[
    {
        "id": "Astronomy_418",
        "problem": "类比是研究问题的常用方法, 科学史上很多重大发现、发明往往发端于类比。\n\n一质量为 $m$ 的人造地球卫星绕地球做匀速圆周运动, 轨道半径为 $r$ 。将地球视为质量均匀分布的球体, 已知地球质量为 $M$, 万有引力常量为 $G$,\n\n求卫星的速度大小 $v$ 和动能 $E_{\\mathrm{k}}$;",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题包含多个待求解的量。\n\n问题:\n类比是研究问题的常用方法, 科学史上很多重大发现、发明往往发端于类比。\n\n一质量为 $m$ 的人造地球卫星绕地球做匀速圆周运动, 轨道半径为 $r$ 。将地球视为质量均匀分布的球体, 已知地球质量为 $M$, 万有引力常量为 $G$,\n\n求卫星的速度大小 $v$ 和动能 $E_{\\mathrm{k}}$;\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你的最终解答的量应该按以下顺序输出：[速度大小, 动能]\n它们的答案类型依次是[表达式, 表达式]\n你需要在输出的最后用以下格式总结答案：“最终答案是\\boxed{ANSWER}”，其中ANSWER应为你的最终答案序列，用英文逗号分隔，例如：5, 7, 2.5",
        "figure_urls": null,
        "answer": [
            "$\\sqrt{\\frac{G M}{r}}$",
            "$\\frac{G M m}{2 r}$"
        ],
        "solution": "根据牛顿第二定律\n\n$$\nG \\frac{M m}{r^{2}}=m \\frac{v^{2}}{r}\n$$\n\n解得卫星的速度大小\n\n$$\nv=\\sqrt{\\frac{G M}{r}}\n$$\n卫星的动能\n\n$$\nE_{\\mathrm{k}}=\\frac{1}{2} m v^{2}=\\frac{G M m}{2 r}\n$$",
        "answer_type": "MPV",
        "unit": [
            null,
            null
        ],
        "answer_sequence": [
            "速度大小",
            "动能"
        ],
        "type_sequence": [
            "EX",
            "EX"
        ],
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_1133",
        "problem": "In the heart of every star, nuclear fusion is taking place. For most stars that involves hydrogen being turned into helium, a process that starts by bringing two protons close enough that the strong nuclear force can act upon them. The smallest stars are the ones that have a core that is only just hot enough for fusion to occur, whilst in the biggest ones the radiation pressure of the photons given out by the fusion reaction pushing on the stellar material can overcome the gravitational forces holding it together.\n[figure1]\n\nFigure 5: Left: The lowest mass star we know of, EBLM J0555-57Ab, was found by von Boetticher et al. (2017) and is about the size of Saturn with a mass of $0.081 M_{\\odot}$. Credit: Amanda Smith, University of Cambridge. Right: The highest mass star we know of, R136a1, is in the centre of the clump of stars on the right of this HST image of the Tarantula Nebula. Schneider et al. (2014) suggest it has a mass of $315 M_{\\odot}$, which is above what stellar evolution models allow. Despite its large mass, other stars have far bigger radii. Credit: NASA \\& ESA.\n\nFor a spherical main sequence star made of a plasma (a fully ionized gas of electrons and nuclei) that is acting like an ideal gas, the temperature at the core can be approximately calculated as\n\n$$\nT_{\\mathrm{int}} \\simeq \\frac{G M \\bar{\\mu}}{k_{\\mathrm{B}} R} \\quad \\text { where } \\quad \\bar{\\mu}=\\frac{m_{\\mathrm{p}}}{2 X+3 Y / 4+Z / 2} .\n$$\n\nIn this equation, $M$ is the mass of the star, $R$ is its radius, $k_{\\mathrm{B}}$ is the Boltzmann constant, and $\\bar{\\mu}$ is the mean mass of the plasma particles (i.e nuclei and electrons) with $m_{\\mathrm{p}}$ the mass of a proton.\n\nClassically, the core of the Sun is not hot enough for fusion, and yet fusion is clearly happening. The key is that it is a fundamentally quantum process, and so protons are able to 'quantum tunnel' through the Coloumb barrier (see Figure 6), allowing fusion to occur at lower temperatures. In quantum mechanics, fusion will happen when $b=\\lambda$ where $\\lambda$ is the de Broglie wavelength of the proton, related to the momentum of the proton by $\\lambda=h / p$.\n\n[figure2]\n\nFigure 6: A diagram showing the way a particle can pass through a classically impenetrable potential barrier due to its wave-like properties on the quantum scale.\n\nCredit: Brooks/Cole - Thomson Learning.\n\nIn the smallest stars, electron degeneracy prevents them from compressing in radius and thus stops the core reaching $T_{\\text {int }} \\gtrsim T_{\\text {quantum }}$. At the limit of electron degeneracy, the number density of electrons $n_{\\mathrm{e}}=1 / \\lambda_{\\mathrm{e}}^{3}$ where $\\lambda_{\\mathrm{e}}$ is the de Broglie wavelength of the electrons.\n\nIn the largest stars, radiation pressure pushes on the outer layers of the star stronger than gravity pulls them in. The brightest luminosity for a star is known as the Eddington luminosity, $L_{\\text {Edd }}$. The acceleration due to radiation pressure can be calculated as\n\n$$\ng_{\\mathrm{rad}}=\\frac{\\kappa_{\\mathrm{e}} I}{c} \\quad \\text { where } \\quad \\kappa_{\\mathrm{e}}=\\frac{\\sigma_{\\mathrm{T}}}{2 m_{\\mathrm{p}}}(1+X)\n$$\n\nand $\\kappa_{\\mathrm{e}}$ is the electron opacity of the stellar material, $\\sigma_{\\mathrm{T}}$ is the Thomson scattering cross-section for electrons $\\left(=66.5 \\mathrm{fm}^{2}\\right.$ ), $X$ is the hydrogen fraction, and $I$ is the intensity of radiation (in $\\mathrm{W} \\mathrm{m}^{-2}$ ). Assuming main-sequence stars follow a mass-luminosity relation of $L \\propto M^{3}$, the maximum mass of a star can be found by considering one that is radiating at $L_{\\text {Edd }}$.c. Given that the proton momentum is related to the average kinetic energy of a particle in the plasma by $E_{K}=p^{2} / 2 m_{p}$ calculate the value of $\\lambda$ and hence calculate $T_{\\text {quantum. }}$. YYou should find that it's below your answer to part a.]",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nIn the heart of every star, nuclear fusion is taking place. For most stars that involves hydrogen being turned into helium, a process that starts by bringing two protons close enough that the strong nuclear force can act upon them. The smallest stars are the ones that have a core that is only just hot enough for fusion to occur, whilst in the biggest ones the radiation pressure of the photons given out by the fusion reaction pushing on the stellar material can overcome the gravitational forces holding it together.\n[figure1]\n\nFigure 5: Left: The lowest mass star we know of, EBLM J0555-57Ab, was found by von Boetticher et al. (2017) and is about the size of Saturn with a mass of $0.081 M_{\\odot}$. Credit: Amanda Smith, University of Cambridge. Right: The highest mass star we know of, R136a1, is in the centre of the clump of stars on the right of this HST image of the Tarantula Nebula. Schneider et al. (2014) suggest it has a mass of $315 M_{\\odot}$, which is above what stellar evolution models allow. Despite its large mass, other stars have far bigger radii. Credit: NASA \\& ESA.\n\nFor a spherical main sequence star made of a plasma (a fully ionized gas of electrons and nuclei) that is acting like an ideal gas, the temperature at the core can be approximately calculated as\n\n$$\nT_{\\mathrm{int}} \\simeq \\frac{G M \\bar{\\mu}}{k_{\\mathrm{B}} R} \\quad \\text { where } \\quad \\bar{\\mu}=\\frac{m_{\\mathrm{p}}}{2 X+3 Y / 4+Z / 2} .\n$$\n\nIn this equation, $M$ is the mass of the star, $R$ is its radius, $k_{\\mathrm{B}}$ is the Boltzmann constant, and $\\bar{\\mu}$ is the mean mass of the plasma particles (i.e nuclei and electrons) with $m_{\\mathrm{p}}$ the mass of a proton.\n\nClassically, the core of the Sun is not hot enough for fusion, and yet fusion is clearly happening. The key is that it is a fundamentally quantum process, and so protons are able to 'quantum tunnel' through the Coloumb barrier (see Figure 6), allowing fusion to occur at lower temperatures. In quantum mechanics, fusion will happen when $b=\\lambda$ where $\\lambda$ is the de Broglie wavelength of the proton, related to the momentum of the proton by $\\lambda=h / p$.\n\n[figure2]\n\nFigure 6: A diagram showing the way a particle can pass through a classically impenetrable potential barrier due to its wave-like properties on the quantum scale.\n\nCredit: Brooks/Cole - Thomson Learning.\n\nIn the smallest stars, electron degeneracy prevents them from compressing in radius and thus stops the core reaching $T_{\\text {int }} \\gtrsim T_{\\text {quantum }}$. At the limit of electron degeneracy, the number density of electrons $n_{\\mathrm{e}}=1 / \\lambda_{\\mathrm{e}}^{3}$ where $\\lambda_{\\mathrm{e}}$ is the de Broglie wavelength of the electrons.\n\nIn the largest stars, radiation pressure pushes on the outer layers of the star stronger than gravity pulls them in. The brightest luminosity for a star is known as the Eddington luminosity, $L_{\\text {Edd }}$. The acceleration due to radiation pressure can be calculated as\n\n$$\ng_{\\mathrm{rad}}=\\frac{\\kappa_{\\mathrm{e}} I}{c} \\quad \\text { where } \\quad \\kappa_{\\mathrm{e}}=\\frac{\\sigma_{\\mathrm{T}}}{2 m_{\\mathrm{p}}}(1+X)\n$$\n\nand $\\kappa_{\\mathrm{e}}$ is the electron opacity of the stellar material, $\\sigma_{\\mathrm{T}}$ is the Thomson scattering cross-section for electrons $\\left(=66.5 \\mathrm{fm}^{2}\\right.$ ), $X$ is the hydrogen fraction, and $I$ is the intensity of radiation (in $\\mathrm{W} \\mathrm{m}^{-2}$ ). Assuming main-sequence stars follow a mass-luminosity relation of $L \\propto M^{3}$, the maximum mass of a star can be found by considering one that is radiating at $L_{\\text {Edd }}$.\n\nproblem:\nc. Given that the proton momentum is related to the average kinetic energy of a particle in the plasma by $E_{K}=p^{2} / 2 m_{p}$ calculate the value of $\\lambda$ and hence calculate $T_{\\text {quantum. }}$. YYou should find that it's below your answer to part a.]\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of \\mathrm{~K}, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_ffe0ae050771e0e3decbg-08.jpg?height=712&width=1508&top_left_y=546&top_left_x=271",
            "https://cdn.mathpix.com/cropped/2024_03_14_ffe0ae050771e0e3decbg-09.jpg?height=514&width=1010&top_left_y=186&top_left_x=523"
        ],
        "answer": [
            "4.86 \\times 10^{6}"
        ],
        "solution": "$$\n\\begin{aligned}\n& \\lambda=\\frac{h}{p} \\therefore p=\\frac{h}{\\lambda} \\text { and } 2 \\times E_{K}=2 \\times \\frac{p^{2}}{2 m_{p}}=E_{P}=\\frac{1}{4 \\pi \\epsilon_{0}} \\frac{e^{2}}{\\lambda} \\\\\n& \\therefore \\frac{h^{2}}{\\lambda^{2} m_{p}}=\\frac{1}{4 \\pi \\epsilon_{0}} \\frac{e^{2}}{\\lambda} \\quad \\therefore \\quad \\lambda=\\frac{4 \\pi \\epsilon_{0} h^{2}}{m_{p} e^{2}} \\\\\n& =\\frac{4 \\pi \\times 8.85 \\times 10^{-12} \\times\\left(6.63 \\times 10^{-34}\\right)^{2}}{1.67 \\times 10^{-27} \\times\\left(1.60 \\times 10^{-19}\\right)^{2}} \\\\\n& =1.14 \\times 10^{-12} \\mathrm{~m}\n\\end{aligned}\n$$\n\n[First mark is for eliminating $p$ and equating with the expression for $E_{\\text {p. }}$. Allow full ecf for this calculation if assuming the stationary proton case, leading to $\\lambda=5.72 \\times 10^{-13} \\mathrm{~m}$ ]\n\n$$\n\\begin{aligned}\n\\frac{3}{2} k_{B} T_{\\text {quantum }}=\\frac{h^{2}}{2 \\lambda^{2} m_{p}} \\therefore T_{\\text {quantum }} & =\\frac{h^{2}}{3 \\lambda^{2} m_{p} k_{B}} \\\\\n& =\\frac{\\left(6.63 \\times 10^{-34}\\right)^{2}}{3 \\times\\left(1.14 \\times 10^{-12}\\right)^{2} \\times 1.67 \\times 10^{-27} \\times 1.38 \\times 10^{-23}} \\\\\n& =4.86 \\times 10^{6} \\mathrm{~K}\n\\end{aligned}\n$$\n\n[First mark is for an expression for $T_{\\text {quantum }}$ in terms of known variables. In the stationary proton case you get $1.95 \\times 10^{7} \\mathrm{~K}$, which is above the value in part a. and so only 2 marks max for this calculation]\n\n[Despite our simplifying assumptions, this is close to the real value of the temperature needed in a star like to Sun to undergo hydrogen fusion, which is $\\approx 3 \\times 10^{6} \\mathrm{~K}$. Deuterium fusion is possible at lower temperatures since each molecule has more mass and hence more momentum at a given temperature, so some very small stars (called brown dwarfs) achieve the required temperature of $\\approx 4.5 \\times 10^{5} \\mathrm{~K}$. The ability to fuse for even a short time is how large gas giants and small stars are distinguished, and is investigated further in the next part of the question]",
        "answer_type": "NV",
        "unit": [
            "\\mathrm{~K}"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_967",
        "problem": "NASA's Mars 2020 mission involves a large rover called Perseverance, and an audacious experiment to see if motor driven flight is possible on the red planet. The helicopter drone is called Ingenuity, shown in Figure 2, and is the first attempt to fly on another world.\n\n[figure1]\n\nFigure 2: The Ingenuity drone with its helicopter rotors ready to take off, with the solar panel used to charge its battery on the very top. In the background you can see a part of the Perseverance rover. Credit: NASA/JPL-Caltech.\n\nAlthough gravity is weaker on Mars, the Martian atmosphere is only about $1 \\%$ the density of that on Earth, which makes it very hard to get lift. To overcome this, the probe has two counterrotating blades (directly above each other) with a tip-to-tip length of $1.2 \\mathrm{~m}$ that will spin at 2400 rpm (about 5 times faster than helicopters on Earth), and is incredibly light with a mass of $1.8 \\mathrm{~kg}$. A consequence of the weight restriction is that the battery is small (and charged by a solar panel on top of the drone, above the blades). Since flying in such a thin atmosphere is a very high power activity the maximum flight duration is therefore short, and given the time needed to recharge, they will be limited to only one flight per day.\n\nThe energy stored in the battery available for flight is $10 \\mathrm{Wh}$ out of a $35 \\mathrm{Wh}$ capacity (the rest is mainly used to keep the drone warm on the cold Martian surface). During flight the drone uses an average of $390 \\mathrm{~W}$. Determine the maximum duration (in seconds) of one flight.",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nNASA's Mars 2020 mission involves a large rover called Perseverance, and an audacious experiment to see if motor driven flight is possible on the red planet. The helicopter drone is called Ingenuity, shown in Figure 2, and is the first attempt to fly on another world.\n\n[figure1]\n\nFigure 2: The Ingenuity drone with its helicopter rotors ready to take off, with the solar panel used to charge its battery on the very top. In the background you can see a part of the Perseverance rover. Credit: NASA/JPL-Caltech.\n\nAlthough gravity is weaker on Mars, the Martian atmosphere is only about $1 \\%$ the density of that on Earth, which makes it very hard to get lift. To overcome this, the probe has two counterrotating blades (directly above each other) with a tip-to-tip length of $1.2 \\mathrm{~m}$ that will spin at 2400 rpm (about 5 times faster than helicopters on Earth), and is incredibly light with a mass of $1.8 \\mathrm{~kg}$. A consequence of the weight restriction is that the battery is small (and charged by a solar panel on top of the drone, above the blades). Since flying in such a thin atmosphere is a very high power activity the maximum flight duration is therefore short, and given the time needed to recharge, they will be limited to only one flight per day.\n\nThe energy stored in the battery available for flight is $10 \\mathrm{Wh}$ out of a $35 \\mathrm{Wh}$ capacity (the rest is mainly used to keep the drone warm on the cold Martian surface). During flight the drone uses an average of $390 \\mathrm{~W}$. Determine the maximum duration (in seconds) of one flight.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of s, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_c3e3c992a9c51eb3e471g-07.jpg?height=731&width=1285&top_left_y=500&top_left_x=385"
        ],
        "answer": [
            "92"
        ],
        "solution": "<br> Recognising that they need to covert from Wh into J, <br> $\\qquad t=\\frac{10 \\times 3600}{390}=92 \\mathrm{~s}$ <br> This is very close to the planned maximum flight time of $90 \\mathrm{~s}$. About $20 \\%$ <br> of the flight will be in high power mode of around $510 \\mathrm{~W}$, during take-off <br> and steering manoeuvres, whilst $80 \\%$ will be travelling at a constant <br> height in a straight line, with a power output of around $360 \\mathrm{~W}$.",
        "answer_type": "NV",
        "unit": [
            "s"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_943",
        "problem": "Meteor showers are caused by\nA: the breakup of asteroids that hit our atmosphere at predictable times\nB: the Earth passing through the debris left behind by a comet as it moves through the inner Solar System\nC: passing asteroids triggering auroral displays\nD: nuclear reactions triggered by an abnormally large meteoritic particle entering the upper atmosphere\n",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nMeteor showers are caused by\n\nA: the breakup of asteroids that hit our atmosphere at predictable times\nB: the Earth passing through the debris left behind by a comet as it moves through the inner Solar System\nC: passing asteroids triggering auroral displays\nD: nuclear reactions triggered by an abnormally large meteoritic particle entering the upper atmosphere\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": [
            "B"
        ],
        "solution": "The dust left by a comet’s tail stays in roughly the same place over short \ntimescales, hence why meteor showers happen at the same time every\nyear.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_855",
        "problem": "An often-repeated fun fact is that humans produce more power per unit volume than stars. If the sun were the same size, but it produced the same amount of power per unit volume as a human, what would its surface temperature be? Assume the \"average human\" produces 100 watts of power and has a volume of 66400 cubic centimeters.\nA: $3500 \\mathrm{~K}$\nB: $10000 \\mathrm{~K}$\nC: $25000 \\mathrm{~K}$\nD: $40000 \\mathrm{~K}$\nE: $50000 \\mathrm{~K}$\n",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn often-repeated fun fact is that humans produce more power per unit volume than stars. If the sun were the same size, but it produced the same amount of power per unit volume as a human, what would its surface temperature be? Assume the \"average human\" produces 100 watts of power and has a volume of 66400 cubic centimeters.\n\nA: $3500 \\mathrm{~K}$\nB: $10000 \\mathrm{~K}$\nC: $25000 \\mathrm{~K}$\nD: $40000 \\mathrm{~K}$\nE: $50000 \\mathrm{~K}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": [
            "E"
        ],
        "solution": "Using the numbers from the problem, the average human produces\n\n$$\nu=\\frac{100}{66400 \\times 10^{-6}}=1506 \\mathrm{~W} / \\mathrm{m}^{3}\n$$\n\nThe volume of the sun is $\\frac{4}{3} \\pi R_{\\odot}^{3}$, so its new power output would be $P=\\frac{4}{3} \\pi R_{\\odot}^{3} u$. To find the equilibrium temperature, we use\n\n$$\nP=\\sigma A T^{4}=4 \\pi \\sigma R_{\\odot}^{2} T^{4}\n$$\n\nSolving for $T$, we get\n\n$$\nT=\\sqrt[4]{\\frac{R_{\\odot} u}{3 \\sigma}}=\\sqrt[4]{\\frac{\\left(6.96 \\times 10^{8}\\right)(1506)}{3\\left(5.67 \\times 10^{-8}\\right)}}=49823 \\approx 50000 \\mathrm{~K} .\n$$",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_215",
        "problem": "质量为 $m$ 的宇宙飞船, 在离月球地面高度 $h$ 处沿圆形轨道绕月球运行。为使飞船到达月球表面 $B$ 点, 喷气发动机在 $A$ 点做一次极短时间的喷气。从喷口射出的气流方向与圆周轨道相切且相对飞船的速度为 $u$, 月球半径为 $R, h=\\frac{R}{16}, A 、 B$ 两点与球心在一直线上, 其速度与飞船到球心的距离成反比。月球表面重力加速度为 $g$ 。若以无穷远处为飞船引力势能的零势能点, 飞船和球心距离为 $r$ 时, 引力势能的表达式为 $E_{\\mathrm{P}}=-\\frac{G M m}{r}$ 。 ( $M$ 是月球质量, 本题中未知)。求:\n\n 飞船刚到 $B$ 点时的速度大小 $v_{B}$;\n\n[图1]",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n\n问题:\n质量为 $m$ 的宇宙飞船, 在离月球地面高度 $h$ 处沿圆形轨道绕月球运行。为使飞船到达月球表面 $B$ 点, 喷气发动机在 $A$ 点做一次极短时间的喷气。从喷口射出的气流方向与圆周轨道相切且相对飞船的速度为 $u$, 月球半径为 $R, h=\\frac{R}{16}, A 、 B$ 两点与球心在一直线上, 其速度与飞船到球心的距离成反比。月球表面重力加速度为 $g$ 。若以无穷远处为飞船引力势能的零势能点, 飞船和球心距离为 $r$ 时, 引力势能的表达式为 $E_{\\mathrm{P}}=-\\frac{G M m}{r}$ 。 ( $M$ 是月球质量, 本题中未知)。求:\n\n 飞船刚到 $B$ 点时的速度大小 $v_{B}$;\n\n[图1]\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-076.jpg?height=271&width=300&top_left_y=1235&top_left_x=341"
        ],
        "answer": [
            "$\\sqrt{\\frac{34 g R}{33}}$"
        ],
        "solution": "设喷气后飞船在 $A$ 点速度变为 $v_{A}^{\\prime}$, 质量变为 $m^{\\prime}$, 飞船从 $A$ 点沿粗圆轨道运动到 $B$点过程机械能守恒, 则有\n\n$$\n-\\frac{G M m^{\\prime}}{R+h}+\\frac{1}{2} m^{\\prime} v_{A}^{\\prime 2}=-\\frac{G M m^{\\prime}}{R}+\\frac{1}{2} m^{\\prime} v_{B}^{2}\n$$\n\n$A 、 B$ 两点与球心在一直线上, 其速度与飞船到球心的距离成反比\n\n$$\n\\frac{v_{A}^{\\prime}}{v_{B}}=\\frac{R}{R+h}\n$$\n\n又因为\n\n$$\nG \\frac{M m}{R^{2}}=m g\n$$\n\n联立可得\n\n$$\nv_{B}=\\sqrt{\\frac{34 g R}{33}}, \\quad v_{A}^{\\prime}=\\frac{16}{17} \\sqrt{\\frac{34 g R}{33}}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_152",
        "problem": "双星的运动是产生引力波的来源之一, 假设宇宙中有一双星系统由 $a 、 b$ 两颗星体组成, 这两颗星绕它们连线的某一点在万有引力作用下做匀速圆周运动, 测得 $a$ 星的周期为 $T, a 、 b$ 两颗星的距离为 $l, a$ 星的轨道半径大于 $b$ 星的轨道半径且 $a 、 b$ 两颗星的轨道半径之差为 $\\Delta r$, 则 $(\\quad)$\nA: $b$ 星的周期为 $\\frac{l-\\Delta r}{l+\\Delta r} T$\nB: $a 、 b$ 两颗星的质量比值为 $\\frac{l-\\Delta r}{l+\\Delta r}$\nC: $a 、 b$ 两颗星的速率之和为 $\\frac{2 \\pi}{T} l$\nD: $a 、 b$ 两颗星的质量之和为 $\\frac{4 \\pi^{2} l^{3}}{G T^{2}}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n双星的运动是产生引力波的来源之一, 假设宇宙中有一双星系统由 $a 、 b$ 两颗星体组成, 这两颗星绕它们连线的某一点在万有引力作用下做匀速圆周运动, 测得 $a$ 星的周期为 $T, a 、 b$ 两颗星的距离为 $l, a$ 星的轨道半径大于 $b$ 星的轨道半径且 $a 、 b$ 两颗星的轨道半径之差为 $\\Delta r$, 则 $(\\quad)$\n\nA: $b$ 星的周期为 $\\frac{l-\\Delta r}{l+\\Delta r} T$\nB: $a 、 b$ 两颗星的质量比值为 $\\frac{l-\\Delta r}{l+\\Delta r}$\nC: $a 、 b$ 两颗星的速率之和为 $\\frac{2 \\pi}{T} l$\nD: $a 、 b$ 两颗星的质量之和为 $\\frac{4 \\pi^{2} l^{3}}{G T^{2}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "B",
            "C",
            "D"
        ],
        "solution": "A. $a 、 b$ 两颗星的周期相同, 所以 $\\mathrm{A}$ 错误;\n\nB. 双星模型中有 $a 、 b$ 两颗星具有相同的角速度, 则对 $a$ 星有\n\n$$\nG \\frac{m_{a} m_{b}}{l^{2}}=m_{a} \\omega^{2} r_{a}\n$$\n\n对 $b$ 星有\n\n$$\nG \\frac{m_{a} m_{b}}{l^{2}}=m_{b} \\omega^{2} r_{b}\n$$\n几何关系可得\n\n$$\n\\begin{gathered}\nr_{a}+r_{b}=l \\\\\nr_{a}-r_{b}=\\Delta r\n\\end{gathered}\n$$\n\n联立解得\n\n$$\n\\frac{m_{a}}{m_{b}}=\\frac{r_{b}}{r_{a}}=\\frac{\\frac{l-\\Delta r}{2}}{\\frac{l+\\Delta r}{2}}=\\frac{l-\\Delta r}{l+\\Delta r}\n$$\n\n所以 $\\mathrm{B}$ 正确;\n\nC. 双星模型中有 $a 、 b$ 两颗星具有相同的角速度, 根据 $v=\\omega r$, 则 $a 、 b$ 两颗星的速率之和为\n\n$$\nv_{a}+v_{b}=\\omega\\left(r_{a}+r_{b}\\right)=\\frac{2 \\pi}{T} l\n$$\n\n所以 $\\mathrm{C}$ 正确;\n\nD. 双星模型中有 $a 、 b$ 两颗星具有相同的角速度, 则对 $a$ 星有\n\n$$\nG \\frac{m_{a} m_{b}}{r_{a}^{2}}=m_{a}\\left(\\frac{2 \\pi}{T}\\right)^{2} r_{a}\n$$\n\n解得\n\n$$\nm_{b}=\\frac{4 \\pi^{2} l^{2} r_{a}}{G T^{2}}\n$$\n\n对 $b$ 星有\n\n$$\nG \\frac{m_{a} m_{b}}{r_{b}^{2}}=m_{b}\\left(\\frac{2 \\pi}{T}\\right)^{2} r_{b}\n$$\n\n解得\n\n$$\nm_{a}=\\frac{4 \\pi^{2} l^{2} r_{b}}{G T^{2}}\n$$\n\n几何关系可得\n\n$$\nr_{a}+r_{b}=l\n$$\n\n联立解得\n\n$$\nm_{a}+m_{b}=\\frac{4 \\pi^{2} l^{2} r_{b}}{G T^{2}}+\\frac{4 \\pi^{2} l^{2} r_{a}}{G T^{2}}=\\frac{4 \\pi^{2} l^{3}}{G T^{2}}\n$$\n\n所以 D 正确;\n\n故选 BCD。",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_58",
        "problem": "在人类的某次星际探测任务中, 发现一个可能宜居的星球。为了更详细的了解该星球的情况, 需要进行登陆探测。在登陆前, 飞船先环绕该星球在半径为 $3 R$ 的圆轨道上做匀速圆周运动, $R$ 为该星球的半径。某一时刻, 飞船运行到 $A$ 点, 此时飞船将登陆器向反方向射出, 但登陆器仍向前运动, 并进入图示的内部栯圆轨道登上该星球表面, 而飞船立即启动发动机进行调整，使其仍保持在圆轨道运行。登陆器在该星球表面探测一段时间后，重新点火加速沿原来的椭圆轨道回到脱离点实现对接。已知该星球的质量为\n\n$M$, 飞船的质量为 $m_{1}$, 登陆器的质量为 $m_{2}$, 且 $m_{1}=2 m_{2}$, 忽略登陆器点火加速的时间,引力常量为 $G$ 。求:飞船在反射出登陆器后, 为保持在原轨道运行, 飞船发动机对飞船做的功 $W$ ?[图1]",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n\n问题:\n在人类的某次星际探测任务中, 发现一个可能宜居的星球。为了更详细的了解该星球的情况, 需要进行登陆探测。在登陆前, 飞船先环绕该星球在半径为 $3 R$ 的圆轨道上做匀速圆周运动, $R$ 为该星球的半径。某一时刻, 飞船运行到 $A$ 点, 此时飞船将登陆器向反方向射出, 但登陆器仍向前运动, 并进入图示的内部栯圆轨道登上该星球表面, 而飞船立即启动发动机进行调整，使其仍保持在圆轨道运行。登陆器在该星球表面探测一段时间后，重新点火加速沿原来的椭圆轨道回到脱离点实现对接。已知该星球的质量为\n\n$M$, 飞船的质量为 $m_{1}$, 登陆器的质量为 $m_{2}$, 且 $m_{1}=2 m_{2}$, 忽略登陆器点火加速的时间,引力常量为 $G$ 。求:飞船在反射出登陆器后, 为保持在原轨道运行, 飞船发动机对飞船做的功 $W$ ?[图1]\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n请记住，你的答案应以[None]为单位计算，但在给出最终答案时，请不要包含单位。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是不含任何单位和等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-001.jpg?height=514&width=371&top_left_y=1439&top_left_x=340"
        ],
        "answer": [
            "$\\frac{6 \\sqrt{2}-11}{48 R} G M m_{1}$"
        ],
        "solution": "登录器在粗圆轨道运动时, 根据几何关系, 长轴、焦距、短轴分别为\n\n$$\n\\begin{gathered}\n2 a=3 R+R=4 R \\\\\n2 c=3 R-R=2 R \\\\\n2 b=\\sqrt{(2 a)^{2}-(2 c)^{2}}=2 \\sqrt{3} R\n\\end{gathered}\n$$\n\n设远“地”点和近“地”点的速度分别为 $v_{A} 、 v_{B}$, 周期为 $T^{\\prime}$, 根据开普勒第二定律\n\n$$\n\\frac{1}{2} \\times 3 R v_{A}=\\frac{1}{2} \\times R v_{B}=\\frac{\\pi a b}{T^{\\prime}}\n$$\n\n根据开普敦第三定律\n\n$$\n\\begin{gathered}\n\\frac{(3 R)^{3}}{T^{2}}=\\frac{\\left(\\frac{R+3 R}{2}\\right)^{3}}{T^{\\prime 2}} \\\\\nT^{\\prime}=4 \\pi R \\sqrt{\\frac{2 R}{G M}}\n\\end{gathered}\n$$\n\n解得\n\n$$\nv_{A}=\\sqrt{\\frac{G M}{6 R}}\n$$\n\n飞船反射出登陆器过程, 根据动量守恒\n\n$$\n\\left(m_{1}+m_{2}\\right) v_{0}=m_{1} v+m_{2} v_{A}\n$$\n\n解得\n\n$$\nv=\\frac{3 \\sqrt{2}-1}{2} \\sqrt{\\frac{G M}{6 R}}\n$$\n\n为保持在原轨道运行, 飞船需要恢复到速度 $v_{0}$, 飞船发动机对飞船做的功\n\n$$\nW=\\frac{1}{2} m_{1} v_{0}^{2}-\\frac{1}{2} m_{1} v^{2}=\\frac{6 \\sqrt{2}-11}{48 R} G M m_{1}\n$$",
        "answer_type": "EX",
        "unit": [
            "[None]"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_273",
        "problem": "宇宙中, 多数恒星是双星系统。我们发现距离地球 149 光年, 有一双星系统 $\\mathrm{AB}$,已知 $\\mathrm{A}$ 和 $\\mathrm{B}$ 的总质量为太阳质量的 1.63 倍, 它们以 156 天的周期互相绕对方公转。则 $\\mathrm{A}$ 和 $\\mathrm{B}$ 的距离约为地球和太阳距离的（）\nA: 1.6 倍\nB: 1 倍\nC: 0.7 倍\nD: 0.3 倍\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n宇宙中, 多数恒星是双星系统。我们发现距离地球 149 光年, 有一双星系统 $\\mathrm{AB}$,已知 $\\mathrm{A}$ 和 $\\mathrm{B}$ 的总质量为太阳质量的 1.63 倍, 它们以 156 天的周期互相绕对方公转。则 $\\mathrm{A}$ 和 $\\mathrm{B}$ 的距离约为地球和太阳距离的（）\n\nA: 1.6 倍\nB: 1 倍\nC: 0.7 倍\nD: 0.3 倍\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "C"
        ],
        "solution": "设 $\\mathrm{A}$ 和 $\\mathrm{B}$ 的距离为 $L$, 它们互绕周期\n\n$$\nT_{A B}=\\frac{156}{365} \\text { 年 } \\approx 0.43 \\text { 年 }\n$$\n\n对 A\n\n$$\nG \\frac{m_{\\mathrm{A}} m_{\\mathrm{B}}}{L^{2}}=m_{\\mathrm{A}} r_{\\mathrm{A}}\\left(\\frac{2 \\pi}{T_{\\mathrm{AB}}}\\right)^{2}\n$$\n\n对 B\n\n$$\nG \\frac{m_{\\mathrm{A}} m_{\\mathrm{B}}}{L^{2}}=m_{\\mathrm{B}} r_{\\mathrm{B}}\\left(\\frac{2 \\pi}{T_{\\mathrm{AB}}}\\right)^{2}\n$$\n\n又设地球绕太阳的周期为 $T$, 地球和太阳距离为 $r$, 地球绕太阳公转, 根据万有引力定律\n\n$$\nG \\frac{M m}{r^{2}}=m r\\left(\\frac{2 \\pi}{T}\\right)^{2}\n$$\n\n由以上各式解得\n\n$$\n\\frac{L}{r}=\\sqrt[3]{\\frac{1.63 M \\cdot T_{A B}^{2}}{M T^{2}}}=\\sqrt[3]{1.63 \\times 0.43 \\times 0.43} \\approx 0.7\n$$\n\nC 正确。\n\n故选 C。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_712",
        "problem": "已知地球的质量为 $M$, 半径为 $R$, 引力常量为 $G$ 。赤道上地球表面附近的重力加速度用 $g_{e}$ 表示, 北极处地球表面附近的重力加速度用 $g_{N}$ 表示, 将地球视为均匀球体。\n用已知量写出 $g_{N}$ 的表达式;\n\n[图1]",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n\n问题:\n已知地球的质量为 $M$, 半径为 $R$, 引力常量为 $G$ 。赤道上地球表面附近的重力加速度用 $g_{e}$ 表示, 北极处地球表面附近的重力加速度用 $g_{N}$ 表示, 将地球视为均匀球体。\n用已知量写出 $g_{N}$ 的表达式;\n\n[图1]\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-093.jpg?height=579&width=599&top_left_y=1735&top_left_x=360"
        ],
        "answer": [
            "$\\frac{G M}{R^{2}}$"
        ],
        "solution": "对北极处地球表面附近的物体, 由\n\n$$\n\\frac{G M m}{R^{2}}=m g_{\\mathrm{N}}\n$$\n\n得\n\n$$\ng_{N}=\\frac{G M}{R^{2}}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_224",
        "problem": "如图所示为三颗卫星 $a 、 b 、 c$ 绕地球做匀速圆周运动的示意图, 其中 $b 、 c$ 是地球同步卫星, $a$ 在半径为 $r$ 的轨道上, 此时 $a 、 b$ 恰好相距最近, 已知地球质量为 $M$, 半径为 $R$, 地球自转的角速度为 $\\omega$, 引力常量为 $G$, 则 ( )\n\n[图1]\nA: 卫星 $b$ 加速一段时间后就可能追上卫星 $\\mathrm{c}$\nB: 卫星 $b$ 和 $c$ 的机械能相等\nC: 到卫星 $a$ 和 $b$ 下一次相距最近，还需经过时间 $t=\\frac{2 \\pi}{\\sqrt{\\frac{G M}{r^{3}}}-\\omega}$\nD: 卫星 $a$ 减速一段时间后就可能追上卫星 $c$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图所示为三颗卫星 $a 、 b 、 c$ 绕地球做匀速圆周运动的示意图, 其中 $b 、 c$ 是地球同步卫星, $a$ 在半径为 $r$ 的轨道上, 此时 $a 、 b$ 恰好相距最近, 已知地球质量为 $M$, 半径为 $R$, 地球自转的角速度为 $\\omega$, 引力常量为 $G$, 则 ( )\n\n[图1]\n\nA: 卫星 $b$ 加速一段时间后就可能追上卫星 $\\mathrm{c}$\nB: 卫星 $b$ 和 $c$ 的机械能相等\nC: 到卫星 $a$ 和 $b$ 下一次相距最近，还需经过时间 $t=\\frac{2 \\pi}{\\sqrt{\\frac{G M}{r^{3}}}-\\omega}$\nD: 卫星 $a$ 减速一段时间后就可能追上卫星 $c$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_6842b9ceb844a90b34c3g-43.jpg?height=391&width=409&top_left_y=173&top_left_x=341"
        ],
        "answer": [
            "C"
        ],
        "solution": "A.卫星 $b$ 加速后将做离心运动, 轨道变高, 不可能追上卫星 $c$, 选项 $\\mathrm{A}$ 错误; B.卫星的机械能等于其动能与势能之和, 因不知道卫星的质量, 故不能确定卫星的机械能大小关系, 选项 B 错误;\n\nC. 对卫星 $a$, 根据万有引力提供向心力有:\n\n$$\nG \\frac{M m}{r^{2}}=m r \\omega_{a}^{2}\n$$\n\n所以卫星 $a$ 的角速度\n\n$$\n\\omega_{a}=\\sqrt{G \\frac{M}{r^{3}}}\n$$\n\n可知半径越大角速度越小, 卫星 $a$ 和 $b$ 由相距最近至再次相距最近时, 圆周运动转过的角度差为 $2 \\pi$, 所以可得经历的时间:\n\n$$\nt=\\frac{2 \\pi}{\\sqrt{\\frac{G M}{r^{3}}}-\\omega}\n$$\n\n选项 C 正确;\n\nD. 卫星 $a$ 减速后将做近心运动, 轨道半径减小, 不可能追上卫星 $c$, 选项 D 错误;故选 C。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_651",
