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38,636 | I am a UI designer working on a sci-fi strategy video game. I was wondering wether anybody here might be able to help? I am looking for a top down projection of the Milky Way that labels regions of space. This is in order to start building a server map of linked 'matches'. The space combat sim Elite:Dangerous has a galactic project split into 42 sectors (see attachment) although I know that most of the regions labelled are made up specifically for the game. Obviously I don't want to copy this, so wondered if there was a more scientific version we could base designs on. Do regional maps of the Milky Way exist in science? Or is it purely a sci-fi thing? | It doesn't exist. It is actually rather difficult to see the shape of the milky way, because we are inside it. The more distant parts are obscured behind the nearer and very much of it is invisible. (We think that a supernova would happen every 50 years or so in a galaxy the size of the Milky Way, but none have been seen in nearly 400 years, so even massive bright events are hidden). There are rough maps based on radio imaging of neutral hydrogen, but these are fuzzy and have artefacts, making part of the map invisible and part of it seems to point strongly towards the Sun. The well-know images of the milky way on Wikipedia are as good an artist's impression as any other. | {
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38,889 | The surface gravity on Mimas is $≈ 0.063\text{ m}/\text{s}^2$ and Saturn's gravitational acceleration at the distance of Mimas' orbit is: $$\frac{{GM}}{{r}^2} = \frac{{6.674 \times 10^{-11} \times 568.34 \times 10^{24}}}{{(185.52 \times 10^{6})}^2} ≈ 1.102 \text{ m}/\text{s}^2$$ How can this be? An object on Mimas' surface would be much more attracted to Saturn than it is to Mimas. Shouldn't Mimas itself get ripped apart or is my math wrong? | An object on Mimas' surface would be much more attracted to Saturn than it is to Mimas. You are missing that Mimas as a whole accelerates gravitationally toward Saturn. What this means is that a point on the surface of the Mimas will feel the acceleration at that point toward Saturn minus the acceleration of Mimas as a whole toward Saturn. This is the tidal acceleration. It is equal to $$a_\text{tidal} = \left|\frac{GM}{(R\pm r)^2}-\frac{GM}{r^2}\right| \approx 2 \frac{GMr}{R^3} = 2\frac{GM}{R^2}\frac{r}{R}$$ where $R$ is Mimas' semi major axis length and $r$ is Mimas' mean radius. The approximation assumes that $r\ll R$ , which most certainly is the case given that Mimas's radius is about 1/1000 of the semi-major axis length of its orbit about Saturn. The result is rather small, about 0.002355 m/s 2 . | {
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38,892 | The famous measurements and calculations done by Eratosthenes around 300 BC are very widely known. He concluded correctly that the circumference of the Earth is about $252\,000$ times the length of an athletic stadium. But what Eratosthenes did would make no sense if the Sun were (for example) only $6000$ miles from the Earth. How did he know it was much farther away than that? | The sun and the moon go around the observer once a day, Eratosthenes knew that the apparent size of moon doesn't change. This must mean that Alexandria is near the centre of the moon's orbit. But the apparent size also doesn't change when viewed from anywhere. So everywhere is close to the centre of the moon's orbit. Thus the moon must be much further than the radius of the Earth. If the moon were 6000 miles from the Earth, then it would seem to grow and shrink in size as it passed by (such an effect can be seen on Mars, where the moon really does orbit close to the planet) And the Sun is further still. At half moon, the sun seems to be at $90^\circ$ to the moon. This is only possible if the sun is much further away than the moon. In conclusion, the distance to the sun must be very very large in comparison to the radius of the Earth, and we can assume that that the rays of light from the sun are parallel. | {
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38,927 | It's easy to find resources stating that the heliocentric model is right and geocentric is wrong . But how wrong was it? Was it correct in any way? It was built on incorrect assumptions, but despite that - was it of any use to describe the apparent motion of celestial bodies? Was it more accurate for some things, but less accurate for others? Or was it altogether a flop and astronomers couldn't get anything out of it either way? I can only find multiple articles proving the heliocentric model, explaining the geocentric one or claiming that it was simply wrong - but I can't find anything about its accuracy and usefulness. (1) (2) (3) Edit 1: I incorrectly used geocentric model when it seems I wanted to say Ptolemaic model - the one with deferents and epicycles, with Earth as its origin. Ptolemaic model (click for full size) Thank you for clarifying, and sorry for the confusion. The answers provided regarding any other geocentric models are still valid and useful, so this is just a minor errata. | Ptolemy's epicyclic, geocentric model, in use until the Renaissance, was very accurate in terms of predicting the positions of planets and the times of eclipses. What it couldn't account for were things like the correlations between apparent size and phase of Venus, or to properly account for the variation in brightness of the planets. Thus the reason for discarding the geocentric model was not really because it lacked precision, but that it failed to explain various other observational facts, especially after the development of telescopes. No doubt you could tune the Ptolemaic system even further (more epicycles?) to iron out some of the small errors that were revealed by Tycho's positional measurements at the turn of the 16th century, which had a precision unavailable to Ptolemy. However, the advent of Kepler's laws and subsequent explanation by Newton, rendered the geocentric model obsolete. As you can judge from (well written) articles like this one , geocentrism is actually quite hard to kill-off observationally, if you are prepared to accept that the universe is arranged "just so". | {
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38,934 | If time passes more slowly, relative to Earth, for a traveler at relativistic speeds, say at .8c, traveling between stars inside this galaxy, does time also pass more slowly for a distant galaxy that is moving away at relativistic speeds, where the speed difference is due to the Hubble expansion of space? | Ptolemy's epicyclic, geocentric model, in use until the Renaissance, was very accurate in terms of predicting the positions of planets and the times of eclipses. What it couldn't account for were things like the correlations between apparent size and phase of Venus, or to properly account for the variation in brightness of the planets. Thus the reason for discarding the geocentric model was not really because it lacked precision, but that it failed to explain various other observational facts, especially after the development of telescopes. No doubt you could tune the Ptolemaic system even further (more epicycles?) to iron out some of the small errors that were revealed by Tycho's positional measurements at the turn of the 16th century, which had a precision unavailable to Ptolemy. However, the advent of Kepler's laws and subsequent explanation by Newton, rendered the geocentric model obsolete. As you can judge from (well written) articles like this one , geocentrism is actually quite hard to kill-off observationally, if you are prepared to accept that the universe is arranged "just so". | {
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39,056 | Background: I'm training to be a geography teacher. Currently I have practice lessons and I'll be discussing solar time and standard time with the class. Now I stumbled over an issue to which I could not find an answer: We teach the students: Sidereal day: In 23 h 56 min the earth rotates 360° Solar day: In 24 h the earth rotates 361° We also teach that in order to construct time zones (as a replacement of solar time), one divided 360° by 24 (to have a zone for each hour of the day), which results in 24 zones with 15° each. Now my question is: Why does one mix the measures for sidereal day and solar day? Or put differently: Why doesn't one calculate 361°/24 or 360° / 23.933? | The Earth takes 23 hours 56 minutes to rotate once. But that is not relevant to most people. Sure, the stars will be in the same position again after 23 hours 56 minutes, but the sun will not be in the same position. It is far more important, for most people, to measure the time from noon to noon. And the average time from noon to noon is 24 hours. This is because the motion of the sun is a combination of both the spinning of the Earth and the orbit of the Earth around the sun. The orbital motion of the Earth adds four minutes. You should also teach the students In twenty-four hours the sun advances 360 degrees. (solar day) Time zones are based on clock time, which is based on the motion of the sun and not the motion of the stars. | {
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39,075 | How do scientists know that distant parts of the universe obey the physical laws exactly as we observe around us? The question might look a bit odd but I am really stuck on my head. We know, scientists (with tools) explored physically only our solar system and some parts of our galaxy which is really a tiny part of the observable universe. And now they are constantly using these knowledge base along with 'tested physical laws' to measure the properties of distant parts of our universe. For example, we tested and found the speed of the light is constant within our local periphery many times (within our Earth and Space around Earth). But yet we presume that the speed of the light is constant even at the farthest part of our universe. Certainly we did not test it in the other distant part of the universe because we have no way until now. Not only light but also the other physical properties like, luminosity, gravity, and etc related physical laws are agreed upon based on tests within our solar system. And based on these laws we deduced the properties of other parts of the universe ( i.e. age, distance, mass, luminosity of stars in millions/billions light years away). My question is, how do we know that these physical laws which we tested within a tiny area of the universe are consistently working in the distant parts of it? Is there any probability that the distant part of our universe obeys physical laws differently and our prediction based on applied physical laws gave us an unreal illusion of the actual reality, yet consistently? | We don't know in general but to the extent we can measure, the laws seem to be the same, even if conditions are not. For example radioactive decay: We know how fast various elements decay, and we can observe the results of radioactive decay in distant supernovae. The conclusion is that, for at least some elements, the rate of radioactive decay is the same on Earth as it is in distant supernovae. After accounting for redshift, spectral emission lines remain unchanged by distance. This implies that the fine-structure constant is indeed constant. Distant galaxies have gravitational fields, and interactions between galaxies proceeds in the same way in distant galaxies as it does in local ones. Eventually, the justification is philosophical: There is no observational reason to believe gravity behaves differently in distant parts of the universe, and so we believe that it does not, In the extreme conditions of the early universe, some physical laws were different. For example, instead of distinct electromagnetic and weak fields, there was a single Electroweak field. But this can be described as single "law" with the electromagnetic and weak interactions being just the low energy approximation of the electroweak interaction. So if it were discovered that Gravity (for example) was working differently in distant parts of the universe, but that there was a consistent pattern or rule for how it varied, then that would simply become the new theory of gravity (with general relativity becoming only the local approximation to this new law). There is a more fundamental assumption: that the behaviour of matter and energy in the universe can be modelled by "laws". There are no angels dancing on pinheads. The justification for this is strictly in the realm of philosophy. | {