        "problem": "近年科学界经过论证认定: 肉眼无法从太空看长城, 但遥感卫星可以“看”到长城。\n\n已知某遥感卫星在离地球高度约为 $300 \\mathrm{~km}$ 的圆轨道上运行, 地球半径约为 $6400 \\mathrm{~km}$, 地球同步卫星离地球高度约为地球半径的 5.6 倍。则以下说法正确的是（）\nA: 遥感卫星的发射速度一定要超过 $7.9 \\mathrm{~km} / \\mathrm{s}$\nB: 遥感卫星运行速度约为 $8.1 \\mathrm{~km} / \\mathrm{s}$\nC: 地球同步卫星运行速度约为 $8.1 \\mathrm{~km} / \\mathrm{s}$\nD: 遥感卫星只需加速, 即可追上同轨道运行的其他卫星\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n近年科学界经过论证认定: 肉眼无法从太空看长城, 但遥感卫星可以“看”到长城。\n\n已知某遥感卫星在离地球高度约为 $300 \\mathrm{~km}$ 的圆轨道上运行, 地球半径约为 $6400 \\mathrm{~km}$, 地球同步卫星离地球高度约为地球半径的 5.6 倍。则以下说法正确的是（）\n\nA: 遥感卫星的发射速度一定要超过 $7.9 \\mathrm{~km} / \\mathrm{s}$\nB: 遥感卫星运行速度约为 $8.1 \\mathrm{~km} / \\mathrm{s}$\nC: 地球同步卫星运行速度约为 $8.1 \\mathrm{~km} / \\mathrm{s}$\nD: 遥感卫星只需加速, 即可追上同轨道运行的其他卫星\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "A"
        ],
        "solution": "A B. 第一宇宙速度 $7.9 \\mathrm{~km} / \\mathrm{s}$ 是最小发射速度也是最大环绕速度, 遥感卫星在离地球高度约为 $300 \\mathrm{~km}$ 的圆轨道上运行, 轨道半径大于近地卫星的轨道半径, 因此发射速度要大于第一宇宙速度; 运行速度小于第一宇宙速度, 则运行速度小于 $7.9 \\mathrm{~km} / \\mathrm{s}$,选项 A 正确, B 错误;\n\nC. 同理, 地球同步卫星的运行速度也要小于第一宇宙速度, 选项 C 错误;\n\nD. 卫星加速后将做离心运动, 偏离原轨道, 不能直接追上同轨道的卫星, 选项 D 错误。故选 A。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_29",
        "problem": "在星球 $\\mathrm{M}$ 上将一轻弹簧坚直固定在水平桌面上, 把物体 $\\mathrm{P}$ 轻放在弹簧上端, $\\mathrm{P}$ 由静止向下运动, 物体的加速度 $a$ 与弹簧的压缩量 $x$ 间的关系如图中实线所示。在另一星球 $\\mathrm{N}$ 上用完全相同的弹簧, 改用物体 $\\mathrm{Q}$ 完成同样的过程, 其 $a-x$ 关系如图中虚线所示,假设两星球均为质量均匀分布的球体。已知星球 $\\mathrm{M}$ 的半径是星球 $\\mathrm{N}$ 的 3 倍, 求:\n\nQ 下落过程中的最大动能与 P 下落过程中的最大动能之比为多少?\n\n[图1]",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题的答案是一个数值。\n\n问题:\n在星球 $\\mathrm{M}$ 上将一轻弹簧坚直固定在水平桌面上, 把物体 $\\mathrm{P}$ 轻放在弹簧上端, $\\mathrm{P}$ 由静止向下运动, 物体的加速度 $a$ 与弹簧的压缩量 $x$ 间的关系如图中实线所示。在另一星球 $\\mathrm{N}$ 上用完全相同的弹簧, 改用物体 $\\mathrm{Q}$ 完成同样的过程, 其 $a-x$ 关系如图中虚线所示,假设两星球均为质量均匀分布的球体。已知星球 $\\mathrm{M}$ 的半径是星球 $\\mathrm{N}$ 的 3 倍, 求:\n\nQ 下落过程中的最大动能与 P 下落过程中的最大动能之比为多少?\n\n[图1]\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是数值。",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-087.jpg?height=300&width=439&top_left_y=330&top_left_x=343"
        ],
        "answer": [
            "4"
        ],
        "solution": "由牛顿第二定律可知\n\n$$\nm g-k x=m a\n$$\n\n得\n\n$$\na=-\\frac{k x}{m}+g\n$$\n\n由图可知, 当 $a=0$ 时, 物体加速度为 0 , 速度达到最大, 在 $\\mathrm{M} 、 \\mathrm{~N}$ 星球分别有\n\n$$\n\\begin{aligned}\n& k \\cdot x_{0}=m_{\\mathrm{P}} \\cdot g_{\\mathrm{M}}=m_{\\mathrm{P}} \\cdot 3 a_{0} \\\\\n& k \\cdot 2 x_{0}=m_{\\mathrm{Q}} \\cdot g_{\\mathrm{N}}=m_{\\mathrm{Q}} \\cdot a_{0}\n\\end{aligned}\n$$\n\n解得\n\n$$\n\\frac{m_{\\mathrm{P}}}{m_{\\mathrm{Q}}}=\\frac{1}{6}\n$$\n\n根据动能定理可得\n\n$$\nF_{\\text {合 }} \\cdot x=m \\cdot a \\cdot x=\\Delta E_{\\mathrm{k}}\n$$\n\n物体合外力是变力, 由图像的面积可得\n\n$$\n\\begin{aligned}\n& E_{\\mathrm{kP}}=\\frac{1}{2} m_{\\mathrm{P}} \\cdot 3 a_{0} \\cdot x_{0} \\\\\n& E_{\\mathrm{kQ}}=\\frac{1}{2} m_{\\mathrm{Q}} \\cdot a_{0} \\cdot 2 x_{0}\n\\end{aligned}\n$$\n\n解得\n\n$$\n\\frac{E_{\\mathrm{kQ}}}{E_{\\mathrm{kP}}}=\\frac{2 m_{\\mathrm{Q}}}{3 m_{\\mathrm{P}}}=4\n$$",
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_614",
        "problem": "如图所示, 卫星先后在近地圆轨道 1 、粗圆形的转移轨道 2 和同步轨道 3 上运动,轨道 $1 、 2$ 相切于 $P$ 点, $2 、 3$ 相切于 $Q$ 点, 则卫星在下列非点火状态下的比较正确的是\n\n[图1]\nA: 在轨道 2 上运行的周期大于在轨道 3 上运行的周期\nB: 在轨道 1 上 $P$ 点的速度大于在轨道 2 上 $P$ 点的速度\nC: 在轨道 1 上 $P$ 点的速度大于在轨道 2 上 $Q$ 点的速度\nD: 在轨道 2 上 $Q$ 点的加速度大于在轨道 3 上 $Q$ 点的加速度\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图所示, 卫星先后在近地圆轨道 1 、粗圆形的转移轨道 2 和同步轨道 3 上运动,轨道 $1 、 2$ 相切于 $P$ 点, $2 、 3$ 相切于 $Q$ 点, 则卫星在下列非点火状态下的比较正确的是\n\n[图1]\n\nA: 在轨道 2 上运行的周期大于在轨道 3 上运行的周期\nB: 在轨道 1 上 $P$ 点的速度大于在轨道 2 上 $P$ 点的速度\nC: 在轨道 1 上 $P$ 点的速度大于在轨道 2 上 $Q$ 点的速度\nD: 在轨道 2 上 $Q$ 点的加速度大于在轨道 3 上 $Q$ 点的加速度\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_a9b05ce8eea7b0e40e5eg-056.jpg?height=380&width=346&top_left_y=1535&top_left_x=341"
        ],
        "answer": [
            "C"
        ],
        "solution": "D. 根据牛顿第二定律有\n\n$$\nG \\frac{M m}{r^{2}}=m a\n$$\n\n解得\n\n$$\na=\\frac{G M}{r^{2}}\n$$\n在轨道 2 上 $Q$ 点的 $r$ 与在轨道 3 上 $Q$ 点的 $r$ 相同, 因此在轨道 2 上 $Q$ 点的加速度等于在轨道 3 上 $Q$ 点的加速度, 选项 $\\mathrm{D}$ 错误;\n\nA. 根据开普勒第三定律有\n\n$$\n\\frac{r_{2}^{3}}{T_{2}^{2}}=\\frac{r_{3}^{3}}{T_{3}^{2}}\n$$\n\n由题意知\n\n$$\nr_{2}<r_{3}\n$$\n\n所以\n\n$$\nT_{2}<T_{3}\n$$\n\n即卫星在轨道 2 上运行的周期小于在轨道 3 上运行的周期, 选项 A 错误;\n\nB. 卫星从轨道 1 上 $P$ 点进入轨道 2 需要加速, 因此卫星在轨道 1 上 $P$ 点的速度小于在轨道 2 上 $P$ 点的速度, 选项 $\\mathrm{B}$ 错误;\n\nC. 根据牛顿第二定律有\n\n$$\nG \\frac{M m}{r^{2}}=m \\frac{v^{2}}{r}\n$$\n\n解得\n\n$$\nv=\\sqrt{\\frac{G M}{r}}\n$$\n\n由题意知\n\n$$\nr_{1}<r_{3}\n$$\n\n所以\n\n$$\nv_{1 P}>v_{3 Q}\n$$\n\n卫星从轨道 2 上 $Q$ 点进入轨道 3 需要加速, 因此有\n\n$$\nv_{3 Q}>v_{2 Q}\n$$\n\n所以\n\n$$\nv_{1 P}>v_{3 Q}>v_{2 Q}\n$$\n\n即卫星在轨道 1 上 $P$ 点的速度大于在轨道 2 上 $Q$ 点的速度, 选项 C 正确。\n\n故选 C。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_445",
        "problem": "2020 年发现的 TOI-1338b 是一颗围绕 TOI-1338 双星系统转动的行星。TOI-1338 双星系统由质量约为 $1.2 \\mathrm{M}$ 的 TOI-1338A 和 $0.3 \\mathrm{M}$ 的 TOI-1338B 组成 $(M$ 为太阳质量), 双星的运动周期约为地球公转周期的 $\\frac{1}{25}$ 。则 TOI-1338A 的线速度与 TOI- $1338 \\mathrm{~B}$线速度的比值以及双星间的距离分别是（ $\\mathrm{AU}$ 为天文单位, $\\mathrm{AU}=$ 日地平均距离）( )\nA: $\\frac{1}{4}, \\frac{\\sqrt[3]{30}}{50} \\mathrm{AU}$\nB: $\\frac{1}{4}, \\frac{\\sqrt[3]{300}}{50} \\mathrm{AU}$\nC: $4, \\frac{\\sqrt{30}}{50} \\mathrm{AU}$\nD: $4, \\frac{\\sqrt[3]{30}}{50} \\mathrm{AU}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n2020 年发现的 TOI-1338b 是一颗围绕 TOI-1338 双星系统转动的行星。TOI-1338 双星系统由质量约为 $1.2 \\mathrm{M}$ 的 TOI-1338A 和 $0.3 \\mathrm{M}$ 的 TOI-1338B 组成 $(M$ 为太阳质量), 双星的运动周期约为地球公转周期的 $\\frac{1}{25}$ 。则 TOI-1338A 的线速度与 TOI- $1338 \\mathrm{~B}$线速度的比值以及双星间的距离分别是（ $\\mathrm{AU}$ 为天文单位, $\\mathrm{AU}=$ 日地平均距离）( )\n\nA: $\\frac{1}{4}, \\frac{\\sqrt[3]{30}}{50} \\mathrm{AU}$\nB: $\\frac{1}{4}, \\frac{\\sqrt[3]{300}}{50} \\mathrm{AU}$\nC: $4, \\frac{\\sqrt{30}}{50} \\mathrm{AU}$\nD: $4, \\frac{\\sqrt[3]{30}}{50} \\mathrm{AU}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "B"
        ],
        "solution": "双星系统转动的角速度相等, 根据 $v=r \\omega$ 可得\n\n$$\n\\frac{v_{1}}{v_{2}}=\\frac{r_{1}}{r_{2}}\n$$\n\n设双星间的距离为 $\\mathrm{L}$ ，根据万有引力提供向心力公式得\n\n$$\nG \\frac{m_{1} m_{2}}{L^{2}}=1.2 M \\frac{v_{1}^{2}}{r_{1}}=0.3 M \\frac{v_{2}^{2}}{r_{2}}\n$$\n\n联立解得\n\n$$\n\\frac{r_{1}}{r_{2}}=\\frac{0.3 M}{1.2 M}=\\frac{1}{4}\n$$\n\n又地球公转有\n\n$$\nG \\frac{M m}{R^{2}}=m R\\left(\\frac{2 \\pi}{T}\\right)^{2}\n$$\n\n解得地球公转周期\n\n$$\nT^{2}=\\frac{4 \\pi^{2} R^{3}}{G M}\n$$\n\n双星的运动周期约为地球公转周期的 $\\frac{1}{25}$, 则有\n\n$$\nG \\frac{m_{1} m_{2}}{L^{2}}=m_{1}\\left(\\frac{2 \\pi}{\\frac{T}{25}}\\right)^{2} r_{1}=m_{2}\\left(\\frac{2 \\pi}{\\frac{T}{25}}\\right)^{2} r_{2}\n$$\n\n解得双星间的距离为\n\n$$\nL=\\frac{\\sqrt[3]{300}}{50} R=\\frac{\\sqrt[3]{300}}{50} \\mathrm{AU}\n$$\n\n故选 B。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_286",
        "problem": "$\\mathrm{A} 、 \\mathrm{~B}$ 两个半径相同的天体各有一个卫星 $\\mathrm{a} 、 \\mathrm{~b}$ 环绕它们做匀速圆周运动, 两个卫星的环绕周期之比为 4; 1, A、B 各自表面重力加速度之比为 4:1 (忽略天体的自转), 则\nA: $a 、 b$ 轨迹半径之比为 $4: 1$\nB: A、B 密度之比为 $4: 1$\nC: $a 、 b$ 扫过相同面积所需时间之比为 $1: 16$\nD: $\\mathrm{a} 、 \\mathrm{~b}$ 所受向心力之比为 $1: 16$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$\\mathrm{A} 、 \\mathrm{~B}$ 两个半径相同的天体各有一个卫星 $\\mathrm{a} 、 \\mathrm{~b}$ 环绕它们做匀速圆周运动, 两个卫星的环绕周期之比为 4; 1, A、B 各自表面重力加速度之比为 4:1 (忽略天体的自转), 则\n\nA: $a 、 b$ 轨迹半径之比为 $4: 1$\nB: A、B 密度之比为 $4: 1$\nC: $a 、 b$ 扫过相同面积所需时间之比为 $1: 16$\nD: $\\mathrm{a} 、 \\mathrm{~b}$ 所受向心力之比为 $1: 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "A",
            "B"
        ],
        "solution": "根据 $G \\frac{M m}{r^{2}}=m\\left(\\frac{2 \\pi}{T}\\right)^{2} r$ 以及 $G \\frac{M m}{R^{2}}=m g$ 可得 $r^{3}=\\frac{G M T^{2}}{4 \\pi^{2}}=\\frac{g R^{2} T^{2}}{4 \\pi^{2}} \\propto g T^{2}$; 可得 $\\mathrm{a} 、 \\mathrm{~b}$ 轨迹半径之比为 $\\frac{r_{a}}{r_{b}}=\\sqrt[3]{\\frac{4}{1} \\times\\left(\\frac{4}{1}\\right)^{2}}=\\frac{4}{1}$, 选项 A 正确; 由 $\\rho=\\frac{M}{\\frac{4}{3} \\pi R^{3}}=\\frac{\\frac{g R^{2}}{G}}{\\frac{4}{3} \\pi R^{3}}=\\frac{3 g}{4 \\pi G R} \\propto g$, 则 $\\mathrm{A} 、 \\mathrm{~B}$ 密度之比为 4:1, 选项 B 正确; 根据 $t=\\frac{r \\theta}{v}$, $\\frac{1}{2} r^{2} \\theta=S$, 即 $t=\\frac{2 S}{r v}=\\frac{2 S}{r \\cdot \\frac{2 \\pi r}{T}}=\\frac{S T}{\\pi r^{2}}$, 当扫过相同面积 $\\mathrm{S}$ 时, 则\n$\\frac{t_{a}}{t_{b}}=\\frac{T_{a}}{T_{b}} \\times \\frac{r_{b}^{2}}{r_{a}^{2}}=\\frac{4}{1} \\times\\left(\\frac{1}{4}\\right)^{2}=\\frac{1}{4}$, 选项 C 错误; 两卫星 $\\mathrm{ab}$ 的质量不确定, 无法比较向心力的大小关系, 选项 D 错误; 故选 AB.",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_382",
        "problem": "火卫一是火星的两颗天然卫星之一, 大小约 $27 \\mathrm{~km} \\times 22 \\mathrm{~km} \\times 18 \\mathrm{~km}$, 运行在距离火星表面约 $6000 \\mathrm{~km}$ 高度的近圆轨道上，周期约为 0.32 个地球日。 2022 年“天问一号”发射两周年之际，国家航天局探月与航天工程中心发布了“天问一号”环绕器拍摄到的火卫一的高清影像图。环绕器运行在近火点距离火星表面高度约 $220 \\mathrm{~km}$ 、远火点距离火星表面高度约 $1.08 \\times 10^{4} \\mathrm{~km}$ 、周期约为 0.29 个地球日的极轨椭圆轨道上。引力常量 $G$ 已知, 根据以上信息可估算出下列哪些物理量（ ）\nA: 火卫一的密度\nB: 火星的密度\nC: 火星的半径\nD: 环绕器在近火点与远火点的速度之比\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n火卫一是火星的两颗天然卫星之一, 大小约 $27 \\mathrm{~km} \\times 22 \\mathrm{~km} \\times 18 \\mathrm{~km}$, 运行在距离火星表面约 $6000 \\mathrm{~km}$ 高度的近圆轨道上，周期约为 0.32 个地球日。 2022 年“天问一号”发射两周年之际，国家航天局探月与航天工程中心发布了“天问一号”环绕器拍摄到的火卫一的高清影像图。环绕器运行在近火点距离火星表面高度约 $220 \\mathrm{~km}$ 、远火点距离火星表面高度约 $1.08 \\times 10^{4} \\mathrm{~km}$ 、周期约为 0.29 个地球日的极轨椭圆轨道上。引力常量 $G$ 已知, 根据以上信息可估算出下列哪些物理量（ ）\n\nA: 火卫一的密度\nB: 火星的密度\nC: 火星的半径\nD: 环绕器在近火点与远火点的速度之比\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "B",
            "C",
            "D"
        ],
        "solution": "A. 由于火卫一为绕火星旋转, 其并不是中心天体, 所以火卫一的质量不可求。则 $\\rho=\\frac{m}{V}$ 中质量未知, 故火卫一的密度不可求, 故 A 错误;\n\nC. 火卫一和天问一号绕同一天体运动, 则\n\n$$\n\\begin{gathered}\nk=\\frac{a_{\\text {卫 }}{ }^{3}}{T_{\\text {卫 }}{ }^{2}}=\\frac{a_{\\text {天 }}^{3}}{T_{\\text {天 }}^{2}} \\\\\na_{\\text {卫 }}=6000 \\mathrm{~km}+R_{\\text {火 }}, \\quad a_{\\text {天 }}=\\frac{220 \\mathrm{~km}+1.08 \\times 10^{4} \\mathrm{~km}+2 R_{\\text {火 }}}{2}\n\\end{gathered}\n$$\n\n其中 $T_{\\text {卫和 }} T_{\\text {天 }}$ 已知, 可求得火星的半径 $R_{\\text {火 }}$, 故 C 正确;\n\nB. 由万有引力提供向心力, 对火卫一可得\n\n$$\nG \\frac{M m}{a_{\\text {卫 }}{ }^{2}}=m\\left(\\frac{2 \\pi}{T_{\\text {卫 }}}\\right)^{2} a_{\\text {卫 }}\n$$\n\n火星的体积\n\n$$\nV=\\frac{4}{3} \\pi R_{\\text {火 }}{ }^{3}\n$$\n\n根据密度的公式 $\\rho=\\frac{M}{V}$ 结合上面两式可求得火星的密度, 故 B 正确;\n\nD. 已知近火点和远火点的长度, 根据开普勒第二定律可知求出环绕器在近火点与远火点的速度之比, 故 D 正确。\n\n故选 BCD。",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_1192",
        "problem": "On $21^{\\text {st }}$ August 2017 the continental United States experienced a total solar eclipse. Dubbed the 'Great American Eclipse', it was estimated to be one of the most watched eclipses in history.\n\n## Total Solar Eclipse of 2017 Aug 21\n\n[figure1]\n\nFigure 3: The path of totality for the Great American Eclipse. The narrow dimension is its width. Credit: Fred Espenak, NASA's GSFC.\n\nThe path of totality (where the Moon completely obscures the Sun) is shown in Figure 3, and the point of greatest eclipse (\"GE\"; where the path was widest since the axis of the cone of the Moon's shadow passed closest to the centre of the Earth) was near the village of Cerulean, Kentucky. The following data can be used for this question:\n\n- The angular radii of the Sun and the Moon (if observed from the centre of the Earth) at the moment of GE are $15^{\\prime} 48.7^{\\prime \\prime}$ and $16^{\\prime} 03.4^{\\prime \\prime}$, respectively, where the notation $x x^{\\prime} y y . y^{\\prime \\prime}$ corresponds to $x x$ arcminutes and $y y . y$ arcseconds ( 60 arcminutes $=1$ degree, and 60 arcseconds $=1$ arcminute $)$\n- The latitude and longitude of the location of GE are $36^{\\circ} 58.0^{\\prime} \\mathrm{N}$ and $87^{\\circ} 40.3^{\\prime} \\mathrm{W}$, respectively\n- Take the mean radii of the Sun, Earth and Moon to be respectively $R_{\\odot}=695700 \\mathrm{~km}, R_{\\oplus}=$ $6371 \\mathrm{~km}$ and $R_{\\text {Moon }}=1737 \\mathrm{~km}$, and a day to be 24 hours\n- Take the semi-major axes of the Sun-Earth and Earth-Moon systems to be $149600000 \\mathrm{~km}$ and $384400 \\mathrm{~km}$, respectively\n- As viewed from a location far above the North Pole, the Moon orbits in an anticlockwise direction around the Earth, and the Earth spins in an anticlockwise direction\n\nFor an ellipse with semi-major axis $a$ it can be shown that the velocity $v$, at a distance $r$ from mass $M$, can be written as:\n\n$$\nv^{2}=G M\\left(\\frac{2}{r}-\\frac{1}{a}\\right)\n$$\n\nThe point of greatest eclipse and greatest duration do not generally coincide, as a more elliptical shadow with a major axis aligned with the path of maximum totality (and thinner path width, equal to the minor axis) can compensate for the shadow moving faster at higher latitudes. For this eclipse the point of greatest duration (\"GD\") was at co-ordinates of $37^{\\circ} 35^{\\prime} \\mathrm{N}$ latitude and $89^{\\circ} 07^{\\prime} \\mathrm{W}$ longitude, reached about 4 minutes before GE, and where totality lasted $0.1 \\mathrm{~s}$ longer than the value calculated in part $\\mathrm{c}$.\n\n[figure2]\n\nFigure 4: The route of the Moon's shadow in the vicinity of the points of greatest duration (GD, near Carbondale) and greatest eclipse (GE, near Hopkinsville). Any places between the two limits on the path of totality will experience at least a very short period of totality - outside that region will only be a partial eclipse and the perpendicular distance between them is the path width. The longest duration of totality at that point of the shadow's journey is indicated as the path of maximum totality, which both GE and GD sit on. The closest part of that path to Carbondale is indicated as CP. Credit: Fred Espenak \\& Google Maps.b. Show that at this point the shadow is moving across the Earth's surface at approximately $0.68 \\mathrm{~km} \\mathrm{~s}^{-1}$. You may use the simplifying approximation that for short time intervals it can be considered as travelling with a constant latitude, and that the apparent movement of the Sun due to the Earth's orbit can be neglected.",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nOn $21^{\\text {st }}$ August 2017 the continental United States experienced a total solar eclipse. Dubbed the 'Great American Eclipse', it was estimated to be one of the most watched eclipses in history.\n\n## Total Solar Eclipse of 2017 Aug 21\n\n[figure1]\n\nFigure 3: The path of totality for the Great American Eclipse. The narrow dimension is its width. Credit: Fred Espenak, NASA's GSFC.\n\nThe path of totality (where the Moon completely obscures the Sun) is shown in Figure 3, and the point of greatest eclipse (\"GE\"; where the path was widest since the axis of the cone of the Moon's shadow passed closest to the centre of the Earth) was near the village of Cerulean, Kentucky. The following data can be used for this question:\n\n- The angular radii of the Sun and the Moon (if observed from the centre of the Earth) at the moment of GE are $15^{\\prime} 48.7^{\\prime \\prime}$ and $16^{\\prime} 03.4^{\\prime \\prime}$, respectively, where the notation $x x^{\\prime} y y . y^{\\prime \\prime}$ corresponds to $x x$ arcminutes and $y y . y$ arcseconds ( 60 arcminutes $=1$ degree, and 60 arcseconds $=1$ arcminute $)$\n- The latitude and longitude of the location of GE are $36^{\\circ} 58.0^{\\prime} \\mathrm{N}$ and $87^{\\circ} 40.3^{\\prime} \\mathrm{W}$, respectively\n- Take the mean radii of the Sun, Earth and Moon to be respectively $R_{\\odot}=695700 \\mathrm{~km}, R_{\\oplus}=$ $6371 \\mathrm{~km}$ and $R_{\\text {Moon }}=1737 \\mathrm{~km}$, and a day to be 24 hours\n- Take the semi-major axes of the Sun-Earth and Earth-Moon systems to be $149600000 \\mathrm{~km}$ and $384400 \\mathrm{~km}$, respectively\n- As viewed from a location far above the North Pole, the Moon orbits in an anticlockwise direction around the Earth, and the Earth spins in an anticlockwise direction\n\nFor an ellipse with semi-major axis $a$ it can be shown that the velocity $v$, at a distance $r$ from mass $M$, can be written as:\n\n$$\nv^{2}=G M\\left(\\frac{2}{r}-\\frac{1}{a}\\right)\n$$\n\nThe point of greatest eclipse and greatest duration do not generally coincide, as a more elliptical shadow with a major axis aligned with the path of maximum totality (and thinner path width, equal to the minor axis) can compensate for the shadow moving faster at higher latitudes. For this eclipse the point of greatest duration (\"GD\") was at co-ordinates of $37^{\\circ} 35^{\\prime} \\mathrm{N}$ latitude and $89^{\\circ} 07^{\\prime} \\mathrm{W}$ longitude, reached about 4 minutes before GE, and where totality lasted $0.1 \\mathrm{~s}$ longer than the value calculated in part $\\mathrm{c}$.\n\n[figure2]\n\nFigure 4: The route of the Moon's shadow in the vicinity of the points of greatest duration (GD, near Carbondale) and greatest eclipse (GE, near Hopkinsville). Any places between the two limits on the path of totality will experience at least a very short period of totality - outside that region will only be a partial eclipse and the perpendicular distance between them is the path width. The longest duration of totality at that point of the shadow's journey is indicated as the path of maximum totality, which both GE and GD sit on. The closest part of that path to Carbondale is indicated as CP. Credit: Fred Espenak \\& Google Maps.\n\nproblem:\nb. Show that at this point the shadow is moving across the Earth's surface at approximately $0.68 \\mathrm{~km} \\mathrm{~s}^{-1}$. You may use the simplifying approximation that for short time intervals it can be considered as travelling with a constant latitude, and that the apparent movement of the Sun due to the Earth's orbit can be neglected.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of \\mathrm{~km} \\mathrm{~s}^{-1}, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_2827c35b7a4e24cd73bcg-08.jpg?height=1011&width=1014&top_left_y=497&top_left_x=521",
            "https://cdn.mathpix.com/cropped/2024_03_14_2827c35b7a4e24cd73bcg-09.jpg?height=859&width=1213&top_left_y=924&top_left_x=410"
        ],
        "answer": [
            "0.681"
        ],
        "solution": "At GE, the latitude is $\\theta=36^{\\circ} 58^{\\prime}=36.967^{\\circ}$\n\n$$\n\\begin{aligned}\nv_{\\mathrm{rot}}=\\frac{2 \\pi R_{E} \\cos \\theta}{T} & =\\frac{2 \\pi \\times 6371 \\times \\cos 36.967}{24 \\times 60 \\times 60} \\\\\n& =0.370 \\mathrm{~km} \\mathrm{~s}^{-1} \\\\\nv^{2}=G M\\left(\\frac{2}{r}-\\frac{1}{a}\\right) \\quad & \\therefore v_{\\text {Moon }}=\\sqrt{G M_{E}\\left(\\frac{2}{d_{\\mathrm{Moon}}}-\\frac{1}{a}\\right)} \\\\\n& =\\sqrt{6.67 \\times 10^{-11} \\times 5.97 \\times 10^{24}\\left(\\frac{2}{371900 \\times 10^{3}}-\\frac{1}{384400 \\times 10^{3}}\\right)} \\\\\n& =1.051 \\mathrm{~km} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$\n\nHence: $v_{\\text {rel }}=v_{\\text {Moon }}-v_{\\text {rot }}=1.051-0.370=0.681 \\mathrm{~km} \\mathrm{~s}^{-1}$\n\n[Since this is a 'show that' question, ensure the student has shown evidence that they have calculated the correct relative velocity (rather than simply quoted it) for the final mark]",
        "answer_type": "NV",
        "unit": [
            "\\mathrm{~km} \\mathrm{~s}^{-1}"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_1052",
        "problem": "In the heart of every star, nuclear fusion is taking place. For most stars that involves hydrogen being turned into helium, a process that starts by bringing two protons close enough that the strong nuclear force can act upon them. The smallest stars are the ones that have a core that is only just hot enough for fusion to occur, whilst in the biggest ones the radiation pressure of the photons given out by the fusion reaction pushing on the stellar material can overcome the gravitational forces holding it together.\n[figure1]\n\nFigure 5: Left: The lowest mass star we know of, EBLM J0555-57Ab, was found by von Boetticher et al. (2017) and is about the size of Saturn with a mass of $0.081 M_{\\odot}$. Credit: Amanda Smith, University of Cambridge. Right: The highest mass star we know of, R136a1, is in the centre of the clump of stars on the right of this HST image of the Tarantula Nebula. Schneider et al. (2014) suggest it has a mass of $315 M_{\\odot}$, which is above what stellar evolution models allow. Despite its large mass, other stars have far bigger radii. Credit: NASA \\& ESA.\n\nFor a spherical main sequence star made of a plasma (a fully ionized gas of electrons and nuclei) that is acting like an ideal gas, the temperature at the core can be approximately calculated as\n\n$$\nT_{\\mathrm{int}} \\simeq \\frac{G M \\bar{\\mu}}{k_{\\mathrm{B}} R} \\quad \\text { where } \\quad \\bar{\\mu}=\\frac{m_{\\mathrm{p}}}{2 X+3 Y / 4+Z / 2} .\n$$\n\nIn this equation, $M$ is the mass of the star, $R$ is its radius, $k_{\\mathrm{B}}$ is the Boltzmann constant, and $\\bar{\\mu}$ is the mean mass of the plasma particles (i.e nuclei and electrons) with $m_{\\mathrm{p}}$ the mass of a proton.\n\nClassically, the core of the Sun is not hot enough for fusion, and yet fusion is clearly happening. The key is that it is a fundamentally quantum process, and so protons are able to 'quantum tunnel' through the Coloumb barrier (see Figure 6), allowing fusion to occur at lower temperatures. In quantum mechanics, fusion will happen when $b=\\lambda$ where $\\lambda$ is the de Broglie wavelength of the proton, related to the momentum of the proton by $\\lambda=h / p$.\n\n[figure2]\n\nFigure 6: A diagram showing the way a particle can pass through a classically impenetrable potential barrier due to its wave-like properties on the quantum scale.\n\nCredit: Brooks/Cole - Thomson Learning.\n\nIn the smallest stars, electron degeneracy prevents them from compressing in radius and thus stops the core reaching $T_{\\text {int }} \\gtrsim T_{\\text {quantum }}$. At the limit of electron degeneracy, the number density of electrons $n_{\\mathrm{e}}=1 / \\lambda_{\\mathrm{e}}^{3}$ where $\\lambda_{\\mathrm{e}}$ is the de Broglie wavelength of the electrons.\n\nIn the largest stars, radiation pressure pushes on the outer layers of the star stronger than gravity pulls them in. The brightest luminosity for a star is known as the Eddington luminosity, $L_{\\text {Edd }}$. The acceleration due to radiation pressure can be calculated as\n\n$$\ng_{\\mathrm{rad}}=\\frac{\\kappa_{\\mathrm{e}} I}{c} \\quad \\text { where } \\quad \\kappa_{\\mathrm{e}}=\\frac{\\sigma_{\\mathrm{T}}}{2 m_{\\mathrm{p}}}(1+X)\n$$\n\nand $\\kappa_{\\mathrm{e}}$ is the electron opacity of the stellar material, $\\sigma_{\\mathrm{T}}$ is the Thomson scattering cross-section for electrons $\\left(=66.5 \\mathrm{fm}^{2}\\right.$ ), $X$ is the hydrogen fraction, and $I$ is the intensity of radiation (in $\\mathrm{W} \\mathrm{m}^{-2}$ ). Assuming main-sequence stars follow a mass-luminosity relation of $L \\propto M^{3}$, the maximum mass of a star can be found by considering one that is radiating at $L_{\\text {Edd }}$.e. By balancing the radiative acceleration with the gravitational acceleration at the surface of a star, derive a formula for $L_{\\text {Edd }}$ in terms of $M$, and hence calculate the maximum mass of a star with a hydrogen fraction like the Sun. Give your answer in $M_{\\odot}$.",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is an expression.\nHere is some context information for this question, which might assist you in solving it:\nIn the heart of every star, nuclear fusion is taking place. For most stars that involves hydrogen being turned into helium, a process that starts by bringing two protons close enough that the strong nuclear force can act upon them. The smallest stars are the ones that have a core that is only just hot enough for fusion to occur, whilst in the biggest ones the radiation pressure of the photons given out by the fusion reaction pushing on the stellar material can overcome the gravitational forces holding it together.\n[figure1]\n\nFigure 5: Left: The lowest mass star we know of, EBLM J0555-57Ab, was found by von Boetticher et al. (2017) and is about the size of Saturn with a mass of $0.081 M_{\\odot}$. Credit: Amanda Smith, University of Cambridge. Right: The highest mass star we know of, R136a1, is in the centre of the clump of stars on the right of this HST image of the Tarantula Nebula. Schneider et al. (2014) suggest it has a mass of $315 M_{\\odot}$, which is above what stellar evolution models allow. Despite its large mass, other stars have far bigger radii. Credit: NASA \\& ESA.\n\nFor a spherical main sequence star made of a plasma (a fully ionized gas of electrons and nuclei) that is acting like an ideal gas, the temperature at the core can be approximately calculated as\n\n$$\nT_{\\mathrm{int}} \\simeq \\frac{G M \\bar{\\mu}}{k_{\\mathrm{B}} R} \\quad \\text { where } \\quad \\bar{\\mu}=\\frac{m_{\\mathrm{p}}}{2 X+3 Y / 4+Z / 2} .\n$$\n\nIn this equation, $M$ is the mass of the star, $R$ is its radius, $k_{\\mathrm{B}}$ is the Boltzmann constant, and $\\bar{\\mu}$ is the mean mass of the plasma particles (i.e nuclei and electrons) with $m_{\\mathrm{p}}$ the mass of a proton.\n\nClassically, the core of the Sun is not hot enough for fusion, and yet fusion is clearly happening. The key is that it is a fundamentally quantum process, and so protons are able to 'quantum tunnel' through the Coloumb barrier (see Figure 6), allowing fusion to occur at lower temperatures. In quantum mechanics, fusion will happen when $b=\\lambda$ where $\\lambda$ is the de Broglie wavelength of the proton, related to the momentum of the proton by $\\lambda=h / p$.\n\n[figure2]\n\nFigure 6: A diagram showing the way a particle can pass through a classically impenetrable potential barrier due to its wave-like properties on the quantum scale.\n\nCredit: Brooks/Cole - Thomson Learning.\n\nIn the smallest stars, electron degeneracy prevents them from compressing in radius and thus stops the core reaching $T_{\\text {int }} \\gtrsim T_{\\text {quantum }}$. At the limit of electron degeneracy, the number density of electrons $n_{\\mathrm{e}}=1 / \\lambda_{\\mathrm{e}}^{3}$ where $\\lambda_{\\mathrm{e}}$ is the de Broglie wavelength of the electrons.\n\nIn the largest stars, radiation pressure pushes on the outer layers of the star stronger than gravity pulls them in. The brightest luminosity for a star is known as the Eddington luminosity, $L_{\\text {Edd }}$. The acceleration due to radiation pressure can be calculated as\n\n$$\ng_{\\mathrm{rad}}=\\frac{\\kappa_{\\mathrm{e}} I}{c} \\quad \\text { where } \\quad \\kappa_{\\mathrm{e}}=\\frac{\\sigma_{\\mathrm{T}}}{2 m_{\\mathrm{p}}}(1+X)\n$$\n\nand $\\kappa_{\\mathrm{e}}$ is the electron opacity of the stellar material, $\\sigma_{\\mathrm{T}}$ is the Thomson scattering cross-section for electrons $\\left(=66.5 \\mathrm{fm}^{2}\\right.$ ), $X$ is the hydrogen fraction, and $I$ is the intensity of radiation (in $\\mathrm{W} \\mathrm{m}^{-2}$ ). Assuming main-sequence stars follow a mass-luminosity relation of $L \\propto M^{3}$, the maximum mass of a star can be found by considering one that is radiating at $L_{\\text {Edd }}$.\n\nproblem:\ne. By balancing the radiative acceleration with the gravitational acceleration at the surface of a star, derive a formula for $L_{\\text {Edd }}$ in terms of $M$, and hence calculate the maximum mass of a star with a hydrogen fraction like the Sun. Give your answer in $M_{\\odot}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_ffe0ae050771e0e3decbg-08.jpg?height=712&width=1508&top_left_y=546&top_left_x=271",
            "https://cdn.mathpix.com/cropped/2024_03_14_ffe0ae050771e0e3decbg-09.jpg?height=514&width=1010&top_left_y=186&top_left_x=523"
        ],
        "answer": [
            "195 M_{\\odot}"
        ],
        "solution": "Balancing the radiative acceleration term with the gravitational one,\n\n$$\n\\begin{aligned}\n& g_{\\text {rad }}=g_{\\text {grav }} \\therefore \\frac{\\kappa_{e} I}{c}=\\frac{G M}{R^{2}} \\\\\n& I=\\frac{L_{E d d}}{4 \\pi R^{2}} \\quad \\therefore L_{E d d}=\\frac{4 \\pi G M c}{\\kappa_{e}}\\left(=\\frac{8 \\pi G M c m_{p}}{\\sigma_{T}(1+X)}\\right)\n\\end{aligned}\n$$\n\n[ $L_{\\text {Edd }}$ can be given in terms of $k$ or in terms of $\\sigma_{T}$ for the second mark]\n\n$$\n\\kappa_{e}=\\frac{\\sigma_{T}}{2 m_{p}}(1+X)=\\frac{66.5 \\times 10^{-30}}{2 \\times 1.67 \\times 10^{-27}}(1+0.72)=0.03425 \\mathrm{~m}^{2} \\mathrm{~kg}^{-1}\n$$\n\nUsing the given mass-luminosity relation,\n\n$$\n\\begin{aligned}\n\\frac{L_{E d d}}{L_{\\odot}}=\\left(\\frac{M_{\\max }}{M_{\\odot}}\\right)^{3} \\therefore \\frac{4 \\pi G M_{\\max } c}{L_{\\odot} \\kappa_{e}}=\\frac{M_{\\max }^{3}}{M_{\\odot}^{3}} \\therefore M_{\\max } & =\\sqrt{\\frac{4 \\pi G c M_{\\odot}^{3}}{L_{\\odot} \\kappa_{e}}} \\\\\n& =\\sqrt{\\frac{4 \\pi \\times 6.67 \\times 10^{-11} \\times 3.00 \\times 10^{8} \\times\\left(1.99 \\times 10^{30}\\right)^{3}}{3.85 \\times 10^{26} \\times 0.03425}} \\\\\n& =3.88 \\times 10^{32} \\mathrm{~kg}=195 M_{\\odot}\n\\end{aligned}\n$$\n\n[The answer must be in $\\mathrm{M}_{\\odot}$ for the final mark. Full marks can be awarded without $k$ being calculated explicitly. Watch for incorrect conversions from $\\mathrm{fm}^{2}$ to $^{2}\\left(1 \\mathrm{fm}^{2}=10^{-30} \\mathrm{~m}^{2}\\right)$ with $\\sigma_{\\mathrm{T}}$ ]\n\n[The real upper limit on stars is about $200 \\mathrm{M}_{\\odot}$, so this simplified model has done rather well]",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_285",