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39,077 | The mass of Venus seems rather complicated to determine to me: Venus doesn't have any satellites, so you can't just apply Kepler's third law (like you would with Jupiter or Saturn for instance) to determine its mass. The gravitational tug of Venus on the Sun is very small compared to that of other Jupiter or Saturn, so it seems like it would be difficult to extract what part of the Sun's proper motion is caused by Venus. There are few asteroids with orbits close to that of Venus, so not many objects who might have their trajectory modified by Venus. Knowing Venus' radius, and assuming it has the same density as the Earth, you can get a pretty close estimate of its mass (85% of Earth's mass with this assumption, when the actual value is 82%). But that's a pretty strong assumption (the density of Earth and Venus only happen to be close by chance) and a rather unsatisfactory "guesstimate". Nowdays, there are a few probes that have flown by Venus, so by looking at their trajectory, you can infer what Venus' gravitational field looks like. But those fly-bys are pretty recent. Did we know about Venus' mass before those fly-bys? How was the mass of Venus measured for the first time? | How was the mass of Venus measured for the first time? In the mid 19 th century, Urbain Le Verrier's predicted of the existence of a then unknown planet beyond the orbit of Uranus. He even predicted this planet's orbit. The discovery of Neptune based on his predictions was perhaps his greatest accomplishment. Le Verrier then went on to investigate Mercury. He used observations of Mercury, Venus, the Sun (as a stand-in for the Earth) and Mars and calculated that Mercury should precess by 532 arc seconds per century based on Newtonian mechanics. Along the way, he had to (and did) estimate the mass of Venus. There was a problem here; the observed precession of Mercury's orbit is 575 arc seconds per century, 43 arc seconds per century greater than his calculated value. This led Le Verrier to conjecture that there was a planet even closer to the Sun than Mercury. Despite the failure to discover the non-existent planet tentatively named Vulcan, Le Verrier's estimate for the mass of Venus was fairly close to the correct figure, within a couple of percent. Once the cause of this 43 arc second per century discrepancy was discovered by Einstein, the mass of Venus was determined with even greater accuracy. Of course, once probes were sent into orbit about Venus, its mass was determined with greater accuracy yet. Reference: Leverrier, M. "On the masses of the planets, and the parallax of the Sun." Monthly Notices of the Royal Astronomical Society 32 (1872): 322. | {
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39,243 | A supernova is the explosion of a single star; so how is it that thousands of stars can "be born of" that one explosion (presumably only using the unspent fuel / lighter elements of the original)? Does that also mean the stars coming from a supernova cannot possibly become supernovae themselves? For example, according to Wikipedia , the Rosetta Nebula has about 2500 including a couple of O-type stars! approximately 2500 young stars lie in this star-forming complex, including the massive O-type stars HD 46223 and HD 46150 Edit to reduce confusion and risk of reinforcing my false premised path of questioning: I somehow had the idea that nebulae came from single supernova explosions and that all nebulae were planetary nebulae. After research, that was an absurd understanding on multiple levels. | Stars don't "come from" a supernova. Stars come from the interstellar gas in the galaxy, particularly where it is more concentrated into nebulae. This gas is mostly hydrogen and helium, but it is "enriched" with heavier elements from old stars, including from stars that have exploded in supernovae. Over the billions of years since the galaxy formed, there have been many supernove which have gradually increased the amount of heavier elements in the interstellar gas. The shockwave from a supernova can cause the gas to become squeezed and more dense, and this can trigger the gravitational collapse that leads to star formation, but the gas for those stars doesn't come from one particular supernova. As a nebula collapses into many stars, some (the largest) will be large enough to supernovae (which can trigger subsequent periods of star formation). | {
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39,401 | I am interested to know whether it's possible for the moon to orbit the earth at such a rate that we would see only one New Moon and one Full Moon a year. If so, what other effects would we experience? I at first assumed this would be a one-year period, and technically I think that would produce one New Moon, but that single New Moon would last all year (i.e., you would never see the moon). So I'm interested to know if there is an orbital period that would give us one New Moon and one Full Moon, as seen from our earth, each year. As I hold my hands up in front of my face and experimentally rotate them around each other, I want to say a six-month period would produce this effect, but I'm not sure that's correct, so I'd appreciate a word from anyone who actually knows what they're talking about. Thanks! (Apologies if this has been asked before; I wasn't able to find it.) | This would naturally be achieved by a quasi-satellite. These are objects with a 1:1 orbital resonance with a different eccentricity to the planet. From the perspective of the planet, this leads to the quasi-satellite appearing to travel around the planet once a year in a retrograde direction. Starting with the quasi-satellite at its maximum distance from the Sun, this would be at full phase. Six months later, the quasi-satellite would be between Earth and the Sun, so would be at new phase. The diagrams below show an idealised version of the configuration based on Keplerian orbits in a non-rotating frame and in a frame where the Earth and the Sun are kept in a fixed position. Real systems will evolve dynamically due to the gravitational interaction between the planet and the quasi-satellite, leading to precession and oscillations of the alignment of the orbits. Earth currently has several known quasi-satellites, the closest of which is (469219) Kamoʻoalewa . The known quasi-satellites in our Solar System do not have long lifetimes: they are relatively recent captures into the quasi-satellite orbit that will eventually escape. Nevertheless, there are some conditions under which quasi-satellites can be stable over long periods , and perhaps some exoplanetary systems have more favourable conditions for quasi-satellites than ours does. | {
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39,561 | In this answer to Is lithium considered a metal in astronomy? this image from Wikimedia of the abundances of elements in the universe was shared: Credit: Wikimedia Commons user 28bytes, under C.C.-by-S.A.-3.0 . I thought it was fascinating that there is a near-constant zig-zag of abundancies. Are these values accurate, and if so, what would cause this zig-zag? | Nuclei are more stable if they have an even number of protons Z, and are also more stable if they have an even number of neutrons N. This is because the particles form pairs. (Almost all nuclei with both N and Z odd are unstable with respect to beta decay.) If a nucleus is going to have significant cosmic abundance, it must be either absolutely stable or at least have an extremely long half-life. This is more likely to happen when Z is even (and also when N is even). Stability also effects probabilities of different types of decay. E.g., if a nucleus can undergo both beta+ and beta- decay, then it is usually more likely to do the decay that results in the more stable daughter. | {
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39,686 | What is the intuition behind why gravity is inversely proportional to exactly square of distance between two object and not cube or not some multiplier. Basically how Newton came up that its exactly square of distance? How it was validated it is in fact square of distance ? | Imagine "gravity" spreading out in a sphere, like light from a bulb. For each doubling of the distance, the sphere has four times the area. The surface area of the sphere is proportional to the square of the radius. If the same gravity is stretched over that sphere, the force of gravity would be inversely proportional to the square of the radius. This gives sufficient intuition for most physicists of the time to believe that gravity was probably inverse square. Newton's real genius was that he was able to prove using mathematics that a planet moving in an inverse square law would obey Kepler's laws, something that his contemporaries had failed to do. A well known story, related by de Moivre: In 1684 Dr Halley came to visit [Newton] at Cambridge. After they had been some time together, the Doctor asked him what he thought the curve would be that would be described by the planets supposing the force of attraction towards the sun to be reciprocal to the square of their distance from it. Sir Isaac replied immediately that it would be an ellipse. The Doctor, struck with joy and amazement, asked him how he knew it. Why, saith he, I have calculated it. | {
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39,697 | Just about every webpage related to asteroids states that, on average, asteroids are about 600,000 miles apart from one another. Have planetary scientists hazarded a guess at how close they might have been 4 billion years ago? 2 billion years ago? 200 million years? I'm not looking for a particular age range, just an understanding of the trend over time. Unfortunately, I'm having trouble looking up an answer to this question on my own because the keywords I've thought of keep grabbing other unrelated things Thank you! | Imagine "gravity" spreading out in a sphere, like light from a bulb. For each doubling of the distance, the sphere has four times the area. The surface area of the sphere is proportional to the square of the radius. If the same gravity is stretched over that sphere, the force of gravity would be inversely proportional to the square of the radius. This gives sufficient intuition for most physicists of the time to believe that gravity was probably inverse square. Newton's real genius was that he was able to prove using mathematics that a planet moving in an inverse square law would obey Kepler's laws, something that his contemporaries had failed to do. A well known story, related by de Moivre: In 1684 Dr Halley came to visit [Newton] at Cambridge. After they had been some time together, the Doctor asked him what he thought the curve would be that would be described by the planets supposing the force of attraction towards the sun to be reciprocal to the square of their distance from it. Sir Isaac replied immediately that it would be an ellipse. The Doctor, struck with joy and amazement, asked him how he knew it. Why, saith he, I have calculated it. | {
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39,709 | Carbon is considered volatile by planetary science, eg Moon lacks volatiles and thus lacks carbon.