        "problem": "2021 年 5 月, “天问一号”探测器成功在火星软着陆, 我国成为世界上第一个首次探测火星就实现“绕、落、巡”三项任务的国家。\n\n 为了简化问题, 可以认为地球和火星在同一平面上绕太阳做匀速圆周运动, 如图 1 所示。已知地球的公转周期为 $T_{1}$, 火星的公转周期为 $T_{2}$ 。\n\n已知地球公转轨道半径为 $r_{1}$, 求火星公转轨道半径 $r_{2}$;\n\n\n[图1]\n\n图1\n\n[图2]\n\n图2",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n\n问题:\n2021 年 5 月, “天问一号”探测器成功在火星软着陆, 我国成为世界上第一个首次探测火星就实现“绕、落、巡”三项任务的国家。\n\n 为了简化问题, 可以认为地球和火星在同一平面上绕太阳做匀速圆周运动, 如图 1 所示。已知地球的公转周期为 $T_{1}$, 火星的公转周期为 $T_{2}$ 。\n\n已知地球公转轨道半径为 $r_{1}$, 求火星公转轨道半径 $r_{2}$;\n\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-035.jpg?height=432&width=508&top_left_y=1937&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-035.jpg?height=417&width=871&top_left_y=1939&top_left_x=844"
        ],
        "answer": [
            "$\\left(\\frac{T_{2}}{T_{1}}\\right)^{\\frac{2}{3}} r_{1}$"
        ],
        "solution": "设太阳质量为 $M_{\\text {太 }}$, 地球质量为 $m_{1}$, 火星质量为 $m_{2}$, 根据万有引力定律结合圆周运动规律，有\n\n$$\nG \\frac{M_{\\text {太 }} m_{1}}{r_{1}^{2}}=m_{1} \\frac{4 \\pi^{2} r_{1}}{T_{1}^{2}}, G \\frac{M_{\\text {太 }} m_{2}}{r_{2}^{2}}=m_{2} \\frac{4 \\pi^{2} r_{2}}{T_{2}^{2}}\n$$\n\n联立可得\n\n$$\nr_{2}=\\left(\\frac{T_{2}}{T_{1}}\\right)^{\\frac{2}{3}} r_{1}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_844",
        "problem": "Let's assume that on an expedition mission to Mars, we take a telescope with 0.01 arcsecond angular resolution from earth. What is the ratio of the number of the stars we can measure the parallax distance to using this telescope on Mars compared to earth? The semimajor axis of Mars is $1.524 \\mathrm{AU}$.\nA: 0.5\nB: 1\nC: 4\nD: 8\n",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nLet's assume that on an expedition mission to Mars, we take a telescope with 0.01 arcsecond angular resolution from earth. What is the ratio of the number of the stars we can measure the parallax distance to using this telescope on Mars compared to earth? The semimajor axis of Mars is $1.524 \\mathrm{AU}$.\n\nA: 0.5\nB: 1\nC: 4\nD: 8\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_7205fccc557018644b5cg-07.jpg?height=216&width=561&top_left_y=968&top_left_x=825"
        ],
        "answer": [
            "D"
        ],
        "solution": "According to the figure below,\n\n[figure1]\n\nfor a given small $\\theta$ we can have: $\\tan (\\theta) \\approx \\theta \\approx \\frac{r}{d} \\rightarrow \\theta=$ const. $=\\frac{r_{e}}{d_{e}}=\\frac{r_{m}}{d_{m}}$\n\n$\\frac{d_{m}}{d_{e}}=\\frac{r_{m}}{r_{e}}$\n\nLet $\\mathrm{N}$ be the number of observed stars:\n\n$N \\propto d^{3} \\rightarrow \\frac{N_{m}}{N_{e}}=\\left(\\frac{d_{m}}{d_{e}}\\right)^{3}=\\left(\\frac{1.5}{1}\\right)^{3} \\approx 4$.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_871",
        "problem": "What is the spectral type of a star with a luminosity of $5.86 * 10^{26} \\mathrm{~W}$ and radius of $8.51 * 10^{8}$ $\\mathrm{m}$ ?\nA: A\nB: F\nC: G\nD: K\nE: M\n",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the spectral type of a star with a luminosity of $5.86 * 10^{26} \\mathrm{~W}$ and radius of $8.51 * 10^{8}$ $\\mathrm{m}$ ?\n\nA: A\nB: F\nC: G\nD: K\nE: M\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": [
            "C"
        ],
        "solution": "Using Stefan-Boltzmann's Law:\n\n$$\n\\begin{gathered}\nL=4 * \\pi * R^{2} * \\sigma * T^{4} \\\\\n5.86 * 10^{26}=4 * \\pi *\\left(8.51 * 10^{8}\\right)^{2} * 5.67 * 10^{-8} * T^{4} \\\\\nT=5805 K\n\\end{gathered}\n$$\n\nThis is extremely close to the temperature of the sun, so the spectral type must be G.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_527",
        "problem": "有 $a 、 b 、 c 、 d$ 四颗地球卫星: $a$ 还未发射, 在地球赤道上随地球表面一起转动; $b$在地球的近地圆轨道上正常运行; $c$ 是地球同步卫星; $d$ 是高空探测卫星。各卫星排列位置如图, 则下列说法正确的是 ( )\n\n[图1]\nA: $a$ 的向心加速度大于 $b$ 的向心加速度\nB: 四颗卫星的速度大小关系是: $v_{a}>v_{b}>v_{c}>v_{d}$\nC: 在相同时间内 $b$ 转过的弧长最长\nD: $d$ 的运动周期可能是 $20 \\mathrm{~h}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n有 $a 、 b 、 c 、 d$ 四颗地球卫星: $a$ 还未发射, 在地球赤道上随地球表面一起转动; $b$在地球的近地圆轨道上正常运行; $c$ 是地球同步卫星; $d$ 是高空探测卫星。各卫星排列位置如图, 则下列说法正确的是 ( )\n\n[图1]\n\nA: $a$ 的向心加速度大于 $b$ 的向心加速度\nB: 四颗卫星的速度大小关系是: $v_{a}>v_{b}>v_{c}>v_{d}$\nC: 在相同时间内 $b$ 转过的弧长最长\nD: $d$ 的运动周期可能是 $20 \\mathrm{~h}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_a9b05ce8eea7b0e40e5eg-020.jpg?height=228&width=853&top_left_y=1665&top_left_x=333"
        ],
        "answer": [
            "C"
        ],
        "solution": "A. 因为 $a 、 c$ 的角速度相同, 根据\n\n$$\na=\\omega^{2} r\n$$\n\n因 $a$ 离地心的距离小于 $c$ 离地心的距离, 所以 $a$ 的向心加速度小于 $c ; b 、 c$ 是围绕地球公转的卫星, 根据万有引力提供向心力\n\n$$\nG \\frac{M m}{r^{2}}=m a\n$$\n解得\n\n$$\na=\\frac{G M}{r^{2}}\n$$\n\n因 $b$ 的轨道半径小于 $c$ 的轨道半径, 所以 $b$ 的向心加速度大于 $c$, 综上分析可知, $a$ 的向心加速度小于 $b$ 的向心加速度, 故 $\\mathrm{A}$ 错误;\n\nB. 因为 $a 、 c$ 的角速度相同, 根据\n\n$$\nv=\\omega r\n$$\n\n知 $a$ 的速度小于 $c ; b 、 c 、 d$ 是围绕地球公转的卫星, 根据万有引力提供向心力\n\n$$\nG \\frac{M m}{r^{2}}=m \\frac{v^{2}}{r}\n$$\n\n解得\n\n$$\nv=\\sqrt{\\frac{G M}{r}}\n$$\n\n因 $b$ 的轨道半径最小, $d$ 的轨道半径最大, 所以 $b$ 的速度大于 $c, c$ 的速度大于 $d$, 则\n\n$$\nv_{b}>v_{c}>v_{d}, \\quad v_{b}>v_{c}>v_{a}\n$$\n\n故 B 错误;\n\nC. 因 $b$ 的线速度最大, 则在相同时间内 $b$ 转过的弧长最长, 故 C 正确;\n\nD. $c 、 d$ 是围绕地球公转的卫星, 根据万有引力提供向心力\n\n$$\nG \\frac{M m}{r^{2}}=m \\frac{4 \\pi^{2}}{T^{2}} r\n$$\n\n解得\n\n$$\nT=2 \\pi \\sqrt{\\frac{r^{3}}{G M}}\n$$\n\n可知因 $d$ 的轨道半径大于 $c$ 的轨道半径, $d$ 的周期大于 $c$, 而 $c$ 的周期是 $24 \\mathrm{~h}$, 则 $d$ 的运动周期大于 $24 \\mathrm{~h}$, 故 D 错误。\n\n故选 C。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_650",
        "problem": "2013 年 6 月 20 日, 航天员王亚平进行了首次太空授课. 如图所示的两组太空实验,下列说法正确的是\n[图1]\nA: 小球呈悬浮状, 说明小球不受重力作用\nB: 小球呈悬浮状, 是因为小球受到的重力太小的缘故\nC: 轻推小球，小球在最低点的速度必须大于某一临界值，才能做完整的圆周运动\nD: 轻推小球, 小球在最低点只要获得速度, 就能做完整的匀速圆周运动\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n2013 年 6 月 20 日, 航天员王亚平进行了首次太空授课. 如图所示的两组太空实验,下列说法正确的是\n[图1]\n\nA: 小球呈悬浮状, 说明小球不受重力作用\nB: 小球呈悬浮状, 是因为小球受到的重力太小的缘故\nC: 轻推小球，小球在最低点的速度必须大于某一临界值，才能做完整的圆周运动\nD: 轻推小球, 小球在最低点只要获得速度, 就能做完整的匀速圆周运动\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_6842b9ceb844a90b34c3g-58.jpg?height=424&width=1442&top_left_y=250&top_left_x=343"
        ],
        "answer": [
            "D"
        ],
        "solution": "$\\mathrm{AB}$. 太空舱中小球呈悬浮状, 是因为完全失重, 而不是不受重力或重力太小,故 $\\mathrm{AB}$ 不符合题意;\n\nCD. 太空舱中小球是因为完全失重轻推小球, 小球在最低点只要获得速度, 就能做完整的匀速圆周运动, 故 C 不符合题意, $\\mathrm{D}$ 合题意;",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_60",
        "problem": "仙王座 VV 是一对双星系统, 分别由一颗红超巨星 VVA (主星) 和一颗蓝矮星 VVB (伴星) 组成, 这是一个食变双星, 两颗星会互相围绕着公共质心 (质量中心, 即质量集中的假想点）公转, 其中 VVA 和 VVB 间的距离为 $25 \\mathrm{AU}$ (1AU 等于日地平均距离),\n它们绕公共质心公转的周期为 20 年, 由于会有物质从 VVA 喷发出去, 最终全部流向 VVB, 导致伴星的质量增大而主星的质量减小, 假定在这个过程中两星间的距离不变,则在 VVA 的质量减小到两星的质量相等的过程中（）\nA: 公共质心离 VVB 越来越远\nB: 两星间的引力将减小\nC: 两星绕公共质心公转的周期大于 20 年\nD: 两星绕公共质心的角速度不变\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n仙王座 VV 是一对双星系统, 分别由一颗红超巨星 VVA (主星) 和一颗蓝矮星 VVB (伴星) 组成, 这是一个食变双星, 两颗星会互相围绕着公共质心 (质量中心, 即质量集中的假想点）公转, 其中 VVA 和 VVB 间的距离为 $25 \\mathrm{AU}$ (1AU 等于日地平均距离),\n它们绕公共质心公转的周期为 20 年, 由于会有物质从 VVA 喷发出去, 最终全部流向 VVB, 导致伴星的质量增大而主星的质量减小, 假定在这个过程中两星间的距离不变,则在 VVA 的质量减小到两星的质量相等的过程中（）\n\nA: 公共质心离 VVB 越来越远\nB: 两星间的引力将减小\nC: 两星绕公共质心公转的周期大于 20 年\nD: 两星绕公共质心的角速度不变\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "D"
        ],
        "solution": "A. 由于 VVA 和 VVB 两星间是由彼此的万有引力来提供向心力, 即两星的向心力大小相等, 且两星做圆周运动的周期、角速度相等, 由\n\n$$\nF_{\\text {向 }}=M \\omega^{2} r\n$$\n\n可知轨道半径与天体的质量成反比, VVB 的质量增大, VVA 的质量减小, 即 VVB 的轨道半径减小, VVA 的轨道半径增大, 公共质心离 VVB 越来越近, A 项错误;\n\nB. 两星的质量之和不变, 且质量之差减小, 由均值不等式知, 当 $M_{\\mathrm{A}}=M_{\\mathrm{B}}$ 时, $M_{\\mathrm{A}} \\cdot M_{\\mathrm{B}}$有最大值，由\n\n$$\nF_{\\text {引引 }}=\\frac{G M_{\\mathrm{A}} M_{\\mathrm{B}}}{L^{2}}\n$$\n\n可知两星间的引力将增大，B 项错误；\n\nCD. 对 VVA 和 VVB 有\n\n$$\n\\begin{gathered}\nF_{\\text {向 }}=\\frac{G M_{\\mathrm{A}} M_{\\mathrm{B}}}{L^{2}}=M_{\\mathrm{A}} \\frac{4 \\pi^{2}}{T^{2}} r_{\\mathrm{A}} \\\\\nF_{\\text {向 }}=\\frac{G M_{\\mathrm{A}} M_{\\mathrm{B}}}{L^{2}}=M_{\\mathrm{B}} \\frac{4 \\pi^{2}}{T^{2}} r_{\\mathrm{B}} \\\\\nr_{\\mathrm{A}}+r_{\\mathrm{B}}=L\n\\end{gathered}\n$$\n\n解得\n\n$$\nT=2 \\pi \\sqrt{\\frac{L^{3}}{G\\left(M_{\\mathrm{A}}+M_{\\mathrm{B}}\\right)}}\n$$\n\n由于 VVA 和 VVB 两星的总质量不变, 所以两星绕公共质心的周期不变, 角速度 $\\omega=\\frac{2 \\pi}{T}$也不变, 故 C 错误, D 正确。\n\n故选 D。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_620",
        "problem": "“嫦娥一号”探月飞行器绕月球做匀速圆周运动, 为保持轨道半径不变, 逐渐消耗所携带的燃料. 若轨道距月球表面的高度为 $h$, 月球质量为 $m$ 、半径为 $r$, 引力常量为 $G$,下列说法正确的是( )\nA: 月球对“嫦娥一号”的万有引力将逐渐减小\nB: “嫦娥一号”绕月球运行的线速度将逐渐减小\nC: “嫦娥一号”绕月球运行的向心加速度为 $\\frac{G m}{(r+h)^{2}}$\nD: “嫦娥一号”绕月球的运行周期为 $2 \\pi \\sqrt{\\frac{r^{3}}{G m}}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n“嫦娥一号”探月飞行器绕月球做匀速圆周运动, 为保持轨道半径不变, 逐渐消耗所携带的燃料. 若轨道距月球表面的高度为 $h$, 月球质量为 $m$ 、半径为 $r$, 引力常量为 $G$,下列说法正确的是( )\n\nA: 月球对“嫦娥一号”的万有引力将逐渐减小\nB: “嫦娥一号”绕月球运行的线速度将逐渐减小\nC: “嫦娥一号”绕月球运行的向心加速度为 $\\frac{G m}{(r+h)^{2}}$\nD: “嫦娥一号”绕月球的运行周期为 $2 \\pi \\sqrt{\\frac{r^{3}}{G m}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "A",
            "C"
        ],
        "solution": "月球对“嫦娥一号”的万物引力\n\n$$\nF=\\frac{G m m^{\\prime}}{(h+r)^{2}}\n$$\n\n“嫦娥一号”燃料持续消耗, $m^{\\prime}$ 将减小, 所以 $F$ 将逐渐减小, A 正确。\n\nB. 万引力提供下心力\n\n$$\n\\frac{G m m^{\\prime}}{(h+r)^{2}}=\\frac{m^{\\prime} v^{2}}{h+r}\n$$\n\n所以\n\n$$\nv=\\sqrt{\\frac{G m}{h+r}}\n$$\n\n$G 、 m 、 h 、 r$ 都不变, 所以 $v$ 不变, B 错误。\n\nC. 万有引力提供向心力\n\n$$\n\\frac{G m m^{\\prime}}{(h+r)^{2}}=m^{\\prime} a\n$$\n\n所以向心加速度 $a=\\frac{G m}{(r+h)^{2}}$, 故 C 正确。\n\nD. 万有引力提供向心力\n\n$$\n\\frac{G m m^{\\prime}}{(h+r)^{2}}=m^{\\prime}\\left(\\frac{2 \\pi}{T}\\right)^{2}(h+r)\n$$\n\n求得 $T=2 \\pi \\sqrt{\\frac{(r+h)^{3}}{G m}}$, 故 D 错误。\n\n故选 $\\mathrm{AC}$ 。",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_947",
        "problem": "One possible theory for why the gas giants have ring systems is that a small moon got too close to the parent planet. When the gravitational tidal forces (due to the difference between the strength of the planet's pull on the near and far sides of the moon) became greater than the gravitational forces holding the moon together, it was ripped apart. This minimum distance is called the \"Roche limit\", named after the French astronomer Edouard Roche who first calculated it. It is defined as when the gravitational force generated by the moon at its surface is equal to the tidal forces it experiences at that distance.\n\n[figure1]\n\nConsider a spherical planet with mass $M$ and radius $R$, and a perfectly rigid spherical moon with mass $m$ and radius $r$, orbiting the planet in a circular orbit of radius $d$. For a small particle of mass $u$ on the surface of the moon, the gravitational and tidal forces it experiences will be\n\n$$\nF_{\\text {grav }}=\\frac{G m u}{r^{2}} \\quad F_{\\text {tidal }}=\\frac{2 G M u r}{d^{3}}\n$$\n\n\nUse your formula to calculate the Roche limit of Saturn for a moon made of water ice $\\left(\\rho_{m}=930 \\mathrm{~kg} \\mathrm{~m}^{-3}\\right)$, given that $M_{\\text {Saturn }}=5.68 \\times 10^{26} \\mathrm{~kg}$ and $R_{\\text {Saturn }}=60270 \\mathrm{~km}$",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nOne possible theory for why the gas giants have ring systems is that a small moon got too close to the parent planet. When the gravitational tidal forces (due to the difference between the strength of the planet's pull on the near and far sides of the moon) became greater than the gravitational forces holding the moon together, it was ripped apart. This minimum distance is called the \"Roche limit\", named after the French astronomer Edouard Roche who first calculated it. It is defined as when the gravitational force generated by the moon at its surface is equal to the tidal forces it experiences at that distance.\n\n[figure1]\n\nConsider a spherical planet with mass $M$ and radius $R$, and a perfectly rigid spherical moon with mass $m$ and radius $r$, orbiting the planet in a circular orbit of radius $d$. For a small particle of mass $u$ on the surface of the moon, the gravitational and tidal forces it experiences will be\n\n$$\nF_{\\text {grav }}=\\frac{G m u}{r^{2}} \\quad F_{\\text {tidal }}=\\frac{2 G M u r}{d^{3}}\n$$\n\n\nUse your formula to calculate the Roche limit of Saturn for a moon made of water ice $\\left(\\rho_{m}=930 \\mathrm{~kg} \\mathrm{~m}^{-3}\\right)$, given that $M_{\\text {Saturn }}=5.68 \\times 10^{26} \\mathrm{~kg}$ and $R_{\\text {Saturn }}=60270 \\mathrm{~km}$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of km, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_6d91a7785df4f4beaa9ag-08.jpg?height=711&width=942&top_left_y=1135&top_left_x=591"
        ],
        "answer": [
            "66300"
        ],
        "solution": "$\\rho_{\\text{Saturn}} = \\frac{M_{\\text{Saturn}}}{\\frac{4}{3} \\pi R^3_{\\text{Saturn}}} = \\frac{5.68 \\times 10^{26}}{\\frac{4}{3} \\pi \\times (6.027 \\times 10^7)^3} = 619 \\text{ kg m}^{-3}$\n\n$d_{RL} = R_{\\text{Saturn}} \\left( \\frac{\\rho_{\\text{Saturn}}}{\\rho_{\\text{ice}}} \\right)^{\\frac{1}{3}} = 60\\,270 \\left( \\frac{619}{930} \\right)^{\\frac{1}{3}} = 66\\,300 \\text{ km} \\quad (= 1.10 R_{\\text{Saturn}})$",
        "answer_type": "NV",
        "unit": [
            "km"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_560",
        "problem": "2017 年 10 月 16 日, 南京紫金山天文台对外发布一项重大发现, 我国南极巡天望远镜追踪探测到首例引力波事件光学信号. 关于引力波, 早在 1916 年爱因斯坦基于广\n义相对论预言了其存在. 1974 年拉塞尔豪尔斯和约瑟夫泰勒发现赫尔斯-泰勒脉冲双星,这双星系统在互相公转时, 由于不断发射引力波而失去能量, 因此逐渐相互靠近, 这现象为引力波的存在提供了首个间接证据. 科学家们猜测该双星系统中体积较小的星体能 “吸食”另一颗体积较大的星球表面的物质, 达到质量转移的目的, 则关于赫尔斯-泰勒脉冲双星周期 $T$ 随双星之间的距离 $L$ 变化关系图正确的是 ( )\n\n[图1]\nA: [图2]\nB: [图3]\nC: [图4]\nD: [图5]\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n2017 年 10 月 16 日, 南京紫金山天文台对外发布一项重大发现, 我国南极巡天望远镜追踪探测到首例引力波事件光学信号. 关于引力波, 早在 1916 年爱因斯坦基于广\n义相对论预言了其存在. 1974 年拉塞尔豪尔斯和约瑟夫泰勒发现赫尔斯-泰勒脉冲双星,这双星系统在互相公转时, 由于不断发射引力波而失去能量, 因此逐渐相互靠近, 这现象为引力波的存在提供了首个间接证据. 科学家们猜测该双星系统中体积较小的星体能 “吸食”另一颗体积较大的星球表面的物质, 达到质量转移的目的, 则关于赫尔斯-泰勒脉冲双星周期 $T$ 随双星之间的距离 $L$ 变化关系图正确的是 ( )\n\n[图1]\n\nA: [图2]\nB: [图3]\nC: [图4]\nD: [图5]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_6842b9ceb844a90b34c3g-79.jpg?height=299&width=531&top_left_y=570&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_04_01_6842b9ceb844a90b34c3g-79.jpg?height=400&width=440&top_left_y=899&top_left_x=454",
            "https://cdn.mathpix.com/cropped/2024_04_01_6842b9ceb844a90b34c3g-79.jpg?height=466&width=448&top_left_y=1332&top_left_x=447",
            "https://cdn.mathpix.com/cropped/2024_04_01_6842b9ceb844a90b34c3g-79.jpg?height=431&width=465&top_left_y=1806&top_left_x=453",
            "https://cdn.mathpix.com/cropped/2024_04_01_6842b9ceb844a90b34c3g-79.jpg?height=420&width=397&top_left_y=2246&top_left_x=453"
        ],
        "answer": [
            "B"
        ],
        "solution": "设脉冲双星的质量, 及轨道半径分别为 $m_{1} 、 m_{2} 、 r_{1} 、 r_{2}$, 间距为 $L=r_{2}+r_{1}$. 由于一对相互作用的万有引力提供向心力且有相同的转动周期, 有 $G \\frac{m_{1} m_{2}}{L^{2}}=m_{1} \\frac{4 \\pi^{2}}{T^{2}} r_{1}=m_{2} \\frac{4 \\pi^{2}}{T^{2}} r_{2}$, 得 $m_{1}=\\frac{4 \\pi^{2} r_{2} L^{2}}{G T^{2}}, m_{2}=\\frac{4 \\pi^{2} r_{1} L^{2}}{G T^{2}}$, 则双星总质量 $\\left(m_{1}+m_{2}\\right)=\\frac{4 \\pi^{2}\\left(r_{1}+r_{2}\\right) L^{2}}{G T^{2}}=\\frac{4 \\pi^{2} L^{3}}{G T^{2}}$, 即 $\\frac{1}{T^{2}}=\\frac{G\\left(m_{1}+m_{2}\\right)}{4 \\pi^{2}} \\cdot \\frac{1}{L^{3}}$, 由于脉冲双星总质量不变, $\\frac{1}{T^{2}}$ 关于 $\\frac{1}{L^{3}}$ 为正比例函数关系; 故选 B.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_429",
        "problem": "2019 年 1 月 3 日, “嫦娥四号”探测器携“玉兔二号”月球车成功着陆在月球背面, 进行科学探测.已知“嫦娥四号”在着陆之前绕月球做圆周运动的半径为 $r_{1}$ 、周期为 $T_{l}$; 月球绕地球做圆周运动的半径为 $r_{2}$ 、周期为 $T_{2}$, 引, 力常量为 $G$. 根据以上条件能得出\nA: 月球与地球的密度\nB: 地球对月球的引力大小\nC: 月球对“嫦娥四号”的引力大小\nD: 关系式 $\\frac{r_{1}^{3}}{T_{1}^{2}}=\\frac{r_{2}^{3}}{T_{2}^{2}}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n2019 年 1 月 3 日, “嫦娥四号”探测器携“玉兔二号”月球车成功着陆在月球背面, 进行科学探测.已知“嫦娥四号”在着陆之前绕月球做圆周运动的半径为 $r_{1}$ 、周期为 $T_{l}$; 月球绕地球做圆周运动的半径为 $r_{2}$ 、周期为 $T_{2}$, 引, 力常量为 $G$. 根据以上条件能得出\n\nA: 月球与地球的密度\nB: 地球对月球的引力大小\nC: 月球对“嫦娥四号”的引力大小\nD: 关系式 $\\frac{r_{1}^{3}}{T_{1}^{2}}=\\frac{r_{2}^{3}}{T_{2}^{2}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "B"
        ],
        "solution": "A. 根据公式:\n\n$$\n\\mathrm{G} \\frac{M m}{r_{2}^{2}}=m \\frac{4 \\pi^{2}}{T_{2}^{2}} r_{2}\n$$\n\n可求出地球质量, 但由于不知道地球半径, 求不出地球体积, 所以算不出地球的密度,同理月球的密度无法求出, $\\mathrm{A}$ 错误;\n\nB. 地球对月球的引力提供月球做圆周运动的向心力:\n\n$$\nF=m_{\\text {月 }} \\frac{4 \\pi^{2}}{T_{2}^{2}} r_{2}\n$$\n\n“嫦娥四号”绕月球做圆周运动, 设“嫦娥四号”探测器的质量为 $m$, 则\n\n$$\nG \\frac{\\mathrm{m}_{\\text {月 }} m}{r_{1}^{2}}=m \\frac{4 \\pi^{2}}{T_{1}^{2}} r_{1}\n$$\n\n求得月球质量:\n\n$$\nm_{\\text {月 }}=\\frac{4 \\pi^{2} r_{1}^{3}}{G T_{1}^{2}}\n$$\n\n所以地球对月球的引力:\n\n$$\nF=m_{\\text {月 }} \\frac{4 \\pi^{2}}{T_{2}^{2}} r_{2}=\\frac{4 \\pi^{2} r_{1}^{3}}{G T_{1}^{2}} \\cdot \\frac{4 \\pi^{2}}{T_{2}^{2}} r_{2}=\\frac{16 \\pi^{4} r_{1}^{3} r_{2}}{G T_{1}^{2} T_{2}^{2}}\n$$\n\nB 正确;\n\nC. “嫦娥四号”绕月球做圆周运动,\n\n$$\nG \\frac{\\mathrm{m}_{\\text {月 }} m}{r_{1}^{2}}=m \\frac{4 \\pi^{2}}{T_{1}^{2}} r_{1}\n$$\n\n可求出月球质量, 不能求出环绕的探测器质量, 无法求月球对“嫦娥四号”的引力大小, C 错误;\n\nD. 开普勒第三定律适用于围绕同一中心天体做圆周运动的卫星, 所以对月球和“嫦娥四号”不适用，D 错误。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_435",
        "problem": "宇宙空间由一种由三颗星体 $A 、 B 、 C$ 组成的三星体系, 它们分别位于等边三角形 $A B C$的三个顶点上, 绕一个固定且共同的圆心 $\\mathrm{O}$ 做匀速圆周运动, 轨道如图中实线所示,\n其轨道半径 $\\mathrm{r}_{\\mathrm{A}}<\\mathrm{r}_{\\mathrm{B}}<\\mathrm{r}_{\\mathrm{C}}$. 忽略其他星体对它们的作用, 可知这三颗星体\n\n[图1]\nA: 线速度大小关系是 $\\mathrm{v}_{\\mathrm{A}}<\\mathrm{v}_{\\mathrm{B}}<\\mathrm{v}_{\\mathrm{C}}$\nB: 加速度大小关系是 $a_{A}>a_{B}>a_{C}$\nC: 质量大小关系是 $\\mathrm{m}_{\\mathrm{A}}>\\mathrm{m}_{\\mathrm{B}}>\\mathrm{m}_{\\mathrm{C}}$\nD: 所受万有引力合力的大小关系是 $\\mathrm{F}_{\\mathrm{A}}>\\mathrm{F}_{\\mathrm{B}}>\\mathrm{F}_{\\mathrm{C}}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n宇宙空间由一种由三颗星体 $A 、 B 、 C$ 组成的三星体系, 它们分别位于等边三角形 $A B C$的三个顶点上, 绕一个固定且共同的圆心 $\\mathrm{O}$ 做匀速圆周运动, 轨道如图中实线所示,\n其轨道半径 $\\mathrm{r}_{\\mathrm{A}}<\\mathrm{r}_{\\mathrm{B}}<\\mathrm{r}_{\\mathrm{C}}$. 忽略其他星体对它们的作用, 可知这三颗星体\n\n[图1]\n\nA: 线速度大小关系是 $\\mathrm{v}_{\\mathrm{A}}<\\mathrm{v}_{\\mathrm{B}}<\\mathrm{v}_{\\mathrm{C}}$\nB: 加速度大小关系是 $a_{A}>a_{B}>a_{C}$\nC: 质量大小关系是 $\\mathrm{m}_{\\mathrm{A}}>\\mathrm{m}_{\\mathrm{B}}>\\mathrm{m}_{\\mathrm{C}}$\nD: 所受万有引力合力的大小关系是 $\\mathrm{F}_{\\mathrm{A}}>\\mathrm{F}_{\\mathrm{B}}>\\mathrm{F}_{\\mathrm{C}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_9938578583ce82f2e878g-44.jpg?height=343&width=331&top_left_y=331&top_left_x=334",
            "https://cdn.mathpix.com/cropped/2024_04_01_9938578583ce82f2e878g-44.jpg?height=374&width=462&top_left_y=1715&top_left_x=320"
        ],
        "answer": [
            "A",
            "C",
            "D"
        ],
        "solution": "三星系统是一种相对稳定的结构, 它们做圆周运动的角速度是相等的, 由 $v=\\omega r$,结合 $r_{A}<r_{B}<r_{C}$. 可知, 线速度大小关系是 $v_{A}<v_{B}<v_{C}$. 故 A 正确; 由 $a=\\omega^{2} r$, 结合 $r_{A}<r_{B}<r_{C}$, 可知加速度大小关系是 $a_{A}<a_{B}<a_{C}$, 故 $\\mathrm{B}$ 错误; 以 $\\mathrm{A}$ 为研究对象, 则受力如图:\n\n[图2]\n\n由于向心力指向圆心, 由矢量关系可知, $\\mathrm{B}$ 对 $\\mathrm{A}$ 的引力大于 $\\mathrm{C}$ 对 $\\mathrm{A}$ 的引力, 结合万有引力定律的表达式: $F=G \\frac{m_{1} m_{2}}{r^{2}}$, 可知 $\\mathrm{B}$ 的质量大于 $\\mathrm{C}$ 的质量. 同理若以 $\\mathrm{C}$ 为研究对象, 可得 $\\mathrm{A}$ 的质量大于 $\\mathrm{B}$ 的质量, 即质量大小关系是 $m_{A}>m_{B}>m_{C}$, 故 $\\mathrm{C}$ 正确. 由于 $m_{A}>m_{B}>m_{C}$, 结合万有引力定律 $F=G \\frac{m_{1} m_{2}}{r^{2}}$ 可知 $\\mathrm{A}$ 与 $\\mathrm{B}$ 之间的引力大于 $\\mathrm{A}$ 与 $\\mathrm{C}$ 之间的引力, 又大于 $\\mathrm{B}$ 与 $\\mathrm{C}$ 之间的引力. 由题可知, $\\mathrm{A} 、 \\mathrm{~B} 、 \\mathrm{C}$ 受到的两个万有引力之间的夹角都是相等的, 根据两个分力的角度一定时, 两个力的大小越大, 合力越大可知 $F_{A}>F_{B}>F_{C}$. 故 D 正确. 故选 ACD.\n【点睛】该题借助于三星模型考查万有引力定律, 解答的过程中向右画出它们的受力的图象，在结合图象中矢量关系，然后和万有引力定律即可正确解答.",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_1058",
        "problem": "The speed of light is considered to be the speed limit of the Universe, however knots of plasma in the jets from active galactic nuclei (AGN) have been observed to be moving with apparent transverse speeds in excess of this, called superluminal speeds. Some of the more extreme examples can be appearing to move at up to 6 times the speed of light (see Figure 7).\n[figure1]\n\nFigure 7: Left: The jet coming from the elliptical galaxy M87 as viewed by the Hubble Space Telescope (HST). Right: Sequence of HST images showing motion at six times the speed of light. The slanting lines track the moving features, and the speeds are given in units of the velocity of light, $c$. Credit: NASA / Space Telescope Science Institute / John Biretta.\n\nThis can be explained by understanding that the jet is offset by an angle $\\theta$ from the sightline to Earth, and that the real speed of the plasma knot, $v$, is less than $c$, and from it we can define the scaled speed $\\beta \\equiv v / c$.\n\nSuperluminal jets are not limited just to AGN, as they have also been observed from systems within our own galaxy. A particularly famous one is the 'microquasar' GRS $1915+105$, which is a low mass X-ray binary consisting of a small star orbiting a black hole. A symmetrical jet with components approaching and receding from us is observed (as expected for jets coming from the poles of the black hole), and the apparent transverse motion of material in those jets has been measured using very high resolution radio imaging. Fender et. al (1999) measure these motions to be $\\mu_{a}=23.6$ mas day $^{-1}$ and $\\mu_{r}=$ 10.0 mas day $^{-1}$ for the approaching and receding jet respectively ( 1 mas $=1$ milliarcsecond, a unit of angle, and there are 3600 arcseconds in a degree) and the distance to the system as $11 \\mathrm{kpc}$.\n\nIn practice, for a given $\\beta_{\\text {app }}$ the values of $\\beta$ and $\\theta$ are degenerate and it is unlikely that the orientation of the jet is such that $\\beta_{\\text {app }}$ has been maximised, so the value in part $\\mathrm{c}$. is just a lower limit. However, since there are two jets then if we assume that they are from the same event (and so equal in speed but opposite in direction) we can break this degeneracy.\n\nSince it is a binary system, we can gain information about the masses of the objects by looking at their period and radial velocity. Formally, the relationship is\n\n$$\n\\frac{\\left(M_{\\mathrm{BH}} \\sin i\\right)^{3}}{\\left(M_{\\mathrm{BH}}+M_{\\star}\\right)^{2}}=\\frac{P_{\\mathrm{orb}} K_{d}^{3}}{2 \\pi G}\n$$\n\nwhere $M_{\\mathrm{BH}}$ is the mass of the black hole, $M_{\\star}$ is the mass of the orbiting star, $i$ is the inclination of the orbit, $P_{\\text {orb }}$ is the orbital period, and $K_{d}$ is the amplitude of the radial velocity curve. Normally the inclination can't be measured, however if we assume that the orbit is perpendicular to the jets then $i=\\theta$ and we can measure the mass of the black hole.f. Greiner et. al (2001) measure $K_{d}=140 \\mathrm{~km} \\mathrm{~s}^{-1}, P_{\\text {orb }}=33.5$ days, and a mass ratio for the two objects of $M_{B H} / M_{*}=12.3$. Using the assumption that $i=\\theta$, calculate $M_{B H}$. Give your answer in $M_{\\odot}$.",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is an expression.\nHere is some context information for this question, which might assist you in solving it:\nThe speed of light is considered to be the speed limit of the Universe, however knots of plasma in the jets from active galactic nuclei (AGN) have been observed to be moving with apparent transverse speeds in excess of this, called superluminal speeds. Some of the more extreme examples can be appearing to move at up to 6 times the speed of light (see Figure 7).\n[figure1]\n\nFigure 7: Left: The jet coming from the elliptical galaxy M87 as viewed by the Hubble Space Telescope (HST). Right: Sequence of HST images showing motion at six times the speed of light. The slanting lines track the moving features, and the speeds are given in units of the velocity of light, $c$. Credit: NASA / Space Telescope Science Institute / John Biretta.\n\nThis can be explained by understanding that the jet is offset by an angle $\\theta$ from the sightline to Earth, and that the real speed of the plasma knot, $v$, is less than $c$, and from it we can define the scaled speed $\\beta \\equiv v / c$.\n\nSuperluminal jets are not limited just to AGN, as they have also been observed from systems within our own galaxy. A particularly famous one is the 'microquasar' GRS $1915+105$, which is a low mass X-ray binary consisting of a small star orbiting a black hole. A symmetrical jet with components approaching and receding from us is observed (as expected for jets coming from the poles of the black hole), and the apparent transverse motion of material in those jets has been measured using very high resolution radio imaging. Fender et. al (1999) measure these motions to be $\\mu_{a}=23.6$ mas day $^{-1}$ and $\\mu_{r}=$ 10.0 mas day $^{-1}$ for the approaching and receding jet respectively ( 1 mas $=1$ milliarcsecond, a unit of angle, and there are 3600 arcseconds in a degree) and the distance to the system as $11 \\mathrm{kpc}$.\n\nIn practice, for a given $\\beta_{\\text {app }}$ the values of $\\beta$ and $\\theta$ are degenerate and it is unlikely that the orientation of the jet is such that $\\beta_{\\text {app }}$ has been maximised, so the value in part $\\mathrm{c}$. is just a lower limit. However, since there are two jets then if we assume that they are from the same event (and so equal in speed but opposite in direction) we can break this degeneracy.\n\nSince it is a binary system, we can gain information about the masses of the objects by looking at their period and radial velocity. Formally, the relationship is\n\n$$\n\\frac{\\left(M_{\\mathrm{BH}} \\sin i\\right)^{3}}{\\left(M_{\\mathrm{BH}}+M_{\\star}\\right)^{2}}=\\frac{P_{\\mathrm{orb}} K_{d}^{3}}{2 \\pi G}\n$$\n\nwhere $M_{\\mathrm{BH}}$ is the mass of the black hole, $M_{\\star}$ is the mass of the orbiting star, $i$ is the inclination of the orbit, $P_{\\text {orb }}$ is the orbital period, and $K_{d}$ is the amplitude of the radial velocity curve. Normally the inclination can't be measured, however if we assume that the orbit is perpendicular to the jets then $i=\\theta$ and we can measure the mass of the black hole.\n\nproblem:\nf. Greiner et. al (2001) measure $K_{d}=140 \\mathrm{~km} \\mathrm{~s}^{-1}, P_{\\text {orb }}=33.5$ days, and a mass ratio for the two objects of $M_{B H} / M_{*}=12.3$. Using the assumption that $i=\\theta$, calculate $M_{B H}$. Give your answer in $M_{\\odot}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_ffe0ae050771e0e3decbg-10.jpg?height=812&width=1458&top_left_y=504&top_left_x=296"
        ],
        "answer": [
            "14.7 M_{\\odot}"
        ],
        "solution": "Rewriting the given equation in terms of the ratio $\\mathrm{M}^{*} / \\mathrm{M}_{\\mathrm{BH}}$,\n\n$$\n\\begin{aligned}\n\\frac{M_{B H}^{3} \\sin ^{3} \\theta}{M_{B H}^{2}\\left(1+\\frac{M_{*}}{M_{B H}}\\right)^{2}}=\\frac{P_{\\text {orb }} K_{d}^{3}}{2 \\pi G} \\quad \\therefore M_{B H} & =\\frac{P_{\\text {orb }} K_{d}^{3}}{2 \\pi G} \\frac{\\left(1+\\frac{M_{*}}{M_{B H}}\\right)^{2}}{\\sin ^{3} \\theta} \\\\\n& =\\frac{(33.5 \\times 24 \\times 3600) \\times\\left(140 \\times 10^{3}\\right)^{3}}{2 \\pi \\times 6.67 \\times 10^{-11}} \\times \\frac{\\left(1+12.3^{-1}\\right)^{2}}{\\sin ^{3} 65.6^{\\circ}} \\\\\n& =2.93 \\times 10^{31} \\mathrm{~kg}=14.7 M_{\\odot}\n\\end{aligned}\n$$\n\n[Must be in $\\mathrm{M}_{\\odot}$ for the final mark]\n\n[At the time of discovery, this meant the black hole in the microquasar was the heaviest stellar black hole known. Subsequent measurements have made the distance a little closer than $11 \\mathrm{kpc}$, and so the mass has been revised down, but until the discovery of the black holes behind detections of gravitational waves (with masses about double this) it was still one of heaviest known]",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_356",
        "problem": "如图所示为一个人造地球卫星沿椭圆轨道绕地球运动的轨迹, 在卫星由近地点 $a$ 运动到远地点 $b$ 的过程中 ( )\n[图1]\nA: 地球引力对卫星不做功\nB: 卫星运行的速率不变\nC: 卫星的重力势能增加\nD: 卫星的动能增加\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图所示为一个人造地球卫星沿椭圆轨道绕地球运动的轨迹, 在卫星由近地点 $a$ 运动到远地点 $b$ 的过程中 ( )\n[图1]\n\nA: 地球引力对卫星不做功\nB: 卫星运行的速率不变\nC: 卫星的重力势能增加\nD: 卫星的动能增加\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/Hkt3gyWv/image.png"
        ],
        "answer": [
            "C"
        ],