However volatiles are defined as "elements or substances with low boiling point", but Carbon boiling point is very high! Its sublimation point is 3900K, so it should be refractory and not volatile.
I am confused why it is volatile with such high sublimation point.
Regards,
Alex | Imagine "gravity" spreading out in a sphere, like light from a bulb. For each doubling of the distance, the sphere has four times the area. The surface area of the sphere is proportional to the square of the radius. If the same gravity is stretched over that sphere, the force of gravity would be inversely proportional to the square of the radius. This gives sufficient intuition for most physicists of the time to believe that gravity was probably inverse square. Newton's real genius was that he was able to prove using mathematics that a planet moving in an inverse square law would obey Kepler's laws, something that his contemporaries had failed to do. A well known story, related by de Moivre: In 1684 Dr Halley came to visit [Newton] at Cambridge. After they had been some time together, the Doctor asked him what he thought the curve would be that would be described by the planets supposing the force of attraction towards the sun to be reciprocal to the square of their distance from it. Sir Isaac replied immediately that it would be an ellipse. The Doctor, struck with joy and amazement, asked him how he knew it. Why, saith he, I have calculated it. | {
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39,749 | This website shows a telescope projecting the sun onto a blackboard: https://astronomyconnect.com/forums/articles/2-three-ways-to-safely-observe-the-sun.21/ Why isn't the board catching fire? You can easily start a fire on a sunny day by targeting the focal point of a magnifying glass onto something flammable. Why isn't the telescope in this picture doing the same thing? Photo by Luis Fernández García | For a magnifying lens or mirror to be able to ignite something with light from the Sun, its surface area must be large relative to the square of the focal length. Solar energy will be spread throughout the projected image, and the size of that image will be essentially proportional to the focal length, making its area proportional to the square of focal length. A typical hand magnifier will have a relatively short focal length, making the projected image quite small. Telescopes, however, are designed to emulate lenses with much longer focal lengths so as to produce larger images. The amount of heating from a telescope will be maximized when it's properly focused, but if the light is spread through a 64mm-diameter image it will be less than 1/1000 as powerful as it would be if it were focused with a shorter lens to produce an image which is only 2mm in diameter. Incidentally, a factor which makes the "Archimedes death ray" improbable as a means of focusing solar energy to directly ignite ships is that the size of the projected image of the Sun would increase with the distance to the enemy ships. On the other hand, the amount of focused solar energy needed to temporarily or permanently blind people is far below the amount required to ignite things. If the crew of a ship had flaming projectiles they wanted to launch at a town, but sunlight focused by the townspeople's shields were to blind some crew members at an inopportune time, it's not hard to imagine that the ship's crew might accidentally set fire to their own ship or other nearby ships. I think it entirely plausible that people witnessing the battle from shore might have observed that solar energy was being focused on ships, and that the ships subsequently ignited; it's not hard to imagine that such people would conclude that the solar energy ignited the ships whether it actually did or not. | {
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39,893 | Planets form from a protoplanetary disk that has been rotating around its star. The initial energy that makes them rotate really matters to me. Why did the protoplanetary disk start rotating around the star? Where did the initial energy to rotate come from? Why hasn't all the disk material been swallowed by the star? | Two rocks placed in space with no relative motion are going to be attracted by gravity, and hit. 3 rocks, placed in space with no carefully rigged symmetry, will likely miss each other, as the gravitational attraction of the additional rock changes their course. Those near misses are the beginning of rotation. Multiply that effect by trillions, and you have a protoplanetary disk. Of course, a lot of those molecules, dust and rock won't even start out with zero relative velocity, so near misses are inevitable. | {
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39,948 | Just a few minutes ago, I got a notification from Space.com stating that the Arecibo Observatory will be, sadly, decommissioned due to extensive damage to its structure. So, with the loss of one of the world's largest telescopes, we will be put at a setback in space exploration and astronomy. What will succeed Arecibo and when will this happen? | There's no simple answer. In the immediate future, different radio telescopes around the world will pick up the slack in various ways; how that happens will depend on the needs of individual observers and collaborations. Unless someone was to build an identical observatory at the same latitude as Arecibo, with the same frequency range, receiver options and field of view . . . we'll have to spread out the observing. Given that we didn't really expect to lose Arecibo for more than about a year $^{\dagger}$ , there are a lot of things up in the air. Insert "probably" and "maybe" into this answer wherever you please. I do pulsar timing, so I'll talk about what a future without Arecibo might look like from our perspective. My collaboration's two main instruments were Arecibo and the 100-meter Green Bank Telescope (GBT), with roughly half of the observing time at Arecibo and half at the GBT. There aren't firm plans in place for the exact strategy we'll take going forward, but we'll have to decrease our observing cadence at the GBT (i.e. how often we observe certain pulsars) so we can use some of the time to observe pulsars we'd normally observe using Arecibo. So the GBT can take some of the load, but as we only have a finite amount of observing time, that impacts our other observations. We'll likely have to decrease the cadence of observations and the number of pulsars we monitor regularly. Pulsar timing arrays will also have options with other existing instruments in the immediate future. A key one is the Canadian Hydrogen Intensity Mapping Experiment (CHIME) in Canada. The folks at CHIME will actually be able to observe all of our northern hemisphere pulsars ( CHIME Pulsar Collaboration, 2017 ), which is excellent. In the short-term, CHIME will slightly lessen the burden of losing Arecibo. We've also done a little bit of timing with the Very Large Array, although I don't think that'll be a major player in the future. That brings us to telescopes currently in the planning stages or under construction. The DSA-2000 ( Hallinan et al. 2019 ) will hopefully see first light in the late 2020s and would likely become a major part of our ongoing observations. The Square Kilometer Array will also be extremely important for radio astronomy as a whole - I don't know whether we in particular will be using it a lot. The SKA should also reach completion in the next decade. These are just the current potential options from my corner of the astronomical world - and they're quite tentative. Arecibo was our most-used screwdriver in the toolbox, so we've certainly been hit. We'll hopefully be able to somewhat supplement the GBT with CHIME and ideally DSA-2000 a decade from now - so in that sense, those will effectively succeed Arecibo for our observations. In reality, all that means is that we won't lose quite as much of our observing capacity - but it doesn't come close to making up for the loss. Other collaborations and astronomers will presumably increase their usage at the observatories that are best equipped for their observations, depending on frontend/backend requirements and field of view. It's not going to be fun. We'll be set back. But science will go on. We'll figure it out. It's worth updating this answer now that the white paper for the proposed Next Generation Arecibo Telescope (NGAT) has been released. The Arecibo Observatory has proposed a phased array of small dishes mounted on a circular platform ~300 meters in diameter (or potentially multiple smaller platforms) which would be angled in various directions to provide reasonable sky coverage. One variant involved 1,112 9-meter dishes, as smaller dishes provide a higher packing efficiency and therefore maximize collecting area. Regardless of the precise configuration, the sinkhole that housed the former 305-meter dish would ideally be the new array's home. The aim is for frequency coverage from 200 MHz to 30 GHz via four broadband receivers and a frequency-dependent field of view ranging from 6 degrees at 300 MHz to 3.5 arcminutes at 30 GHz, a significant field of view increase over its predecessor by a factor of 500. The dishes would also be equipped with transmitters for radar observations. Whether the NGAT will be built is a question for which nobody has any answers just yet. It would cost roughly $450 million, which could take money away from other projects unless it received special Congressional funding or support from private sources. Additionally, the setup has not been tested on such a large scale, so there are some technical and engineering concerns (the group did consider but reject single-dish and classical array designs). All that said, the design was put together in an impressively short amount of time, and even if the NGAT isn't build, the scientific considerations that went into it may inspire future designs for replacement telescopes at Arecibo. $^{\dagger}$ This isn't totally true - a scenario where we lose Arecibo had already been studied in detail. But . . . I think a lot of folks felt that it was more likely that it would be out of commission for a bit but could be salvaged in some way, and hopefully brought back online before too long. This was not the Thursday we were expecting. | {
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39,952 | As a black hole shrinks in volume and mass, shouldn't its temperature get lower? Shouldn't it evaporate more slowly? Naively, (very naively), I think that with a smaller surface area (as per its event horizon), it would have fewer and fewer random quantum foam particles popping up just outside its surface.... | There's no simple answer. In the immediate future, different radio telescopes around the world will pick up the slack in various ways; how that happens will depend on the needs of individual observers and collaborations. Unless someone was to build an identical observatory at the same latitude as Arecibo, with the same frequency range, receiver options and field of view . . . we'll have to spread out the observing. Given that we didn't really expect to lose Arecibo for more than about a year $^{\dagger}$ , there are a lot of things up in the air. Insert "probably" and "maybe" into this answer wherever you please. I do pulsar timing, so I'll talk about what a future without Arecibo might look like from our perspective. My collaboration's two main instruments were Arecibo and the 100-meter Green Bank Telescope (GBT), with roughly half of the observing time at Arecibo and half at the GBT. There aren't firm plans in place for the exact strategy we'll take going forward, but we'll have to decrease our observing cadence at the GBT (i.e. how often we observe certain pulsars) so we can use some of the time to observe pulsars we'd normally observe using Arecibo. So the GBT can take some of the load, but as we only have a finite amount of observing time, that impacts our other observations. We'll likely have to decrease the cadence of observations and the number of pulsars we monitor regularly. Pulsar timing arrays will also have options with other existing instruments in the immediate future. A key one is the Canadian Hydrogen Intensity Mapping Experiment (CHIME) in Canada. The folks at CHIME will actually be able to observe all of our northern hemisphere pulsars ( CHIME Pulsar Collaboration, 2017 ), which is excellent. In the short-term, CHIME will slightly lessen the burden of losing Arecibo. We've also done a little bit of timing with the Very Large Array, although I don't think that'll be a major player in the future. That brings us to telescopes currently in the planning stages or under construction. The DSA-2000 ( Hallinan et al. 2019 ) will hopefully see first light in the late 2020s and would likely become a major part of our ongoing observations. The Square Kilometer Array will also be extremely important for radio astronomy as a whole - I don't know whether we in particular will be using it a lot. The SKA should also reach completion in the next decade. These are just the current potential options from my corner of the astronomical world - and they're quite tentative. Arecibo was our most-used screwdriver in the toolbox, so we've certainly been hit. We'll hopefully be able to somewhat supplement the GBT with CHIME and ideally DSA-2000 a decade from now - so in that sense, those will effectively succeed Arecibo for our observations. In reality, all that means is that we won't lose quite as much of our observing capacity - but it doesn't come close to making up for the loss. Other collaborations and astronomers will presumably increase their usage at the observatories that are best equipped for their observations, depending on frontend/backend requirements and field of view. It's not going to be fun. We'll be set back. But science will go on. We'll figure it out. It's worth updating this answer now that the white paper for the proposed Next Generation Arecibo Telescope (NGAT) has been released. The Arecibo Observatory has proposed a phased array of small dishes mounted on a circular platform ~300 meters in diameter (or potentially multiple smaller platforms) which would be angled in various directions to provide reasonable sky coverage. One variant involved 1,112 9-meter dishes, as smaller dishes provide a higher packing efficiency and therefore maximize collecting area. Regardless of the precise configuration, the sinkhole that housed the former 305-meter dish would ideally be the new array's home. The aim is for frequency coverage from 200 MHz to 30 GHz via four broadband receivers and a frequency-dependent field of view ranging from 6 degrees at 300 MHz to 3.5 arcminutes at 30 GHz, a significant field of view increase over its predecessor by a factor of 500. The dishes would also be equipped with transmitters for radar observations. Whether the NGAT will be built is a question for which nobody has any answers just yet. It would cost roughly $450 million, which could take money away from other projects unless it received special Congressional funding or support from private sources. Additionally, the setup has not been tested on such a large scale, so there are some technical and engineering concerns (the group did consider but reject single-dish and classical array designs). All that said, the design was put together in an impressively short amount of time, and even if the NGAT isn't build, the scientific considerations that went into it may inspire future designs for replacement telescopes at Arecibo. $^{\dagger}$ This isn't totally true - a scenario where we lose Arecibo had already been studied in detail. But . . . I think a lot of folks felt that it was more likely that it would be out of commission for a bit but could be salvaged in some way, and hopefully brought back online before too long. This was not the Thursday we were expecting. | {
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40,047 | Since atmospheres don't end abruptly but gradually get thinner the higher you go, I wonder how we can get the total mass of an atmosphere if we don't know where exactly it ends. E.g. the Earth's atmosphere's mass is defined as 5.1480 × 10 18 kg. Does this value include the exosphere (which doesn't have an abrupt end either)? Or is it up to the exobase only? Or is it even the significant part only, up to the mesopause or to the Kármán line or something? Also, if we mention the mass of a celestial body, does this value (e.g. in case of Venus 4.867 × 10 24 kg) include its atmosphere's mass or not? | There is a simple $^*$ way to know the total mass of the atmosphere: measuring the pressure it exerts on the surface, which necessarily integrate all of the atmosphere above ground level. If you take an atmospheric pressure of $1\cdot10^5$ Pa, it is equivalent to a force of $1\cdot10^5$ newton over one square meter. Multiply by the area of the planet in square meters, you get the total weight of the atmosphere: $1\cdot10^5 \times 5.1\cdot10^{14} = 5.1\cdot10^{19}$ N. Divide by the acceleration of gravity to convert this weight to a mass: $\frac{5.1\cdot10^{19}}{9.8} = 5.2\cdot10^{18}$ kg. There you go! $^*$ Well, I guess it is simple on Earth, but could be more challenging on other planets... | {
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40,110 | We seem to have named every moon orbiting other planets. Why haven't we named our own moon? And for that matter, why doesn't our sun have a name since we name or number stars? | What is a name? A name is a word, that is reasonably unique, that is used to identify a person or thing. When a child is born there is no word that identifies it, and so its parents have to choose a "name". Similarly when a new astronomical object, such as an asteroid, is discovered, it has no word that identifies it and so it is assigned first a number and later a name. But there is already a word that identifies our sun and moon: "the Sun" and "the Moon". There is no need for a new name because they already have a name. Notice that English (unlike some other languages) gives us a typographical clue that these are names, because the first letter is capitalised. The Moon is a moon of the Earth and the Earth is a planet orbiting the Sun. Earth, Moon and Sun are the correct names in English. | {
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40,219 | In 2017 a series of stellar occultations by asteroid 2014 MU69 "Ultima Thule" now officially named 486958 Arrokoth were timed in order to obtain better orbital information before the New Horizons flyby and to look for additional large chunks of debris that might be of scientific interest and pose a danger of collision thereby warranting a more distant flyby. There was a pleasant surprise in the shape of the asteroid determined from an array of portable telescopes deployed across the occultation paths. The asteroid appeared to be binary, which was puzzling because there were no obvious oscillations in its light curve, but that's a different story. Below is an assemblage of three such occultation measurements. We can see the asteroid's double circle shape nicely. But what interests me here is the single trace for July 10th. It seems to show an occulting object clearly distinct from yet incredibly close to Arrokoth. A pure coincidence would be more than extremely unlikely as the time window is seconds and the distance less than 100 km at ~45 AU! Question: But I don't remember reading about a separate secondary body orbiting Arrokoth. So what caused this apparent occultation on July 10, 2017? From this answer to Will the upcoming observations of occultation by Arrokoth (2014 MU69) be of a single object, or two? | There were three attempts to measure Arrokoth by occultation, and the June 3rd attempt didn't detect anything. The July 10th attempt had a tiny blip, that appeared to be in the "wrong place", well away from the location that astrometry had predicted. The July 17th occultation was successful, it determined the shape and location well. Some thought that the July 10th "blip" might have been evidence of a moon . It was not, it turned out that there was an error in the software that was used to produce the astronometry (the author of the software describes one as "a case of me failing to read my own documentation, another one a sign error") If those bugs are fixed, then the blip on July 10th is moved to the same place as the blips on July 17th. That blip is Arrokoth. | {
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40,362 | Keep in mind I'm just a newbie. So.... I have a new Newtonian telescope. It has 150mm opening and 1400mm Focal Distance on top of an equatorial mount. I have 25 and 10 mm eyepieces... and a 2x Barlow. I sat down to do observation yesterday and was able to see Saturn and Jupiter (warming up for the alignment on the 21th). Now, I did notice that as they sank closer to the horizon, the effect of the atmosphere was more apparent (and they were less sharp) so I'd like to see if it is possible to see them earlier when they are higher in the sky to try to minimize the effect of the atmosphere... but that might mean to watch them when the sun hasn't set yet. I know that we can't see them with the naked eye when it is daytime but given that the telescope is letting a lot more light in, then it should be possible to see them if I point in the right direction? I intend to use the sun to get RA/DEC alignment (using the telescope shadow, don't intent to look at the sun with the telescope.... yet! :-D) so that then I can point the telescope in the direction of the planets (and I am using Stellarium on my Debian box to get the numbers and stuff). | Jupiter can be seen during the day. This image is by Philip Crude. Philip is an experienced astro-photographer. On his webpage http://www.billionplanetsquest.com/p/planets.html he gives details of the equipment and settings used: This image was captured through my [i.e., Philip Crude's] Celestron CGEM-800 using a ZWO ASI120MC camera at prime focus. Resolution 640 x 480, 35fps best of 900, Brightness @ 1, Gamma @ 48, Gain @ 11. The image was acquistioned with SharpCap and processed with RegiStax and CS6. Philip also tweeted the image here . The main difficulty is pointing the telescope in the right direction, as you won't be able to align it against the background of stars, so you need to predict the position and point the telescope accordingly. | {
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40,433 | Mercury 's orbital period around the Sun is about 88 days. Comets and other things have gotten closer to the Sun than Mercury does. But has there ever been an asteroid or some other body discovered that has a shorter orbital period? Are there theoretical constraints on if such a body can exist? The IAU defines a planet as one that has swept out its orbit . Does the neighborhood of Mercury extend to the Sun? | The recently discovered asteroids 2019 LF6 and 2020 AV2 , each taking 151 days to orbit the Sun, have the shortest periods currently listed in the JPL Small Body Database . Vulcanoids are difficult to detect from Earth; none are known yet.