        "solution": "A. 卫星由近地点和远地点移动时, 地球引力对卫星做负功; 故 A 错误;\n\n$\\mathrm{BCD}$. 由于卫星高度增加, 重力重力势能增加, 动能减小; 故速率减小; 故 BD 错误, C 正确;\n\n故选 C。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_1061",
        "problem": "Recent years have seen an explosion in the discovery of new exoplanets. About $85 \\%$ of transiting exoplanets discovered by the NASA Kepler telescope have radii less than Neptune ( $\\sim 4 R_{\\oplus}$ ), meaning we are improving our understanding of what the transition between rocky Earth-size planets and gaseous Neptune-size planets looks like.\n\nGiven how common these \"super-Earths\" and \"gas dwarfs\" seem to be, it was odd that we didn't have any in our own Solar System. However, Batygin \\& Brown (2016) suggested that a hypothetical ninth planet (called 'Planet Nine') could explain some of the unusual properties of the orbits of objects in the Kuiper Belt. This planet is inferred to have a mass of $10 M_{\\oplus}$, and so would be an example of a super-Earth.\n\n[figure1]\n\nFigure 5: A plot of planet density versus radius for 33 extrasolar planets (circles) and the planets in our solar system (diamonds).\n\nCredit: Marcy et al. (2014).\n\nAnalysing exoplanets discovered by Kepler, Marcy et al. (2014) used a piecewise function to describe their planetary density data such that:\n\n$$\n\\begin{aligned}\n\\text { For } R_{\\mathrm{P}} \\leq 1.5 R_{\\oplus} & \\rho & =2.32+3.18 \\frac{R_{\\mathrm{P}}}{R_{\\oplus}}\\left[\\mathrm{g} \\mathrm{cm}^{-3}\\right] \\\\\n\\text { For } 1.5 R_{\\oplus}<R_{\\mathrm{P}} \\leq 4.2 R_{\\oplus} & \\frac{M_{\\mathrm{P}}}{M_{\\oplus}} & =2.69\\left(\\frac{R_{\\mathrm{P}}}{R_{\\oplus}}\\right)^{0.93}\n\\end{aligned}\n$$\n\nwhere $R_{\\mathrm{P}}$ is the radius of the planet, $M_{\\mathrm{P}}$ is the mass of the planet, and the model's transition between rocky super-Earth and non-rocky gas dwarf occurs at $R_{\\mathrm{P}}=1.5 R_{\\oplus}$.\n\nThe minimum speed necessary to fully escape a planet's gravity (rather than be put into an elliptical orbit) is called the escape velocity and is calculated as\n\n$$\nv_{\\mathrm{esc}}=\\sqrt{\\frac{2 G M_{\\mathrm{P}}}{R_{\\mathrm{P}}}}\n$$\n\nwhere $G$ is the universal gravitational constant.\n\nIn contrast, the maximum speed a rocket can provide is determined by the ejection velocity, $v_{\\mathrm{e}}$, of the gas used, as determined by the chemical energy stored in the bonds of the fuel used, and the fraction of the rocket that is fuel. Since the rocket gets lighter as it burns its fuel the final speed can be bigger than $v_{\\mathrm{e}}$. The rocket equation is\n\n$$\nv_{\\max }=v_{\\mathrm{e}} \\ln \\frac{m_{0}}{m_{1}}\n$$\n\nwhere $m_{0}$ is the mass at launch and $m_{1}$ is the mass once all the fuel has been burnt. The most energetic chemical reaction we can use in a rocket is hydrogen-oxygen, which gives $v_{\\mathrm{e}}=4.46 \\mathrm{~km} \\mathrm{~s}^{-1}$, and engineering limits us to a rocket design with a maximum of $96 \\%$ of launch mass being fuel (as was used with the solid rockets that launched the space shuttle).c. Although In practice the transition between rocky and gaseous planets is not likely to be sharply at $1.5 R_{E}$, with some planets of both types existing above and below the limit. Calculate $R_{P}$ (in units of $\\left.R_{E}\\right)$ for Planet Nine if it was a rocky super-Earth. How does its average density compare to an Earth-sized rocky exoplanet?",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nRecent years have seen an explosion in the discovery of new exoplanets. About $85 \\%$ of transiting exoplanets discovered by the NASA Kepler telescope have radii less than Neptune ( $\\sim 4 R_{\\oplus}$ ), meaning we are improving our understanding of what the transition between rocky Earth-size planets and gaseous Neptune-size planets looks like.\n\nGiven how common these \"super-Earths\" and \"gas dwarfs\" seem to be, it was odd that we didn't have any in our own Solar System. However, Batygin \\& Brown (2016) suggested that a hypothetical ninth planet (called 'Planet Nine') could explain some of the unusual properties of the orbits of objects in the Kuiper Belt. This planet is inferred to have a mass of $10 M_{\\oplus}$, and so would be an example of a super-Earth.\n\n[figure1]\n\nFigure 5: A plot of planet density versus radius for 33 extrasolar planets (circles) and the planets in our solar system (diamonds).\n\nCredit: Marcy et al. (2014).\n\nAnalysing exoplanets discovered by Kepler, Marcy et al. (2014) used a piecewise function to describe their planetary density data such that:\n\n$$\n\\begin{aligned}\n\\text { For } R_{\\mathrm{P}} \\leq 1.5 R_{\\oplus} & \\rho & =2.32+3.18 \\frac{R_{\\mathrm{P}}}{R_{\\oplus}}\\left[\\mathrm{g} \\mathrm{cm}^{-3}\\right] \\\\\n\\text { For } 1.5 R_{\\oplus}<R_{\\mathrm{P}} \\leq 4.2 R_{\\oplus} & \\frac{M_{\\mathrm{P}}}{M_{\\oplus}} & =2.69\\left(\\frac{R_{\\mathrm{P}}}{R_{\\oplus}}\\right)^{0.93}\n\\end{aligned}\n$$\n\nwhere $R_{\\mathrm{P}}$ is the radius of the planet, $M_{\\mathrm{P}}$ is the mass of the planet, and the model's transition between rocky super-Earth and non-rocky gas dwarf occurs at $R_{\\mathrm{P}}=1.5 R_{\\oplus}$.\n\nThe minimum speed necessary to fully escape a planet's gravity (rather than be put into an elliptical orbit) is called the escape velocity and is calculated as\n\n$$\nv_{\\mathrm{esc}}=\\sqrt{\\frac{2 G M_{\\mathrm{P}}}{R_{\\mathrm{P}}}}\n$$\n\nwhere $G$ is the universal gravitational constant.\n\nIn contrast, the maximum speed a rocket can provide is determined by the ejection velocity, $v_{\\mathrm{e}}$, of the gas used, as determined by the chemical energy stored in the bonds of the fuel used, and the fraction of the rocket that is fuel. Since the rocket gets lighter as it burns its fuel the final speed can be bigger than $v_{\\mathrm{e}}$. The rocket equation is\n\n$$\nv_{\\max }=v_{\\mathrm{e}} \\ln \\frac{m_{0}}{m_{1}}\n$$\n\nwhere $m_{0}$ is the mass at launch and $m_{1}$ is the mass once all the fuel has been burnt. The most energetic chemical reaction we can use in a rocket is hydrogen-oxygen, which gives $v_{\\mathrm{e}}=4.46 \\mathrm{~km} \\mathrm{~s}^{-1}$, and engineering limits us to a rocket design with a maximum of $96 \\%$ of launch mass being fuel (as was used with the solid rockets that launched the space shuttle).\n\nproblem:\nc. Although In practice the transition between rocky and gaseous planets is not likely to be sharply at $1.5 R_{E}$, with some planets of both types existing above and below the limit. Calculate $R_{P}$ (in units of $\\left.R_{E}\\right)$ for Planet Nine if it was a rocky super-Earth. How does its average density compare to an Earth-sized rocky exoplanet?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of \\rho_{\\text {Earth }}, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_2827c35b7a4e24cd73bcg-10.jpg?height=691&width=922&top_left_y=797&top_left_x=567"
        ],
        "answer": [
            "1.51"
        ],
        "solution": "$$\n\\begin{aligned}\n& \\rho=1000\\left[2.32+3.18\\left(\\frac{R_{P}}{R_{E}}\\right)\\right]=\\frac{M_{P}}{\\frac{4}{3} \\pi R_{P}^{3}} \\\\\n& \\therefore \\frac{3180}{R_{E}} R_{P}^{4}+2320 R_{P}^{3}-\\frac{M_{P}}{\\frac{4}{3} \\pi}=0\n\\end{aligned}\n$$\n\nOnly positive real root is $R_{P}=1.20 \\times 10^{7} \\mathrm{~m}=1.88 R_{E}$\n\n$$\n\\begin{aligned}\n& \\rho_{\\text {Earth }}=2.32+3.18=5.50 \\mathrm{~g} \\mathrm{~cm}^{-3} \\\\\n& \\rho_{\\text {av }}=2.32+(3.18 \\times 1.88)=8.30 \\mathrm{~g} \\mathrm{~cm}^{-3}=1.51 \\rho_{\\text {Earth }}\n\\end{aligned}\n$$\n\n[Allow any valid method to solve the quartic, such as using their graphical calculator to plot the graph and find roots, or even a trial and error iteration. If when constructing the quartic they forget to convert the densities into $\\mathrm{kg} \\mathrm{m}^{-3}$ (leading to $R_{P}=7.20 \\times 10^{7} \\mathrm{~m}=11.3 R_{E}$ ) then only lose 1 mark]",
        "answer_type": "NV",
        "unit": [
            "\\rho_{\\text {Earth }}"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_887",
        "problem": "The following series of photos were taken on March 17 around noon time (each photo taken about 3 minutes apart from the previous one). In each photo you can see a white vertical pole (about 1 foot high), and a piece of paper at the base of the pole. Notice how the shadow cast by the pole on the piece of paper is slowly disappearing. Based on the date and time, and the length of the pole's shadow, where were the photos taken?\n[figure1]\n\n(photo for problem 12. Copyright: Ana-Roxana Pop. Ask for author's permission before using.)\nA: Seattle, USA $\\left(47.61^{\\circ} \\mathrm{N}, 122.33^{\\circ} \\mathrm{W}\\right)$\nB: Los Angeles, USA $\\left(34.05^{\\circ} \\mathrm{N}, 118.24^{\\circ} \\mathrm{W}\\right)$\nC: Cancun, Mexico $\\left(21.16^{\\circ} \\mathrm{N}, 86.85^{\\circ} \\mathrm{W}\\right)$\nD: Galapagos Islands, Ecuador $\\left(0.74^{\\circ} \\mathrm{S}, 90.32^{\\circ} \\mathrm{W}\\right.$ )\nE: Sydney, Australia $\\left(33.87^{\\circ} \\mathrm{S}, 151.21^{\\circ} \\mathrm{E}\\right)$\n",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following series of photos were taken on March 17 around noon time (each photo taken about 3 minutes apart from the previous one). In each photo you can see a white vertical pole (about 1 foot high), and a piece of paper at the base of the pole. Notice how the shadow cast by the pole on the piece of paper is slowly disappearing. Based on the date and time, and the length of the pole's shadow, where were the photos taken?\n[figure1]\n\n(photo for problem 12. Copyright: Ana-Roxana Pop. Ask for author's permission before using.)\n\nA: Seattle, USA $\\left(47.61^{\\circ} \\mathrm{N}, 122.33^{\\circ} \\mathrm{W}\\right)$\nB: Los Angeles, USA $\\left(34.05^{\\circ} \\mathrm{N}, 118.24^{\\circ} \\mathrm{W}\\right)$\nC: Cancun, Mexico $\\left(21.16^{\\circ} \\mathrm{N}, 86.85^{\\circ} \\mathrm{W}\\right)$\nD: Galapagos Islands, Ecuador $\\left(0.74^{\\circ} \\mathrm{S}, 90.32^{\\circ} \\mathrm{W}\\right.$ )\nE: Sydney, Australia $\\left(33.87^{\\circ} \\mathrm{S}, 151.21^{\\circ} \\mathrm{E}\\right)$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_56d1b5239b3c83be7aceg-08.jpg?height=919&width=1647&top_left_y=744&top_left_x=239"
        ],
        "answer": [
            "D"
        ],
        "solution": "on the Vernal Equinox (March 21), the plane of Earth's equator passes through the center of the Sun. On this day, at a point on the equatorial line, the Sun is exactly overhead. Therefore, since the photos were taken close to March 21st, they should have been taken somewhere near the Equator, i.e. in the Galapagos Islands.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_698",
        "problem": "如图所示, $O_{1}$ 是一个半径为 $2 R$, 质量为 $M$ 的密度均匀球体的球心, 现在其内以 $O_{2}$为球心挖去一个半径为 $R$ 的球, 并在空心球内某点 $P$ 放置一个质量为 $m$ 的质点。若已知质量分布均匀的薄球壳对壳内物体的引力为零, 则 $O_{1}$ 球剩余部分对该质点的万有引力\n\n[图1]\nA: 方向由 $P$ 点指向 $O_{1} O_{2}$ 连线上某点\nB: 方向与 $O_{1} O_{2}$ 连线平行\nC: 大小为 $\\frac{G M m}{8 R^{2}}$\nD: 大小为 $\\frac{G M m}{2 R^{2}}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n如图所示, $O_{1}$ 是一个半径为 $2 R$, 质量为 $M$ 的密度均匀球体的球心, 现在其内以 $O_{2}$为球心挖去一个半径为 $R$ 的球, 并在空心球内某点 $P$ 放置一个质量为 $m$ 的质点。若已知质量分布均匀的薄球壳对壳内物体的引力为零, 则 $O_{1}$ 球剩余部分对该质点的万有引力\n\n[图1]\n\nA: 方向由 $P$ 点指向 $O_{1} O_{2}$ 连线上某点\nB: 方向与 $O_{1} O_{2}$ 连线平行\nC: 大小为 $\\frac{G M m}{8 R^{2}}$\nD: 大小为 $\\frac{G M m}{2 R^{2}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_86694e5d1e9acbe7af1ag-057.jpg?height=451&width=463&top_left_y=1936&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_04_01_86694e5d1e9acbe7af1ag-058.jpg?height=568&width=583&top_left_y=1658&top_left_x=334"
        ],
        "answer": [
            "B",
            "C"
        ],
        "solution": "$\\mathrm{AB}$. 设球的密度为 $\\rho$, 则实心大球质量可表示为\n\n$$\nM=\\rho \\cdot \\frac{4}{3} \\pi(2 R)^{3}\n$$\n\n由题意可知质量分布均匀的薄球壳对壳内物体的引力为零, 故实心大球对 $P$ 点的引力等于以 $O_{1} P$ 为半径的实心小球对质点 $P$ 的引力。以 $O_{1} P$ 为半径的实心小球的质量为\n\n$$\nM_{1}=\\rho \\cdot \\frac{4}{3} \\pi \\cdot O_{1} P^{3}\n$$\n\n故以 $O_{1} P$ 为半径的实心小球对 $P$ 点的引力为\n\n$$\nF_{1}=G \\frac{M_{1} m}{O_{1} P^{2}}=G \\rho \\cdot \\frac{4}{3} \\pi m \\cdot O_{1} P\n$$\n\n同理, 以 $O_{2} P$ 为半径的实心小球的质量为\n\n$$\nM_{2}=\\rho \\cdot \\frac{4}{3} \\pi \\cdot O_{2} P^{3}\n$$\n\n则以 $O_{2} P$ 为半径的实心小球对 $P$ 点的引力为\n\n$$\nF_{2}=G \\frac{M_{2} m}{O_{2} P^{2}}=G \\rho \\cdot \\frac{4}{3} \\pi m \\cdot O_{2} P\n$$\n\n受力情况如图\n\n[图2]\n\n由几何关系可知\n\n$$\n\\frac{F_{1}}{O_{1} P}=\\frac{F_{2}}{O_{2} P}\n$$\n\n故\n\n$$\nF_{\\text {合 }} \\| O_{1} O_{2}\n$$\n\n$\\mathrm{A}$ 错误, B 正确;\n\nCD. 由上述分析可知\n\n$$\n\\frac{F_{1}}{O_{1} P}=\\frac{F_{\\text {合 }}}{O_{2} O_{1}}\n$$\n\n解得\n\n$$\nF_{\\text {合 }}=\\frac{G M m}{8 R^{2}}\n$$\n\nC 正确, D 错误。\n\n故选 BC。",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_436",
        "problem": "如图所示的卫星轨道, 其中圆 $a$ 是地球同步轨道, 现在有 $\\mathrm{A} 、 \\mathrm{~B}$ 两颗卫星分别位于 $a 、 b$ 轨道运行, 且卫星 $\\mathrm{A}$ 的运行方向与地球自转方向相反, 已知 $\\mathrm{A} 、 \\mathrm{~B}$ 的运行周期分别为 $T_{1} 、 T_{2}$, 地球自转周期为 $T_{0}, P$ 为轨道的近地点, 则有 $(\\quad)$\n\n[图1]\nA: 卫星 $\\mathrm{A}$ 是地球同步卫星\nB: 卫星 $\\mathrm{B}$ 在 $P$ 点时速度最大\nC: $T_{1}>T_{2}$\nD: $T_{0}=T_{l}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n如图所示的卫星轨道, 其中圆 $a$ 是地球同步轨道, 现在有 $\\mathrm{A} 、 \\mathrm{~B}$ 两颗卫星分别位于 $a 、 b$ 轨道运行, 且卫星 $\\mathrm{A}$ 的运行方向与地球自转方向相反, 已知 $\\mathrm{A} 、 \\mathrm{~B}$ 的运行周期分别为 $T_{1} 、 T_{2}$, 地球自转周期为 $T_{0}, P$ 为轨道的近地点, 则有 $(\\quad)$\n\n[图1]\n\nA: 卫星 $\\mathrm{A}$ 是地球同步卫星\nB: 卫星 $\\mathrm{B}$ 在 $P$ 点时速度最大\nC: $T_{1}>T_{2}$\nD: $T_{0}=T_{l}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_cbd0a496f6e2fb8d7781g-029.jpg?height=368&width=388&top_left_y=1758&top_left_x=340"
        ],
        "answer": [
            "B",
            "C"
        ],
        "solution": "A. 圆 $a$ 是地球同步轨道, 但卫星 A 的运行方向与地球自转方向相反, 所以不是地球同步卫星, 选项 $\\mathrm{A}$ 错误;\n\nB. 卫星 B 在椭圆轨道上, 根据开普勒第二定律可知在近地点 $P$ 速度最大, 选项 $\\mathrm{B}$ 正确 C. 根据万有引力提供向心力可得\n\n$$\nG \\frac{M m}{r^{2}}=m \\frac{4 \\pi^{2}}{T^{2}} r\n$$\n\n得出周期\n\n$$\nT=\\sqrt{\\frac{4 \\pi^{2} r^{3}}{G M}}\n$$\n\n因为 $r_{A}>r_{B}$, 所以 $\\mathrm{A} 、 \\mathrm{~B}$ 的运行周期 $T_{1}>T_{2}$, 选项 C 正确;\n\nD. 卫星 A 与地球同步卫星的半径相同, 则周期相同, 而地球同步卫星的周期与地球自转周期为 $T_{0}$, 所以 $T_{0}=T_{1}$, 选项 D 正确。\n\n故选 BCD。",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_461",
        "problem": "中国科幻电影《流浪地球》讲述了地球逃离太阳系的故事, 假设人们在逃离过程中发现一种三星组成的孤立系统, 三星的质量相等、半径均为 $R$, 稳定分布在等边三角形的三个顶点上, 三角形的边长为 $d$, 三星绕 $O$ 点做周期为 $T$ 的匀速圆周运动。已知万有引力常量为 $G$, 忽略星体的自转, 下列说法正确的是 ( )\n\n[图1]\nA: 匀速圆周运动的半径为 $\\frac{\\sqrt{3}}{2} d$\nB: 每个星球的质量为 $\\frac{4 \\pi^{2} d^{3}}{3 G T^{2}}$\nC: 每个星球表面的重力加速度大小为 $\\frac{\\pi^{2} d}{T^{2}}$\nD: 每个星球的第一宇宙速度大小为 $\\frac{2 \\pi d}{T} \\sqrt{\\frac{d}{3 R}}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n中国科幻电影《流浪地球》讲述了地球逃离太阳系的故事, 假设人们在逃离过程中发现一种三星组成的孤立系统, 三星的质量相等、半径均为 $R$, 稳定分布在等边三角形的三个顶点上, 三角形的边长为 $d$, 三星绕 $O$ 点做周期为 $T$ 的匀速圆周运动。已知万有引力常量为 $G$, 忽略星体的自转, 下列说法正确的是 ( )\n\n[图1]\n\nA: 匀速圆周运动的半径为 $\\frac{\\sqrt{3}}{2} d$\nB: 每个星球的质量为 $\\frac{4 \\pi^{2} d^{3}}{3 G T^{2}}$\nC: 每个星球表面的重力加速度大小为 $\\frac{\\pi^{2} d}{T^{2}}$\nD: 每个星球的第一宇宙速度大小为 $\\frac{2 \\pi d}{T} \\sqrt{\\frac{d}{3 R}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_cbd0a496f6e2fb8d7781g-003.jpg?height=283&width=279&top_left_y=772&top_left_x=343",
            "https://cdn.mathpix.com/cropped/2024_04_01_cbd0a496f6e2fb8d7781g-003.jpg?height=320&width=325&top_left_y=2053&top_left_x=340"
        ],
        "answer": [
            "B",
            "D"
        ],
        "solution": "A. 如图所示, 三星均围绕边长为 $d$ 的等边三角形的中心 $O$ 做匀速圆周运动,由几何关系, 匀速圆周运动的半径\n\n$$\nr=\\frac{d}{2 \\cos 30^{\\circ}}=\\frac{\\sqrt{3}}{3} d\n$$\n\n故 A 错误;\n\nB. 设星球的质量为 $M$, 星球间的万有引力 $F_{\\text {万 }}=\\frac{G M^{2}}{d^{2}}$, 如下图所示\n\n[图2]\n\n星球做匀速圆周运动的向心力\n\n$$\nF_{\\mathrm{n}}=2 F_{\\text {万 }} \\cos 30^{\\circ}\n$$\n\n由牛顿第二定律\n\n$$\nF_{\\mathrm{n}}=M\\left(\\frac{2 \\pi}{T}\\right)^{2} r\n$$\n\n综合可得\n\n$$\nM=\\frac{4 \\pi^{2} d^{3}}{3 G T^{2}}\n$$\n\n故 B 正确;\n\nC. 星球表面重力近似等于万有引力\n\n$$\nm g=\\frac{G M m}{R^{2}}\n$$\n\n结合\n\n$$\nM=\\frac{4 \\pi^{2} d^{3}}{3 G T^{2}}\n$$\n\n可得星球表面的重力加速度\n\n$$\ng=\\frac{4 \\pi^{2} d^{3}}{3 R^{2} T^{2}}\n$$\n\n故 C 错误;\n\nD. 由\n\n$$\n\\frac{G M m}{R^{2}}=\\frac{m v^{2}}{R}\n$$\n\n结合\n\n$$\nM=\\frac{4 \\pi^{2} d^{3}}{3 G T^{2}}\n$$\n\n可得星球的第一宇宙速度\n\n$$\nv=\\frac{2 \\pi d}{T} \\sqrt{\\frac{d}{3 R}}\n$$\n\n故 D 正确。\n\n故选 BD。",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_1215",
        "problem": "In order to move a spacecraft between orbits we must apply a thrust using rockets, which changes the velocity of the spacecraft by $\\Delta v$. In this question we will ignore changes in the mass of the spacecraft due to the burning of fuel.\n\nFor an object of mass $m$ in a circular orbit of radius $r$ around an object with mass $M$ (where $m \\ll M$ ) the orbital velocity, $v_{\\text {orb }}$, is given by the formula $v_{\\text {orb }}=\\sqrt{\\frac{G M}As part of their plan to rule the galaxy the First Order has created the Starkiller Base. Built within an ice planet and with a superweapon capable of destroying entire star systems, it is charged using the power of stars. The Starkiller Base has moved into the solar system and seeks to use the Sun to power its weapon to destroy the Earth.\n\n[figure1]\n\nFigure 3: The Starkiller Base charging its superweapon by draining energy from the local star. Credit: Star Wars: The Force Awakens, Lucasfilm.\n\nFor this question you will need that the gravitational binding energy, $U$, of a uniform density spherical object with mass $M$ and radius $R$ is given by\n\n$$\nU=\\frac{3 G M^{2}}{5 R}\n$$\n\nand that the mass-luminosity relation of low-mass main sequence stars is given by $L \\propto M^{4}$.{r}}$.f. Assume that at the moment of destruction of the Starkiller Base the mass of the new star formed is equal to the mass drained from the Sun $\\left(0.10 M_{\\odot}\\right)$. Derive an expression for the main sequence lifetime in terms of stellar mass, and hence calculate the main sequence lifetime of this new star.",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nIn order to move a spacecraft between orbits we must apply a thrust using rockets, which changes the velocity of the spacecraft by $\\Delta v$. In this question we will ignore changes in the mass of the spacecraft due to the burning of fuel.\n\nFor an object of mass $m$ in a circular orbit of radius $r$ around an object with mass $M$ (where $m \\ll M$ ) the orbital velocity, $v_{\\text {orb }}$, is given by the formula $v_{\\text {orb }}=\\sqrt{\\frac{G M}As part of their plan to rule the galaxy the First Order has created the Starkiller Base. Built within an ice planet and with a superweapon capable of destroying entire star systems, it is charged using the power of stars. The Starkiller Base has moved into the solar system and seeks to use the Sun to power its weapon to destroy the Earth.\n\n[figure1]\n\nFigure 3: The Starkiller Base charging its superweapon by draining energy from the local star. Credit: Star Wars: The Force Awakens, Lucasfilm.\n\nFor this question you will need that the gravitational binding energy, $U$, of a uniform density spherical object with mass $M$ and radius $R$ is given by\n\n$$\nU=\\frac{3 G M^{2}}{5 R}\n$$\n\nand that the mass-luminosity relation of low-mass main sequence stars is given by $L \\propto M^{4}$.{r}}$.\n\nproblem:\nf. Assume that at the moment of destruction of the Starkiller Base the mass of the new star formed is equal to the mass drained from the Sun $\\left(0.10 M_{\\odot}\\right)$. Derive an expression for the main sequence lifetime in terms of stellar mass, and hence calculate the main sequence lifetime of this new star.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of years, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_204b2e236273ea30e8d2g-06.jpg?height=611&width=1448&top_left_y=505&top_left_x=310"
        ],
        "answer": [
            "1 \\times 10^{13}"
        ],
        "solution": "We can combine the mass-luminosity relation with the expression we used in part a.\n\n$$\n\\begin{aligned}\n& t_{\\mathrm{MS}} \\propto \\frac{M}{L} \\quad \\text { but } L \\propto M^{4} \\quad \\therefore \\quad t_{\\mathrm{MS}} \\propto \\frac{M}{M^{4}} \\propto M^{-3} \\\\\n& \\therefore t_{\\mathrm{MS}, \\text { new }}=\\left(\\frac{M_{\\text {new }}}{\\mathrm{M}_{\\odot}}\\right)^{-3} t_{\\mathrm{MS}, \\odot}=0.1^{-3} \\times 10^{10} \\text { years } \\\\\n& =10^{13} \\text { years }\n\\end{aligned}\n$$",
        "answer_type": "NV",
        "unit": [
            "years"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_960",
        "problem": "Main sequence stars fuse hydrogen atoms to form helium in their cores. About $90 \\%$ of the stars in the Universe, including the Sun, are main sequence stars. These stars can range from about a tenth of the mass of the Sun to up to 200 times as massive. The main source of energy in main sequence stars is from nuclear fusion. The mass of one hydrogen nucleus is $m_{\\mathrm{H}}=1.674 \\times 10^{-27} \\mathrm{~kg}$, and the mass of one helium nucleus is $m_{\\mathrm{He}}=6.649 \\times 10^{-27} \\mathrm{~kg}$.\n[figure1]\n\nFigure 3: Left: The proton-proton chain reaction is one of two known sets of nuclear fusion reactions by which stars convert hydrogen to helium. It dominates in stars with masses less than or equal to that of the Sun's, and involves a net change of four hydrogen nuclei becoming one helium nucleus.\n\nRight: Only the core of a main sequence star will undergo nuclear fusion due to the higher temperature than the surrounding hydrogen shell.\nCalculate the number of hydrogen nuclei that must fuse to form helium every second to provide the measured solar luminosity, and the percentage of hydrogen mass per reaction that is converted into energy via fusion.",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nMain sequence stars fuse hydrogen atoms to form helium in their cores. About $90 \\%$ of the stars in the Universe, including the Sun, are main sequence stars. These stars can range from about a tenth of the mass of the Sun to up to 200 times as massive. The main source of energy in main sequence stars is from nuclear fusion. The mass of one hydrogen nucleus is $m_{\\mathrm{H}}=1.674 \\times 10^{-27} \\mathrm{~kg}$, and the mass of one helium nucleus is $m_{\\mathrm{He}}=6.649 \\times 10^{-27} \\mathrm{~kg}$.\n[figure1]\n\nFigure 3: Left: The proton-proton chain reaction is one of two known sets of nuclear fusion reactions by which stars convert hydrogen to helium. It dominates in stars with masses less than or equal to that of the Sun's, and involves a net change of four hydrogen nuclei becoming one helium nucleus.\n\nRight: Only the core of a main sequence star will undergo nuclear fusion due to the higher temperature than the surrounding hydrogen shell.\nCalculate the number of hydrogen nuclei that must fuse to form helium every second to provide the measured solar luminosity, and the percentage of hydrogen mass per reaction that is converted into energy via fusion.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %(percentage), but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_13148c5721a741e30941g-08.jpg?height=606&width=1400&top_left_y=785&top_left_x=356"
        ],
        "answer": [
            "0.702"
        ],
        "solution": "First, we calculate the number of reactions per second, $\\mathrm{N}$ : <br> $\\qquad N=\\frac{L_{\\odot}}{E_{\\mathrm{p}-\\mathrm{p}}}=\\frac{3.85 \\times 10^{26}}{4.23 \\times 10^{-12}}=9.10 \\times 10^{37}$ <br> Since there are four hydrogen nuclei used per reaction, the total number <br> of hydrogen nuclei fusing per second is $3.64 \\times 10^{38}$ <br> The percentage of hydrogen mass converted into energy is simply the <br> change in mass in the p-p chain divided by the total mass of hydrogen <br> used in the p-p chain: <br> $\\qquad \\frac{\\Delta m}{4 m_{\\mathrm{H}}}=\\frac{4.7 \\times 10^{-29}}{4 \\times 1.674 \\times 10^{-27}}=0.702 \\%",
        "answer_type": "NV",
        "unit": [
            "%(percentage)"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_730",
        "problem": "如图所示, $A$ 为静止于地球赤道上的物体, $B$ 为绕地球沿粗圆轨道运行的卫星, $C$为绕地球做圆周运动的卫星, $P$ 为 $B 、 C$ 两卫星轨道的交点. 已知 $A 、 B 、 C$ 绕地心运动的周期相同，下列说法中正确的是\n\n[图1]\nA: 物体 $A$ 的速度小于第一宇宙速度\nB: 物体 $A$ 的速度小于卫星 $C$ 的运行速度\nC: 物体 $A$ 和卫星 $C$ 具有相同大小的加速度\nD: 卫星 $B$ 在 $P$ 点的加速度与卫星 $C$ 在 $P$ 点的加速度大小不相等\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n如图所示, $A$ 为静止于地球赤道上的物体, $B$ 为绕地球沿粗圆轨道运行的卫星, $C$为绕地球做圆周运动的卫星, $P$ 为 $B 、 C$ 两卫星轨道的交点. 已知 $A 、 B 、 C$ 绕地心运动的周期相同，下列说法中正确的是\n\n[图1]\n\nA: 物体 $A$ 的速度小于第一宇宙速度\nB: 物体 $A$ 的速度小于卫星 $C$ 的运行速度\nC: 物体 $A$ 和卫星 $C$ 具有相同大小的加速度\nD: 卫星 $B$ 在 $P$ 点的加速度与卫星 $C$ 在 $P$ 点的加速度大小不相等\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_9938578583ce82f2e878g-39.jpg?height=300&width=231&top_left_y=2117&top_left_x=336"
        ],
        "answer": [
            "A",
            "B"
        ],
        "solution": "$\\mathrm{B} 、$ 物体 $A$ 和卫星 $C$ 的周期相等, 则角速度相等, 根据 $v=r \\omega$ 知, 半径越大,线速度越大, 所以卫星 $\\mathrm{C}$ 的运行速度大于物体 $A$ 的速度; 故 B 正确.\n\nA、第一宇宙速度是最大的环绕速度, 根据 $v=\\sqrt{\\frac{G M}{r}}$, 卫星 $C$ 的运行速度小于第一宇宙速度, 卫星 $C$ 的运行速度大于物体 $A$ 的速度, 所以物体 $A$ 的运行速度一定小于第一宇宙速度; A 正确.\n\n$\\mathrm{C}$ 、物体 $\\mathrm{A}$ 静止于地球赤道上随地球一起自转, 卫星 $C$ 为绕地球做圆周运动, 它们绕地心运动的周期相同, 根据向心加速度的公式 $a=\\left(\\frac{2 \\pi}{T}\\right)^{2} r$, 得卫星 $C$ 的加速度较大; $\\mathrm{C}$ 错误.\n\nD、卫星做圆周或椭圆都是受万有引力产生加速度, $\\frac{G M m}{r^{2}}=m a$, 可得 $a=\\frac{G M}{r^{2}}$, 则两卫星距离地心的距离相等时加速度相等, D 错误.\n\n故选 AB.",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_664",
        "problem": "2013 年 12 月 10 日晚上九点二十分, 在太空飞行了九天的“嫦娥三号”飞船再次成功变轨, 从 $100 \\mathrm{~km} \\times 100 \\mathrm{~km}$ 的环月圆轨道 I 降低到椭圆轨道II(近月点 $15 \\mathrm{~km}$ 、远月点 $100 \\mathrm{~km})$, 两轨道相交于点 $P$, 如图所示. 关于“嫦娥三号”飞船, 以下说法正确的是 ( )\n\n[图1]\nA: 飞船在轨道 $\\mathrm{I}$ 上运动到 $\\mathrm{P}$ 点的速度比在轨道II上运动到 $\\mathrm{P}$ 点的速度大\nB: 飞船在轨道 $I$ 上运动到 $p$ 点的向心加速度比在轨道II上运动到 $P$ 点的向心加速度小\nC: 飞船在轨道 I 上的引力势能与动能之和比在轨道II上的引力势能与动能之和\nD: 飞船在轨道II上运动的周期大于在轨道 I 上运动的周期\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n2013 年 12 月 10 日晚上九点二十分, 在太空飞行了九天的“嫦娥三号”飞船再次成功变轨, 从 $100 \\mathrm{~km} \\times 100 \\mathrm{~km}$ 的环月圆轨道 I 降低到椭圆轨道II(近月点 $15 \\mathrm{~km}$ 、远月点 $100 \\mathrm{~km})$, 两轨道相交于点 $P$, 如图所示. 关于“嫦娥三号”飞船, 以下说法正确的是 ( )\n\n[图1]\n\nA: 飞船在轨道 $\\mathrm{I}$ 上运动到 $\\mathrm{P}$ 点的速度比在轨道II上运动到 $\\mathrm{P}$ 点的速度大\nB: 飞船在轨道 $I$ 上运动到 $p$ 点的向心加速度比在轨道II上运动到 $P$ 点的向心加速度小\nC: 飞船在轨道 I 上的引力势能与动能之和比在轨道II上的引力势能与动能之和\nD: 飞船在轨道II上运动的周期大于在轨道 I 上运动的周期\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_9938578583ce82f2e878g-27.jpg?height=346&width=325&top_left_y=2094&top_left_x=337"
        ],
        "answer": [
            "A",
            "C"
        ],
        "solution": "$\\mathrm{A}$ 、沿轨道I运动至 $P$ 时, 制动减速, 万有引力大于向心力做向心运动, 才能进入轨道II, 故在轨道I上运动到 $P$ 点的速度比在轨道II上运动到 $P$ 点的速度大;故 A 正确. B、“嫦娥三号”卫星变轨前通过椭圆轨道远地点时只有万有引力来提供加速度, 变轨后沿圆轨道运动也是只有万有引力来提供加速度, 同一地点万有引力相同, 所以加速度相等;故 B 错误.\n\nC、变轨的时候点火, 发动机做功, 从轨道I进入轨道II, 发动机要做功使卫星减速, 故在轨道I上的势能与动能之和比在轨道II上的势能与动能之和大;故 C 正确.\n\nD、根据开普勒第三定律 $\\frac{a^{3}}{T^{2}}$ 为常数, 可得半长轴 $a$ 越大, 运动周期越大, 显然轨道I的半长轴 (半径) 大于轨道II的半长轴, 故沿轨道II运动的周期小于沿轨道 I 运动的周期;故 D 错误.\n\n故选 AC.",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_140",
        "problem": "利用金星凌日现象, 我们可以估算出地球与太阳之间的平均距离。日地平均距离也被定义为 1 个天文单位 (1A.U.), 是天文学中常用的距离单位。\n\n金星轨道在地球轨道内侧, 某些特殊时刻, 地球、金星、太阳恰在一条直线上, 这时从地球上可以看到金星就像一个小黑点一样在太阳表面缓慢移动, 如图甲所示, 天文学称之为“金星凌日”。在地球上的不同地点, 比如图乙中的 $A 、 B$ 两点, 它们在同一时刻观察到的金星在日面上的位置是不同的，我们分别记为 $A^{\\prime} 、 B^{\\prime}$ 。\n\n人类在此之前就观察到金星绕日公转的周期是 0.62 年。据此估算金星与太阳的平均距离大约是多少个天文单位;\n\n[图1]\n\n甲\n\n[图2]\n\n丙\n\n[图3]\n\n乙",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题的答案是一个数值。\n\n问题:\n利用金星凌日现象, 我们可以估算出地球与太阳之间的平均距离。日地平均距离也被定义为 1 个天文单位 (1A.U.), 是天文学中常用的距离单位。\n\n金星轨道在地球轨道内侧, 某些特殊时刻, 地球、金星、太阳恰在一条直线上, 这时从地球上可以看到金星就像一个小黑点一样在太阳表面缓慢移动, 如图甲所示, 天文学称之为“金星凌日”。在地球上的不同地点, 比如图乙中的 $A 、 B$ 两点, 它们在同一时刻观察到的金星在日面上的位置是不同的，我们分别记为 $A^{\\prime} 、 B^{\\prime}$ 。\n\n人类在此之前就观察到金星绕日公转的周期是 0.62 年。据此估算金星与太阳的平均距离大约是多少个天文单位;\n\n[图1]\n\n甲\n\n[图2]\n\n丙\n\n[图3]\n\n乙\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n请记住，你的答案应以A.U.为单位计算，但在给出最终答案时，请不要包含单位。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是不包含任何单位的数值。",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-038.jpg?height=214&width=240&top_left_y=201&top_left_x=474",
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-038.jpg?height=249&width=280&top_left_y=178&top_left_x=751",
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-038.jpg?height=205&width=853&top_left_y=480&top_left_x=333"
        ],
        "answer": [
            "0.727"
        ],
        "solution": "由开普勒第三定律\n\n$$\n\\frac{r_{\\text {地 }}^{3}}{T_{\\text {地 }}^{2}}=\\frac{r_{\\text {金 }}^{3}}{T_{\\text {金 }}^{2}}\n$$\n\n金星与太阳的距离为\n\n$$\nr_{\\text {金 }}=\\sqrt[3]{0.62^{2}} \\text { A.U. }=0.727 \\text { A.U. }\n$$",
        "answer_type": "NV",
        "unit": [
            "A.U."