To remain in such an orbit, Evans and Tabachnik 1999 estimate a minimum diameter of 100 m and a semimajor axis between 0.09 and 0.21 au. There are various competing definitions for a planet's neighborhood. Soter 2006 says: Two bodies share an "orbital zone" if their orbits cross a common radial distance from the primary and their periods are nonresonant and differ by less than an order of magnitude. A Sun-grazing asteroid with perihelion 0.01 au and aphelion 0.31 au (Mercury's perihelion) would orbit in 23 days, well above the 9 day minimum for that definition. | {
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40,445 | There are two clusters of stars that I always thought were the Big Dipper and Little Dipper. But after looking at images of the Little Dipper and Big Dipper online, I am not too sure if that's what they are. Also, I found out that the front edge of the Big Dipper is supposed to be aligned with Polaris (or the North Star), which is part of the Little Dipper, but that's not the case with these clusters of stars that I see. So last night I took this photo: I circled the clusters of stars that I thought were the Little Dipper and Big Dipper. I would like to know if either of these clusters are indeed the Little Dipper or Big Dipper. If either of them are not the Little Dipper or Big Dipper, what are they? And where should I look in relation to these two clusters to find the Little Dipper and Big Dipper? | As mentioned, these are the Pleiades, and the belt of Orion. These are visible in the South at this time of year. The Big and Little Dippers are in the North, so turn around. The best way to find them is a map: I've marked the approximate edge of your photo, with Orion, the V of the Hyades and the small cluster called the Pleiades. The big dipper is much bigger. It is usually quite easy to find, in the North East, In this map it is labeled "Ursa Major" which means "great bear". The little dipper is a rather faint constellation. Even its brightest star, Polaris, is only second magnitude. You can find it by tracing from the pointers in Ursa Major (I've coloured them green). It is the only moderately bright star in that region of the sky. Once you have found the Pole star, it is possible to make out the faint stars that make up the rest of the Ursa Minor, the little dipper. | {
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40,703 | We know that the universe has more dark and anti matter as compared to normal matter. Can there be dark matter galaxies or antimatter galaxies? | Dark matter galaxies are possible but very speculative. On a theoretical level, they are hard to form because dark matter interacts only gravitationally (see Anders Sandberg's answer), which makes it hard to lose energy and become bound structures. On an observational level, they would be hard to detect. Gravitational lensing can do something, but since one cannot actually see the galaxy, it's also hard to say where the dark galaxy is -- if there is one at all. Still, people have studied the idea , so it's not impossible. Antimatter galaxies : At some level the idea that there are antimatter galaxies out here is appealing. First it can solve the baryon asymmetry problem at a stroke. It's also the case that an antimatter star would shine. From long distance, it would also be practically indistinguishable from a "normal" star. However , there are strong reasons to believe that there are no antimatter galaxies. That's because antimatter annihilates with normal matter, which leaves experimental signatures. If any part of the Earth were made of antimatter, it would immediately vanish in a flash, so we can be sure that the Earth is mostly matter. Similarly, if the Sun were made of antimatter, we would be quickly annihilated (thanks to the antimatter solar wind radiating from the anti-Sun), so we can be sure the Sun is also mostly matter. Similar arguments allow us to conclude that the Milky Way is almost entirely matter, the Local Group is almost entirely matter, etc, all the way up to the largest structures in the sky . If antimatter galaxies exist, they are probably outside our observable universe, at which point some will argue it's no longer science. | {
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40,801 | Is it possible for planetary rings to be perpendicular (or near perpendicular) to the planet's orbit around the host star? | Yes, the plane of the rings of Uranus are at about 98 degrees to the plane of its orbit around the Sun. This means that the ring system looks as in your picture twice per orbit. As the planet orbits the Sun, the rings, although still inclined at 98 degrees to the orbital plane gradually become "face-on" when viewed from the Sun. This will happen about quarter of an orbital period after the configuration illustrated in the picture. Then another quarter of an orbital period later, Uranus will be on the other side of the Sun, but with its rings tilted as shown. What cannot happen is that the rings are oriented as shown throughout the entirety of a planet's orbit. Conservation of angular momentum demands that the plane of the rings (or the axis of rotation of the ring material) does not vary, unless some external torque were brought to bear in order to change it. Therefore after a quarter of an orbit, the rings in your picture would be face-on to the star. | {
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40,837 | Gas giants like Jupiter and Saturn have bands of different colors in their atmosphere. These are due to the rotation of the planets. Stars rotate too, so why do most stars have patches/blotches of color rather than having latitudinal stripes? | Just rotation is the wrong tree to bark up on. You see color variations on gas giants due to differences in composition, i.e. ammonia vs. sulfuric acid clouds on Jupiter, which are transported differently on the rotating planet in the up/downwelling bands. Spots on stars originate due to very different physics. At the temperatures which are prevalent on stellar surfaces, molecules are mostly dissociated and ionized, we call this state a plasma, so no more colour effects due to ammonia or others. Star spots are concentrations of the local magnetic field, which is coupled to the plasma dynamics. The magnetic field pushes the gas aside and lets the surface cool locally. This makes dark spots that you see on stars like our sun. There is a transition region in terms of surface temperature of 2000-3000K, at the masses of brown dwarves. Those failed stars seem to exhibit dark bands on their surfaces, which are thought to be due to absorption from exotic, high-temperature molecules like TiO and VO (Titanium and Vanadium oxide). | {
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40,871 | Most stars orbit in the Milky Way's galactic disc. But is it possible for one to orbit perpendicular to it? Here on Earth since we're inside the galactic plane we can't get a good view of what the Milky Way looks like. But would the whole Milky Way be visible from a planet orbiting such a star? | The Sun and most of the other stars are in the bulging disk of the Milky Way galaxy, but about 1% of the galaxy's stellar mass is in the galactic halo. The halo also includes 50 globular clusters and about 20 satellite galaxies according to Helmi 2008 : The stellar halo of the Galaxy . Here is a nice graphic: Note that the Sagittarius Stream of stars is extremely close to passing the galactic poles. If you were on a planet next to one of these stars as it passed a pole, you would probably have a glorious view of the Milky Way stellar disk. Bodies in the galactic halo don't necessarily follow the elliptical paths predicted by Kepler, so their orbits may not be consistently perpendicular to the galactic plane. Some stellar streams have extremely odd orbital paths like the Phlegethon stellar stream, referenced in this answer : There is some speculation that the Halo stellar streams are the remnants of dwarf galaxies the Milky Way has absorbed. Indeed, Liang et al. say that: the Galactic halo has complicated assembly history and it not only
interacts but also [is] strongly mixed with other components of the Galaxy
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40,930 | When Neptune and Pluto are closest, about 100 million mi (160 million km) from each other, would an observer on Neptune (or rather on one of its moons, since Neptune is gaseous) be able to see Pluto, and maybe even Charon, with the naked eye? If not, could Pluto be seen in average binoculars? I think Pluto would appear a bit smaller than Mercury from Earth (but at much lower apparent brightness because of the distance from the Sun). | No, it cannot. Far from it. The closest approach between both planets is roughly 16 AU due to the 3:2 orbit resonance. Pluto will even then be a tiny dot among many with a brightness around 14 mag. You can try that with Stellarium yourself, placing the observer on Neptune and looking for Pluto. You just have to find the right time. One such time is approx. in the year 2877. | {
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40,993 | If stars are primarily made of hydrogen, which is then burned to helium, and then on and on and on down the chain until you either reach iron, or in extreme cases much heavier metals, this then implies that as time goes on, more and more heavy elements will disperse into the universe. Now, the universe is huge, and there's still a lot of hydrogen left to make stars out of.