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_1210",
        "problem": "On $21^{\\text {st }}$ December 2020, Jupiter and Saturn formed a true spectacle in the southwestern sky just after sunset in the UK, separated by only $0.102^{\\circ}$. This is close enough that when viewed through a telescope both planets and their moons could be seen in the same field of view (see Fig 4).\n\nWhen two planets occupy the same piece of sky it is known as a conjunction, and when it is Jupiter and Saturn it is known as a great conjunction (so named because they are the rarest of the naked-eye planet conjunctions). The reason it happens is because Jupiter's orbital velocity is higher than Saturn's and so as time goes on Jupiter catches up with and overtakes Saturn (at least as viewed from Earth), with the moment of overtaking corresponding to the conjunction. The process of the two getting closer and closer together has been seen in sky throughout the year (see Fig 5).\n[figure1]\n\nFigure 4: Left: The view of the planets the day before their closest approach, as captured by the 16 \" telescope at the Institute of Astronomy, Cambridge. Credit: Robin Catchpole.\n\nRight: The view from Arizona on the day of the closest approach as viewed with the Lowell Discovery\n\nTelescope. Credit: Levine / Elbert / Bosh / Lowell Observatory.\n\n[figure2]\n\nFigure 5: Left: Demonstrating how Jupiter and Saturn have been getting closer and closer together over the last few months. Separations are given in degrees and arcminutes $\\left(1 / 60^{\\text {th }}\\right.$ of a degree). Credit: Pete Lawrence.\n\nRight: The positions of the Earth, Jupiter and Saturn that were responsible for the 2020 great conjunction. The precise timing of the apparent alignment is clearly sensitive to where Earth is in its orbit. Credit:\n\ntimeanddate.com.\n\nThe time between conjunctions is known as the synodic period. Although this period will change slightly from conjunction to conjunction due to different lines of perspective as viewed from Earth (see Fig 5), we can work out the average time between great conjunctions by considering both planets travelling on circular coplanar orbits and ignoring the position of the Earth.\n\nFor circular coplanar orbits the centre of Jupiter's disc would pass in front of the centre of Saturn's disc every conjunction, and hence have an angular separation of $\\theta=0^{\\circ}$ (it is measured from the centre of each disc). In practice, the planets follow elliptical orbits that are in planes inclined at different angles to each other.\n\nFig 6 shows how this affects the real values for over 8000 years' worth of data, which along with different synodic periods between conjunctions makes it a difficult problem to solve precisely without a computer. However, after one synodic period Saturn has moved about $2 / 3$ of the way around its orbit, and so roughly every 3 synodic periods it is in a similar part of the sky. Consequently every third great conjunction follows a reasonably regular pattern which can be fit with a sinusoidal function.\n\n[figure3]\n\nFigure 6: Top: All great conjunctions from 1800 to 2300, calculated for the real celestial mechanics of the Solar System. We can see that each great conjunction belongs to one of three different series or tracks, with Track A indicated with orange circles, Track B with green squares, and Track C with blue triangles.\n\nBottom: The same idea but extended over a much larger date range, up to $10000 \\mathrm{AD}$. It is clear the distinct series form broadly sinusoidal patterns which can be used with the average synodic period to give rough predictions for the separations of great conjunctions. The opacity of points is related to each conjunction's angular separation from the Sun (where low opacity means close to the Sun, so it is harder for any observers to see). Credit: Nick Koukoufilippas, but inspired by the work of Steffen Thorsen and Graham Jones / Sky \\& Telescope.b. Jupiter has a period of 4332.589 days and Saturn has a period of 10759.22 days (where 1 day $=24$ hours). Note: be careful as your calculations will be very sensitive to rounding errors.\n\nii. Use your answer to predict the date (to the nearest day) of the next great conjunction.",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is an expression.\nHere is some context information for this question, which might assist you in solving it:\nOn $21^{\\text {st }}$ December 2020, Jupiter and Saturn formed a true spectacle in the southwestern sky just after sunset in the UK, separated by only $0.102^{\\circ}$. This is close enough that when viewed through a telescope both planets and their moons could be seen in the same field of view (see Fig 4).\n\nWhen two planets occupy the same piece of sky it is known as a conjunction, and when it is Jupiter and Saturn it is known as a great conjunction (so named because they are the rarest of the naked-eye planet conjunctions). The reason it happens is because Jupiter's orbital velocity is higher than Saturn's and so as time goes on Jupiter catches up with and overtakes Saturn (at least as viewed from Earth), with the moment of overtaking corresponding to the conjunction. The process of the two getting closer and closer together has been seen in sky throughout the year (see Fig 5).\n[figure1]\n\nFigure 4: Left: The view of the planets the day before their closest approach, as captured by the 16 \" telescope at the Institute of Astronomy, Cambridge. Credit: Robin Catchpole.\n\nRight: The view from Arizona on the day of the closest approach as viewed with the Lowell Discovery\n\nTelescope. Credit: Levine / Elbert / Bosh / Lowell Observatory.\n\n[figure2]\n\nFigure 5: Left: Demonstrating how Jupiter and Saturn have been getting closer and closer together over the last few months. Separations are given in degrees and arcminutes $\\left(1 / 60^{\\text {th }}\\right.$ of a degree). Credit: Pete Lawrence.\n\nRight: The positions of the Earth, Jupiter and Saturn that were responsible for the 2020 great conjunction. The precise timing of the apparent alignment is clearly sensitive to where Earth is in its orbit. Credit:\n\ntimeanddate.com.\n\nThe time between conjunctions is known as the synodic period. Although this period will change slightly from conjunction to conjunction due to different lines of perspective as viewed from Earth (see Fig 5), we can work out the average time between great conjunctions by considering both planets travelling on circular coplanar orbits and ignoring the position of the Earth.\n\nFor circular coplanar orbits the centre of Jupiter's disc would pass in front of the centre of Saturn's disc every conjunction, and hence have an angular separation of $\\theta=0^{\\circ}$ (it is measured from the centre of each disc). In practice, the planets follow elliptical orbits that are in planes inclined at different angles to each other.\n\nFig 6 shows how this affects the real values for over 8000 years' worth of data, which along with different synodic periods between conjunctions makes it a difficult problem to solve precisely without a computer. However, after one synodic period Saturn has moved about $2 / 3$ of the way around its orbit, and so roughly every 3 synodic periods it is in a similar part of the sky. Consequently every third great conjunction follows a reasonably regular pattern which can be fit with a sinusoidal function.\n\n[figure3]\n\nFigure 6: Top: All great conjunctions from 1800 to 2300, calculated for the real celestial mechanics of the Solar System. We can see that each great conjunction belongs to one of three different series or tracks, with Track A indicated with orange circles, Track B with green squares, and Track C with blue triangles.\n\nBottom: The same idea but extended over a much larger date range, up to $10000 \\mathrm{AD}$. It is clear the distinct series form broadly sinusoidal patterns which can be used with the average synodic period to give rough predictions for the separations of great conjunctions. The opacity of points is related to each conjunction's angular separation from the Sun (where low opacity means close to the Sun, so it is harder for any observers to see). Credit: Nick Koukoufilippas, but inspired by the work of Steffen Thorsen and Graham Jones / Sky \\& Telescope.\n\nproblem:\nb. Jupiter has a period of 4332.589 days and Saturn has a period of 10759.22 days (where 1 day $=24$ hours). Note: be careful as your calculations will be very sensitive to rounding errors.\n\nii. Use your answer to predict the date (to the nearest day) of the next great conjunction.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_f4dc8cb2d9258a843a19g-07.jpg?height=706&width=1564&top_left_y=834&top_left_x=244",
            "https://cdn.mathpix.com/cropped/2024_03_14_f4dc8cb2d9258a843a19g-08.jpg?height=578&width=1566&top_left_y=196&top_left_x=242",
            "https://cdn.mathpix.com/cropped/2024_03_14_f4dc8cb2d9258a843a19g-09.jpg?height=1072&width=1564&top_left_y=1191&top_left_x=246"
        ],
        "answer": [
            "30^{\\text {th }}$ Oct 2040"
        ],
        "solution": "Rewriting $T_{J+S}$ into something more convenient\n\n$T_{J+S}=20$ years -51.545 days\n\n$\\therefore$ next conjunction 52 days before $21^{\\text {st }}$ Dec\n\n52 days $=21$ in Dec +30 in Nov +1 in Oct $\\therefore 30^{\\text {th }}$ Oct 2040\n\n[One mark for the year (2040), one for the date. Allow $\\pm 1$ day. Alternatively, some students may look for the $304^{\\text {th }}$ day in 2040]\n\n[The real date is $31^{\\text {st }}$ Oct 2040 so this model has done remarkably well - due to the dependence on the Earth's position in its orbit the gap between conjunctions can vary up to $\\sim 320$ days above or below the average value]",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_816",
        "problem": "Samvit observes a binary star system of masses $M_{1}$ and $M_{2}$. Unfortunately, the star with mass $M_{2}$ is too dim for him to observe it, leading to the following snapshot below.\n\n[figure1]\n\nWhat could Samvit hypothesize to be the position and mass of the other star at this instant that would be consistent with the laws of physics and the orbit snapshot that he sees? To be clear, he has no knowledge of the value of $M_{2}$ or the period of the binary system.\nA: At $\\left(\\sqrt{a^{2}-b^{2}}, 0\\right)$ with mass $M_{1}+M_{2}$\nB: At $(0,-b)$ with mass $M_{1}$\nC: At $\\left(\\sqrt{a^{2}-b^{2}}, 0\\right)$ with mass $\\frac{M_{1} M_{2}}{M_{1}+M_{2}}$\nD: At $\\left(-2 \\sqrt{a^{2}-b^{2}},-b\\right)$ with mass $M_{1}$\nE: At $\\left(\\frac{M_{2}}{M_{1}} \\sqrt{a^{2}-b^{2}}, 0\\right)$ with mass $M_{2}$\n",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSamvit observes a binary star system of masses $M_{1}$ and $M_{2}$. Unfortunately, the star with mass $M_{2}$ is too dim for him to observe it, leading to the following snapshot below.\n\n[figure1]\n\nWhat could Samvit hypothesize to be the position and mass of the other star at this instant that would be consistent with the laws of physics and the orbit snapshot that he sees? To be clear, he has no knowledge of the value of $M_{2}$ or the period of the binary system.\n\nA: At $\\left(\\sqrt{a^{2}-b^{2}}, 0\\right)$ with mass $M_{1}+M_{2}$\nB: At $(0,-b)$ with mass $M_{1}$\nC: At $\\left(\\sqrt{a^{2}-b^{2}}, 0\\right)$ with mass $\\frac{M_{1} M_{2}}{M_{1}+M_{2}}$\nD: At $\\left(-2 \\sqrt{a^{2}-b^{2}},-b\\right)$ with mass $M_{1}$\nE: At $\\left(\\frac{M_{2}}{M_{1}} \\sqrt{a^{2}-b^{2}}, 0\\right)$ with mass $M_{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_7205fccc557018644b5cg-09.jpg?height=412&width=631&top_left_y=1252&top_left_x=774"
        ],
        "answer": [
            "D"
        ],
        "solution": "The trick to this question is that the center of mass of the system has to lie at the focus of the ellipse, and this is the only thing that matters if you have no knowledge of the other parameters in the system. In fact, if you had the period of the binary system, you could fully determine the value of $M_{2}$ in terms of $M_{1}$ and the other parameters of the problem. So, we can simply go through the choices and see which choice places the center of mass of the system at one of the foci, $\\left( \\pm \\sqrt{a^{2}-b^{2}}, 0\\right)$, and choice (d) does.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_88",
        "problem": "百余年前, 爱因斯坦的广义相对论率先对黑洞作出预言。2019 年 4 月 10 日 21 点整, 天文学家召开全球新闻发布会, 宣布首次直接拍摄到黑洞的照片。若认为黑洞为一个密度极大的球形天体, 质量为 $M$, 半径为 $R$, 吸引光绕黑洞做匀速圆周运动。已知光速为 $c$, 以黑洞中心为起点, 到黑洞外圈视界边缘的长度为临界半径, 称为史瓦西半径。下面说法正确的是（）\n\n[图1]\nA: 史瓦西半径为 $\\frac{G M}{c^{2}}$\nB: 史瓦西半径为 $2 \\frac{G M}{c^{2}}$\nC: 黑洞密度为 $\\frac{3 c}{4 G \\pi R^{3}}$\nD: 黑洞密度为 $\\frac{3 c^{2}}{4 G \\pi R^{3}}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n百余年前, 爱因斯坦的广义相对论率先对黑洞作出预言。2019 年 4 月 10 日 21 点整, 天文学家召开全球新闻发布会, 宣布首次直接拍摄到黑洞的照片。若认为黑洞为一个密度极大的球形天体, 质量为 $M$, 半径为 $R$, 吸引光绕黑洞做匀速圆周运动。已知光速为 $c$, 以黑洞中心为起点, 到黑洞外圈视界边缘的长度为临界半径, 称为史瓦西半径。下面说法正确的是（）\n\n[图1]\n\nA: 史瓦西半径为 $\\frac{G M}{c^{2}}$\nB: 史瓦西半径为 $2 \\frac{G M}{c^{2}}$\nC: 黑洞密度为 $\\frac{3 c}{4 G \\pi R^{3}}$\nD: 黑洞密度为 $\\frac{3 c^{2}}{4 G \\pi R^{3}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_6842b9ceb844a90b34c3g-40.jpg?height=323&width=491&top_left_y=1186&top_left_x=337"
        ],
        "answer": [
            "B"
        ],
        "solution": "AB．逃逸速度\n\n$$\nv=\\sqrt{\\frac{2 G M}{r}}\n$$\n\n此为脱离中心天体吸引的临界速度, 代入光速可知临界半径为\n\n$$\nr_{\\text {临 }}=2 \\frac{G M}{c^{2}}\n$$\n\nA 错误, B 正确;\n\nCD. 若光绕黑洞表面做匀速圆周运动, 轨道半径等于黑洞半径, 由\n\n$$\nG \\frac{M m}{R^{2}}=m \\frac{c^{2}}{R}\n$$\n\n可知\n\n$$\nM=\\frac{R c^{2}}{G}\n$$\n\n密度为\n\n$$\n\\rho=\\frac{M}{\\frac{4}{3} \\pi R^{3}}=\\frac{3 c^{2}}{4 G \\pi R^{2}}\n$$\n\n$\\mathrm{CD}$ 错误。\n\n故选 B。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_666",
        "problem": "物体在引力场中具有的势能叫做引力势能, 取无穷远处为引力势能零点。质量为 $m$\n的物体在地球引力场中具有的引力势能 $E_{\\mathrm{p}}=-\\frac{G M m}{r_{0}}$ (式中 $G$ 为引力常量, $M$ 为地球的质量, $r_{0}$ 为物体到地心的距离), 如果用 $R$ 表示地球的半径, $g$ 表示地球表面重力加速度。则下列说法正确的是 ( )\nA: 质量为 $m$ 的人造地球卫星在半径为 $r$ 的圆形轨道上运行时, 其机械能为 $-\\frac{G M m}{2 r}$\nB: 如果地球的第一宇宙速度为 $v_{1}$, 则将质量为 $m$ 的卫星从地球表面发射到半径为 $r$的轨道上运行时至少需要的能量 $E=\\frac{1}{2} m v_{1}^{2}+m g(r-R)$\nC: 由于受高空稀薄空气的阻力作用, 质量为 $m$ 的卫星从半径为 $r_{1}$ 的圆轨道缓慢减小到半径为 $r_{2}$ 的圆轨道的过程中克服空气阻力做的功为 $\\frac{C M m}{2}\\left(\\frac{1}{r_{1}}-\\frac{1}{r_{2}}\\right)$\nD: 位于赤道上的物体引力势能为零\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n物体在引力场中具有的势能叫做引力势能, 取无穷远处为引力势能零点。质量为 $m$\n的物体在地球引力场中具有的引力势能 $E_{\\mathrm{p}}=-\\frac{G M m}{r_{0}}$ (式中 $G$ 为引力常量, $M$ 为地球的质量, $r_{0}$ 为物体到地心的距离), 如果用 $R$ 表示地球的半径, $g$ 表示地球表面重力加速度。则下列说法正确的是 ( )\n\nA: 质量为 $m$ 的人造地球卫星在半径为 $r$ 的圆形轨道上运行时, 其机械能为 $-\\frac{G M m}{2 r}$\nB: 如果地球的第一宇宙速度为 $v_{1}$, 则将质量为 $m$ 的卫星从地球表面发射到半径为 $r$的轨道上运行时至少需要的能量 $E=\\frac{1}{2} m v_{1}^{2}+m g(r-R)$\nC: 由于受高空稀薄空气的阻力作用, 质量为 $m$ 的卫星从半径为 $r_{1}$ 的圆轨道缓慢减小到半径为 $r_{2}$ 的圆轨道的过程中克服空气阻力做的功为 $\\frac{C M m}{2}\\left(\\frac{1}{r_{1}}-\\frac{1}{r_{2}}\\right)$\nD: 位于赤道上的物体引力势能为零\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "A"
        ],
        "solution": "A. 在半径为 $r$ 的圆形轨道上运行的质量为 $m$ 的人造地球卫星有\n\n$$\nG \\frac{M m}{r^{2}}=m \\frac{v_{1}^{2}}{r}\n$$\n\n动能\n\n$$\nE_{\\mathrm{k}}=\\frac{1}{2} m v_{1}^{2}=\\frac{G M m}{2 r}\n$$\n\n引力势能为\n\n$$\nE_{\\mathrm{p}}=-\\frac{G M m}{r}\n$$\n\n则机械能为\n\n$$\nE=E_{\\mathrm{k}}+E_{\\mathrm{p}}=\\frac{G M m}{2 r}-\\frac{G M m}{r}=-\\frac{G M m}{2 r}\n$$\n\n选项 A 正确;\n\nB. 将质量为 $m$ 的卫星从地球表面发射到半径为 $r$ 的轨道上运行时至少需要的能量\n\n$$\nE=\\frac{1}{2} m v_{1}^{2}+\\left(E_{r}-E_{R}\\right)=\\frac{1}{2} m v_{1}^{2}+\\frac{G M m}{2} \\times \\frac{r-R}{R r}\n$$\n\n在地球表面上有\n\n$$\nG \\frac{M m^{\\prime}}{R^{2}}=m^{\\prime} g\n$$\n\n所以\n\n$$\nE=\\frac{1}{2} m v_{1}^{2}+\\frac{m g R(r-R)}{2 r}\n$$\n\n选项 B 错误;\n\nC. 由于受高空稀薄空气的阻力作用, 质量为 $m$ 的卫星从半径为 $r_{1}$ 的圆轨道缓慢减小到\n半径为 $r_{2}$ 的圆轨道的过程中克服空气阻力做的功为\n\n$$\nW_{\\mathrm{f}}=E_{1}-E_{2}=-\\frac{G M m}{2 r_{1}}-\\left(-\\frac{G M m}{2 r_{2}}\\right)=\\frac{G M m}{2}\\left(\\frac{1}{r_{2}}-\\frac{1}{r_{1}}\\right)\n$$\n\n选项 C 错误;\n\nD. 位于赤道上的物体引力势能为\n\n$$\nE_{\\mathrm{p}}^{\\prime}=-\\frac{G M m}{R}\n$$\n\n选项 D 错误。\n\n故选 A。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_576",
        "problem": "如图所示, 飞行器 $P$ 绕某星球做匀速圆周运动, 下列说法不正确的是\n\n[图1]\nA: 轨道半径越大, 周期越长\nB: 张角越大, 速度越大\nC: 若测得周期和星球相对飞行器的张角, 则可得到星球的平均密度\nD: 若测得周期和轨道半径, 则可得到星球的平均密度\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图所示, 飞行器 $P$ 绕某星球做匀速圆周运动, 下列说法不正确的是\n\n[图1]\n\nA: 轨道半径越大, 周期越长\nB: 张角越大, 速度越大\nC: 若测得周期和星球相对飞行器的张角, 则可得到星球的平均密度\nD: 若测得周期和轨道半径, 则可得到星球的平均密度\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_6842b9ceb844a90b34c3g-13.jpg?height=483&width=437&top_left_y=181&top_left_x=341"
        ],
        "answer": [
            "D"
        ],
        "solution": "A. 根据开普勒第三定律 $\\frac{r^{3}}{T^{2}}=k$, 可知轨道半径越大, 飞行器的周期越长, $\\mathrm{A}$正确;\n\nB. 根据卫星的速度公式 $v=\\sqrt{\\frac{G M}{r}}$, 可知张角越大, 轨道半径越小, 速度越大, B 正确 C. 根据公式 $G \\frac{M m}{r^{2}}=m \\frac{4 \\pi^{2} r}{T^{2}}$ 可得\n\n$$\nM=\\frac{4 \\pi^{2} r^{3}}{G T^{2}}\n$$\n\n设星球的质量为 $M$, 半径为 $R$, 平均密度为 $\\rho$, 飞行器的质量为 $m$, 轨道半径为 $r$, 周期为 $T$ ，对于飞行器，由几何关系得\n\n$$\nR=r \\sin \\frac{\\theta}{2}\n$$\n\n星球的平均密度为\n\n$$\n\\rho=\\frac{M}{\\frac{4}{3} \\pi R^{3}}\n$$\n\n由以上三式知, 测得周期和张角, 就可得到星球的平均密度, C 正确;\n\nD. 由 $G \\frac{M m}{r^{2}}=m \\frac{4 \\pi^{2} r}{T^{2}}$ 可得\n\n$$\nM=\\frac{4 \\pi^{2} r^{3}}{G T^{2}}\n$$\n\n星球的平均密度为\n\n$$\n\\rho=\\frac{M}{\\frac{4}{3} \\pi R^{3}}\n$$\n\n可知若测得周期和轨道半径, 可得到星球的质量, 但星球的半径未知, 不能求出星球的\n体积, 故不能求出平均密度, $\\mathrm{D}$ 错误。\n\n故选 D。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_1153",
        "problem": "Recently a group of researchers announced that they had discovered an Earth-sized exoplanet around our nearest star, Proxima Centauri. Its closeness raises an intriguing possibility about whether or not we might be able to image it directly using telescopes. The difficulty comes from the small angular scales that need to be resolved and the extreme differences in brightness between the reflected light from the planet and the light given out by the star.\n\n[figure1]\n\nFigure 6: Artist's impression of the view from the surface of Proxima Centauri b. Credit: ESO / M. Kornmesser\n\nData about the star and the planet are summarised below:\n\n| Proxima Centauri (star) |  | Proxima Centauri b (planet) |  |\n| :--- | :--- | :--- | :--- |\n| Distance | $1.295 \\mathrm{pc}$ | Orbital period | 11.186 days |\n| Mass | $0.123 \\mathrm{M}_{\\odot}$ | Mass $(\\mathrm{min})$ | $\\approx 1.27 \\mathrm{M}_{\\oplus}$ |\n| Radius | $0.141 \\mathrm{R}_{\\odot}$ | Radius $(\\mathrm{min})$ | $\\approx 1.1 \\mathrm{R}_{\\oplus}$ |\n| Surface temperature | $3042 \\mathrm{~K}$ |  |  |\n| Apparent magnitude | 11.13 |  |  |\n\nThe following formulae may also be helpful:\n\n$$\nm-\\mathcal{M}=5 \\log \\left(\\frac{d}{10}\\right) \\quad \\mathcal{M}-\\mathcal{M}_{\\odot}=-2.5 \\log \\left(\\frac{L}{\\mathrm{~L}_{\\odot}}\\right) \\quad \\Delta m=2.5 \\log C R\n$$\n\nwhere $m$ is the apparent magnitude, $\\mathcal{M}$ is the absolute magnitude, $d$ is the distance in parsecs, and the contrast ratio $(C R)$ is defined as the ratio of fluxes from the star and planet, $C R=\\frac{f_{\\text {star }}}{f_{\\text {planet }}}$.e. How long an exposure would JWST need in order to get the same SNR as Keck II, again if observed at the longest wavelength for which the planet can still be resolved from the star by the telescope? (Make similar assumptions about the received flux and use the same value of $\\varepsilon$ ).",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nRecently a group of researchers announced that they had discovered an Earth-sized exoplanet around our nearest star, Proxima Centauri. Its closeness raises an intriguing possibility about whether or not we might be able to image it directly using telescopes. The difficulty comes from the small angular scales that need to be resolved and the extreme differences in brightness between the reflected light from the planet and the light given out by the star.\n\n[figure1]\n\nFigure 6: Artist's impression of the view from the surface of Proxima Centauri b. Credit: ESO / M. Kornmesser\n\nData about the star and the planet are summarised below:\n\n| Proxima Centauri (star) |  | Proxima Centauri b (planet) |  |\n| :--- | :--- | :--- | :--- |\n| Distance | $1.295 \\mathrm{pc}$ | Orbital period | 11.186 days |\n| Mass | $0.123 \\mathrm{M}_{\\odot}$ | Mass $(\\mathrm{min})$ | $\\approx 1.27 \\mathrm{M}_{\\oplus}$ |\n| Radius | $0.141 \\mathrm{R}_{\\odot}$ | Radius $(\\mathrm{min})$ | $\\approx 1.1 \\mathrm{R}_{\\oplus}$ |\n| Surface temperature | $3042 \\mathrm{~K}$ |  |  |\n| Apparent magnitude | 11.13 |  |  |\n\nThe following formulae may also be helpful:\n\n$$\nm-\\mathcal{M}=5 \\log \\left(\\frac{d}{10}\\right) \\quad \\mathcal{M}-\\mathcal{M}_{\\odot}=-2.5 \\log \\left(\\frac{L}{\\mathrm{~L}_{\\odot}}\\right) \\quad \\Delta m=2.5 \\log C R\n$$\n\nwhere $m$ is the apparent magnitude, $\\mathcal{M}$ is the absolute magnitude, $d$ is the distance in parsecs, and the contrast ratio $(C R)$ is defined as the ratio of fluxes from the star and planet, $C R=\\frac{f_{\\text {star }}}{f_{\\text {planet }}}$.\n\nproblem:\ne. How long an exposure would JWST need in order to get the same SNR as Keck II, again if observed at the longest wavelength for which the planet can still be resolved from the star by the telescope? (Make similar assumptions about the received flux and use the same value of $\\varepsilon$ ).\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of \\mathrm{~s}, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_204b2e236273ea30e8d2g-10.jpg?height=708&width=1082&top_left_y=551&top_left_x=493"
        ],
        "answer": [
            "4.96"
        ],
        "solution": "(using similar reasoning to the previous part of the question)\n\n$$\n\\begin{aligned}\n& \\lambda_{\\text {JWST }}=9.70 \\times 10^{-7} \\mathrm{~m} \\\\\n& f=0.55 \\text { photons } \\mathrm{m}^{-2} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$\n\n(alternative method gives $\\mathrm{f}=16.5$ photons $\\mathrm{m}^{-2} \\mathrm{~s}^{-1}$ )\n\nSince $b<<f$ we can simplify the denominator of the SNR formula by ignoring the second term, so\n\n$$\nt=\\frac{S N R^{2}}{f A \\epsilon}=\\frac{3^{2}}{0.55 \\times\\left(\\pi \\times 3.25^{2}\\right) \\times 0.1}=4.96 \\mathrm{~s}\n$$\n\n(alternative method gives $t=0.16 \\mathrm{~s}$, and day side reflection decreases $t$ by a factor of 2 )\n\n[This is much more reasonable - the optimistic assumptions we have made throughout this question mean this is just a lower limit and so the actual exposure time will be longer, perhaps several minutes to get a much higher SNR and to compensate for the fact for that the sensitivity is lower at $970 \\mathrm{~nm}$ than we have quoted since it varies with wavelength. The main thing that would prevent direct observation of the exoplanet by JWST would be if the contrast ratio was much higher than calculated here as that would mean the star's light would need to be blocked out by the Near Infrared Camera's coronagraph - unfortunately this can only be used at wavelengths too long to resolve the system.]",
        "answer_type": "NV",
        "unit": [
            "\\mathrm{~s}"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_7",
        "problem": "宇宙中存在着由四颗星组成的孤立星系。如图所示, 一颗母星处在正三角形的中心,三角形的顶点各有一个质量相等的小星围绕母星做圆周运动。如果两颗小星间的万有引力为 $F$, 母星与任意一颗小星间的万有引力为 $9 F$, 则 ( )\n\n[图1]\nA: 每颗小星受到的万有引力为 $(\\sqrt{3}+9) F$\nB: 每颗小星受到的万有引力为 $\\left(\\frac{\\sqrt{3}}{2}+9\\right) F$\nC: 母星的质量是每颗小星质量的 9 倍\nD: 母星的质量是每颗小星质量的 $3 \\sqrt{3}$ 倍\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n宇宙中存在着由四颗星组成的孤立星系。如图所示, 一颗母星处在正三角形的中心,三角形的顶点各有一个质量相等的小星围绕母星做圆周运动。如果两颗小星间的万有引力为 $F$, 母星与任意一颗小星间的万有引力为 $9 F$, 则 ( )\n\n[图1]\n\nA: 每颗小星受到的万有引力为 $(\\sqrt{3}+9) F$\nB: 每颗小星受到的万有引力为 $\\left(\\frac{\\sqrt{3}}{2}+9\\right) F$\nC: 母星的质量是每颗小星质量的 9 倍\nD: 母星的质量是每颗小星质量的 $3 \\sqrt{3}$ 倍\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_ef01104c57d69d8b0f5ag-007.jpg?height=328&width=379&top_left_y=156&top_left_x=336"
        ],
        "answer": [
            "A"
        ],
        "solution": "CD. 假设每颗小星的质量为 $m$, 母星的质量为 $M$, 等边三角形的边长为 $a$,则小星绕母星运动轨道半径为\n\n$$\nr=\\frac{\\sqrt{3}}{3} a\n$$\n\n根据万有引力定律，两颗小星间的万有引\n\n$$\nF=\\frac{G m m}{a^{2}}\n$$\n\n母星与任意一颗小星间的万有引力为\n\n$$\n9 F=G \\frac{M m}{\\left(\\frac{\\sqrt{3}}{3} a\\right)^{2}}\n$$\n\n解得\n\n$$\nM=3 m\n$$\n\n故 CD 错误;\n\nAB. 根据受力分析可知, 每颗小星受到其余两颗小星和一颗母星的引力, 其合力指向母星以提供向心力, 即每颗小星受到的万有引力为\n\n$$\nF=2 \\frac{G m^{2}}{a^{2}} \\cos 30^{\\circ}+G \\frac{M m}{\\left(\\frac{\\sqrt{3}}{3} a\\right)^{2}}=(\\sqrt{3}+9) F\n$$\n\n故 A 正确, B 错误。\n\n故选 A。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_330",
        "problem": "“嫦娥二号”进入环月轨道后, 在距月球表面一定高度的轨道上做匀速圆周运动, 此高度远小于月球的半径, 设“嫦娥二号”绕月与月绕地的转动方向同向. 已知地球的质量为月球质量的 $k$ 倍, 月球绕地球运行的轨道半径为月球的半径的 $n$ 倍, 月球绕地球运行的周期为 $T$. 若某时刻“嫦娥二号”距地球最远，经 $\\Delta t$ 时间“嫦娥二号”距地球最近, 则 $\\Delta t$最短为 ( )\nA: $\\frac{T}{2\\left(\\sqrt{\\frac{n^{3}}{k}}-1\\right)}$\nB: $\\frac{T}{2\\left(1-\\sqrt{\\frac{k}{n^{3}}}\\right)}$\nC: $\\frac{T}{2\\left(\\sqrt{\\frac{k}{n^{3}}}-1\\right)}$\nD: $\\frac{T}{2\\left(1-\\sqrt{\\frac{n^{3}}{k}}\\right)}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n“嫦娥二号”进入环月轨道后, 在距月球表面一定高度的轨道上做匀速圆周运动, 此高度远小于月球的半径, 设“嫦娥二号”绕月与月绕地的转动方向同向. 已知地球的质量为月球质量的 $k$ 倍, 月球绕地球运行的轨道半径为月球的半径的 $n$ 倍, 月球绕地球运行的周期为 $T$. 若某时刻“嫦娥二号”距地球最远，经 $\\Delta t$ 时间“嫦娥二号”距地球最近, 则 $\\Delta t$最短为 ( )\n\nA: $\\frac{T}{2\\left(\\sqrt{\\frac{n^{3}}{k}}-1\\right)}$\nB: $\\frac{T}{2\\left(1-\\sqrt{\\frac{k}{n^{3}}}\\right)}$\nC: $\\frac{T}{2\\left(\\sqrt{\\frac{k}{n^{3}}}-1\\right)}$\nD: $\\frac{T}{2\\left(1-\\sqrt{\\frac{n^{3}}{k}}\\right)}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "A"
        ],
        "solution": "对于月球绕地球运动, 根据万有引力提供向心力 $G \\frac{M_{\\text {地 }} M_{\\text {月 }}}{\\left(n R_{\\text {月 }}\\right)^{2}}=M_{\\text {月 }} \\frac{4 \\pi^{2}}{T^{2}}\\left(n R_{\\text {月 }}\\right)$,即 $G M_{\\text {地 }}=\\frac{4 \\pi^{2} n^{3} R_{\\text {月 }}{ }^{3}}{T^{2}}$\n\n又因为已知地球的质量为月球质量的 $\\mathrm{k}$ 倍, 所以 $G k M_{\\text {月 }}=\\frac{4 \\pi^{2} n^{3} R_{\\text {月 }}{ }^{3}}{T^{2}}$\n\n对于“嫦娥二号”绕月球运动, $G \\frac{M_{\\text {月 }} m}{R_{\\text {月 }}{ }^{2}}=m \\frac{4 \\pi^{2}}{T^{\\prime 2}} R_{\\text {月 }}$, 即 $G M_{\\text {月 }}=\\frac{4 \\pi^{2} R_{\\text {月 }}{ }^{3}}{T^{\\prime 2}}$\n\n所以 $\\mathrm{T}^{\\prime}=\\sqrt{\\frac{k}{n^{3}}} T$\n\n所以月球的角速度 $\\omega=\\frac{2 \\pi}{T}$, “嫦娥二号”的角速度 $\\omega^{\\prime}=\\frac{2 \\pi}{T^{\\prime}}=\\frac{2 \\pi}{T \\sqrt{\\frac{k}{n^{3}}}}$\n由于“嫦娥二号”绕月与月绕地的转动方向同向, 所以有 $\\pi=\\left(\\omega^{\\prime}-\\omega\\right) \\cdot \\triangle \\mathrm{t}$\n\n即 $\\pi=\\left(\\frac{2 \\pi}{T \\sqrt{\\frac{k}{n^{3}}}}-\\frac{2 \\pi}{T}\\right) \\Delta t$; 所以 $\\Delta t=\\frac{T}{2\\left(\\sqrt{\\frac{n^{3}}{k}}-1\\right)}$\n\n故 A 正确, BCD 错误. 故选 A.",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_557",
        "problem": "2021 年 2 月, 我国“天问一号”火星探测器到达火星附近, 进入环火轨道, 进行火星\n科学探测。探测器着陆前运行的轨道分别如图中 $1 、 2 、 3$ 所示, 则探测器 ( )\n\n[图1]\nA: 沿不同轨道运动, 经过 $P$ 点时的加速度相同\nB: 由轨道 2 变到轨道 3 , 需要在 $P$ 点加速\nC: 在轨道 1 上运行的周期比在轨道 2 上运行的周期小\nD: 沿轨道 3 运动时的机械能大于沿轨道 2 运动时的机械能\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n2021 年 2 月, 我国“天问一号”火星探测器到达火星附近, 进入环火轨道, 进行火星\n科学探测。探测器着陆前运行的轨道分别如图中 $1 、 2 、 3$ 所示, 则探测器 ( )\n\n[图1]\n\nA: 沿不同轨道运动, 经过 $P$ 点时的加速度相同\nB: 由轨道 2 变到轨道 3 , 需要在 $P$ 点加速\nC: 在轨道 1 上运行的周期比在轨道 2 上运行的周期小\nD: 沿轨道 3 运动时的机械能大于沿轨道 2 运动时的机械能\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_ef01104c57d69d8b0f5ag-045.jpg?height=389&width=768&top_left_y=228&top_left_x=333"
        ],
        "answer": [
            "A"
        ],
        "solution": "A. 根据牛顿第二定律，有\n\n$$\nG \\frac{M m}{r^{2}}=m a\n$$\n\n由于经过 $P$ 点的距离 $r$ 相同, 所以沿不同轨道运动, 经过 $P$ 点时的加速度相同, 选项 A 正确;\n\nB. 由轨道 2 变到轨道 3 , 半径减小, 做向心运动, 因此需要在 $P$ 点减速, 选项 $\\mathrm{B}$ 错误 C. 根据万有引力提供向心力, 有\n\n$$\nG \\frac{M m}{r^{2}}=m r \\frac{4 \\pi^{2}}{T^{2}}\n$$\n\n可得\n\n$$\nT=\\sqrt{\\frac{4 \\pi^{2} r^{3}}{G M}}\n$$\n\n在轨道 1 上运行的半径比在轨道 2 上运行的半径大, 所以在轨道 1 上运行的周期比在轨道 2 上运行的周期大，选项 C 错误;\n\nD. 由轨道 3 变到轨道 2, 需要点火加速, 因此沿轨道 3 运动时的机械能小于沿轨道 2 运动时的机械能, 选项 D 错误。\n\n故选 A。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_2",
        "problem": "在地球上观测, 太阳与地内行星 (金星、水星) 可视为质点, 它们与眼睛连线的夹角有最大值时叫大距。地内行星在太阳东边时为东大距, 在太阳西边时为西大距, 如图所示。已知水星到太阳的平均距离约为 0.4 天文单位 (1 天文单位约为太阳与地球间的平均距离), 金星到太阳的平均距离约为 0.7 天文单位, 地内行星与地球可认为在同一平面内的圆轨道上运动, 地球的自转方向与公转方向相同, 取 $\\sqrt{0.7} \\approx 0.8, \\sqrt{0.4} \\approx 0.6$,则下列说法中正确的是（ ）\n\n[图1]\nA: 水星的公转周期为 0.4 年\nB: 水星的线速度大约为金星线速度的 1.3 倍\nC: 水星两次东大距的间隔时间大约 $\\frac{6}{19}$ 年\nD: 金星两次东大距的间隔时间比水星短\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n在地球上观测, 太阳与地内行星 (金星、水星) 可视为质点, 它们与眼睛连线的夹角有最大值时叫大距。地内行星在太阳东边时为东大距, 在太阳西边时为西大距, 如图所示。已知水星到太阳的平均距离约为 0.4 天文单位 (1 天文单位约为太阳与地球间的平均距离), 金星到太阳的平均距离约为 0.7 天文单位, 地内行星与地球可认为在同一平面内的圆轨道上运动, 地球的自转方向与公转方向相同, 取 $\\sqrt{0.7} \\approx 0.8, \\sqrt{0.4} \\approx 0.6$,则下列说法中正确的是（ ）\n\n[图1]\n\nA: 水星的公转周期为 0.4 年\nB: 水星的线速度大约为金星线速度的 1.3 倍\nC: 水星两次东大距的间隔时间大约 $\\frac{6}{19}$ 年\nD: 金星两次东大距的间隔时间比水星短\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_cbd0a496f6e2fb8d7781g-083.jpg?height=474&width=445&top_left_y=1028&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_04_01_cbd0a496f6e2fb8d7781g-083.jpg?height=133&width=464&top_left_y=2549&top_left_x=796"
        ],
        "answer": [
            "B",
            "C"
        ],
        "solution": "A. 行星绕太阳公转时, 由万有引力提供向心力, 则得\n\n$$\nG \\frac{M m}{r^{2}}=m \\frac{4 \\pi^{2}}{T^{2}} r\n$$\n\n可得行星公转周期为\n\n$$\nT=2 \\pi \\sqrt{\\frac{r^{3}}{G M}}\n$$\n\n式中 $M$ 是太阳的质量, $r$ 是行星的公转轨道半径。\n\n则水星与地球公转周期之比\n\n[图2]\n\n所以水星的公转周期为\n\n$$\nT_{\\text {水 }}=0.4 \\sqrt{0.4} \\text { 年 }\n$$\n\n故 A 错误\n\nB. 由万有引力提供向心力得\n\n$$\nG \\frac{M m}{r^{2}}=m \\frac{v^{2}}{r}\n$$\n\n得\n\n$$\nv=\\sqrt{\\frac{G M}{r}}\n$$\n\n则水星的线速度与金星线速度之比\n\n$$\n\\frac{v_{\\text {水 }}}{v_{\\text {金 }}}=\\sqrt{\\frac{r_{\\text {金 }}}{r_{\\text {水 }}}}=\\sqrt{\\frac{0.7}{0.4}} \\approx 1.3\n$$\n\n则 B 正确。\n\nC. 设水星两次东大距的间隔时间为 $t$ 。则\n\n$$\n2 \\pi=\\frac{2 \\pi}{T_{\\text {水 }}} t-\\frac{2 \\pi}{T_{\\text {地 }}} t\n$$\n\n得\n\n$$\nt=\\frac{T_{\\text {地 }} T_{\\text {水 }}}{T_{\\text {地 }}-T_{\\text {水 }}}=\\frac{1 \\times 0.4 \\sqrt{0.4}}{1-0.4 \\sqrt{0.4}} \\text { 年 } \\approx \\frac{6}{19} \\text { 年 }\n$$\n\n则 C 正确;\n\nD. 因金星的周期长, 则金星两次东大距的间隔时间比水星长, 则 $\\mathrm{D}$ 错误。\n\n故选 BC。",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_16",
        "problem": "太阳主要是由 e、 ${ }_{1}^{1} \\mathrm{H}$ 和 ${ }_{2}^{4} \\mathrm{He}$ 等粒子组成的。维持太阳辐射的是其内部的核聚变反应,\n\n核反应方程是 $2 \\mathrm{e}+4{ }_{1}^{1} \\mathrm{H} \\rightarrow{ }_{2}^{4} \\mathrm{He}$, 该核反应产生的核能最后转化为辐射能。根据目前关于\n恒星演化的理论, 若由于聚变反应而使太阳中的 ${ }_{1}^{1} \\mathrm{H}$ 核数目从现有数减少 $10 \\%$, 太阳将离开主序星阶段而转入红巨星的演化阶段。为了简化模型, 假定目前太阳全部由 $\\mathrm{e}$ 和 ${ }_{1}{ }^{1} \\mathrm{H}$核组成, 并据此回答下列问题。\n\n已知地球半径 $R=6.4 \\times 10^{6} \\mathrm{~m}$, 地球质量 $m=6.0 \\times 10^{24} \\mathrm{~kg}$, 日地中心的距离 $r=1.5 \\times 10^{11} \\mathrm{~m}$, 地球表面处的重力加速度 $g=10 \\mathrm{~m} / \\mathrm{s}^{2}, 1$ 年约为 $3.2 \\times 10^{7} \\mathrm{~s}$, \n\n已知质子质量 $m_{\\mathrm{p}}=1.6726 \\times 10^{-27} \\mathrm{~kg},{ }_{2}^{4} \\mathrm{He}$ 质量 $m_{\\alpha}=6.6458 \\times 10^{-27} \\mathrm{~kg}$, 电子质量 $m_{\\mathrm{e}}=9.1 \\times 10^{-31} \\mathrm{~kg}$, 光速 $c=3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}$ \n已知地球上与太阳垂直的每平方米截面上, 每秒通过的太阳辐射能 $w=1.35 \\times 10^{3} \\mathrm{~W} / \\mathrm{m}^{2}$ 。试估算太阳继续保持在主序星阶段的寿命。（结果保留一位有效数字）",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题的答案是一个数值。\n\n问题:\n太阳主要是由 e、 ${ }_{1}^{1} \\mathrm{H}$ 和 ${ }_{2}^{4} \\mathrm{He}$ 等粒子组成的。维持太阳辐射的是其内部的核聚变反应,\n\n核反应方程是 $2 \\mathrm{e}+4{ }_{1}^{1} \\mathrm{H} \\rightarrow{ }_{2}^{4} \\mathrm{He}$, 该核反应产生的核能最后转化为辐射能。根据目前关于\n恒星演化的理论, 若由于聚变反应而使太阳中的 ${ }_{1}^{1} \\mathrm{H}$ 核数目从现有数减少 $10 \\%$, 太阳将离开主序星阶段而转入红巨星的演化阶段。为了简化模型, 假定目前太阳全部由 $\\mathrm{e}$ 和 ${ }_{1}{ }^{1} \\mathrm{H}$核组成, 并据此回答下列问题。\n\n已知地球半径 $R=6.4 \\times 10^{6} \\mathrm{~m}$, 地球质量 $m=6.0 \\times 10^{24} \\mathrm{~kg}$, 日地中心的距离 $r=1.5 \\times 10^{11} \\mathrm{~m}$, 地球表面处的重力加速度 $g=10 \\mathrm{~m} / \\mathrm{s}^{2}, 1$ 年约为 $3.2 \\times 10^{7} \\mathrm{~s}$, \n\n已知质子质量 $m_{\\mathrm{p}}=1.6726 \\times 10^{-27} \\mathrm{~kg},{ }_{2}^{4} \\mathrm{He}$ 质量 $m_{\\alpha}=6.6458 \\times 10^{-27} \\mathrm{~kg}$, 电子质量 $m_{\\mathrm{e}}=9.1 \\times 10^{-31} \\mathrm{~kg}$, 光速 $c=3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}$ \n已知地球上与太阳垂直的每平方米截面上, 每秒通过的太阳辐射能 $w=1.35 \\times 10^{3} \\mathrm{~W} / \\mathrm{m}^{2}$ 。试估算太阳继续保持在主序星阶段的寿命。（结果保留一位有效数字）\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n请记住，你的答案应以年为单位计算，但在给出最终答案时，请不要包含单位。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是不包含任何单位的数值。",
        "figure_urls": null,
        "answer": [
            "$1 \\times 10^{10}$"
        ],
        "solution": "根据题给假设, 在太阳继续保持在主序星阶段的时间内, 发生题中所述的核聚变反应的次数为\n\n$$\nN=\\frac{M}{4 m_{\\mathrm{p}}} \\times 10 \\%\n$$\n\n因此, 太阳总共辐射出的能量为\n\n$$\nE=N \\cdot \\Delta E\n$$\n\n设太阳辐射是各向同性的, 则每秒内太阳向外放出的辐射能为\n\n$$\n\\varepsilon=4 \\pi r^{2} w\n$$\n\n所以太阳继续保持在主序星的时间为\n\n$$\nt=\\frac{E}{\\varepsilon}\n$$\n\n由以上各式解得\n\n$$\nt=1 \\times 10^{10} \\text { 年 }\n$$",
        "answer_type": "NV",
        "unit": [
            "年"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_5",
        "problem": "人造地球卫星是发射数量最多, 用途最广, 发展最快的航天器。已知引力常量为 $G$, 地球半径为 $R$, 地球表面的重力加速度为 $\\mathrm{g}$, 地球自转的周期为 $T$ 。\n一颗在赤道上空运行的人造地球卫星，其轨道半径为 $2 R$, 卫星的运行方向与地球的自转方向相同, 若某时刻卫星通过某建筑物的正上方, 求它再次通过该建筑物正上方所需的时间。(可以不进行分母有理化)",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n\n问题:\n人造地球卫星是发射数量最多, 用途最广, 发展最快的航天器。已知引力常量为 $G$, 地球半径为 $R$, 地球表面的重力加速度为 $\\mathrm{g}$, 地球自转的周期为 $T$ 。\n一颗在赤道上空运行的人造地球卫星，其轨道半径为 $2 R$, 卫星的运行方向与地球的自转方向相同, 若某时刻卫星通过某建筑物的正上方, 求它再次通过该建筑物正上方所需的时间。(可以不进行分母有理化)\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": [
            "$\\frac{2 \\pi}{\\sqrt{\\frac{g}{8 R}}-\\frac{2 \\pi}{T}}$"
        ],
        "solution": "由万有引力提供向心力得\n\n$$\n\\begin{gathered}\nG \\frac{M m}{(2 R)^{2}}=m \\omega^{2} \\cdot 2 R \\\\\nG \\frac{M m}{R^{2}}=m g\n\\end{gathered}\n$$\n\n联立解得得卫星运行的角速度\n\n$$\n\\omega=\\sqrt{\\frac{g}{8 R}}\n$$\n\n根据\n\n$$\n\\omega \\Delta t-\\omega_{0} \\Delta t=2 \\pi\n$$\n\n解得\n\n$$\n\\Delta t=\\frac{2 \\pi}{\\omega-\\omega_{0}}=\\frac{2 \\pi}{\\sqrt{\\frac{g}{8 R}}-\\frac{2 \\pi}{T}}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_600",
        "problem": "“嫦娥四号”已成功降落月球背面, 未来中国还将建立绕月轨道空间站。如图所示,关闭动力的宇宙飞船在月球引力作用下沿地一月转移轨道向月球靠近, 并将与空间站在 $A$ 处对接。已知空间站绕月轨道半径为 $r$, 周期为 $T$, 万有引力常量为 $G$, 月球的半径为 $R$, 下列说法正确的是 ( )\n\n[图1]\nA: 地一月转移轨道的周期小于 $T$\nB: 宇宙飞船在 $A$ 处由椭圆轨道进入空间站轨道必须点火减速\nC: 宇宙飞船飞向 $A$ 的过程中加速度逐渐减小\nD: 月球的质量为 $M=\\frac{4 \\pi^{2} R^{2}}{G T^{2}}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n“嫦娥四号”已成功降落月球背面, 未来中国还将建立绕月轨道空间站。如图所示,关闭动力的宇宙飞船在月球引力作用下沿地一月转移轨道向月球靠近, 并将与空间站在 $A$ 处对接。已知空间站绕月轨道半径为 $r$, 周期为 $T$, 万有引力常量为 $G$, 月球的半径为 $R$, 下列说法正确的是 ( )\n\n[图1]\n\nA: 地一月转移轨道的周期小于 $T$\nB: 宇宙飞船在 $A$ 处由椭圆轨道进入空间站轨道必须点火减速\nC: 宇宙飞船飞向 $A$ 的过程中加速度逐渐减小\nD: 月球的质量为 $M=\\frac{4 \\pi^{2} R^{2}}{G T^{2}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_6842b9ceb844a90b34c3g-28.jpg?height=409&width=451&top_left_y=164&top_left_x=323"
        ],
        "answer": [
            "B"
        ],
        "solution": "A. 根据开普勒第三定律可知, 飞船在粗圆轨道的半长轴大于圆轨道的半径,所以地-月转移轨道的周期大于 $T$ ，选项 A 错误;\n\nB. 宇宙飞船在椭圆轨道的 $A$ 点做离心运动, 只有在点火减速后, 才能进入圆轨道的空间站轨道, 选项 B 正确;\n\nC. 宇宙飞船飞向 $A$ 的过程中, 根据\n\n$$\nG \\frac{M m}{r^{2}}=m a\n$$\n\n知半径越来越小, 加速度越来越大, 选项 C 错误;\n\nD. 对空间站, 根据万有引力提供向心力有\n\n$$\nG \\frac{M m}{r^{2}}=m \\frac{4 \\pi^{2}}{T^{2}} r\n$$\n\n解得\n\n$$\nM=\\frac{4 \\pi^{2} r^{3}}{G T^{2}}\n$$\n\n其中 $r$ 为空间站的轨道半径, 选项 D 错误。\n\n故选 B。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_164",
        "problem": "嫦娥四号是我国探月工程二期发射的月球探测器, 也是人类第一个着陆月球背面的探测器。嫦娥四号于 2018 年 12 月 12 日完成近月制动被月球捕获。若探测器近月环绕周期为 $T$, 探测器降落到月球表面后, 从距月球表面高度为 $h$ 处由静止释放一物体, 测出物体落到月球表面的时间为 $t$, 已知月球半径为 $R$, 引力常量为 $G$, 将月球视为质量均匀分布的球体，通过以上物理量可求得月球的平均密度为（）\nA: $\\frac{h}{2 \\pi G R t^{2}}$\nB: $\\frac{\\pi G R t^{2}}{6 h}$\nC: $\\frac{3 \\pi}{G T}$\nD: $\\frac{3 h}{2 \\pi G R t^{2}}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n嫦娥四号是我国探月工程二期发射的月球探测器, 也是人类第一个着陆月球背面的探测器。嫦娥四号于 2018 年 12 月 12 日完成近月制动被月球捕获。若探测器近月环绕周期为 $T$, 探测器降落到月球表面后, 从距月球表面高度为 $h$ 处由静止释放一物体, 测出物体落到月球表面的时间为 $t$, 已知月球半径为 $R$, 引力常量为 $G$, 将月球视为质量均匀分布的球体，通过以上物理量可求得月球的平均密度为（）\n\nA: $\\frac{h}{2 \\pi G R t^{2}}$\nB: $\\frac{\\pi G R t^{2}}{6 h}$\nC: $\\frac{3 \\pi}{G T}$\nD: $\\frac{3 h}{2 \\pi G R t^{2}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "D"
        ],
        "solution": "C. 探测器近月环绕运动, 根据万有引力提供向心力, 有\n\n$$\nG \\frac{M m}{R^{2}}=m \\frac{4 \\pi^{2} R}{T^{2}}\n$$\n\n解得\n\n$$\nM=\\frac{4 \\pi^{2} R^{3}}{G T^{2}}\n$$\n\n根据密度公式 $\\rho=\\frac{M}{V}$ 和体积公式 $V=\\frac{4}{3} \\pi R^{3}$ 解得\n\n$$\n\\rho=\\frac{3 \\pi}{G T^{2}}\n$$\n\n选项 C 错误;\n\n$\\mathrm{ABD}$. 物体落到月球表面有\n\n$$\nh=\\frac{1}{2} g_{\\text {月 }} t^{2}\n$$\n\n月球表面有\n\n$$\nG \\frac{M m}{R^{2}}=m g_{\\text {月 }}\n$$\n\n联立密度公式 $\\rho=\\frac{M}{V}$ 和体积公式 $V=\\frac{4}{3} \\pi R^{3}$ 解得\n\n$$\n\\rho=\\frac{3 h}{2 \\pi G R t^{2}}\n$$\n\n选项 D 正确, $\\mathrm{AB}$ 错误。\n\n故选 D。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_332",
        "problem": "如图, $\\mathrm{A}$ 为太阳系中的天王星, 它绕太阳 $O$ 运行的轨道视为圆时, 运动的轨道半径为 $R_{0}$, 周期为 $T_{0}$ 的匀速圆周运动。天文学家长期观测发现, 天王星实际运动的轨道与圆轨道总有一些偏离, 且每隔 $t_{0}$ 时间发生一次最大偏离, 形成这种现象的原因可能是天王星外侧还存在着一颗未知的行星 $\\mathrm{B}$, 假设行星 $\\mathrm{B}$ 与 $\\mathrm{A}$ 在同一平面内, 且与 $\\mathrm{A}$ 的绕行方向相同, 它对天王星的万有引力引起天王星轨道的偏离, 由此可推测未知行星的运动轨道半径是 ( )\n\n[图1]\nA: $\\frac{t_{0}}{t_{0}-T_{0}} R_{0}$\nB: $R_{0} \\sqrt{\\left(\\frac{t_{0}}{t_{0}-T_{0}}\\right)^{3}}$\nC: $R_{0} \\sqrt[3]{\\left(\\frac{t_{0}-T_{0}}{t_{0}}\\right)^{2}}$\nD: $R_{0} \\sqrt[3]{\\left(\\frac{t_{0}}{t_{0}-T_{0}}\\right)^{2}}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图, $\\mathrm{A}$ 为太阳系中的天王星, 它绕太阳 $O$ 运行的轨道视为圆时, 运动的轨道半径为 $R_{0}$, 周期为 $T_{0}$ 的匀速圆周运动。天文学家长期观测发现, 天王星实际运动的轨道与圆轨道总有一些偏离, 且每隔 $t_{0}$ 时间发生一次最大偏离, 形成这种现象的原因可能是天王星外侧还存在着一颗未知的行星 $\\mathrm{B}$, 假设行星 $\\mathrm{B}$ 与 $\\mathrm{A}$ 在同一平面内, 且与 $\\mathrm{A}$ 的绕行方向相同, 它对天王星的万有引力引起天王星轨道的偏离, 由此可推测未知行星的运动轨道半径是 ( )\n\n[图1]\n\nA: $\\frac{t_{0}}{t_{0}-T_{0}} R_{0}$\nB: $R_{0} \\sqrt{\\left(\\frac{t_{0}}{t_{0}-T_{0}}\\right)^{3}}$\nC: $R_{0} \\sqrt[3]{\\left(\\frac{t_{0}-T_{0}}{t_{0}}\\right)^{2}}$\nD: $R_{0} \\sqrt[3]{\\left(\\frac{t_{0}}{t_{0}-T_{0}}\\right)^{2}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_86694e5d1e9acbe7af1ag-018.jpg?height=317&width=340&top_left_y=2029&top_left_x=344"
        ],
        "answer": [
            "D"
        ],
        "solution": "$\\mathrm{A} 、 \\mathrm{~B}$ 行星发生最大偏离时, 两行星在同一直线上且位于太阳的同一侧, 则有\n\n$$\n\\frac{2 \\pi}{T_{0}} t-\\frac{2 \\pi}{T} t=2 \\pi\n$$\n\n解得未知行星的运行周期\n\n$$\nT=\\frac{t_{0}}{t_{0}-T_{0}} T_{0}\n$$\n\n由开普勒第三定律有\n\n$$\n\\frac{R_{0}^{3}}{T_{0}^{2}}=\\frac{R^{3}}{T^{2}}\n$$\n\n解得\n\n$$\nR=R_{0} \\sqrt[3]{\\left(\\frac{t_{0}}{t_{0}-T_{0}}\\right)^{2}}\n$$\n\n故 D 正确。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_66",
        "problem": "当某一地外行星 (火星、木星、土星、天王星、海王星) 于绕日公转过程中运行到试卷第 48 页，共 150 页\n与地球、太阳成一直线的状态, 且地球恰好位于太阳和外行星之间的这种天文现象叫“冲日”, 冲日前后是观测地外行星的好时机。如图所示是土星冲日示意图, 已知地球质量为 $M$, 半径为 $R$, 公转周期是 1 年, 公转半径为 $r$, 土星质量是地球的 95 倍, 土星半径是地球的 9.5 倍, 土星的公转半径是地球的 9.5 倍。求: $\\left(\\sqrt{9.5^{3}} \\approx 29\\right)$\n土星冲日平均多长时间出现一次？\n\n[图1]",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题的答案是一个数值。\n\n问题:\n当某一地外行星 (火星、木星、土星、天王星、海王星) 于绕日公转过程中运行到试卷第 48 页，共 150 页\n与地球、太阳成一直线的状态, 且地球恰好位于太阳和外行星之间的这种天文现象叫“冲日”, 冲日前后是观测地外行星的好时机。如图所示是土星冲日示意图, 已知地球质量为 $M$, 半径为 $R$, 公转周期是 1 年, 公转半径为 $r$, 土星质量是地球的 95 倍, 土星半径是地球的 9.5 倍, 土星的公转半径是地球的 9.5 倍。求: $\\left(\\sqrt{9.5^{3}} \\approx 29\\right)$\n土星冲日平均多长时间出现一次？\n\n[图1]\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n请记住，你的答案应以年为单位计算，但在给出最终答案时，请不要包含单位。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是不包含任何单位的数值。",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-049.jpg?height=286&width=491&top_left_y=848&top_left_x=337"
        ],
        "answer": [
            "1.04"
        ],
        "solution": "设行星周期为 $T$ ，有\n\n$$\nG \\frac{M m}{r^{2}}=m\\left(\\frac{2 \\pi}{T}\\right)^{2} r\n$$\n\n得\n\n$$\nT=\\sqrt{\\frac{4 \\pi^{2} r^{3}}{G M}}\n$$\n\n所以土星的周期为\n\n$$\nT_{\\text {土 }}=\\sqrt{9.5^{3}} \\text { 年 }=29 \\text { 年 }\n$$\n\n由题设每经 $t$ 年出现土星冲日, 有\n\n$$\n\\frac{t}{T_{\\text {地 }}}-\\frac{t}{T_{ \\pm}}=1\n$$\n\n得\n\n$$\nt=1.04 \\text { 年 }\n$$",
        "answer_type": "NV",
        "unit": [
            "年"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_441",
        "problem": "已知金星绕太阳公转的周期小于 1 年, 则可判定（）\nA: 金星到太阳的距离小于地球到太阳的距离\nB: 金星的质量大于地球的质量\nC: 金星的密度大于地球的密度\nD: 金星的向心加速度大于地球的向心加速度\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n已知金星绕太阳公转的周期小于 1 年, 则可判定（）\n\nA: 金星到太阳的距离小于地球到太阳的距离\nB: 金星的质量大于地球的质量\nC: 金星的密度大于地球的密度\nD: 金星的向心加速度大于地球的向心加速度\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "A",
            "D"
        ],
        "solution": "根据 $G \\frac{m M}{r^{2}}=m \\frac{4 \\pi^{2}}{T^{2}} r$ 得, 周期 $T=\\sqrt{\\frac{4 \\pi^{2} r^{3}}{G M}}$, 由于地球公转的周期大于金星公转周期, 则地球的轨道半径大于金星的轨道半径, 故 A 正确. 金星和地球属于环绕天体, 根据万有引力定律理论无法求出环绕天体的质量, 则无法比较金星的质量和地球质量的大小关系, 故 B 错误. 由于金星和地球的质量无法比较, 半径也未知, 无法比较密度的大小关系, 故 C 错误. 根据 $G \\frac{m M}{r^{2}}=m a$ 得, $a=\\frac{G M}{r^{2}}$, 由于地球的轨道半径大,则地球的向心加速度小, 故 D 正确. 故选 AD.",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_193",
        "problem": "有人提出了一种不用火箭发射人造地球卫星的设想。其设想如下: 沿地球的一条弦挖一通道, 如图乙所示. 在通道的两个出口处 $A$ 和 $B$, 分别将质量为 $M$ 的物体和质量为 $m$ 的待发射卫星同时自由释放, 只要 $M$ 比 $m$ 足够大, 碰撞后, 质量为 $m$ 的物体, 即待发射的卫星就会从通道口 $B$ 冲出通道。(已知地球表面的重力加速度为 $g$, 地球半径为 $\\left.R_{0}\\right)$\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n如图甲所示, 一个质量为 $m$ 的物体(可视为质点)在一个质量均匀分布的大球(半径为 $R$, 密度为 $\\rho)$ 的内部距球心 $r$ 处, 那么 $m$ 与大球之间的万有引力就是 $F=G \\frac{m M^{\\prime}}{r^{2}}$, 式中 $M^{\\prime}$ 是以 $r$ 为半径的球体的质量。试根据所给条件求出物体 $m$ 所受到的万有引力多大?",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n\n问题:\n有人提出了一种不用火箭发射人造地球卫星的设想。其设想如下: 沿地球的一条弦挖一通道, 如图乙所示. 在通道的两个出口处 $A$ 和 $B$, 分别将质量为 $M$ 的物体和质量为 $m$ 的待发射卫星同时自由释放, 只要 $M$ 比 $m$ 足够大, 碰撞后, 质量为 $m$ 的物体, 即待发射的卫星就会从通道口 $B$ 冲出通道。(已知地球表面的重力加速度为 $g$, 地球半径为 $\\left.R_{0}\\right)$\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n如图甲所示, 一个质量为 $m$ 的物体(可视为质点)在一个质量均匀分布的大球(半径为 $R$, 密度为 $\\rho)$ 的内部距球心 $r$ 处, 那么 $m$ 与大球之间的万有引力就是 $F=G \\frac{m M^{\\prime}}{r^{2}}$, 式中 $M^{\\prime}$ 是以 $r$ 为半径的球体的质量。试根据所给条件求出物体 $m$ 所受到的万有引力多大?\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-024.jpg?height=229&width=232&top_left_y=682&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-024.jpg?height=237&width=368&top_left_y=681&top_left_x=570"
        ],
        "answer": [
            "$\\frac{4}{3} \\pi \\rho G m r$"
        ],
        "solution": "$F=\\frac{4}{3} \\pi \\rho G m r$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_30",
        "problem": "某颗地球同步卫星正下方的地球表面上有一观察者, 他用天文望远镜观察被太阳光照射的此卫星, 春分那天 (太阳光直射赤道) 在日落 12 小时内有 $t_{1}$ 时间该观察者看不见此卫星。已知地球半径为 $R$, 地球表面处的重力加速度为 $g$, 地球自转周期为 $T$, 卫星的绕地方向与地球转动方向相同, 不考虑大气对光的折射。下列说法中正确的是 ( )\nA: 同步卫星离地高度为 $\\sqrt[3]{\\frac{g R^{2} T^{2}}{4 \\pi^{2}}}$\nB: 同步卫星加速度小于赤道上物体向心加速度\nC: $t_{1}=\\frac{T \\sin ^{-1} \\frac{R}{\\sqrt[3]{\\frac{g R^{2} T^{2}}{4 \\pi^{2}}}}}{\\pi}$\nD: 同步卫星的加速度大于近地卫星的加速度\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某颗地球同步卫星正下方的地球表面上有一观察者, 他用天文望远镜观察被太阳光照射的此卫星, 春分那天 (太阳光直射赤道) 在日落 12 小时内有 $t_{1}$ 时间该观察者看不见此卫星。已知地球半径为 $R$, 地球表面处的重力加速度为 $g$, 地球自转周期为 $T$, 卫星的绕地方向与地球转动方向相同, 不考虑大气对光的折射。下列说法中正确的是 ( )\n\nA: 同步卫星离地高度为 $\\sqrt[3]{\\frac{g R^{2} T^{2}}{4 \\pi^{2}}}$\nB: 同步卫星加速度小于赤道上物体向心加速度\nC: $t_{1}=\\frac{T \\sin ^{-1} \\frac{R}{\\sqrt[3]{\\frac{g R^{2} T^{2}}{4 \\pi^{2}}}}}{\\pi}$\nD: 同步卫星的加速度大于近地卫星的加速度\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_a9b05ce8eea7b0e40e5eg-005.jpg?height=320&width=394&top_left_y=243&top_left_x=340"
        ],
        "answer": [
            "C"
        ],
        "solution": "A. 设地球同步卫星的轨道半径为 $r$, 其受到的地球万有引力提供向心力, 即\n\n$$\n\\frac{G M m}{r^{2}}=m \\frac{4 \\pi^{2} r}{T^{2}}\n$$\n\n对地面上的物体万有引力等于重力有\n\n$$\n\\frac{G M m}{R^{2}}=m g\n$$\n\n由以上两式可得:\n\n$$\nr=\\sqrt[3]{\\frac{g R^{2} T^{2}}{4 \\pi^{2}}}\n$$\n\n所以同步卫星离地高度为\n\n$$\nh=\\sqrt[3]{\\frac{g R^{2} T^{2}}{4 \\pi^{2}}}-R\n$$\n\n故 A 错误;\n\nB. 根据\n\n$$\na=\\omega^{2} r\n$$\n\n因为同步卫星与赤道上物体转动角速度相同, 同步卫星离地心距离较大, 同步卫星加速度大于赤道上物体向心加速度, 故 B 错误;\n\nC. 根据光的直线传播规律, 日落 12 小时内有 $t_{1}$ 时间该观察者看不见此卫星图示如下图\n所示\n\n[图1]\n\n同步卫星相对地心转过角度为\n\n$$\n\\begin{gathered}\n\\theta=2 \\alpha \\\\\n\\sin \\alpha=\\frac{R}{r}\n\\end{gathered}\n$$\n\n结合\n\n$$\n\\theta=\\omega t=\\frac{2 \\pi}{T} t\n$$\n\n计算得出\n\n$$\nt_{1}=\\frac{T \\sin ^{-1} \\frac{R}{\\sqrt[3]{\\frac{g R^{2} T^{2}}{4 \\pi^{2}}}}}{\\pi}\n$$\n\n所以 $\\mathrm{C}$ 选项是正确的;\n\nD. 根据\n\n$$\nG \\frac{M m}{r^{2}}=m a\n$$\n\n得\n\n$$\na=\\frac{G M}{r^{2}}\n$$\n\n同步卫星的轨道半径比近地卫星轨道半径大, 同步卫星的加速度小于近地卫星的加速度,故 D 错误。\n\n故选 C。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_403",
        "problem": "如图所示是“嫦娥三号”着陆器携“玉兔号”奔月过程中某阶段的运动示意图, 关闭动力的“嫦娥三号”着陆器在月球引力作用下向月球靠近, 并将沿椭圆轨道在 $P$ 处轨迹入圆轨道。已知着陆器绕月做圆周运动的轨道半径为 $r$, 周期为 $T$, 引力常量为 $G$, 下列说法中正确的是 ( )\n\n[图1]\nA: “嫦娥三号”经陏圆轨道到 $P$ 处时的线速度大于经圆形轨道到 $P$ 处时的线速度\nB: “嫦娥三号”经椭圆轨道到 $P$ 处时的加速度和经圆形轨道到 $P$ 处时的加速度不等\nC: “嫦娥三号”携“玉兔号”绕月球做圆周运动的过程中, “玉兔号”所受重力为零\nD: 图中“嫦娥三号”着陆器在 $P$ 处由栯圆轨道进入圆轨道前后机械能守恒\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图所示是“嫦娥三号”着陆器携“玉兔号”奔月过程中某阶段的运动示意图, 关闭动力的“嫦娥三号”着陆器在月球引力作用下向月球靠近, 并将沿椭圆轨道在 $P$ 处轨迹入圆轨道。已知着陆器绕月做圆周运动的轨道半径为 $r$, 周期为 $T$, 引力常量为 $G$, 下列说法中正确的是 ( )\n\n[图1]\n\nA: “嫦娥三号”经陏圆轨道到 $P$ 处时的线速度大于经圆形轨道到 $P$ 处时的线速度\nB: “嫦娥三号”经椭圆轨道到 $P$ 处时的加速度和经圆形轨道到 $P$ 处时的加速度不等\nC: “嫦娥三号”携“玉兔号”绕月球做圆周运动的过程中, “玉兔号”所受重力为零\nD: 图中“嫦娥三号”着陆器在 $P$ 处由栯圆轨道进入圆轨道前后机械能守恒\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_6842b9ceb844a90b34c3g-23.jpg?height=377&width=329&top_left_y=1590&top_left_x=338"
        ],
        "answer": [
            "A"
        ],
        "solution": "$\\mathrm{AD}$. 将运动倒过来分析, 如果从圆轨道向粗圆轨道变轨, 一定要加速做离心运动, 因此在圆轨道上运动的线速度小于椭圆轨道上 $P$ 点的线速度, 故从椭圆轨道进入圆形轨道, 一定进行“刹车”, 机械能一定减小, A 正确, D 错误;\n\nB. “嫦娥三号”在两个轨道上的 $P$ 点受到的万有引力相等, 根据牛顿第二定律, 加速度相等，B 错误;\n\nC. “嫦娥三号”携“玉兔号”绕月球做圆周运动的过程中，“玉兔号”处于失重状态，但不是不受重力, 而是重力全部用来提供“玉兔号”做圆周运动的向心力, $\\mathrm{C}$ 错误。故选 A。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_362",
        "problem": "为简化“天问一号”探测器在火星软着陆的问题, 可以认为地球和火星在同一平面上绕太阳做匀速圆周运动, 如图 1 所示。火星探测器在火星附近的 $A$ 点减速后, 被火星捕获进入了 1 号椭圆轨道, 紧接着在 $B$ 点进行了一次“侧手翻”, 即从与火星赤道平行的 1 号轨道, 调整为经过火星两极的 2 号轨道, 将探测器绕火星飞行的路线从“横着绕”变成\n\n“坚着绕”, 从而实现对火星表面的全面扫描, 如图 2 所示。以火星为参考系, 质量为 $M_{1}$的探测器沿 1 号轨道到达 $B$ 点时速度为 $v_{1}$, 为了实现“侧手翻”, 此时启动发动机, 在极短的时间内喷出部分气体, 假设气体为一次性喷出, 喷气后探测器质量变为 $M_{2}$ 、速度变为与 $v_{1}$ 垂直的 $v_{2}$ 。已知地球的公转周期为 $T_{1}$, 火星的公转周期为 $T_{2}$, 地球公转轨道半径为 $r_{1}$, 以下说法正确的是 ( )\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 火星公转轨道半径为 $r_{2}=\\left(\\frac{T_{2}}{T_{1}}\\right)^{\\frac{3}{2}} r_{1}$\nB: 喷出气体速度 $u$ 的大小为 $\\frac{\\sqrt{\\left(M_{1} v_{1}\\right)^{2}-\\left(M_{2} v_{2}\\right)^{2}}}{M_{1}-M_{2}}$\nC: 假设实现“侧手翻”的能量全部来源于化学能, 化学能向动能转化比例为 $k(k<1)$, 此次“侧手翻”消耗的化学能 $\\Delta E=\\frac{M_{1} M_{2}\\left(v_{1}^{2}+v_{2}^{2}\\right)}{2 k\\left(M_{1}-M_{2}\\right)}$\nD: 考虑到飞行时间和节省燃料, 地球和火星处于图 1 中相对位置时是在地球上发射火星探测器的最佳时机, 则在地球上相邻两次发射火星探测器最佳时机的时间间隔为 $\\Delta t=\\frac{T_{1} T_{2}}{T_{2}-T_{1}}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n为简化“天问一号”探测器在火星软着陆的问题, 可以认为地球和火星在同一平面上绕太阳做匀速圆周运动, 如图 1 所示。火星探测器在火星附近的 $A$ 点减速后, 被火星捕获进入了 1 号椭圆轨道, 紧接着在 $B$ 点进行了一次“侧手翻”, 即从与火星赤道平行的 1 号轨道, 调整为经过火星两极的 2 号轨道, 将探测器绕火星飞行的路线从“横着绕”变成\n\n“坚着绕”, 从而实现对火星表面的全面扫描, 如图 2 所示。以火星为参考系, 质量为 $M_{1}$的探测器沿 1 号轨道到达 $B$ 点时速度为 $v_{1}$, 为了实现“侧手翻”, 此时启动发动机, 在极短的时间内喷出部分气体, 假设气体为一次性喷出, 喷气后探测器质量变为 $M_{2}$ 、速度变为与 $v_{1}$ 垂直的 $v_{2}$ 。已知地球的公转周期为 $T_{1}$, 火星的公转周期为 $T_{2}$, 地球公转轨道半径为 $r_{1}$, 以下说法正确的是 ( )\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 火星公转轨道半径为 $r_{2}=\\left(\\frac{T_{2}}{T_{1}}\\right)^{\\frac{3}{2}} r_{1}$\nB: 喷出气体速度 $u$ 的大小为 $\\frac{\\sqrt{\\left(M_{1} v_{1}\\right)^{2}-\\left(M_{2} v_{2}\\right)^{2}}}{M_{1}-M_{2}}$\nC: 假设实现“侧手翻”的能量全部来源于化学能, 化学能向动能转化比例为 $k(k<1)$, 此次“侧手翻”消耗的化学能 $\\Delta E=\\frac{M_{1} M_{2}\\left(v_{1}^{2}+v_{2}^{2}\\right)}{2 k\\left(M_{1}-M_{2}\\right)}$\nD: 考虑到飞行时间和节省燃料, 地球和火星处于图 1 中相对位置时是在地球上发射火星探测器的最佳时机, 则在地球上相邻两次发射火星探测器最佳时机的时间间隔为 $\\Delta t=\\frac{T_{1} T_{2}}{T_{2}-T_{1}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_86694e5d1e9acbe7af1ag-040.jpg?height=411&width=506&top_left_y=680&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_04_01_86694e5d1e9acbe7af1ag-040.jpg?height=425&width=873&top_left_y=667&top_left_x=840"
        ],
        "answer": [
            "C",
            "D"
        ],
        "solution": "A. 根据开普勒第三定律有\n\n$$\n\\frac{r_{1}^{3}}{T_{1}^{2}}=\\frac{r_{2}^{3}}{T_{2}^{2}}\n$$\n\n可得\n\n$$\nr_{2}=\\left(\\frac{T_{2}}{T_{1}}\\right)^{\\frac{2}{3}} r_{1}\n$$\n\n故 A 错误;\n\nB. 根据动量守恒定律, 在 $v_{1}$ 方向上有\n\n$$\nM_{1} v_{1}=\\left(M_{1}-M_{2}\\right) u_{1}\n$$\n\n在与 $v_{1}$ 垂直的方向上有\n\n$$\nM_{2} v_{2}=\\left(M_{1}-M_{2}\\right) u_{2}\n$$\n\n解得\n\n$$\n\\begin{aligned}\n& u_{1}=\\frac{M_{1} v_{1}}{M_{1}-M_{2}} \\\\\n& u_{2}=\\frac{M_{2} v_{2}}{M_{1}-M_{2}}\n\\end{aligned}\n$$\n\n喷出气体速度 $u$ 的大小为\n\n$$\nu=\\sqrt{u_{1}^{2}+u_{2}^{2}}=\\frac{\\sqrt{\\left(M_{1} v_{1}\\right)^{2}+\\left(M_{2} v_{2}\\right)^{2}}}{M_{1}-M_{2}}\n$$\n\n故 B 错误;\n\nC. 以探测器和喷出气体组成的系统为研究对象, 喷气前的总动能为\n\n$$\nE_{\\mathrm{k} 1}=\\frac{1}{2} M v_{1}^{2}\n$$\n\n喷气后的总动能为\n\n$$\nE_{\\mathrm{k} 2}=\\frac{1}{2} M_{2} v_{2}^{2}+\\frac{1}{2}\\left(M_{1}-M_{2}\\right) u^{2}\n$$\n\n可得消耗的化学能为\n\n$$\n\\Delta E=\\frac{E_{\\mathrm{k} 2}-E_{\\mathrm{k} 1}}{k}=\\frac{M_{1} M_{2}\\left(v_{1}^{2}+v_{2}^{2}\\right)}{2 k\\left(M_{1}-M_{2}\\right)}\n$$\n\n故 C 正确;\n\nD. 在地球上相邻两次发射火星探测器最佳时机的时间间隔 $\\Delta t$ 内, 地球比火星多转一周,则有\n\n$$\n\\frac{\\Delta t}{T_{1}}-\\frac{\\Delta t}{T_{2}}=1\n$$\n\n解得\n\n$$\n\\Delta t=\\frac{T_{1} T_{2}}{T_{2}-T_{1}}\n$$\n\n故 D 正确。\n\n故选 CD。",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_146",
        "problem": "电影中的太空电梯非常吸引人。现假设已经建成了如图所示的太空电梯, 其通过超级缆绳将地球赤道上的固定基地、同步空间站和配重空间站连接在一起, 它们随地球同步旋转。图中配重空间站比同步空间站更高, $\\mathrm{P}$ 是缆绳上的一个平台。则下列说法正确的是 ( )\n\n[图1]\nA: 太空电梯上各点线速度的平方与该点离地球球心的距离成反比\nB: 超级缆绳对 $\\mathrm{P}$ 平台的作用力方向指向地心\nC: 若从 $\\mathrm{P}$ 平台向外自由释放一个小物块, 则小物块会一边朝 $\\mathrm{P}$ 点转动的方向向前运动一边落向地球\nD: 若两空间站之间缆绳断裂, 配重空间站将绕地球做椭圆运动, 且断裂处为椭圆的远地点\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n电影中的太空电梯非常吸引人。现假设已经建成了如图所示的太空电梯, 其通过超级缆绳将地球赤道上的固定基地、同步空间站和配重空间站连接在一起, 它们随地球同步旋转。图中配重空间站比同步空间站更高, $\\mathrm{P}$ 是缆绳上的一个平台。则下列说法正确的是 ( )\n\n[图1]\n\nA: 太空电梯上各点线速度的平方与该点离地球球心的距离成反比\nB: 超级缆绳对 $\\mathrm{P}$ 平台的作用力方向指向地心\nC: 若从 $\\mathrm{P}$ 平台向外自由释放一个小物块, 则小物块会一边朝 $\\mathrm{P}$ 点转动的方向向前运动一边落向地球\nD: 若两空间站之间缆绳断裂, 配重空间站将绕地球做椭圆运动, 且断裂处为椭圆的远地点\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_a9b05ce8eea7b0e40e5eg-022.jpg?height=443&width=947&top_left_y=144&top_left_x=332"
        ],
        "answer": [
            "C"
        ],
        "solution": "A. 太空电梯上各点具有相同的角速度, 根据 $v=\\omega r$ 可知, 太空电梯上各点线速度与该点离地球球心的距离成正比, 故 A 错误;\n\nB. P 平台如果只受地球万有引力, 则圆周运动角速度比同步空间站要快, 而实际圆周运动角速度等于同步空间站角速度, 则在万有引力之外, $\\mathrm{P}$ 平台还受到缆绳拉力, 故地球的引力与缆绳拉力提供 $\\mathrm{P}$ 平台做圆周运动所需的向心力, $\\mathrm{P}$ 平台做圆周运动所需的向心力小于地球对它的万有引力, 所以超级缆绳对 $\\mathrm{P}$ 平台的作用力方向背离地心, 故 $\\mathrm{B}$错误;\n\nC. 若从 $\\mathrm{P}$ 平台向外自由释放一个小物块, 则小物块会一边朝 $\\mathrm{P}$ 点转动的方向向前运动一边落向地球，做近心运动，故 C 正确;\n\nD. 若两空间站之间缆绳断裂, 配重空间站将绕地球做椭圆运动, 其断裂处为椭圆的近地点, 因为在近地点线速度较大, 半径较小, 需要的向心力更大, 故 D 错误。故选 C。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_408",
        "problem": "如图所示, 横截面积为 $A$ 、质量为 $m$ 的柱状飞行器沿半径为 $R$ 的圆形轨道在高空绕地球做无动力运行。将地球看作质量为 $M$ 的均匀球体。万有引力常量为 $G$ 。\n\n在飞行器运行轨道附近范围内有密度为 $\\rho$ (恒量) 的稀薄空气。稀薄空气可看成是由彼此没有相互作用的均匀小颗粒组成，所有小颗粒原来都静止。假设每个小颗粒与飞行器碰撞后具有与飞行器相同的速度, 且碰撞时间很短。频繁碰撞会对飞行器产生持续阻力, 飞行器的轨道高度会逐渐降低。观察发现飞行器绕地球运行很多圈之后, 其轨道高度下降了 $\\Delta H$ 。由于 $\\Delta H \\leq R$, 可将飞行器绕地球运动的每一圈运动均视为匀速圆周运动。已知当飞行器到地球球心距离为 $r$ 时, 飞行器与地球组成的系统具有的引力势能 $E_{\\mathrm{p}}=-\\frac{G M m}{r}$ 。请根据上述条件推导:\n\n飞行器由半径为 $R$ 的轨道下降 $\\Delta H$ 的过程中, 飞行器绕地球运动圈数 $n$ 的表达式。\n\n[图1]",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n\n问题:\n如图所示, 横截面积为 $A$ 、质量为 $m$ 的柱状飞行器沿半径为 $R$ 的圆形轨道在高空绕地球做无动力运行。将地球看作质量为 $M$ 的均匀球体。万有引力常量为 $G$ 。\n\n在飞行器运行轨道附近范围内有密度为 $\\rho$ (恒量) 的稀薄空气。稀薄空气可看成是由彼此没有相互作用的均匀小颗粒组成，所有小颗粒原来都静止。假设每个小颗粒与飞行器碰撞后具有与飞行器相同的速度, 且碰撞时间很短。频繁碰撞会对飞行器产生持续阻力, 飞行器的轨道高度会逐渐降低。观察发现飞行器绕地球运行很多圈之后, 其轨道高度下降了 $\\Delta H$ 。由于 $\\Delta H \\leq R$, 可将飞行器绕地球运动的每一圈运动均视为匀速圆周运动。已知当飞行器到地球球心距离为 $r$ 时, 飞行器与地球组成的系统具有的引力势能 $E_{\\mathrm{p}}=-\\frac{G M m}{r}$ 。请根据上述条件推导:\n\n飞行器由半径为 $R$ 的轨道下降 $\\Delta H$ 的过程中, 飞行器绕地球运动圈数 $n$ 的表达式。\n\n[图1]\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-058.jpg?height=445&width=314&top_left_y=1391&top_left_x=354"
        ],
        "answer": [
            "$\\frac{m \\Delta H}{4 \\pi \\rho A(R-\\Delta H) R}$"
        ],
        "solution": "设飞行器在半径为 $R$ 的轨道上运行时的动能为 $E_{\\mathrm{k} 1}$ 、势能为 $E_{\\mathrm{p} 1}$ 、机械能为 $E_{1}$ 。则有\n\n$$\nE_{\\mathrm{k} 1}=\\frac{1}{2} m v^{2}, \\quad E_{\\mathrm{p} 1}=-\\frac{G M m}{R}\n$$\n\n解得\n\n$$\nE_{1}=-\\frac{G M m}{2 R}\n$$\n\n飞行器高度下降 $\\Delta H$, 其运行的轨道半径变为 $(R-\\Delta H)$, 设在该该轨道上飞行器的机械能为 $E_{2}$, 同理可知其机械能为\n\n$$\nE_{2}=-\\frac{G M m}{2(R-\\Delta H)}\n$$\n\n可知在该过程中飞行器机械能的改变量为\n\n$$\n\\Delta E=\\frac{G M m}{2 R}-\\frac{G M m}{2(R-\\Delta H)}\n$$\n\n飞行器机械能减小是因为克服大气阻力做了功, 设卫星在沿轨道半径为 $R$ 的轨道上运行一周的过程中, 稀薄空气颗粒作用于飞行器上的阻力所做的功为 $W_{0}$, 利用累计法可知\n\n$$\nW_{0}=-F \\cdot 2 \\pi R=-2 \\pi \\rho A G M\n$$\n\n上式表明, 卫星在绕不同轨道运行一周的过程中, 稀薄空气颗粒施加的阻力所做的功为恒量, 与轨道半径无关, 则有\n\n$$\n\\Delta E=n W_{0}\n$$\n\n解得\n\n$$\nn=\\frac{m \\Delta H}{4 \\pi \\rho A(R-\\Delta H) R}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_1",
        "problem": "如图所示。甲、乙为地球赤道面内围绕地球运转的通讯卫星。已知甲是与地面相对静止的同步卫星; 乙的运转方向与地球自转方向相反, 轨道半径为地球半径的 2 倍, 周期为 $T$, 在地球赤道上的 $P$ 点有一位观测者，观测者始终相对于地面静止。若地球半径为 $R$, 地球的自转周期为 $T_{0}$ 。求:\n\n卫星甲距离地面的高度;\n\n[图1]",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n\n问题:\n如图所示。甲、乙为地球赤道面内围绕地球运转的通讯卫星。已知甲是与地面相对静止的同步卫星; 乙的运转方向与地球自转方向相反, 轨道半径为地球半径的 2 倍, 周期为 $T$, 在地球赤道上的 $P$ 点有一位观测者，观测者始终相对于地面静止。若地球半径为 $R$, 地球的自转周期为 $T_{0}$ 。求:\n\n卫星甲距离地面的高度;\n\n[图1]\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-092.jpg?height=440&width=397&top_left_y=1096&top_left_x=338"
        ],
        "answer": [
            "$2 R \\cdot \\sqrt[3]{\\frac{T_{0}^{2}}{T^{2}}}-R$"
        ],
        "solution": "甲乙两卫星均受万有引力提供向心力, 由开普勒第三定律有\n\n$$\n\\frac{T^{2}}{(2 R)^{3}}=\\frac{T_{0}^{2}}{(h+R)^{3}}\n$$\n\n解得卫星甲距离地面的高度为\n\n$$\nh=2 R \\cdot \\sqrt[3]{\\frac{T_{0}^{2}}{T^{2}}}-R\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_809",
        "problem": "$H_{\\alpha}$ is a prominent absorption line in the sun's spectrum with $\\lambda=656.281$. For an observer on $\\alpha$-Centauri, assuming the Jupiter-Sun orbital plane is in the line of sight, calculate the magnitude of the periodic shift in wavelength ( $\\mathrm{nm}$ ) for the $H_{\\alpha}$ line.\nA: $2.7 \\times 10^{-5} \\mathrm{~nm}$\nB: $2.7 \\times 10^{-4} \\mathrm{~nm}$\nC: $2.7 \\times 10^{-3} \\mathrm{~nm}$\nD: $2.7 \\times 10^{-2} \\mathrm{~nm}$\n",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n$H_{\\alpha}$ is a prominent absorption line in the sun's spectrum with $\\lambda=656.281$. For an observer on $\\alpha$-Centauri, assuming the Jupiter-Sun orbital plane is in the line of sight, calculate the magnitude of the periodic shift in wavelength ( $\\mathrm{nm}$ ) for the $H_{\\alpha}$ line.\n\nA: $2.7 \\times 10^{-5} \\mathrm{~nm}$\nB: $2.7 \\times 10^{-4} \\mathrm{~nm}$\nC: $2.7 \\times 10^{-3} \\mathrm{~nm}$\nD: $2.7 \\times 10^{-2} \\mathrm{~nm}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": [
            "A"
        ],
        "solution": "$H_{\\alpha}$ is a prominent absorption line in the sun's spectrum with $\\lambda=656.281$. For an observer on $\\alpha$-Centauri, assuming the Jupiter-Sun orbital plane is in the line of sight, calculate the magnitude of the periodic shift in wavelength $(\\mathrm{nm})$ for the $H_{\\alpha}$ line.\n\nSolution: A: $2.7 \\times 10^{-5} \\mathrm{~nm}$\n\nUse\n\n$$\n\\frac{\\Delta_{\\lambda}}{\\lambda} \\approx \\frac{v}{c}\n$$",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_938",
        "problem": "Which of these stars are closest to the Sun in October 2021?\nA: Rigel (Right ascension $=05^{\\mathrm{h}} 15^{\\mathrm{m}}$, declination $\\left.=-8.20^{\\circ}\\right)$\nB: Regulus (Right ascension $=10^{\\mathrm{h}} 08^{\\mathrm{m}}$, declination $=+11.97^{\\circ}$ )\nC: Spica (Right ascension $=13^{\\mathrm{h}} 25^{\\mathrm{m}}$, declination $=-11.16^{\\circ}$ )\nD: Vega $\\left(\\right.$ Right ascension $=18^{\\mathrm{h}} 37^{\\mathrm{m}}$, declination $\\left.=+38.78^{\\circ}\\right)$\n",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of these stars are closest to the Sun in October 2021?\n\nA: Rigel (Right ascension $=05^{\\mathrm{h}} 15^{\\mathrm{m}}$, declination $\\left.=-8.20^{\\circ}\\right)$\nB: Regulus (Right ascension $=10^{\\mathrm{h}} 08^{\\mathrm{m}}$, declination $=+11.97^{\\circ}$ )\nC: Spica (Right ascension $=13^{\\mathrm{h}} 25^{\\mathrm{m}}$, declination $=-11.16^{\\circ}$ )\nD: Vega $\\left(\\right.$ Right ascension $=18^{\\mathrm{h}} 37^{\\mathrm{m}}$, declination $\\left.=+38.78^{\\circ}\\right)$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": [
            "C"
        ],
        "solution": "<br> The right ascension (RA) of the Sun is $0 \\mathrm{~h} 00 \\mathrm{~m}$ at the vernal equinox in <br> March. October is 7 months later, so the expected RA is roughly <br> $\\frac{7}{12} \\times 24^{\\mathrm{h}}=14 \\mathrm{~h}$. Spica is the closest to that, whilst also having small <br> declination (so is close to the ecliptic)",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_492",
        "problem": "三颗人造卫星 $\\mathrm{A} 、 \\mathrm{~B} 、 \\mathrm{C}$ 都在赤道正上方同方向绕地球做匀速圆周运动, $\\mathrm{A} 、 \\mathrm{C}$ 为地球同步卫星, 某时刻 $\\mathrm{A} 、 \\mathrm{~B}$ 相距最近, 如图所示。已知地球自转周期为 $T_{1}, \\mathrm{~B}$ 的运行周期为 $T_{2}$, 则下列说法正确的是（）\n\n[图1]\nA: A、C 受到的万有引力大小相等\nB: 经过时间 $\\frac{T_{1} T_{2}}{T_{1}+T_{2}}, \\mathrm{~A} 、 \\mathrm{~B}$ 再次相距最近\nC: 经过时间 $\\frac{T_{1} T_{2}}{2\\left(T_{1}-T_{2}\\right)}, \\mathrm{A} 、 \\mathrm{~B}$ 相距最远\nD: 在相同时间内, $\\mathrm{A}$ 与地心连线扫过的面积大于 B 与地心连线扫过的面积\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n三颗人造卫星 $\\mathrm{A} 、 \\mathrm{~B} 、 \\mathrm{C}$ 都在赤道正上方同方向绕地球做匀速圆周运动, $\\mathrm{A} 、 \\mathrm{C}$ 为地球同步卫星, 某时刻 $\\mathrm{A} 、 \\mathrm{~B}$ 相距最近, 如图所示。已知地球自转周期为 $T_{1}, \\mathrm{~B}$ 的运行周期为 $T_{2}$, 则下列说法正确的是（）\n\n[图1]\n\nA: A、C 受到的万有引力大小相等\nB: 经过时间 $\\frac{T_{1} T_{2}}{T_{1}+T_{2}}, \\mathrm{~A} 、 \\mathrm{~B}$ 再次相距最近\nC: 经过时间 $\\frac{T_{1} T_{2}}{2\\left(T_{1}-T_{2}\\right)}, \\mathrm{A} 、 \\mathrm{~B}$ 相距最远\nD: 在相同时间内, $\\mathrm{A}$ 与地心连线扫过的面积大于 B 与地心连线扫过的面积\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_86694e5d1e9acbe7af1ag-106.jpg?height=329&width=329&top_left_y=1640&top_left_x=338"
        ],
        "answer": [
            "C",
            "D"
        ],
        "solution": "$\\mathrm{A}$. 卫星 $\\mathrm{A}$ 和 $\\mathrm{C}$ 的角速度与地球赤道物体自转的角速度相等, 轨道半径大于地球半径，根据\n\n$$\na_{n}=r \\omega^{2}\n$$\n可知, 卫星 $\\mathrm{A}$ 和 $\\mathrm{C}$ 的向心加速度相等, 但 $\\mathrm{A} 、 \\mathrm{C}$ 质量未知, 无法得出向心力大小, 故 $\\mathrm{A}$错误;\n\nB. 卫星 A、B 由相距最近到再次相距最近, 圆周运动转过的角度差为 $2 \\pi$, 所以可得\n\n$$\n\\omega_{B} t-\\omega_{A} t=2 \\pi\n$$\n\n其中\n\n$$\n\\begin{aligned}\n\\omega_{B} & =\\frac{2 \\pi}{T_{2}} \\\\\n\\omega_{A} & =\\frac{2 \\pi}{T_{1}}\n\\end{aligned}\n$$\n\n则经历的时间\n\n$$\nt=\\frac{T_{1} T_{2}}{T_{1}-T_{2}}\n$$\n\n故 B 错误;\n\nC. 卫星 A、B 由相距最近到相距最远, 圆周运动转过的角度差为 $\\pi$, 所以可得\n\n$$\n\\omega_{B} t-\\omega_{A} t=\\pi\n$$\n\n其中\n\n$$\n\\begin{aligned}\n& \\omega_{B}=\\frac{2 \\pi}{T_{2}} \\\\\n& \\omega_{A}=\\frac{2 \\pi}{T_{1}}\n\\end{aligned}\n$$\n\n则经历的时间\n\n$$\nt^{\\prime}=\\frac{T_{1} T_{2}}{2\\left(T_{1}-T_{2}\\right)}\n$$\n\n故 C 正确;\n\nD. 绕地球运动的卫星与地心的连线在相同时间 $t$ 内扫过的面积\n\n$$\nS=\\frac{1}{2} v t r\n$$\n\n由万有引力提供向心力, 可知\n\n$$\nG \\frac{M m}{r^{2}}=m \\frac{v^{2}}{r}\n$$\n\n整理解得\n\n$$\nS=\\frac{1}{2} v t r=\\frac{t}{2} \\sqrt{G M r}\n$$\n\n可知, 在相同时间内, $\\mathrm{A}$ 与地心连线扫过的面积大于 $\\mathrm{B}$ 与地心连线扫过的面积, 故 $\\mathrm{D}$正确。\n\n故选 CD。",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_144",
        "problem": "“天问一号”探测器需要通过霍曼转移轨道从地球发送到火星, 地球轨道和火星轨道看成圆形轨道, 此时霍曼转移轨道是一个近日点 $M$ 和远日点 $P$ 都与地球轨道、火星轨道相切的粗圆轨道 (如图所示)。在近日点短暂点火后“天问一号”进入霍曼转移轨道,接着“天问一号”沿着这个轨道直至抵达远日点, 然后再次点火进入火星轨道。已知万有引力常量为 $G$, 太阳质量为 $m$, 地球轨道和火星轨道半径分别为 $r$ 和 $R$, 地球、火星、 “天问一号”运行方向都为逆时针方向。下列说法正确的是（）\n\n[图1]\nA: 两次点火方向都与运动方向相同\nB: 两次点火之间的时间为 $\\frac{\\pi}{2 \\sqrt{2}} \\sqrt{\\frac{(r+R)^{3}}{G m}}$\nC: “天问一号”在地球轨道上的角速度小于在火星轨道上的角速度\nD: “天问一号”在转移轨道上近日点的速度大小等于地球公转速度大小\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n“天问一号”探测器需要通过霍曼转移轨道从地球发送到火星, 地球轨道和火星轨道看成圆形轨道, 此时霍曼转移轨道是一个近日点 $M$ 和远日点 $P$ 都与地球轨道、火星轨道相切的粗圆轨道 (如图所示)。在近日点短暂点火后“天问一号”进入霍曼转移轨道,接着“天问一号”沿着这个轨道直至抵达远日点, 然后再次点火进入火星轨道。已知万有引力常量为 $G$, 太阳质量为 $m$, 地球轨道和火星轨道半径分别为 $r$ 和 $R$, 地球、火星、 “天问一号”运行方向都为逆时针方向。下列说法正确的是（）\n\n[图1]\n\nA: 两次点火方向都与运动方向相同\nB: 两次点火之间的时间为 $\\frac{\\pi}{2 \\sqrt{2}} \\sqrt{\\frac{(r+R)^{3}}{G m}}$\nC: “天问一号”在地球轨道上的角速度小于在火星轨道上的角速度\nD: “天问一号”在转移轨道上近日点的速度大小等于地球公转速度大小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_ef01104c57d69d8b0f5ag-075.jpg?height=466&width=488&top_left_y=1055&top_left_x=336"
        ],
        "answer": [
            "B"
        ],
        "solution": "A. 两次点火都让探测器做加速运动, 因此点火的方向都与运动方向相反, $\\mathrm{A}$错误;\n\nB. 地球绕太阳运行时, 根据牛顿第二定律\n\n$$\n\\frac{G m}{r^{2}}=\\frac{4 \\pi^{2}}{T_{1}^{2}} r\n$$\n\n由根据开普勒第三定律\n\n$$\n\\frac{r^{3}}{T_{1}^{2}}=\\frac{\\left(\\frac{R+r}{2}\\right)^{3}}{T_{2}^{2}}\n$$\n\n可得“天问一号”探测器沿椭圆轨道运行的周期\n\n$$\nT_{2}=\\frac{\\pi}{\\sqrt{2}} \\sqrt{\\frac{(r+R)^{3}}{G m}}\n$$\n\n而两次点火之间的时间\n\n$$\nt=\\frac{1}{2} T_{2}=\\frac{\\pi}{2 \\sqrt{2}} \\sqrt{\\frac{(r+R)^{3}}{G m}}\n$$\n\nB 正确;\n\nC. 绕太阳运行时, 根据\n\n$$\nG \\frac{m m_{0}}{r^{2}}=m_{0} \\omega^{2} r\n$$\n\n解得 $\\omega=\\sqrt{\\frac{G m}{r^{3}}}$, 可得轨道半径越大, 角速度越小, 因此“天问一号”在地球轨道上的角速度大于在火星轨道上的角速度, $\\mathrm{C}$ 错误;\n\nD. 由于“天问一号”通过近日点后将做离心运动, 因此在转移轨道上近日点的速度大于地球公转速度, D 错误。\n\n故选 B。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_242",
        "problem": "科学家在银河系发现一颗类地行星, 半径是地球半径的两倍, 质量是地球质量的三倍。卫星 $a 、 b$ 分别绕地球、类地行星做匀速圆周运动, 它们距中心天体表面的高度均\n等于地球的半径。则卫星 $a 、 b$ 的（）\n\n[图1]\nA: 线速度之比为 $1: \\sqrt{2}$\nB: 角速度之比为 $3: 2 \\sqrt{2}$\nC: 周期之比为 $2 \\sqrt{2}: \\sqrt{3}$\nD: 加速度之比为 4: 3\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n科学家在银河系发现一颗类地行星, 半径是地球半径的两倍, 质量是地球质量的三倍。卫星 $a 、 b$ 分别绕地球、类地行星做匀速圆周运动, 它们距中心天体表面的高度均\n等于地球的半径。则卫星 $a 、 b$ 的（）\n\n[图1]\n\nA: 线速度之比为 $1: \\sqrt{2}$\nB: 角速度之比为 $3: 2 \\sqrt{2}$\nC: 周期之比为 $2 \\sqrt{2}: \\sqrt{3}$\nD: 加速度之比为 4: 3\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_cbd0a496f6e2fb8d7781g-032.jpg?height=360&width=680&top_left_y=248&top_left_x=334"
        ],
        "answer": [
            "A",
            "B"
        ],
        "solution": "由万有引力提供相信力可得\n\n$$\nG \\frac{M m}{R^{2}}=m \\frac{v^{2}}{R}=m \\omega^{2} R=m \\frac{4 \\pi^{2} R}{T^{2}}=m a\n$$\n\n可得\n\n$$\n\\begin{gathered}\nv=\\sqrt{\\frac{G M}{R}} \\\\\n\\omega=\\sqrt{\\frac{G M}{R^{3}}} \\\\\nT=2 \\pi \\sqrt{\\frac{R^{3}}{G M}} \\\\\na=\\frac{G M}{R^{2}}\n\\end{gathered}\n$$\n\nA. 卫星 $a 、 b$ 的线速度之比\n\n$$\n\\frac{v_{a}}{v_{b}}=\\frac{\\sqrt{\\frac{G M}{2 R}}}{\\sqrt{\\frac{G \\cdot 3 M}{3 R}}}=\\frac{1}{\\sqrt{2}}\n$$\n\nA 正确;\n\nB. 卫星 $a 、 b$ 的角速度之比\n\n$$\n\\frac{\\omega_{a}}{\\omega_{b}}=\\frac{\\sqrt{\\frac{G M}{(2 R)^{3}}}}{\\sqrt{\\frac{G \\cdot 3 M}{(3 R)^{3}}}}=\\frac{3}{2 \\sqrt{2}}\n$$\n\nB 正确;\n\nC. 卫星 $a 、 b$ 的周期之比\n\n$$\n\\frac{T_{a}}{T_{b}}=\\frac{2 \\pi \\sqrt{\\frac{(2 R)^{3}}{G M}}}{2 \\pi \\sqrt{\\frac{(3 R)^{3}}{G \\cdot 3 M}}}=\\frac{2 \\sqrt{2}}{3}\n$$\n\nC 错误;\n\nD. 卫星 $a 、 b$ 的向心加速度之比\n\n$$\n\\frac{a_{a}}{a_{b}}=\\frac{\\frac{G M}{(2 R)^{2}}}{\\frac{G \\cdot 3 M}{(3 R)^{2}}}=\\frac{3}{4}\n$$\n\n$\\mathrm{D}$ 错误。\n\n故选 $\\mathrm{AB}$ 。",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_852",
        "problem": "TESS Object of Interest (TOI) 402.01 has an orbital period of $4.756 \\pm 0.000023$ (days) and was last observed to transit on $2139.1 \\pm 0.0027008$ (in TESS Julian days, i.e. BJD - 2457000). For follow-up observation, we would like to predict the next transit - this would be the 23rd transit since the last observation. In TESS Julian days, when will the next transit occur?\nA: $2243.732 \\pm 0.0027238$ TESS JD\nB: $2248.488 \\pm 0.000529$ TESS JD\nC: $2248.488 \\pm 0.0027238$ TESS JD\nD: $2248.488 \\pm 0.0032298$ TESS JD\n",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTESS Object of Interest (TOI) 402.01 has an orbital period of $4.756 \\pm 0.000023$ (days) and was last observed to transit on $2139.1 \\pm 0.0027008$ (in TESS Julian days, i.e. BJD - 2457000). For follow-up observation, we would like to predict the next transit - this would be the 23rd transit since the last observation. In TESS Julian days, when will the next transit occur?\n\nA: $2243.732 \\pm 0.0027238$ TESS JD\nB: $2248.488 \\pm 0.000529$ TESS JD\nC: $2248.488 \\pm 0.0027238$ TESS JD\nD: $2248.488 \\pm 0.0032298$ TESS JD\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": [