That being said, in denser star-forming regions, much of this gas is available for recycling as the remnants of old stars get turned into new ones. Doesn't this all this imply that as time goes on, more and more heavy metals will be included within young stars from the start? It may be a very small percentile, sure, but it I assume it would increase slowly as the generations go by. Furthermore, I assume many rocks get swallowed up by a star during the formation of the planetary disc, which likely adds to the impurity. Does any of this have any adverse effect on the lifespan of the sun? Or are we talking quantities so small in objects so huge in time-spans so large, that even a dip of a hundred years is completely meaningless? | The chemical enrichment of the Universe over time is indeed a thing. The plot below ( source ) shows observational measurements of the cosmic density of ionised Carbon in the Universe against redshift (higher redshift -> further back in time). The abundance of other heavy elements over time shows a similar trend. Stars are explicitly classified based on their metallicity, and its direct relation to which generation of stars they are part of. This is known as the Baade chemical classification . The first stars to form from primordial gas are confusingly known as Population III. These stars have yet to be observed, but it's hoped that the James Webb Space Telescope may glimpse them . They are theorised to be very massive and short lived, hence the difficulty in detecting them. The next generation, formed from the enriched material from this first generation of stars are known as Population II. And the subsequent generation, of which our own Sun is a member, are Population I stars. Population I stars have the highest metallicity, due to the fact they formed from material that has already participated in Population III and II star formation. The metallicity of a star is almost entirely determined by the metallicity of its birth cloud. Stars do swallow up small amounts of material over their lifetimes, but this does not significantly affect the overall metallicity of a star. A stars lifetime is mostly set by it's mass, and whilst metallicity does have a small effect, it doesn't change significantly due to outside effects over a stars lifetime. | {
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41,127 | My understanding is that the moon was created a long time ago when Earth was hit by a big asteroid. The debris then agglomerated into the Moon, which happens to be orbiting at the exact speed required to neither crash back into the Earth, nor escape into space. Having the exact correct speed seems extremely unlikely. Yet, our moon is there, and many other planets have moons. Are these just the few survivors out of thousands of events that didnt have the « goldilock » speed? 2022 Edit: I got my "ah HA!" moment where everything makes sense after playing 10 minutes of the tutorial of the "Kerbal Space Program" game. Highly recommended. | There isn't a "Goldilocks speed" for orbit. If you put two objects in space, and give them a velocity relative to each other, then provided that velocity is less than the escape velocity (at their relative distance) the two objects will orbit each other. Those orbits will be elliptical, and it is possible that the ellipse is skinny and "eccentric" enough for the two bodies to collide when they are closest to each other. But for an object that is several hundred thousand km from Earth, there is a quite a wide range of possible elliptical orbits. So when (and if) the grand collision happened, there was a huge amount of matter that was ejected up into space. Some probably was moving so fast that it escaped, Some certainly went into orbits that didn't have enough energy and so were small skinny ellipses and the matter fell back to Earth. But there was a lot that ended up in some kind of elliptical orbit. This matter was not all in the same orbit, but it started to coalesce, and form into a single ball, under its own gravity. Other moons weren't formed like this, they either formed at the same time as their planets as a "mini solar system" (such as the four major moons of Jupiter) or they were captured from the asteroid or Kuiper belts). Initially, the captured moons may have had rather elliptical orbits. But most moons are in rather circular orbits. Even if the moon was originally in an elliptical orbit, tidal effects will tend to make the orbit more circular. A planet and moon system has a certain amount of angular momentum and a certain amount of energy. The angular momentum can't change, but energy can be converted into heat and since tides dissipate some energy as heat, the orbit will tend to change to a shape that minimizes energy, for a given amount of angular momentum. That shape is a circle. (See Is the moon's orbit circularizing? Why does tidal heating circularize orbits? ) So the effect of tides is to give moons the "Goldilocks speed" that keeps them in a circular orbit. | {
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41,131 | I came up with this confusing theory a few years ago, but did not ask it on this site. If the Big Rip and the Multiverse theories are true, then when the Big Rip occurs, the universe's size will expand to infinity. However, the size of the Multiverse is also infinite (supposedly), and there are an infinite number of universes (also supposedly), then there is a paradox of infinities. I am still rather confused about this paradox, and is there a work-around to it, or is this completely invalid? | There isn't a "Goldilocks speed" for orbit. If you put two objects in space, and give them a velocity relative to each other, then provided that velocity is less than the escape velocity (at their relative distance) the two objects will orbit each other. Those orbits will be elliptical, and it is possible that the ellipse is skinny and "eccentric" enough for the two bodies to collide when they are closest to each other. But for an object that is several hundred thousand km from Earth, there is a quite a wide range of possible elliptical orbits. So when (and if) the grand collision happened, there was a huge amount of matter that was ejected up into space. Some probably was moving so fast that it escaped, Some certainly went into orbits that didn't have enough energy and so were small skinny ellipses and the matter fell back to Earth. But there was a lot that ended up in some kind of elliptical orbit. This matter was not all in the same orbit, but it started to coalesce, and form into a single ball, under its own gravity. Other moons weren't formed like this, they either formed at the same time as their planets as a "mini solar system" (such as the four major moons of Jupiter) or they were captured from the asteroid or Kuiper belts). Initially, the captured moons may have had rather elliptical orbits. But most moons are in rather circular orbits. Even if the moon was originally in an elliptical orbit, tidal effects will tend to make the orbit more circular. A planet and moon system has a certain amount of angular momentum and a certain amount of energy. The angular momentum can't change, but energy can be converted into heat and since tides dissipate some energy as heat, the orbit will tend to change to a shape that minimizes energy, for a given amount of angular momentum. That shape is a circle. (See Is the moon's orbit circularizing? Why does tidal heating circularize orbits? ) So the effect of tides is to give moons the "Goldilocks speed" that keeps them in a circular orbit. | {
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41,264 | Somebody suggested that the side of the moon that always faces the earth is as dark as the far side of the moon, but is that really the case? Doesn’t earth-shine make the earth-facing side of the moon overall brighter? Or does the absence of a moon atmosphere negate any possible gain from earth-shine? | Somebody suggested that the side of the moon that always faces the earth is as dark as the far side of the moon, but is that really the case? That is not the case. I'll look at two different wavelengths: visible and radio. Visible Doing astronomy from the surface of the Earth when the Moon is full is much more difficult than is doing astronomy from the surface of the Earth when the Moon is new. While much of this difficulty arises due to the Earth's atmosphere, some is the direct result of the presence of a largish, well-lit object that has same angular size as does the Sun. The full Earth as seen from the Moon is, on average, over 40 times brighter than is the full Moon as seen from the Earth. Reading a newspaper on the Moon at night under a full Earth would be a piece of cake compared to reading a newspaper on the Earth at night under a full Moon. Radio (and microwave) These are the wavelengths where an observatory on the far side of the Moon would truly shine. The Earth radiates significant amounts of electromagnetic radiation in the radio and microwave wavelengths. There are a few places on the Earth where radio towers and microwave towers are absolutely forbidden, such as the area around the Murchison Radio-Astronomy Observatory in Australia, the area around the Itapeting Radio Observatory in Brazil, and the area around the National Radio Astronomy Observatory (Green Banks, West Virginia) in the US. These Earth-bound radio quiet zones are still plagued with radio and microwave radiation reflected by the atmosphere and by radiation from satellites orbiting the Earth. Radio telescopes on the far side of the Moon would be shielded from the large amounts of electromagnetic radiation in those wavelengths emitted by the Earth, and possibly even from the electromagnetic radiation in those wavelengths emitted by satellites orbiting the Earth. Moreover, radio telescopes on the far side of the Moon would not be hindered by the Earth's atmosphere. The transparency of the Earth's atmosphere at radio wavelengths ends at the very long wavelengths that would be useful in studying the early universe, and it ends at the upper microwave that would be useful in studying molecular gas clouds. | {