            "D"
        ],
        "solution": "The predicted epoch of the next transit is $(2139.1 \\pm 0.0027008)+23 *(4.756 \\pm 0.000023)=$ $2248.488 \\pm 0.0032298$. (The error on the last observation gets added to 23 times the error on the period to get the final period.)",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_473",
        "problem": "土卫十和土卫十一是土星的两颗卫星, 都沿近似为圆周的轨道线土星运动. 其参数如表, 两卫星相比土卫十\n\n|  | 卫星半径 $(\\mathrm{m})$ | 卫星质量 $(\\mathrm{kg})$ | 轨道半径 $(\\mathrm{m})$ |\n| :--- | :--- | :--- | :--- |\n\n\n| 土卫十 | $8.90 \\times 10^{4}$ | $2.01 \\times 10^{18}$ | $1.51 \\times 10^{18}$ |\n| :--- | :--- | :--- | :--- |\n| 土卫十一 | $5.70 \\times 10^{4}$ | $5.60 \\times 10^{17}$ | $1.51 \\times 10^{18}$ |\nA: 受土星的万有引力较大\nB: 绕土星的圆周运动的周期较大\nC: 绕土星做圆周运动的向心加速度较大\nD: 动能较大\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n土卫十和土卫十一是土星的两颗卫星, 都沿近似为圆周的轨道线土星运动. 其参数如表, 两卫星相比土卫十\n\n|  | 卫星半径 $(\\mathrm{m})$ | 卫星质量 $(\\mathrm{kg})$ | 轨道半径 $(\\mathrm{m})$ |\n| :--- | :--- | :--- | :--- |\n\n\n| 土卫十 | $8.90 \\times 10^{4}$ | $2.01 \\times 10^{18}$ | $1.51 \\times 10^{18}$ |\n| :--- | :--- | :--- | :--- |\n| 土卫十一 | $5.70 \\times 10^{4}$ | $5.60 \\times 10^{17}$ | $1.51 \\times 10^{18}$ |\n\nA: 受土星的万有引力较大\nB: 绕土星的圆周运动的周期较大\nC: 绕土星做圆周运动的向心加速度较大\nD: 动能较大\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "A",
            "D"
        ],
        "solution": "根据 $G M m / r^{2}=m a$ 向 $=m v^{2} / r=m r \\omega^{2}=m r(2 \\pi / T)^{2}$, 两卫星相比, 土卫十受土星的万有引力较大, 速度较大, 动能也较大.\n\n根据根据 $F=G \\frac{M m}{R^{2}}=m \\frac{v^{2}}{R}=m \\frac{4 \\pi^{2}}{T^{2}} r=m \\omega^{2} R=m a$ 进行比较, 可知, 土卫十受引力较\n\n大. 轨道半径一样, 速度、周期和向心加速度大小相等, 但是土卫十质量较大, 动能较大, 故 $\\mathrm{AD}$ 正确 $\\mathrm{BC}$ 错误.",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_498",
        "problem": "在人类的某次星际探测任务中, 发现一个可能宜居的星球。为了更详细的了解该星球的情况, 需要进行登陆探测。在登陆前, 飞船先环绕该星球在半径为 $3 R$ 的圆轨道上做匀速圆周运动, $R$ 为该星球的半径。某一时刻, 飞船运行到 $A$ 点, 此时飞船将登陆器向反方向射出, 但登陆器仍向前运动, 并进入图示的内部栯圆轨道登上该星球表面, 而飞船立即启动发动机进行调整，使其仍保持在圆轨道运行。登陆器在该星球表面探测一段时间后，重新点火加速沿原来的椭圆轨道回到脱离点实现对接。已知该星球的质量为\n\n$M$, 飞船的质量为 $m_{1}$, 登陆器的质量为 $m_{2}$, 且 $m_{1}=2 m_{2}$, 忽略登陆器点火加速的时间,引力常量为 $G$ 。求:飞船在圆轨道上运行时的速度 $v_{0}$ 与周期 $T$ 的大小?[图1]",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题包含多个待求解的量。\n\n问题:\n在人类的某次星际探测任务中, 发现一个可能宜居的星球。为了更详细的了解该星球的情况, 需要进行登陆探测。在登陆前, 飞船先环绕该星球在半径为 $3 R$ 的圆轨道上做匀速圆周运动, $R$ 为该星球的半径。某一时刻, 飞船运行到 $A$ 点, 此时飞船将登陆器向反方向射出, 但登陆器仍向前运动, 并进入图示的内部栯圆轨道登上该星球表面, 而飞船立即启动发动机进行调整，使其仍保持在圆轨道运行。登陆器在该星球表面探测一段时间后，重新点火加速沿原来的椭圆轨道回到脱离点实现对接。已知该星球的质量为\n\n$M$, 飞船的质量为 $m_{1}$, 登陆器的质量为 $m_{2}$, 且 $m_{1}=2 m_{2}$, 忽略登陆器点火加速的时间,引力常量为 $G$ 。求:飞船在圆轨道上运行时的速度 $v_{0}$ 与周期 $T$ 的大小?[图1]\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你的最终解答的量应该按以下顺序输出：[飞船在圆轨道上运行时的速度 $v_{0}$, 飞船在圆轨道上运行时的周期 $T$ ]\n它们的答案类型依次是[数值, 数值]\n你需要在输出的最后用以下格式总结答案：“最终答案是\\boxed{ANSWER}”，其中ANSWER应为你的最终答案序列，用英文逗号分隔，例如：5, 7, 2.5",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-001.jpg?height=514&width=371&top_left_y=1439&top_left_x=340"
        ],
        "answer": [
            "$v_{0}=\\sqrt{\\frac{G M}{3 R}}$",
            "$T=6 \\pi R \\sqrt{\\frac{3 R}{G M}}$"
        ],
        "solution": "(1) 飞船在圆轨道上运行时, 根据万有引力提供向心力\n\n$$\n\\frac{G M m_{1}}{9 R^{2}}=\\frac{m_{1} v_{0}^{2}}{3 R}\n$$\n\n解得\n\n$$\nv_{0}=\\sqrt{\\frac{G M}{3 R}}\n$$\n\n周期\n\n$$\nT=\\frac{2 \\pi \\times 3 R}{v_{0}}=6 \\pi R \\sqrt{\\frac{3 R}{G M}}\n$$",
        "answer_type": "MPV",
        "unit": [
            null,
            null
        ],
        "answer_sequence": [
            "飞船在圆轨道上运行时的速度 $v_{0}$",
            "飞船在圆轨道上运行时的周期 $T$ "
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_379",
        "problem": "中国科学家提出的“以石击石”改变小行星飞行轨迹的方案为人类应对小行星的潜在威胁提供了新的思路。如图所示, 已知天体在轨道上的运动方向均为顺时针, 地球公转圆轨道的半径为 $R$ 、周期为 $T_{0}$, 偏转圆轨道的半径为 $4 R$, 若小行星与地球预计在 $A$ 点撞击, 设想在地球运行到 $B$ 点时发射无人飞行器, 飞行器在太空中捕获百吨级质量的岩石后构成质量为 $m_{0}$ 的组合撞击体, 该撞击体在小行星椭圆轨道的远日点 $C$ 沿切线与质量为 $3 m_{0}$ 的小行星发生完全非弹性碰撞, 从而使小行星改变飞行轨迹, 到达偏转圆轨道稳定运行。组合撞击体与小行星撞击前瞬间的速度大小是地球公转线速度的 $\\frac{3}{4}$, 则小行星撞击前在椭圆轨道的远日点 $C$ 的速度大小为 ( )\n\n[图1]\nA: $\\frac{6 \\pi R}{7 T_{0}}$\nB: $\\frac{5 \\pi R}{6 T_{0}}$\nC: $\\frac{3 \\pi R}{4 T_{0}}$\nD: $\\frac{2 \\pi R}{3 T_{0}}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n中国科学家提出的“以石击石”改变小行星飞行轨迹的方案为人类应对小行星的潜在威胁提供了新的思路。如图所示, 已知天体在轨道上的运动方向均为顺时针, 地球公转圆轨道的半径为 $R$ 、周期为 $T_{0}$, 偏转圆轨道的半径为 $4 R$, 若小行星与地球预计在 $A$ 点撞击, 设想在地球运行到 $B$ 点时发射无人飞行器, 飞行器在太空中捕获百吨级质量的岩石后构成质量为 $m_{0}$ 的组合撞击体, 该撞击体在小行星椭圆轨道的远日点 $C$ 沿切线与质量为 $3 m_{0}$ 的小行星发生完全非弹性碰撞, 从而使小行星改变飞行轨迹, 到达偏转圆轨道稳定运行。组合撞击体与小行星撞击前瞬间的速度大小是地球公转线速度的 $\\frac{3}{4}$, 则小行星撞击前在椭圆轨道的远日点 $C$ 的速度大小为 ( )\n\n[图1]\n\nA: $\\frac{6 \\pi R}{7 T_{0}}$\nB: $\\frac{5 \\pi R}{6 T_{0}}$\nC: $\\frac{3 \\pi R}{4 T_{0}}$\nD: $\\frac{2 \\pi R}{3 T_{0}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_a9b05ce8eea7b0e40e5eg-042.jpg?height=414&width=648&top_left_y=1438&top_left_x=336"
        ],
        "answer": [
            "B"
        ],
        "solution": "地球公转的线速度大小\n\n$$\nv_{0}=\\frac{2 \\pi R}{T_{0}}\n$$\n\n设小行星在偏转圆轨道上运行的周期为 $T$, 根据开普勒第三定律有\n\n$$\n\\frac{R^{3}}{T_{0}^{3}}=\\frac{(4 R)^{3}}{T^{2}}\n$$\n\n则小行星在偏转圆轨道上运行的线速度大小\n\n$$\nv=\\frac{2 \\pi \\times 4 R}{T}\n$$\n\n组合撞击体与小行星发生完全非弹性碰撞, 根据动量守恒定律有\n\n$$\nm_{0} \\times \\frac{3}{4} v_{0}+3 m_{0} v_{C}=\\left(m_{0}+3 m_{0}\\right) v\n$$\n\n联立解得\n\n$$\nv_{C}=\\frac{5 \\pi R}{6 T_{0}}\n$$\n\n故选 B。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_520",
        "problem": "牛顿发现了万有引力定律以后, 还设想了发射人造卫星的情景, 若要发射人造卫星并将卫星以一定的速度送入预定轨道。发射场一般选择在尽可能靠近赤道的地方, 如图这样选址的优点是, 在赤道附近 ( )\n\n[图1]\nA: 地球的引力较大\nB: 地球自转角速度较大\nC: 重力加速度较大\nD: 地球自转线速度较大\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n牛顿发现了万有引力定律以后, 还设想了发射人造卫星的情景, 若要发射人造卫星并将卫星以一定的速度送入预定轨道。发射场一般选择在尽可能靠近赤道的地方, 如图这样选址的优点是, 在赤道附近 ( )\n\n[图1]\n\nA: 地球的引力较大\nB: 地球自转角速度较大\nC: 重力加速度较大\nD: 地球自转线速度较大\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_6842b9ceb844a90b34c3g-19.jpg?height=326&width=346&top_left_y=585&top_left_x=341"
        ],
        "answer": [
            "D"
        ],
        "solution": "A. 由万有引力定律可知物体在地球表面各点所受的引力大小相等, 故 A 错误\n\nB. 在地球上各点具有相同的角速度, 故 B 错误;\n\nC. 赤道处重力加速度最小. 故 C 错误;\n\nD. 相对于地心的发射速度等于相对于地面的发射速度加上地球自转的线速度. 地球自转的线速度越大, 相对于地心的发射速度越大, 卫星越容易发射出去. 赤道处, 半径最大, 所以自转线速度最大. 故 D 正确.\n\n故选 D。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_1145",
        "problem": "On $21^{\\text {st }}$ December 2020, Jupiter and Saturn formed a true spectacle in the southwestern sky just after sunset in the UK, separated by only $0.102^{\\circ}$. This is close enough that when viewed through a telescope both planets and their moons could be seen in the same field of view (see Fig 4).\n\nWhen two planets occupy the same piece of sky it is known as a conjunction, and when it is Jupiter and Saturn it is known as a great conjunction (so named because they are the rarest of the naked-eye planet conjunctions). The reason it happens is because Jupiter's orbital velocity is higher than Saturn's and so as time goes on Jupiter catches up with and overtakes Saturn (at least as viewed from Earth), with the moment of overtaking corresponding to the conjunction. The process of the two getting closer and closer together has been seen in sky throughout the year (see Fig 5).\n[figure1]\n\nFigure 4: Left: The view of the planets the day before their closest approach, as captured by the 16 \" telescope at the Institute of Astronomy, Cambridge. Credit: Robin Catchpole.\n\nRight: The view from Arizona on the day of the closest approach as viewed with the Lowell Discovery\n\nTelescope. Credit: Levine / Elbert / Bosh / Lowell Observatory.\n\n[figure2]\n\nFigure 5: Left: Demonstrating how Jupiter and Saturn have been getting closer and closer together over the last few months. Separations are given in degrees and arcminutes $\\left(1 / 60^{\\text {th }}\\right.$ of a degree). Credit: Pete Lawrence.\n\nRight: The positions of the Earth, Jupiter and Saturn that were responsible for the 2020 great conjunction. The precise timing of the apparent alignment is clearly sensitive to where Earth is in its orbit. Credit:\n\ntimeanddate.com.\n\nThe time between conjunctions is known as the synodic period. Although this period will change slightly from conjunction to conjunction due to different lines of perspective as viewed from Earth (see Fig 5), we can work out the average time between great conjunctions by considering both planets travelling on circular coplanar orbits and ignoring the position of the Earth.\n\nFor circular coplanar orbits the centre of Jupiter's disc would pass in front of the centre of Saturn's disc every conjunction, and hence have an angular separation of $\\theta=0^{\\circ}$ (it is measured from the centre of each disc). In practice, the planets follow elliptical orbits that are in planes inclined at different angles to each other.\n\nFig 6 shows how this affects the real values for over 8000 years' worth of data, which along with different synodic periods between conjunctions makes it a difficult problem to solve precisely without a computer. However, after one synodic period Saturn has moved about $2 / 3$ of the way around its orbit, and so roughly every 3 synodic periods it is in a similar part of the sky. Consequently every third great conjunction follows a reasonably regular pattern which can be fit with a sinusoidal function.\n\n[figure3]\n\nFigure 6: Top: All great conjunctions from 1800 to 2300, calculated for the real celestial mechanics of the Solar System. We can see that each great conjunction belongs to one of three different series or tracks, with Track A indicated with orange circles, Track B with green squares, and Track C with blue triangles.\n\nBottom: The same idea but extended over a much larger date range, up to $10000 \\mathrm{AD}$. It is clear the distinct series form broadly sinusoidal patterns which can be used with the average synodic period to give rough predictions for the separations of great conjunctions. The opacity of points is related to each conjunction's angular separation from the Sun (where low opacity means close to the Sun, so it is harder for any observers to see). Credit: Nick Koukoufilippas, but inspired by the work of Steffen Thorsen and Graham Jones / Sky \\& Telescope.c. By empirically fitting a sinusoidal function (which is assumed to be the same for each track, just with a fixed phase difference between them) and assuming all conjunctions are separated by the average synodic period, we can give rough estimations for the separations of any given great conjunction. Note: be careful as your calculations will be very sensitive to rounding errors.\n\nv. Similarly, when was the last great conjunction at least as close as the 2020 one? Give its year and the value of $\\theta$.",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nOn $21^{\\text {st }}$ December 2020, Jupiter and Saturn formed a true spectacle in the southwestern sky just after sunset in the UK, separated by only $0.102^{\\circ}$. This is close enough that when viewed through a telescope both planets and their moons could be seen in the same field of view (see Fig 4).\n\nWhen two planets occupy the same piece of sky it is known as a conjunction, and when it is Jupiter and Saturn it is known as a great conjunction (so named because they are the rarest of the naked-eye planet conjunctions). The reason it happens is because Jupiter's orbital velocity is higher than Saturn's and so as time goes on Jupiter catches up with and overtakes Saturn (at least as viewed from Earth), with the moment of overtaking corresponding to the conjunction. The process of the two getting closer and closer together has been seen in sky throughout the year (see Fig 5).\n[figure1]\n\nFigure 4: Left: The view of the planets the day before their closest approach, as captured by the 16 \" telescope at the Institute of Astronomy, Cambridge. Credit: Robin Catchpole.\n\nRight: The view from Arizona on the day of the closest approach as viewed with the Lowell Discovery\n\nTelescope. Credit: Levine / Elbert / Bosh / Lowell Observatory.\n\n[figure2]\n\nFigure 5: Left: Demonstrating how Jupiter and Saturn have been getting closer and closer together over the last few months. Separations are given in degrees and arcminutes $\\left(1 / 60^{\\text {th }}\\right.$ of a degree). Credit: Pete Lawrence.\n\nRight: The positions of the Earth, Jupiter and Saturn that were responsible for the 2020 great conjunction. The precise timing of the apparent alignment is clearly sensitive to where Earth is in its orbit. Credit:\n\ntimeanddate.com.\n\nThe time between conjunctions is known as the synodic period. Although this period will change slightly from conjunction to conjunction due to different lines of perspective as viewed from Earth (see Fig 5), we can work out the average time between great conjunctions by considering both planets travelling on circular coplanar orbits and ignoring the position of the Earth.\n\nFor circular coplanar orbits the centre of Jupiter's disc would pass in front of the centre of Saturn's disc every conjunction, and hence have an angular separation of $\\theta=0^{\\circ}$ (it is measured from the centre of each disc). In practice, the planets follow elliptical orbits that are in planes inclined at different angles to each other.\n\nFig 6 shows how this affects the real values for over 8000 years' worth of data, which along with different synodic periods between conjunctions makes it a difficult problem to solve precisely without a computer. However, after one synodic period Saturn has moved about $2 / 3$ of the way around its orbit, and so roughly every 3 synodic periods it is in a similar part of the sky. Consequently every third great conjunction follows a reasonably regular pattern which can be fit with a sinusoidal function.\n\n[figure3]\n\nFigure 6: Top: All great conjunctions from 1800 to 2300, calculated for the real celestial mechanics of the Solar System. We can see that each great conjunction belongs to one of three different series or tracks, with Track A indicated with orange circles, Track B with green squares, and Track C with blue triangles.\n\nBottom: The same idea but extended over a much larger date range, up to $10000 \\mathrm{AD}$. It is clear the distinct series form broadly sinusoidal patterns which can be used with the average synodic period to give rough predictions for the separations of great conjunctions. The opacity of points is related to each conjunction's angular separation from the Sun (where low opacity means close to the Sun, so it is harder for any observers to see). Credit: Nick Koukoufilippas, but inspired by the work of Steffen Thorsen and Graham Jones / Sky \\& Telescope.\n\nproblem:\nc. By empirically fitting a sinusoidal function (which is assumed to be the same for each track, just with a fixed phase difference between them) and assuming all conjunctions are separated by the average synodic period, we can give rough estimations for the separations of any given great conjunction. Note: be careful as your calculations will be very sensitive to rounding errors.\n\nv. Similarly, when was the last great conjunction at least as close as the 2020 one? Give its year and the value of $\\theta$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of percentage, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_f4dc8cb2d9258a843a19g-07.jpg?height=706&width=1564&top_left_y=834&top_left_x=244",
            "https://cdn.mathpix.com/cropped/2024_03_14_f4dc8cb2d9258a843a19g-08.jpg?height=578&width=1566&top_left_y=196&top_left_x=242",
            "https://cdn.mathpix.com/cropped/2024_03_14_f4dc8cb2d9258a843a19g-09.jpg?height=1072&width=1564&top_left_y=1191&top_left_x=246"
        ],
        "answer": [
            "5.42"
        ],
        "solution": "Again, to avoid brute force, the graph suggests we see whether there is anything suitable around Track B's last x-intecept\n\n$$\n0=1.20 \\sin \\left(\\frac{2 \\pi t}{2500}-0.98\\right) \\therefore t=\\frac{2500 \\times 0.98}{2 \\pi}=388\n$$\n\nThese repeat every $1 / 2 \\lambda$ so the last one was $388+\\frac{2500}{2}=1638$\n\nLooking at the number of track intervals you need to go back,\n\n$$\n\\begin{aligned}\n& n=\\frac{(2020.975+19.86)-1638}{3 \\times 19.86}=6.76 \\therefore \\text { try } 7 \\text { intervals back from } 2040 \\\\\n& \\therefore t=(2020.975+19.86)-(7 \\times 3 \\times 19.86)=1623.80 \\\\\n& \\quad \\therefore \\text { year }=1623 \\\\\n& \\therefore \\theta=\\left|1.20 \\sin \\left(\\frac{2 \\pi \\times(2020.975+19.86)}{2500}-0.98\\right)\\right|=0.043^{\\circ} \\text { (good guess) }\n\\end{aligned}\n$$\n\n[Another example of intuition saving us lots of time. We have added the 19.86 to the 2020.975 to move from Track A to Track B. The real one was on $16^{\\text {th }}$ July 1623 (our model predicts $19^{\\text {th }}$ October) and had a separation of $0.086^{\\circ}$. Some closer separations do happen, including transits (where the disc of Jupiter overlaps with the disc of Saturn), the next of which are due to happen in 7541 and 8674, and occultations (where Jupiter completely blocks out Saturn) in 7541, 13340 and 13738. (Note: 7541 sees part of a sequence called a triple conjunction, where retrograde motion means that Jupiter and Saturn meet three times in a short period. The transit is in February and the occultation is in June)]",
        "answer_type": "NV",
        "unit": [
            "percentage"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_352",
        "problem": "2020 年 7 月 23 日, 我国首次火星探测任务“天问一号”探测器, 在中国文昌航天发射场, 应用长征五号运载火箭送入地火转移轨道。火星距离地球最远时有 4 亿公里, 最近时大约 0.55 亿公里。由于距离遥远, 地球与火星之间的信号传输会有长时间的时延。当火星离我们最远时, 从地球发出一个指令, 约 22 分钟才能到达火星。为了节省燃料,我们要等火星与地球之间相对位置合适的时候发射探测器。受天体运行规律的影响, 这样的发射机会很少。为简化计算, 已知火星的公转周期约是地球公转周期的 1.9 倍,认为地球和火星在同一平面上、沿同一方向绕太阳做匀速圆周运动，如图所示。根据上述材料, 结合所学知识, 判断下列说法正确的是 ( )\n\n[图1]\nA: 当探测器加速后刚离开 $A$ 处的加速度与速度均比火星在轨时的要大\nB: 当火星离地球最近时, 地球上发出的指令需要约 10 分钟到达火星\nC: 如果火星运动到 $B$ 点, 地球恰好在 $A$ 点时发射探测器, 那么探测器将沿轨迹 $A C$ 运动到 $C$ 点时, 恰好与火星相遇\nD: 下一个发射时机需要再等约 2.7 年\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n2020 年 7 月 23 日, 我国首次火星探测任务“天问一号”探测器, 在中国文昌航天发射场, 应用长征五号运载火箭送入地火转移轨道。火星距离地球最远时有 4 亿公里, 最近时大约 0.55 亿公里。由于距离遥远, 地球与火星之间的信号传输会有长时间的时延。当火星离我们最远时, 从地球发出一个指令, 约 22 分钟才能到达火星。为了节省燃料,我们要等火星与地球之间相对位置合适的时候发射探测器。受天体运行规律的影响, 这样的发射机会很少。为简化计算, 已知火星的公转周期约是地球公转周期的 1.9 倍,认为地球和火星在同一平面上、沿同一方向绕太阳做匀速圆周运动，如图所示。根据上述材料, 结合所学知识, 判断下列说法正确的是 ( )\n\n[图1]\n\nA: 当探测器加速后刚离开 $A$ 处的加速度与速度均比火星在轨时的要大\nB: 当火星离地球最近时, 地球上发出的指令需要约 10 分钟到达火星\nC: 如果火星运动到 $B$ 点, 地球恰好在 $A$ 点时发射探测器, 那么探测器将沿轨迹 $A C$ 运动到 $C$ 点时, 恰好与火星相遇\nD: 下一个发射时机需要再等约 2.7 年\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_a9b05ce8eea7b0e40e5eg-110.jpg?height=505&width=417&top_left_y=176&top_left_x=337"
        ],
        "answer": [
            "A"
        ],
        "solution": "A. 当探测器加速后刚离开 $A$ 处, 根据万有引力提供加速度可知\n\n$$\nG \\frac{M m}{r^{2}}=m a\n$$\n\n解得\n\n$$\na=G \\frac{M}{r^{2}}\n$$\n\n探测器 $A$ 处距太阳距离较小, 则加速度较大, 探测器在 $A$ 处做圆周运动的线速度 $v=\\sqrt{\\frac{G M}{r}}$\n\n探测器 $A$ 处距太阳距离较小,探测器在 $A$ 处做圆周运动的线速度 $v$ 比火星在轨的线速度大, 探测器加速后刚离开 $A$ 处速度比探测器在 $A$ 处做圆周运动的线速度 $v$ 大,因此当探测器加速后刚离开 $A$ 处的速度均比火星在轨时的要大, $\\mathrm{A}$ 正确;\n\nB. 火星距离地球最远时有 4 亿公里, 从地球发出一个指令, 约 22 分钟才能到达火星,最近时大约 0.55 亿公里, 因为指令传播速度相同, 则时间为\n\n$$\nt=\\frac{22 \\times 0.55}{4} \\mathrm{~s}=3.025 \\mathrm{~s}\n$$\n\nB 错误;\n\nC. 根据开普勒第三定律, 火星与探测器的公转半径不同, 则公转周期不相同, 因此探测器与火星不能在 $C$ 点相遇, $C$ 错误;\n\nD . 地球的公转周期为 1 年, 火星的公转周期约是地球公转周期的 1.9 倍, 两者的角速度之差为\n\n$$\n\\Delta \\omega=\\frac{2 \\pi}{1}-\\frac{2 \\pi}{1.9}=\\frac{1.8 \\pi}{1.9}\n$$\n\n则地球再一次追上火星的用时为\n\n$$\nt=\\frac{2 \\pi}{\\Delta \\omega}=2.1 \\text { 年 }\n$$\n\nD 错误;\n\n故选 A。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_1176",
        "problem": "e. Calculate the projected physical separation, $r_{p}$, between the galaxy and the Voorwerp.g. High precision measurements showed that the Voorwerp is slightly further away than the galaxy, and so $\\theta=125^{\\circ}$. Use this with your expression from the previous part of the question to estimate an upper limit for the number of years that have passed since the quasar was last active.",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\ne. Calculate the projected physical separation, $r_{p}$, between the galaxy and the Voorwerp.\n\nproblem:\ng. High precision measurements showed that the Voorwerp is slightly further away than the galaxy, and so $\\theta=125^{\\circ}$. Use this with your expression from the previous part of the question to estimate an upper limit for the number of years that have passed since the quasar was last active.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of  \\mathrm{~s}, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": [
            "4.11 \\times 10^{12}"
        ],
        "solution": "$$\n\\begin{aligned}\n\\Delta t=\\frac{r_{p}}{c \\sin \\theta}(1-\\cos \\theta) \\quad & =\\frac{20.8 \\times 10^{3} \\times 3.09 \\times 10^{16}}{3.00 \\times 10^{8} \\times \\sin 125^{\\circ}}\\left(1-\\cos 125^{\\circ}\\right) \\\\\n& =4.11 \\times 10^{12} \\mathrm{~s} \\\\\n& \\approx 130000 \\text { years }\n\\end{aligned}\n$$\n\n(This is remarkably recent on astronomical timescales!)",
        "answer_type": "NV",
        "unit": [
            " \\mathrm{~s}"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_93",
        "problem": "人类对来知事物的好奇和科学家们的不懈努力, 使人类对宇宙的认识越来越丰富。\n在研究匀变速直线运动的位移时，我们常用“以恒代变”的思想：在研究曲线运动的“瞬时速度”时，又常用“化曲为直”的思想，而在研究一般的曲线运动时，我们用的更多的是一种“化曲为圆”的思想, 即对于一般的曲线运动, 尽管曲线各个位置的弯曲程度不一样, 但在研究时, 可以将曲线分割为许多很短的小段, 质点在每小段的运动都可以看做半径为某个合适值 $\\rho$ 的圆周运动的一部分, 进而采用圆周运动的分析方法来进行研究、 $\\rho$ 叫做曲率半径, 如图 2 所示, 试据此分析图 3 所示的斜抛运动中, 轨迹最高点处的曲率半径 $\\rho$ 。\n\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n[图3]\n\n图3",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n\n问题:\n人类对来知事物的好奇和科学家们的不懈努力, 使人类对宇宙的认识越来越丰富。\n在研究匀变速直线运动的位移时，我们常用“以恒代变”的思想：在研究曲线运动的“瞬时速度”时，又常用“化曲为直”的思想，而在研究一般的曲线运动时，我们用的更多的是一种“化曲为圆”的思想, 即对于一般的曲线运动, 尽管曲线各个位置的弯曲程度不一样, 但在研究时, 可以将曲线分割为许多很短的小段, 质点在每小段的运动都可以看做半径为某个合适值 $\\rho$ 的圆周运动的一部分, 进而采用圆周运动的分析方法来进行研究、 $\\rho$ 叫做曲率半径, 如图 2 所示, 试据此分析图 3 所示的斜抛运动中, 轨迹最高点处的曲率半径 $\\rho$ 。\n\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n[图3]\n\n图3\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-070.jpg?height=460&width=625&top_left_y=1897&top_left_x=336",
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-070.jpg?height=272&width=468&top_left_y=2074&top_left_x=994",
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-071.jpg?height=614&width=568&top_left_y=156&top_left_x=390"
        ],
        "answer": [
            "$\\frac{v_{0}^{2} \\cos ^{2} \\theta}{g}$"
        ],
        "solution": "在斜抛运动的最高点, 质点的速度为\n\n$$\nv=v_{0} \\cos \\theta\n$$\n\n在最高点可以把质点的运动看作半径为 $\\rho$ 的圆周运动, 质点受到重力提供了向心力, 根据牛顿第二定律有\n\n$$\nm g=m \\frac{v^{2}}{\\rho}\n$$\n\n联立解得\n\n$$\n\\rho=\\frac{v_{0}^{2} \\cos ^{2} \\theta}{g}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_697",
        "problem": "2020 年 11 月 24 日 4 点 30 分, 嫦娥五号探测器成功发射升空。若嫦娥五号在距月球表面高度分别为 $h_{1} 、 h_{2}$ 的轨道 I、II上运行, 均可视为匀速圆周运动, 则在轨道 I、II 上运行时, 嫦娥五号与月球中心连线扫过相同面积所用的时间之比为 (月球看成半径为 $R$ 、质量均匀分布的球体）（）\nA: $\\sqrt{\\frac{h_{1}}{h_{2}}}$\nB: $\\frac{R+h_{1}}{R+h_{2}}$\nC: $\\sqrt{\\frac{R+h_{1}}{R+h_{2}}}$\nD: $\\sqrt{\\frac{R+h_{2}}{R+h_{1}}}$\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n2020 年 11 月 24 日 4 点 30 分, 嫦娥五号探测器成功发射升空。若嫦娥五号在距月球表面高度分别为 $h_{1} 、 h_{2}$ 的轨道 I、II上运行, 均可视为匀速圆周运动, 则在轨道 I、II 上运行时, 嫦娥五号与月球中心连线扫过相同面积所用的时间之比为 (月球看成半径为 $R$ 、质量均匀分布的球体）（）\n\nA: $\\sqrt{\\frac{h_{1}}{h_{2}}}$\nB: $\\frac{R+h_{1}}{R+h_{2}}$\nC: $\\sqrt{\\frac{R+h_{1}}{R+h_{2}}}$\nD: $\\sqrt{\\frac{R+h_{2}}{R+h_{1}}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": [
            "D"
        ],
        "solution": "根据万有引力提供向心力\n\n$$\nG \\frac{M m}{(R+h)^{2}}=m \\omega^{2}(R+h)\n$$\n\n可知嫦娥五号在距月球表面高度为 $h_{1} 、 h_{2}$ 的轨道 I、II上的角速度分别为\n\n$$\n\\omega_{1}=\\sqrt{\\frac{G M}{\\left(R+h_{1}\\right)^{3}}}, \\omega_{2}=\\sqrt{\\frac{G M}{\\left(R+h_{2}\\right)^{3}}}\n$$\n\n又因为嫦娥五号与月球中心连线扫过的面积为\n\n$$\ns=\\frac{1}{2} \\omega t(R+h)^{2}\n$$\n\n当扫过面积相等时，有\n\n$$\n\\frac{1}{2} \\omega_{1} t_{1}\\left(R+h_{1}\\right)^{2}=\\frac{1}{2} \\omega_{2} t_{2}\\left(R+h_{2}\\right)^{2}\n$$\n\n解得\n\n$$\n\\frac{t_{1}}{t_{2}}=\\sqrt{\\frac{R+h_{2}}{R+h_{1}}}\n$$\n\n故选 D。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Astronomy_1163",
        "problem": "The Event Horizon Telescope (EHT) is a project to use many widely-spaced radio telescopes as a Very Long Baseline Interferometer (VBLI) to create a virtual telescope as big as the Earth. This extraordinary size allows sufficient angular resolution to be able to image the space close to the event horizon of a super massive black hole (SMBH), and provide an opportunity to test the predictions of Einstein's theory of General Relativity (GR) in a very strong gravitational field. In April 2017 the EHT collaboration managed to co-ordinate time on all of the telescopes in the array so that they could observe the SMBH (called M87*) at the centre of the Virgo galaxy, M87, and they plan to also image the SMBH at the centre of our galaxy (called Sgr A*).\n[figure1]\n\nFigure 3: Left: The locations of all the telescopes used during the April 2017 observing run. The solid lines correspond to baselines used for observing M87, whilst the dashed lines were the baselines used for the calibration source. Credit: EHT Collaboration.\n\nRight: A simulated model of what the region near an SMBH could look like, modelled at much higher resolution than the EHT can achieve. The light comes from the accretion disc, but the paths of the photons are bent into a characteristic shape by the extreme gravity, leading to a 'shadow' in middle of the disc - this is what the EHT is trying to image. The left side of the image is brighter than the right side as light emitted from a substance moving towards an observer is brighter than that of one moving away. Credit: Hotaka Shiokawa.\n\nSome data about the locations of the eight telescopes in the array are given below in 3-D cartesian geocentric coordinates with $X$ pointing to the Greenwich meridian, $Y$ pointing $90^{\\circ}$ away in the equatorial plane (eastern longitudes have positive $Y$ ), and positive $Z$ pointing in the direction of the North Pole. This is a left-handed coordinate system.\n\n| Facility | Location | $X(\\mathrm{~m})$ | $Y(\\mathrm{~m})$ | $Z(\\mathrm{~m})$ |\n| :--- | :--- | :---: | :---: | :---: |\n| ALMA | Chile | 2225061.3 | -5440061.7 | -2481681.2 |\n| APEX | Chile | 2225039.5 | -5441197.6 | -2479303.4 |\n| JCMT | Hawaii, USA | -5464584.7 | -2493001.2 | 2150654.0 |\n| LMT | Mexico | -768715.6 | -5988507.1 | 2063354.9 |\n| PV | Spain | 5088967.8 | -301681.2 | 3825012.2 |\n| SMA | Hawaii, USA | -5464555.5 | -2492928.0 | 2150797.2 |\n| SMT | Arizona, USA | -1828796.2 | -5054406.8 | 3427865.2 |\n| SPT | Antarctica | 809.8 | -816.9 | -6359568.7 |\n\nThe minimum angle, $\\theta_{\\min }$ (in radians) that can be resolved by a VLBI array is given by the equation\n\n$$\n\\theta_{\\min }=\\frac{\\lambda_{\\mathrm{obs}}}{d_{\\max }},\n$$\n\nwhere $\\lambda_{\\text {obs }}$ is the observing wavelength and $d_{\\max }$ is the longest straight line distance between two telescopes used (called the baseline), assumed perpendicular to the line of sight during the observation.\n\nAn important length scale when discussing black holes is the gravitational radius, $r_{g}=\\frac{G M}{c^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the black hole and $c$ is the speed of light. The familiar event horizon of a non-rotating black hole is called the Schwartzschild radius, $r_{S} \\equiv 2 r_{g}$, however this is not what the EHT is able to observe - instead the closest it can see to a black hole is called the photon sphere, where photons orbit in the black hole in unstable circular orbits. On top of this the image of the black hole is gravitationally lensed by the black hole itself magnifying the apparent radius of the photon sphere to be between $(2 \\sqrt{3+2 \\sqrt{2}}) r_{g}$ and $(3 \\sqrt{3}) r_{g}$, determined by spin and inclination; the latter corresponds to a perfectly spherical non-spinning black hole. The area within this lensed image will appear almost black and is the 'shadow' the EHT is looking for.\n\n[figure2]\n\nFigure 4: Four nights of data were taken for M87* during the observing window of the EHT, and whilst the diameter of the disk stayed relatively constant the location of bright spots moved, possibly indicating gas that is orbiting the black hole. Credit: EHT Collaboration.\n\nThe EHT observed M87* on four separate occasions during the observing window (see Fig 4), and the team saw that some of the bright spots changed in that time, suggesting they may be associated with orbiting gas close to the black hole. The Innermost Stable Circular Orbit (ISCO) is the equivalent of the photon sphere but for particles with mass (and is also stable). The total conserved energy of a circular orbit close to a non-spinning black hole is given by\n\n$$\nE=m c^{2}\\left(\\frac{1-\\frac{2 r_{g}}{r}}{\\sqrt{1-\\frac{3 r_{g}}{r}}}\\right)\n$$\n\nand the radius of the ISCO, $r_{\\mathrm{ISCO}}$, is the value of $r$ for which $E$ is minimised.\n\nWe expect that most black holes are in fact spinning (since most stars are spinning) and the spin of a black hole is quantified with the spin parameter $a \\equiv J / J_{\\max }$ where $J$ is the angular momentum of the black hole and $J_{\\max }=G M^{2} / c$ is the maximum possible angular momentum it can have. The value of $a$ varies from $-1 \\leq a \\leq 1$, where negative spins correspond to the black hole rotating in the opposite direction to its accretion disk, and positive spins in the same direction. If $a=1$ then $r_{\\text {ISCO }}=r_{g}$, whilst if $a=-1$ then $r_{\\text {ISCO }}=9 r_{g}$. The angular velocity of a particle in the ISCO is given by\n\n$$\n\\omega^{2}=\\frac{G M}{\\left(r_{\\text {ISCO }}^{3 / 2}+a r_{g}^{3 / 2}\\right)^{2}}\n$$\n\n[figure3]\n\nFigure 5: Due to the curvature of spacetime, the real distance travelled by a particle moving from the ISCO to the photon sphere (indicated with the solid red arrow) is longer than you would get purely from subtracting the radial co-ordinates of those orbits (indicated with the dashed blue arrow), which would be valid for a flat spacetime. Relations between these distances are not to scale in this diagram. Credit: Modified from Bardeen et al. (1972).\n\nThe spacetime near a black hole is curved, as described by the equations of GR. This means that the distance between two points can be substantially different to the distance you would expect if spacetime was flat. GR tells us that the proper distance travelled by a particle moving from radius $r_{1}$ to radius $r_{2}$ around a black hole of mass $M$ (with $r_{1}>r_{2}$ ) is given by\n\n$$\n\\Delta l=\\int_{r_{2}}^{r_{1}}\\left(1-\\frac{2 r_{g}}{r}\\right)^{-1 / 2} \\mathrm{~d} r\n$$a. Determine $\\mathrm{d}_{\\max }$ for the observations of M87 (i.e. the solid lines in Fig 3) and hence $\\theta_{\\min }$ if the EHT uses radio waves of frequency $230 \\mathrm{GHz}$. Give your answer for $\\mathrm{d}_{\\max }$ in $\\mathrm{km}$ and $\\theta_{\\min }$ in microarcseconds ( 1 degree $=3600$ arcseconds).",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is an expression.\nHere is some context information for this question, which might assist you in solving it:\nThe Event Horizon Telescope (EHT) is a project to use many widely-spaced radio telescopes as a Very Long Baseline Interferometer (VBLI) to create a virtual telescope as big as the Earth. This extraordinary size allows sufficient angular resolution to be able to image the space close to the event horizon of a super massive black hole (SMBH), and provide an opportunity to test the predictions of Einstein's theory of General Relativity (GR) in a very strong gravitational field. In April 2017 the EHT collaboration managed to co-ordinate time on all of the telescopes in the array so that they could observe the SMBH (called M87*) at the centre of the Virgo galaxy, M87, and they plan to also image the SMBH at the centre of our galaxy (called Sgr A*).