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41,345 | I was looking at a map of our local stellar neighborhood, and it occured to me, the stars are really close, if one compares them to the size of some nebulae. So can it be, that the Sun, Alpha Centauri system, and all the other stars in the neighborhood actually formed from the same nebula? Have we discovered any information about this? | There are three main reasons why we can tell that local stars did not, for the most part, form from the same molecular cloud that the Sun formed from. The first is that unless stars are born in a very tightly bound system such as a globular cluster (which the Sun is definitely not in), they will drift apart from their birth companions over time in slightly different orbits in the Milky Way galaxy. The Sun has gone around the center of the galaxy about twenty times since it was born; this is more than enough time for its initial cluster to become completely scrambled in azimuth (the angle in the plane of the galaxy). To see how this would work, imagine two stars with very similar orbits, one with a period of 200 million years and the other with a period of 210 million years. If they start off right next to each other, then after 2 billion years, the first star has made 10 complete orbits, while the second has made about 9.5 -- meaning it will now be on the other side of the galaxy from the first star. You can also take the current velocities of nearby stars, and see that they can be rather different, which tells you they are not on very similar orbits. Alpha Centauri is moving about 20 km/s in our direction, while Barnard's star has a velocity in our direction of over 100 km/s (in both cases, they also have movement in perpendicular directions; they're not moving directly toward us). So these are stars which happen to be near each other now , but are on different orbits and therefore were almost certainly not near each other at any given time in the past. The second reason is the one that David Hammen pointed out: stars born from the same clouds should have very similar chemical compositions ("metallicities" in astronomical parlance). The third reason is that stars born from the same cloud should have the same age (to within a few million years, anyway), since molecular clouds tend to be disrupted by the star-formation process and don't keep forming stars for hundreds of millions or billions of years. Looking at some of the very nearest stars: the Sun has what we define as "solar metallicity" and is about 4.6 billion years old. The stars of Alpha Centauri are slightly more metal-rich than the Sun and probably several hundred million years older. Barnard's Star is more metal-poor than the Sun (maybe only 1/3 the iron fraction), and about 5 billion years older. Sirius is significantly more metal-rich than the Sun (its iron fraction is about three times that of the Sun) and is only about 250 million years old. So you can see that the Sun and the nearby stars are a heterogeneous group, and clearly not all born at the same time from the same patch of interstellar gas. | {
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41,351 | I found this infographic that seems to say that oxygen is the most abundant element on the surface of the Moon. Is this really the case? If so, under what form is this oxygen? | Yes, that's correct ; it's also true for the Earth's crust . The reason is that "rocks" are typically made up of components containing combinations of silicon or one or more metals (e.g., magnesium, aluminum, iron) and oxygen , such as silica ( $\mathrm{SiO}_{2}$ ); alumina ( $\mathrm{Al}_{2}\mathrm{O}_{3}$ ); lime ( $\mathrm{CaO}$ ); iron oxide ( $\mathrm{FeO}$ ); and magnesium oxide ( $\mathrm{MgO}$ ). Examples of common lunar minerals formed from these components includes plagioclase feldspars (mixtures of NaAlSi $_{3}$ O $_{8}$ and CaAl $_{2}$ Si $_{2}$ O $_{8}$ ), pyroxene (typically XYSi $_{2}$ O $_{6}$ , where X and Y are metals such as calcium, sodium, iron, magnesium, and aluminum), and olivine (made up of Mg $_{2}$ SiO $_{4}$ and Fe $_{2}$ SiO $_{4}$ ), along with oxide minerals like ilmenite (FeTiO $_{3}$ ). ( Source ) Since in all these cases you have between one and two oxygen atoms for every non-oxygen atom, you end up with oxygen as the most abundant single element. | {
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41,377 | To my understanding, the Earth's precession causes an approximately 20-minute difference between the sidereal year and the tropical year. Also to my understanding we use the tropical year for our calendar which means we ignore those 20-minutes. These 20-minutes will cause the calendar to "shift" one day every 72 years so if we don't do anything about it in a few centuries winter and summer will switch places. So every century they add an extra day to the calendar to counter this. This I understand. What I don't understand is that this 20-minute difference means that the Sun's position in the sky will vary by 20 minutes every year. Meaning if the Sun rises at the right ascension of 0 this year, next year it should rise at the right ascension of 20 minutes, and so on. But each year the Sun will actually rise at the same spot in a particular day. How is this possible? Where do these 20 minutes go? | Yes, that's correct ; it's also true for the Earth's crust . The reason is that "rocks" are typically made up of components containing combinations of silicon or one or more metals (e.g., magnesium, aluminum, iron) and oxygen , such as silica ( $\mathrm{SiO}_{2}$ ); alumina ( $\mathrm{Al}_{2}\mathrm{O}_{3}$ ); lime ( $\mathrm{CaO}$ ); iron oxide ( $\mathrm{FeO}$ ); and magnesium oxide ( $\mathrm{MgO}$ ). Examples of common lunar minerals formed from these components includes plagioclase feldspars (mixtures of NaAlSi $_{3}$ O $_{8}$ and CaAl $_{2}$ Si $_{2}$ O $_{8}$ ), pyroxene (typically XYSi $_{2}$ O $_{6}$ , where X and Y are metals such as calcium, sodium, iron, magnesium, and aluminum), and olivine (made up of Mg $_{2}$ SiO $_{4}$ and Fe $_{2}$ SiO $_{4}$ ), along with oxide minerals like ilmenite (FeTiO $_{3}$ ). ( Source ) Since in all these cases you have between one and two oxygen atoms for every non-oxygen atom, you end up with oxygen as the most abundant single element. | {
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41,601 | If the sun were to suddenly disappear, the planets would continue to travel tangentially to their former orbits. (This I know from this answer to a somewhat related question here.) In such a scenario, is it possible (however unlikely) that a new orbital system would form from two or more of the now-rogue planets? For example, is it physically and mathematically possible that Earth might end up getting captured by Jupiter and thereby become one of its moons? Or do the size and speed of the planets completely rule out such a scenario, even if the sun disappeared at such a time that the planets' new trajectories would cross? | The issue here is whether pairs of planets can become gravitationally bound to each other. In the two-body problem the trajectories or orbits are ellipses (bound orbits), parabolas and hyperbolas (unbound). For all practical purposes, an encounter looks like they start out at infinity with some finite speed, approach each other, and then maybe fly away or remain bound. This situation (finite speed when far away) corresponds to a hyperbola, so the generic case will not form a new bound system: some of that energy needs to be shed somehow. One way this can happen is an inelastic collision, i.e. a merger. A less extreme case is a close encounter where tidal forces dissipate some of the kinetic energy. But the kinetic energy of planets is large, so a single encounter is unlikely to dissipate enough to make them bound. Another important possibility is a three-body encounter. This is how comets get into short-periodic orbits when falling into the solar system: an encounter with Jupiter (or some other planet) leads to a gravitational "assist" that makes it bound with the sun. In your scenario where the rogue planets fly off along their tangents it is very unlikely that any start out bound to each other since planetary orbital velocities around the sun are much bigger than the orbital velocities of a bound planet-planet pair at a large distance. Hence they will just fly apart. In theory they might have an encounter before leaving the solar system vicinity (e.g. Mercury by sheer chance sweeping past Jupiter) and that leaves a tiny chance of a three-body interaction with moons or a merger, but the chances are truly tiny. The generic case is just scattering. Addendum: Some estimates For a body to be bound to another body of mass $M$ it need to move slower relative to it than escape velocity $v_{esc}=\sqrt{2GM/r}$ . If we assume planets move in circles we can then take their orbital velocities for all possible positions along the orbit, calculate the relative velocity, and compare it to the escape velocity. The result is this diagram, where I have plotted the separation on the x-axis and how many times escape velocity the relative velocity is on the y-axis. The numbers denote which pair of planets I am comparing. The blue horizontal line is $\Delta v=v_{esc}$ . Looking just at the lowest relative velocities shows that the closest to being potentially bound is Saturn to Jupiter were the sun vanish just as they are moving in parallel, but even in this case Saturn has several times Jupiter escape velocity. One could complicate it with inclinations and elliptic orbits, but it will not change the qualitative picture: the planets are not even close to being bound to each other in the absence of the sun. | {
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41,722 | The super massive black hole at the centre of Milky Way has a mass of about 4 million times that of our sun. Is this enough to keep the entire galaxy together, is it this black hole which keeps the galaxy together? I did read that there are also galaxies that don't have a super massive blackhole, what keeps those galaxies together? I did ask this question at space stackexchange but it got closed because it was posted at wrong place. | The galaxy is kept together by the combined mass of the matter in the galaxy, of which the supermassive black hole is a negligible part. There are galaxies that don't have a central black hole (such as the Triangulum galaxy), and they are also held together by their combined mass. In particular, the dark matter of the galaxy is what provides most of the mass that holds the visible matter together. Galaxies usually have far more dark matter than visible matter. This mass is distributed in a roughly spherical shape, with the greatest density towards the centre of the galaxy, and dropping off further from the centre. This "dark matter halo" is substantially larger than the visible disc. | {