\n[figure1]\n\nFigure 3: Left: The locations of all the telescopes used during the April 2017 observing run. The solid lines correspond to baselines used for observing M87, whilst the dashed lines were the baselines used for the calibration source. Credit: EHT Collaboration.\n\nRight: A simulated model of what the region near an SMBH could look like, modelled at much higher resolution than the EHT can achieve. The light comes from the accretion disc, but the paths of the photons are bent into a characteristic shape by the extreme gravity, leading to a 'shadow' in middle of the disc - this is what the EHT is trying to image. The left side of the image is brighter than the right side as light emitted from a substance moving towards an observer is brighter than that of one moving away. Credit: Hotaka Shiokawa.\n\nSome data about the locations of the eight telescopes in the array are given below in 3-D cartesian geocentric coordinates with $X$ pointing to the Greenwich meridian, $Y$ pointing $90^{\\circ}$ away in the equatorial plane (eastern longitudes have positive $Y$ ), and positive $Z$ pointing in the direction of the North Pole. This is a left-handed coordinate system.\n\n| Facility | Location | $X(\\mathrm{~m})$ | $Y(\\mathrm{~m})$ | $Z(\\mathrm{~m})$ |\n| :--- | :--- | :---: | :---: | :---: |\n| ALMA | Chile | 2225061.3 | -5440061.7 | -2481681.2 |\n| APEX | Chile | 2225039.5 | -5441197.6 | -2479303.4 |\n| JCMT | Hawaii, USA | -5464584.7 | -2493001.2 | 2150654.0 |\n| LMT | Mexico | -768715.6 | -5988507.1 | 2063354.9 |\n| PV | Spain | 5088967.8 | -301681.2 | 3825012.2 |\n| SMA | Hawaii, USA | -5464555.5 | -2492928.0 | 2150797.2 |\n| SMT | Arizona, USA | -1828796.2 | -5054406.8 | 3427865.2 |\n| SPT | Antarctica | 809.8 | -816.9 | -6359568.7 |\n\nThe minimum angle, $\\theta_{\\min }$ (in radians) that can be resolved by a VLBI array is given by the equation\n\n$$\n\\theta_{\\min }=\\frac{\\lambda_{\\mathrm{obs}}}{d_{\\max }},\n$$\n\nwhere $\\lambda_{\\text {obs }}$ is the observing wavelength and $d_{\\max }$ is the longest straight line distance between two telescopes used (called the baseline), assumed perpendicular to the line of sight during the observation.\n\nAn important length scale when discussing black holes is the gravitational radius, $r_{g}=\\frac{G M}{c^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the black hole and $c$ is the speed of light. The familiar event horizon of a non-rotating black hole is called the Schwartzschild radius, $r_{S} \\equiv 2 r_{g}$, however this is not what the EHT is able to observe - instead the closest it can see to a black hole is called the photon sphere, where photons orbit in the black hole in unstable circular orbits. On top of this the image of the black hole is gravitationally lensed by the black hole itself magnifying the apparent radius of the photon sphere to be between $(2 \\sqrt{3+2 \\sqrt{2}}) r_{g}$ and $(3 \\sqrt{3}) r_{g}$, determined by spin and inclination; the latter corresponds to a perfectly spherical non-spinning black hole. The area within this lensed image will appear almost black and is the 'shadow' the EHT is looking for.\n\n[figure2]\n\nFigure 4: Four nights of data were taken for M87* during the observing window of the EHT, and whilst the diameter of the disk stayed relatively constant the location of bright spots moved, possibly indicating gas that is orbiting the black hole. Credit: EHT Collaboration.\n\nThe EHT observed M87* on four separate occasions during the observing window (see Fig 4), and the team saw that some of the bright spots changed in that time, suggesting they may be associated with orbiting gas close to the black hole. The Innermost Stable Circular Orbit (ISCO) is the equivalent of the photon sphere but for particles with mass (and is also stable). The total conserved energy of a circular orbit close to a non-spinning black hole is given by\n\n$$\nE=m c^{2}\\left(\\frac{1-\\frac{2 r_{g}}{r}}{\\sqrt{1-\\frac{3 r_{g}}{r}}}\\right)\n$$\n\nand the radius of the ISCO, $r_{\\mathrm{ISCO}}$, is the value of $r$ for which $E$ is minimised.\n\nWe expect that most black holes are in fact spinning (since most stars are spinning) and the spin of a black hole is quantified with the spin parameter $a \\equiv J / J_{\\max }$ where $J$ is the angular momentum of the black hole and $J_{\\max }=G M^{2} / c$ is the maximum possible angular momentum it can have. The value of $a$ varies from $-1 \\leq a \\leq 1$, where negative spins correspond to the black hole rotating in the opposite direction to its accretion disk, and positive spins in the same direction. If $a=1$ then $r_{\\text {ISCO }}=r_{g}$, whilst if $a=-1$ then $r_{\\text {ISCO }}=9 r_{g}$. The angular velocity of a particle in the ISCO is given by\n\n$$\n\\omega^{2}=\\frac{G M}{\\left(r_{\\text {ISCO }}^{3 / 2}+a r_{g}^{3 / 2}\\right)^{2}}\n$$\n\n[figure3]\n\nFigure 5: Due to the curvature of spacetime, the real distance travelled by a particle moving from the ISCO to the photon sphere (indicated with the solid red arrow) is longer than you would get purely from subtracting the radial co-ordinates of those orbits (indicated with the dashed blue arrow), which would be valid for a flat spacetime. Relations between these distances are not to scale in this diagram. Credit: Modified from Bardeen et al. (1972).\n\nThe spacetime near a black hole is curved, as described by the equations of GR. This means that the distance between two points can be substantially different to the distance you would expect if spacetime was flat. GR tells us that the proper distance travelled by a particle moving from radius $r_{1}$ to radius $r_{2}$ around a black hole of mass $M$ (with $r_{1}>r_{2}$ ) is given by\n\n$$\n\\Delta l=\\int_{r_{2}}^{r_{1}}\\left(1-\\frac{2 r_{g}}{r}\\right)^{-1 / 2} \\mathrm{~d} r\n$$\n\nproblem:\na. Determine $\\mathrm{d}_{\\max }$ for the observations of M87 (i.e. the solid lines in Fig 3) and hence $\\theta_{\\min }$ if the EHT uses radio waves of frequency $230 \\mathrm{GHz}$. Give your answer for $\\mathrm{d}_{\\max }$ in $\\mathrm{km}$ and $\\theta_{\\min }$ in microarcseconds ( 1 degree $=3600$ arcseconds).\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_0117b7b4f76996307b50g-07.jpg?height=704&width=1414&top_left_y=698&top_left_x=331",
            "https://cdn.mathpix.com/cropped/2024_03_14_0117b7b4f76996307b50g-08.jpg?height=374&width=1562&top_left_y=1698&top_left_x=263",
            "https://cdn.mathpix.com/cropped/2024_03_14_0117b7b4f76996307b50g-09.jpg?height=468&width=686&top_left_y=1388&top_left_x=705"
        ],
        "answer": [
            "24.7 \\mu \\text { as }"
        ],
        "solution": "Longest (solid) baseline from Fig. 3 is JCMT -> PV, so using 3-D Pythagoras,\n\n$$\n\\begin{aligned}\nd_{\\max } & =\\sqrt{\\left(x_{J C M T}-x_{P V}\\right)^{2}+\\left(y_{J C M T}-y_{P V}\\right)^{2}+\\left(z_{J C M T}-z_{P V}\\right)^{2}} \\\\\n& =10907.93 \\mathrm{~km}\n\\end{aligned}\n$$\n\n[Accept using SMA -> PV (giving $\\mathrm{d}_{\\max }=10907.86 \\mathrm{~km}$ ) for full marks, and $\\mathrm{d}_{\\max }$ should be at least 3 s.f. and in $\\mathrm{km}$ for the final mark. If the student uses PV -> SPT (the longest baseline, but a dashed line so not used for the M87 observation) giving $d_{\\max }=11388.83 \\mathrm{~km}$ then they lose the final mark only]\n\nWavelength being used,\n\n$$\n\\begin{aligned}\n& \\lambda_{\\text {obs }}=\\frac{c}{f}=\\frac{3.00 \\times 10^{8}}{230 \\times 10^{9}}=1.30 \\mathrm{~mm} \\\\\n& \\theta_{\\text {min }}=\\frac{\\lambda_{\\text {obs }}}{d_{\\max }}=\\frac{1.30 \\times 10^{-3}}{10907.93 \\times 10^{3}}=1.20 \\times 10^{-10} \\mathrm{rad}=24.7 \\mu \\text { as }\n\\end{aligned}\n$$\n\n[Must be in microacrseconds ( $\\mu$ as) for the final mark]",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_1000",
        "problem": "Neso is the outermost moon of Neptune, and is notable for being the moon with the greatest known orbital distance from its planet, corresponding to a period of over 26 Earth years. At its furthest point (apoapsis), it is further from Neptune than Mercury gets from the Sun, meaning it is close to the limit of being held by Neptune's gravity (see Figure 1).\n\n[figure1]\n\nFigure 1: Plan view of Neptune's outermost satellite orbits. Neptune is represented by the black dot at the centre. The green orbit corresponds to Nereid, the two in blue are the prograde orbits of Sao and Laomedeia, and the red are the retrograde orbits of Halimede, Psamathe and Neso (bottom right). The dotted circle shows the theoretical outer limit of stability for Neptune satellites. Taken from Shepperd et al. (2006).\n\nOrbital data about Neso and Mercury are summarised below:\n\n|  | Mercury | Neso |\n| :--- | :---: | :---: |\n| Semi-major axis, $a\\left(\\times 10^{9} \\mathrm{~m}\\right)$ | 57.909 | 49.285 |\n| Eccentricity, $e$ | 0.2056 | 0.5714 |\n| Orbital period, $T$ (Earth days) | 87.97 | 9740.73 |\n\nUse the orbital data to work out the ratio of Neptune's mass to the Sun's, $M_{\\mathrm{Nep}} / M_{\\odot}$.",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nNeso is the outermost moon of Neptune, and is notable for being the moon with the greatest known orbital distance from its planet, corresponding to a period of over 26 Earth years. At its furthest point (apoapsis), it is further from Neptune than Mercury gets from the Sun, meaning it is close to the limit of being held by Neptune's gravity (see Figure 1).\n\n[figure1]\n\nFigure 1: Plan view of Neptune's outermost satellite orbits. Neptune is represented by the black dot at the centre. The green orbit corresponds to Nereid, the two in blue are the prograde orbits of Sao and Laomedeia, and the red are the retrograde orbits of Halimede, Psamathe and Neso (bottom right). The dotted circle shows the theoretical outer limit of stability for Neptune satellites. Taken from Shepperd et al. (2006).\n\nOrbital data about Neso and Mercury are summarised below:\n\n|  | Mercury | Neso |\n| :--- | :---: | :---: |\n| Semi-major axis, $a\\left(\\times 10^{9} \\mathrm{~m}\\right)$ | 57.909 | 49.285 |\n| Eccentricity, $e$ | 0.2056 | 0.5714 |\n| Orbital period, $T$ (Earth days) | 87.97 | 9740.73 |\n\nUse the orbital data to work out the ratio of Neptune's mass to the Sun's, $M_{\\mathrm{Nep}} / M_{\\odot}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3776e2d93eca0bbf48b9g-07.jpg?height=991&width=985&top_left_y=727&top_left_x=547"
        ],
        "answer": [
            "$5.028 \\times 10^{-5}$"
        ],
        "solution": "Kepler's Third Law (on page 2 of the paper) states that $\\frac{a^{3}}{T^{2}}=\\frac{G M}{4 \\pi^{2}}$ so <br> $\\qquad \\frac{M_{\\mathrm{Nep}}}{M_{\\odot}}=\\frac{\\left(\\frac{a_{\\mathrm{Neso}}}{a_{\\mathrm{Mer}}}\\right)^{3}}{\\left(\\frac{T_{\\mathrm{Neso}}}{T_{\\mathrm{Mer}}}\\right)^{2}}$ <br> $\\qquad \\therefore \\frac{M_{\\mathrm{Nep}}}{M_{\\odot}}=\\frac{\\left(\\frac{49.285}{57.909}\\right)^{3}}{\\left(\\frac{9740.73}{87.97}\\right)^{2}}$ <br> $\\qquad=5.028 \\times 10^{-5}$ <br>",
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_737",
        "problem": "图为“嫦娥一号” 某次在近地点 $A$ 由轨道 1 变轨为轨道 2 的示意图, 其中 $B 、 C$ 分别为两个轨道的远地点。关于上述变轨过程及“嫦娥一号”在两个轨道上运动的情况，下列说法中正确的是 ( )\n\n[图1]\nA: “嫦娥一号”在轨道 1 的 $A$ 点处应点火加速\nB: “嫦娥一号”在轨道 1 的 $A$ 点处的速度比在轨道 2 的 $A$ 点处的速度大\nC: “嫦娥一号”在轨道 1 的 $B$ 点处的速度比在轨道 2 的 $C$ 点处的速度小\nD: “嫦娥一号”在轨道 1 的 $B$ 点处的加速度比在轨道 2 的 $C$ 点处的加速度小\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图为“嫦娥一号” 某次在近地点 $A$ 由轨道 1 变轨为轨道 2 的示意图, 其中 $B 、 C$ 分别为两个轨道的远地点。关于上述变轨过程及“嫦娥一号”在两个轨道上运动的情况，下列说法中正确的是 ( )\n\n[图1]\n\nA: “嫦娥一号”在轨道 1 的 $A$ 点处应点火加速\nB: “嫦娥一号”在轨道 1 的 $A$ 点处的速度比在轨道 2 的 $A$ 点处的速度大\nC: “嫦娥一号”在轨道 1 的 $B$ 点处的速度比在轨道 2 的 $C$ 点处的速度小\nD: “嫦娥一号”在轨道 1 的 $B$ 点处的加速度比在轨道 2 的 $C$ 点处的加速度小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_6842b9ceb844a90b34c3g-38.jpg?height=391&width=579&top_left_y=2323&top_left_x=356"
        ],
        "answer": [
            "A"
        ],
        "solution": "$\\mathrm{AB}$. 要想使“嫦娥一号”在近地点 $A$ 由轨道 1 变轨为轨道 2 , 需要加速做离心运动, 故应在 $A$ 点加速, $\\mathrm{A}$ 正确, $\\mathrm{B}$ 错误;\n\nC. “嫦娥一号” 从 $B$ 点运动至 $C$ 点, 万有引力做负功, 动能减小, 引力势能增大, 故嫦娥一号”在轨道 1 的 $B$ 点处的速度比在轨道 2 的 $C$ 点处的速度大, 故 C 错误;\n\nD. 根据牛顿第二定律\n\n$$\na=\\frac{\\frac{G M m}{r^{2}}}{m}=\\frac{G M}{r^{2}}\n$$\n\n其中 $r$ 为“嫦娥一号”到地心的距离。由图知 $r_{\\mathrm{B}}<r_{\\mathrm{C}}$, 所以 $a_{\\mathrm{B}}>a_{\\mathrm{C}}$, 故 $\\mathrm{D}$ 错误。故选 A。",
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_309",
        "problem": "已知某卫星在赤道上空的圆形轨道运行, 轨道半径为 $r_{1}$, 运行周期为 $T$, 卫星运动方向与地球自转方向相同, 不计空气阻力, 万有引力常量为 $G \\circ$ 。求:\n\n如图所示, 假设某时刻, 该卫星在 $A$ 点变轨进入粗圆轨道, 近地点 $B$ 到地心距离为 $r_{2}$, 求卫星在椭圆轨道上的周期 $T_{1}$;\n\n[图1]",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n\n问题:\n已知某卫星在赤道上空的圆形轨道运行, 轨道半径为 $r_{1}$, 运行周期为 $T$, 卫星运动方向与地球自转方向相同, 不计空气阻力, 万有引力常量为 $G \\circ$ 。求:\n\n如图所示, 假设某时刻, 该卫星在 $A$ 点变轨进入粗圆轨道, 近地点 $B$ 到地心距离为 $r_{2}$, 求卫星在椭圆轨道上的周期 $T_{1}$;\n\n[图1]\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-136.jpg?height=514&width=531&top_left_y=654&top_left_x=337"
        ],
        "answer": [
            "$\\left(\\frac{r_{1}+r_{2}}{2 r_{1}}\\right)^{\\frac{3}{2}} \\cdot T$"
        ],
        "solution": "根据开普勒第三定律\n\n$$\n\\frac{\\left(\\frac{r_{1}+r_{2}}{2}\\right)^{3}}{T_{1}^{2}}=\\frac{r_{1}^{3}}{T^{2}}\n$$\n\n得\n\n$$\nT_{1}=\\left(\\frac{r_{1}+r_{2}}{2 r_{1}}\\right)^{\\frac{3}{2}} \\cdot T\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_903",
        "problem": "The NEOWISE telescope discovered a new comet on $27^{\\text {th }}$ March 2020, later given the official designation C/2020 F3 NEOWISE. Although when first discovered it only had an apparent magnitude of 18.0, it would become sufficiently bright that it could be seen with the naked eye by observers throughout the northern hemisphere, and was one of the brightest comets since Hale-Bopp in 1997.\n\n[figure1]\n\nFigure 4: The comet C/2020 F3 NEOWISE as seen from the UK in late July. Credit: Alex Calverley.\n\nOn its discovery date the comet was 1.702 au from the Earth and 2.089 au from the Sun, and at perihelion (when it was closest to the Sun) on $3^{\\text {rd }}$ July 2020 it was only 0.294649 au from the Sun.\n
Treating the comet nucleus as a sphere of fixed radius as it orbits, estimate the comet's apparent magnitude when it is at aphelion (when it is furthest from the Sun), assuming the phase angle is zero when it is observed from the Earth. Comment on whether the Hubble Space Telescope (limiting magnitude $\\sim 30$ ) would be able to see it.
",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe NEOWISE telescope discovered a new comet on $27^{\\text {th }}$ March 2020, later given the official designation C/2020 F3 NEOWISE. Although when first discovered it only had an apparent magnitude of 18.0, it would become sufficiently bright that it could be seen with the naked eye by observers throughout the northern hemisphere, and was one of the brightest comets since Hale-Bopp in 1997.\n\n[figure1]\n\nFigure 4: The comet C/2020 F3 NEOWISE as seen from the UK in late July. Credit: Alex Calverley.\n\nOn its discovery date the comet was 1.702 au from the Earth and 2.089 au from the Sun, and at perihelion (when it was closest to the Sun) on $3^{\\text {rd }}$ July 2020 it was only 0.294649 au from the Sun.\n
Treating the comet nucleus as a sphere of fixed radius as it orbits, estimate the comet's apparent magnitude when it is at aphelion (when it is furthest from the Sun), assuming the phase angle is zero when it is observed from the Earth. Comment on whether the Hubble Space Telescope (limiting magnitude $\\sim 30$ ) would be able to see it.
\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_c3e3c992a9c51eb3e471g-10.jpg?height=859&width=1285&top_left_y=593&top_left_x=385",
            "https://i.postimg.cc/PJ98BF6j/Screenshot-2024-04-06-at-22-11-52.png"
        ],
        "answer": [
            "43.7"
        ],
        "solution": "Calculating the aphelion distance using the formula from page 2 : <br> $\\qquad r_{a p h}=a(1+e)=362.9 \\times(1+0.999188)=725.4 \\mathrm{au}$ <br> Calculating the phase angle on the discovery date using the cosine rule: <br> $\\qquad \\therefore \\theta=E^{2}=S C^{2}+E C^{2}-2 \\times S C \\times E C \\cos \\theta$ <br> $\\therefore \\theta=\\cos ^{-1}\\left(\\frac{S E^{2}-S C^{2}-E C^{2}}{-2 \\times S C \\times E C}\\right)$ <br> Evaluating $p(\\theta)$ at both aphelion and in the discovery location: <br> $p(\\theta)_{\\text {aph }}=p\\left(0^{\\circ}\\right)=B$\n[figure2]\nAssuming a cross-sectional area for the comet of $A$, we can calculate the <br> total power incident for a given distance from the Sun, $d_{S}$ : <br> $\\qquad P_{\\text {inc }}=b_{\\text {inc }} \\times A=\\frac{L_{\\odot}}{4 \\pi d_{S}^{2}} \\times A$ <br> We can now use this to work out the brightness (i.e. intensity) of the light <br> reflected from the comet as observed a given distance from the Earth, $d_{E}$ : <br> $\\qquad b_{r e f}=\\frac{P_{r e f}}{4 \\pi d_{E}^{2}}=\\frac{\\frac{L_{\\odot}}{4 \\pi d_{S}^{2}} \\times A \\times p(\\theta)}{4 \\pi d_{E}^{2}}=\\frac{L_{\\odot} A p(\\theta)}{16 \\pi d_{S}^{2} d_{E}^{2}}$ <br> Considering the ratio of the reflected brightness for the two locations: <br> $\\qquad \\frac{b_{\\text {aph }}}{b_{\\text {disc }}}=\\frac{p(\\theta)_{\\text {aph }} d_{S, \\text { disc }}^{2} d_{E, \\text { disc }}^{2}}{p(\\theta)_{\\text {disc }} d_{S, a p h}^{2} d_{E, \\text { aph }}^{2}}$ <br> $\\qquad \\frac{B .089^{2} \\times 1.702^{2}}{0.893 B \\times 725.4^{2} \\times 724.4^{2}}=5.13 \\times 10^{-11}$ <br> [Allow $d_{E \\text { aph }}=726.4$ au too giving $5.10 \\times 10^{-11}$, but penalise by 0.5 <br> marks any other value] <br> We can now use this flux ratio to work out the magnitude at aphelion: <br> $m_{\\text {aph }}=m_{\\text {disc }}-2.5 \\log \\left(\\frac{b_{a p h}}{b_{\\text {disc }}}\\right)=18.0-2.5 \\log \\left(5.13 \\times 10^{-11}\\right)=43.7$ <br> This is much fainter than the Hubble Space Telescope's limiting magnitude <br> so would not be visible",
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_653",
        "problem": "如图所示，宇宙飞船绕地球做圆周运动时，由于地球遮挡阳光，会经历“日全食”过\n\n程, 太阳光可看作平行光, 宇航员在 $A$ 点测出地球的张角为 $\\alpha$ 。已知地球半径为 $R$, 地\n球质量为 $\\mathrm{M}$, 引力常量为 $G$, 不考虑地球公转的影响。求:\n\n飞船绕地球一周经历“日全食”过程的时间 $t$ 。\n\n[图1]",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这个问题的答案是一个表达式。\n\n问题:\n如图所示，宇宙飞船绕地球做圆周运动时，由于地球遮挡阳光，会经历“日全食”过\n\n程, 太阳光可看作平行光, 宇航员在 $A$ 点测出地球的张角为 $\\alpha$ 。已知地球半径为 $R$, 地\n球质量为 $\\mathrm{M}$, 引力常量为 $G$, 不考虑地球公转的影响。求:\n\n飞船绕地球一周经历“日全食”过程的时间 $t$ 。\n\n[图1]\n\n你输出的所有数学公式和符号应该使用LaTeX表示！\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER是一个不含等号的表达式，例如ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_29925d26250e50e92016g-086.jpg?height=277&width=514&top_left_y=393&top_left_x=334"
        ],
        "answer": [
            "$\\alpha \\sqrt{\\frac{R^{3}}{G M \\sin ^{3} \\frac{\\alpha}{2}}}$"
        ],
        "solution": "飞船绕地球做圆周运动\n\n$$\nG \\frac{M m}{r^{2}}=m\\left(\\frac{2 \\pi}{T}\\right)^{2} r\n$$\n\n每次“日全食”对应的圆心角\n\n$$\n\\beta=\\alpha\n$$\n\n又\n\n$$\n\\begin{aligned}\n\\beta & =\\omega t \\\\\n2 \\pi & =\\omega T\n\\end{aligned}\n$$\n\n解得\n\n$$\nt=\\alpha \\sqrt{\\frac{R^{3}}{G M \\sin ^{3} \\frac{\\alpha}{2}}}\n$$",
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_1150",
        "problem": "The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on $12^{\\text {th }}$ August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $r_{\\text {peri }}=9.86 R_{\\odot}$, about 7 times closer than any previous probe, the first of which is due on $24^{\\text {th }}$ December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.\n[figure1]\n\nFigure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.\n\n$$\nv^{2}=G M\\left(\\frac{2}{r}-\\frac{1}{a}\\right)\n$$\n\nGiven that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in $\\mathrm{km} \\mathrm{s}^{-1}$.\n\nClose to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.\n\n[figure2]\n\nFigure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.\n\nWhen considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called 'anomalies') necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, $E$, is then the angle between the major axis and the perpendicular projection of the object (some time $t$ after perihelion) onto the circle as measured from the centre of the ellipse ( $\\angle x c z$ in the figure). The mean anomaly, $M$, is the angle between the major axis and where the object would have been at time $t$ if it was indeed on the circular orbit ( $\\angle y c z$ in the figure, such that the shaded areas are the same).\n\n[figure3]\n\nFigure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled $S$ and the probe $P . M$ and $E$ are the mean and eccentric anomalies respectively. The angle $\\theta$ is called the true anomaly and is not needed for this question. Credit: Wikipedia.\n\n\nThe eccentric anomaly can be related to the mean anomaly through Kepler's Equation,\n\n$$\nM=E-e \\sin E \\text {. }\n$$e. Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on $12^{\\text {th }}$ August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $r_{\\text {peri }}=9.86 R_{\\odot}$, about 7 times closer than any previous probe, the first of which is due on $24^{\\text {th }}$ December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.\n[figure1]\n\nFigure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.\n\n$$\nv^{2}=G M\\left(\\frac{2}{r}-\\frac{1}{a}\\right)\n$$\n\nGiven that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in $\\mathrm{km} \\mathrm{s}^{-1}$.\n\nClose to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.\n\n[figure2]\n\nFigure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.\n\nWhen considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called 'anomalies') necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, $E$, is then the angle between the major axis and the perpendicular projection of the object (some time $t$ after perihelion) onto the circle as measured from the centre of the ellipse ( $\\angle x c z$ in the figure). The mean anomaly, $M$, is the angle between the major axis and where the object would have been at time $t$ if it was indeed on the circular orbit ( $\\angle y c z$ in the figure, such that the shaded areas are the same).\n\n[figure3]\n\nFigure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled $S$ and the probe $P . M$ and $E$ are the mean and eccentric anomalies respectively. The angle $\\theta$ is called the true anomaly and is not needed for this question. Credit: Wikipedia.\n\n\nThe eccentric anomaly can be related to the mean anomaly through Kepler's Equation,\n\n$$\nM=E-e \\sin E \\text {. }\n$$\n\nproblem:\ne. Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of days, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_ffe0ae050771e0e3decbg-04.jpg?height=708&width=1438&top_left_y=694&top_left_x=318",
            "https://cdn.mathpix.com/cropped/2024_03_14_ffe0ae050771e0e3decbg-05.jpg?height=411&width=1539&top_left_y=383&top_left_x=264",
            "https://cdn.mathpix.com/cropped/2024_03_14_ffe0ae050771e0e3decbg-05.jpg?height=603&width=714&top_left_y=1429&top_left_x=677"
        ],
        "answer": [
            "9.84"
        ],
        "solution": "With the perihelion distance we can find the eccentricity,\n\n$$\nr_{\\text {peri }}=a(1-e) \\therefore \\quad e=1-\\frac{r_{\\text {peri }}}{a}=1-\\frac{9.86 \\times 6.96 \\times 10^{8}}{5.79 \\times 10^{10}}=0.882\n$$\n\nUsing the formula just derived we can find $E$ for $r=0.25$ au,\n\n$$\nE=\\cos ^{-1}\\left(\\frac{1}{e}\\left(1-\\frac{r}{a}\\right)\\right)=\\cos ^{-1}\\left(\\frac{1}{0.882}\\left(1-\\frac{0.25 \\times 1.50 \\times 10^{11}}{5.79 \\times 10^{10}}\\right)\\right)=1.16 \\mathrm{rad}\\left(=66.4^{\\circ}\\right)[1]\n$$\n\nUsing the formula given we can find $\\mathrm{M}$,\n\n$$\nM=E-e \\sin E=1.16-0.882 \\times \\sin 1.16=0.351 \\mathrm{rad}\\left(=20.1^{\\circ}\\right)\n$$\n\nAllowing for both sides of the ellipse, $\\Delta M=0.703 \\mathrm{rad}\\left(=40.3^{\\circ}\\right)$\n\nMovement through the circular orbit has constant angular velocity, meaning\n\n$$\n\\frac{\\Delta M}{\\Delta t}=\\frac{2 \\pi}{T} \\quad \\therefore \\Delta t=\\frac{T \\Delta M}{2 \\pi}=\\frac{88 \\times 0.703}{2 \\pi}=9.84 \\text { days ( }=9 \\text { days } 20 \\text { hours } 12 \\text { mins) [1] }\n$$\n\n[If they forget the factor of two (so $\\Delta t=4.92$ days) allow 4 marks. Allow $\\pm 1$ hour on the final answer to account for intermediate rounding errors. Allow full ecf for using their value of the semi-major axis from part $b$. Must be given in days for the final mark]",
        "answer_type": "NV",
        "unit": [
            "days"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_1075",
        "problem": "Hanny's Voorwerp (Dutch for 'object') is a rare type of astronomical object discovered in 2007 by the school teacher Hanny van Arkel whilst participating as a volunteer in the Galaxy Zoo project. When inspecting the image of the galaxy IC 2497 in the constellation Leo Minor, she observed a bright green blob close to the galaxy.\n\n[figure1]\n\nFigure 5: HST image of galaxy IC 2497 and the glowing Voorwerp below it.\n\nCredit: Keel et al. (2012) \\& Galaxy Zoo.\n\nSubsequent observations have shown that the galaxy IC 2497 is at a redshift of $z=0.05$, with the Voorwerp at a similar distance and with a projected angular separation of 20 arcseconds from the centre of the galaxy $\\left(3600\\right.$ arcseconds $\\left.=1^{\\circ}\\right)$. Radio observations suggest that the Voorwerp is a massive cloud of gas, made of ionized hydrogen, with a size of $10 \\mathrm{kpc}$ and a mass of $10^{11} \\mathrm{M}_{\\odot}$. It is probably a cloud of gas that was stripped from the galaxy during a merger with another nearby galaxy.\n\nIn this question you will explore the cause of the 'glow' of the Voorwerp and will learn about a new type of an astronomical object; a quasar.b. Calculate the power (luminosity) of the source required to completely ionize the Voorwerp (assumed to be spherical), given that the mass of a hydrogen atom is $1.67 \\times 10^{-27} \\mathrm{~kg}$ and the ionization energy of hydrogen is $13.6 \\mathrm{eV}$, where $1 \\mathrm{eV}=1.60 \\times 10^{-19} \\mathrm{~J}$.",
        "prompt": "You are participating in an international Astronomy competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nHanny's Voorwerp (Dutch for 'object') is a rare type of astronomical object discovered in 2007 by the school teacher Hanny van Arkel whilst participating as a volunteer in the Galaxy Zoo project. When inspecting the image of the galaxy IC 2497 in the constellation Leo Minor, she observed a bright green blob close to the galaxy.\n\n[figure1]\n\nFigure 5: HST image of galaxy IC 2497 and the glowing Voorwerp below it.\n\nCredit: Keel et al. (2012) \\& Galaxy Zoo.\n\nSubsequent observations have shown that the galaxy IC 2497 is at a redshift of $z=0.05$, with the Voorwerp at a similar distance and with a projected angular separation of 20 arcseconds from the centre of the galaxy $\\left(3600\\right.$ arcseconds $\\left.=1^{\\circ}\\right)$. Radio observations suggest that the Voorwerp is a massive cloud of gas, made of ionized hydrogen, with a size of $10 \\mathrm{kpc}$ and a mass of $10^{11} \\mathrm{M}_{\\odot}$. It is probably a cloud of gas that was stripped from the galaxy during a merger with another nearby galaxy.\n\nIn this question you will explore the cause of the 'glow' of the Voorwerp and will learn about a new type of an astronomical object; a quasar.\n\nproblem:\nb. Calculate the power (luminosity) of the source required to completely ionize the Voorwerp (assumed to be spherical), given that the mass of a hydrogen atom is $1.67 \\times 10^{-27} \\mathrm{~kg}$ and the ionization energy of hydrogen is $13.6 \\mathrm{eV}$, where $1 \\mathrm{eV}=1.60 \\times 10^{-19} \\mathrm{~J}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of \\mathrm{~W}, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_204b2e236273ea30e8d2g-08.jpg?height=800&width=577&top_left_y=508&top_left_x=745"
        ],
        "answer": [
            "6.50 \\times 10^{37}"
        ],
        "solution": "Given we know the radius of the cloud $(10 \\mathrm{kpc})$ and the mass $\\left(10^{11} \\mathrm{M}_{\\odot}\\right)$ we can work out the number density of hydrogen atoms\n\n$$\n\\begin{aligned}\n& S_{*}=V n^{2} \\alpha \\quad \\text { and } \\quad n=\\frac{M / m_{\\mathrm{H}}}{V} \\quad \\text { and } \\quad V=\\frac{4}{3} \\pi R^{3} \\\\\n& \\therefore S_{*}=\\left(\\frac{M}{m_{\\mathrm{H}}}\\right)^{2} \\frac{\\alpha}{V} \\\\\n& =\\left(\\frac{10^{11} \\mathrm{M}_{\\odot}}{m_{\\mathrm{H}}}\\right)^{2} \\frac{\\alpha}{\\frac{4}{3} \\pi R^{3}}=\\left(\\frac{10^{11} \\times 1.99 \\times 10^{30}}{1.67 \\times 10^{-27}}\\right)^{2} \\frac{2.6 \\times 10^{-19}}{\\frac{4}{3} \\pi \\times\\left(10 \\times 10^{3} \\times 3.09 \\times 10^{16}\\right)^{3}} \\\\\n& =2.99 \\times 10^{55} \\text { photons s }^{-1}\n\\end{aligned}\n$$\n\n(watch that the units of $\\alpha$ are converted correctly to SI)\n\nThe luminosity can then be calculated as we know the energy of each photon\n\n$$\n\\begin{aligned}\nL=S_{*} E_{\\text {photon }} & =2.99 \\times 10^{55} \\times 13.6 \\times 1.60 \\times 10^{-19} \\\\\n& =6.50 \\times 10^{37} \\mathrm{~W}\n\\end{aligned}\n$$\n\n(Allow full credit for interpreting the $10 \\mathrm{kpc}$ 'size' of the cloud to mean its diameter rather than its radius, giving $\\mathrm{S}_{*}=2.39 \\times 10^{56}$ photons $\\mathrm{s}^{-1}$ and $\\mathrm{L}=5.20 \\times 10^{38} \\mathrm{~W}$ )\n\n[Working out $S_{*}$ directly may prove difficult for some calculators as $\\left(M / m_{H}\\right)^{2}$ may exceed their largest power of ten, in which case students should work out $\\mathrm{VS} *$ and then square it later.]",
        "answer_type": "NV",
        "unit": [
            "\\mathrm{~W}"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Astronomy_49",
        "problem": "如图所示, $A 、 B$ 两卫星绕地球运行, 运动方向相同, 此时两卫星距离最近, 其中 $A$是地球同步卫星, 轨道半径为 $r$. 地球可看成质量均匀分布的球体, 其半径为 $R$, 自转周期为 $T$. 若经过时间 $t$ 后, $A 、 B$ 第一次相距最远, 下列说法正确的有\n\n[图1]\nA: 在地球两极, 地表重力加速度是 $\\frac{4 \\pi^{2} r^{3}}{T^{2} R^{2}}$\nB: 卫星 $B$ 的运行周期是 $\\frac{2 T t}{T+t}$\nC: 卫星 $B$ 的轨道半径为是 $r \\sqrt[3]{\\left(\\frac{2 t}{2 t+T}\\right)^{2}}$\nD: 若卫星 $B$ 通过变轨与 $\\mathrm{A}$ 对接之后, $\\mathrm{B}$ 的机械能可能不变\n",
        "prompt": "你正在参加一个国际天文竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n如图所示, $A 、 B$ 两卫星绕地球运行, 运动方向相同, 此时两卫星距离最近, 其中 $A$是地球同步卫星, 轨道半径为 $r$. 地球可看成质量均匀分布的球体, 其半径为 $R$, 自转周期为 $T$. 若经过时间 $t$ 后, $A 、 B$ 第一次相距最远, 下列说法正确的有\n\n[图1]\n\nA: 在地球两极, 地表重力加速度是 $\\frac{4 \\pi^{2} r^{3}}{T^{2} R^{2}}$\nB: 卫星 $B$ 的运行周期是 $\\frac{2 T t}{T+t}$\nC: 卫星 $B$ 的轨道半径为是 $r \\sqrt[3]{\\left(\\frac{2 t}{2 t+T}\\right)^{2}}$\nD: 若卫星 $B$ 通过变轨与 $\\mathrm{A}$ 对接之后, $\\mathrm{B}$ 的机械能可能不变\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_04_01_9938578583ce82f2e878g-16.jpg?height=434&width=485&top_left_y=1887&top_left_x=337"
        ],
        "answer": [
            "A",
            "C"
        ],
        "solution": "$\\mathrm{A}$ 、对于卫星 $\\mathrm{A}$, 根据万有引力提供向心力, 可得: $G \\frac{M m}{r^{2}}=m \\frac{4 \\pi^{2}}{T^{2}} r$, 可得地球的质量: $M=\\frac{4 \\pi^{2} r^{3}}{G T^{2}}$, 在地球两极, 据万有引力等于重力, 可得: $m^{\\prime} g=G \\frac{M m^{\\prime}}{R^{2}}$, 联立解得: $g=\\frac{4 \\pi^{2} r^{3}}{R^{2} T^{2}}$, 故 A 正确;\n\n$\\mathrm{B} 、$ 卫星 $\\mathrm{A}$ 的运行周期等于地球自转周期 $\\mathrm{T}$. 设卫星 $\\mathrm{B}$ 的周期为 $T^{\\prime}$. 当卫星卫星 $\\mathrm{B}$ 比 $\\mathrm{A}$多转半周时, $\\mathrm{A} 、 \\mathrm{~B}$ 第一次相距最远, 则有: $\\frac{2 \\pi}{T^{\\prime}} t-\\frac{2 \\pi}{T} t=\\pi$, 解得: $T^{\\prime}=\\frac{2 T t}{T+2 t}$, 故 B 错误;\n\nC、根据开普勒第三定律得: $\\frac{r^{3}}{r_{B}^{3}}=\\frac{T^{2}}{T^{\\prime 2}}$, 解得: $r B=r \\sqrt[3]{\\left(\\frac{2 t}{2 t+T}\\right)^{2}}$, 故 C 正确;\n\n$\\mathrm{D}$ 、卫星 $B$ 通过变轨与 $\\mathrm{A}$ 对接, 则需要在原轨道上对卫星 $\\mathrm{B}$ 加速, 使万有引力不足以提供向心力, 做离心运动, 最后与 $\\mathrm{A}$ 对接, 则卫星 $\\mathrm{B}$ 的机械能要增大, 故 $\\mathrm{D}$ 错误.",
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Astronomy",
        "language": "ZH",
        "modality": "multi-modal"
    }
]