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41,936 | Could we carve a large radio dish in the Antarctic ice that could replace the Arecibo Observatory? Could this work, and what would be some potential limitations? | Great question! There are many open and active antarctic permanent bases and several antarctic astronomical observatories . There are even major large projects that involve substantial drilling and removal of ice such as the IceCube Neutrino Observatory . The biggest construction project is probably the Amundsen–Scott South Pole Station . This is certainly a valid idea to consider and should not be dismissed out of hand by "Antarctica is hard". There's no bedrock on which to affix your dish; it will be floating in creeping ice and subject to drifting snow. From Amundsen–Scott South Pole Station : The new station included a modular design, to accommodate rises in population, and an adjustable elevation to prevent it from being buried in snow. Since roughly 20 centimetres (8 in) of snow accumulates every year without ever thawing the building's designers included rounded corners and edges around the structure to help reduce snow drifts. Basically I think @ PeterErwin's comment summarizes the problem nicely. For more details see their link and also Challenges That Face South Pole Architecture which mentions the following: WIND Constant winds result in snow pile-up on buildings. The new station faces into the wind, and is airfoil-shaped. The airfoil forces air into a compressed space where it accelerates. The fast wind scours out built-up snow. Years later, if snow still builds up, the building can be lifted two more
stories on its columns. ICE FLOW The station sits atop a 2-mile deep layer of ice. Each year, the geographic South Pole is marked. Ice (cold water), slowly drips down to the ocean with gravity. The trail of yearly South Pole markers shows that the ice moves 33 feet per year. ICE CREEP The weight of the building also causes the ice to move locally. Ice compresses and shifts away from sources of pressure. Resulting variable rates of sinking make keeping the building level a challenge. Architectural elements built into the design will help meet that challenge Basically everything is built up on stilts above the surface. This lets wind blow under things and prevents snow drifting which would quickly burry a surface structure. The South Pole Telescope consisting of a 10 meter point-able dish is also built on an elevated platform sitting on stilts to prevent drifting (see below). You could propose heating the entire dish to de-ice and de-snow it regularly, and pumping out the melt water, but power is at a premium at the South Pole. Fuel is flown in on airplanes during a short summer season and burned in electrical generators for power. Building something that requires electrical power to constantly melt drifting snow has serious practical hurdles. Radio telescope dishes built into fixed surface depressions At least two very large, single dish telescopes have been built into naturally occurring depressions in local rock: Arecibo at 18.3° N and FAST at 25.7° N latitudes. Both telescope designs have large primary focus structures (small buildings where people can work) suspended high above the ground at the primary focus of the dish, and in order to view the celestial equator and to track objects as the Earth rotates, moveable cables suspended from very tall towers were designed allow these structures to move around by about in two directions by roughly +/-19° and +/-26°. For more on that see this answer to Is there an advantage to the equatorial region of the far side of the moon for a radio telescope or would any crater on the far side work? Arecibo succumbed to deterioration of its cables caused by environmental factors and fell down. Is there any role today that would justify building a large single dish radio telescope to replace Arecibo? What will succeed the Arecibo Observatory? How could the loss of the Arecibo Observatory have been prevented? Building a floating, ocean-going giant radio telescope? Maintaining these kinds of necessary tall structures in the high winds and snow storms at the South Pole would be quite a challenge as well. Sources: nsf.gov news and South Pole Station Destination Alpha Source: nsf.gov news Sources left and right Source Sources for both: Antarctic Bases and Buildings - 3 Building at The South Pole via @PeterErwin's comment. left: "The South Pole radio telescope, the very clear and dry air at the pole make it the best place on earth to have a space telescope.
Photo Christopher Michel from San Francisco Creative Commons Attribution 2 Generic license" right: "In the 6 months of night time of the South Polar winter with the Aurora Australis overhead, July 2008 Photo: Patrick Cullis - National Science Foundation" | {
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42,048 | I just learned that the Fraunhofer lines of the Sun's spectrum indicate that the Sun contains various elements other than just hydrogen and helium (for example, Na and Fe) but don't the Sun's p-p chain reactions only involve hydrogen and helium? So where do these other elements came from? | The Sun is currently turning hydrogen into helium. There are no other nuclear reactions taking place at any significant rate in the Sun. The Sun will not start to make heavier elements until it reaches the tip of the red giant branch in about 7 billion years time. The elements heavier than helium that are present in the Sun were almost all made inside other stars. These stars lived and died before the Sun was born 4.5 billion years ago. The nucleosynthesis products from these previous stars were recycled in stellar winds, supernovae, kilonovae and novae, and then mixed into the turbulent interstellar medium. | {
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42,076 | Correct me if I am wrong, but if we count sunsets by the center of the Sun apparently crossing the horizon then the Sun is supposed to set every day at latitudes under the arctic circle. (Yes if you count in the disc then adjust by 0.27 degrees.) I was playing around with PyEphem (a python library that claims an approximately 1 arcsecond accuracy), and found that according to it the sun stays up for a couple of days even under it ( $66.2<90-23.4$ ). Can someone explain what is going on? I created this simple plot to illustrate the issue. The left panel shows the whole period while the right panel shows that the problem is not that the data is under sampled. The plot goes from 2021-06-13 00:12:58.085383+00:00 to 2021-06-28 23:54:05.880661+00:00 . And one more example: Kuusamo is a city at 65°58′N 29°11′E and apparently it also has a few days when the sun doesn't set: LINK1 LINK2 | From the PyEphem Quick Reference Guide : Rising and setting are sensitive to atmospheric refraction at the
horizon, and therefore to the observer’s temp and pressure; set the
pressure to zero to turn off refraction. It seems likely that, if you're using the default settings, the result returned is including atmospheric refraction, giving the results you would expect to see at the location, where the sun is visible even though a direct line to the disc of the sun would put it below the horizon. Image is from " What is Refraction of Light " by By Konstantin Bikos and Aparna Kher at timeanddate.com | {
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43,299 | I'm doing some research on black holes for a science video contest . I want to explain the physics of how they work, but also want to have a little background on how they're formed. As far as I've searched, black holes are predicted by general relativity (GR). But I saw this site that said 'general relativity is inaccurate at very small sizes' and that it kind deviates from GR. So I want to confirm whether black holes really are a prediction of GR or an inconsistency with GR. Can someone please help? Thanks. | Well, yes, but we must be careful with the meaning of "predict". The Schwarzschild solution, developed by Karl Schwarzschild in 1916 [1], is the first closed-form, explicit solution of Einstein's field equations for gravitation. It describes a spherically symmetric, static, vacuum spacetime. The solution goes singular at a specific radius (the Schwarzschild radius). In the weak field limit, it correctly replicates the Newtonian gravitational field of a compact object. Though Birkhoff's theorem (general relativity's version of the shell theorem) was not yet known in 1916, the Schwarzschild solution was nonetheless recognized as the general relativistic description of the gravitational field outside a compact gravitating body. The fact that it went singular at the Schwarzschild radius was either ignored or taken to imply that objects that are as small as, or smaller than, this radius cannot exist. In any case, just because a solution exists in general relativity does not mean that objects described by that solution exist in Nature. (E.g., the field equations admit solutions that blatantly violate causality; yet I don't see any time machines out there.) For all they knew, the Schwarzschild solution was nothing more than a mathematical curiosity, an idealized case that does not describe reality. For this reason, I'd suggest that a much more groundbreaking paper is that of Oppenheimer and Snyder from 1939 [2]. This paper demonstrates that a spherical cloud of dust initially at rest will undergo "continued gravitational contraction". An observer that is falling with the collapsing matter would see total collapse in a finite amount of time, but to a distant observer, the collapse will continue forever, the object asymptotically approaching, but never quite reaching, its Schwarzschild radius. And it was in 1957 I believe that Regge and Wheeler first demonstrated that a Schwarzschild singularity is stable under small perturbations [3], i.e., perfect symmetry is not required. (Wheeler, of course, was also the first to popularize the name, "black hole".) Lastly, as I was reminded in a comment, we should not forget Penrose's 1965 singularity theorem [4], which introduces the concept of a trapped surface and shows that gravitational collapse is indeed a very generic feature of general relativity. After all, this is the result that earned Penrose the 2020 Nobel prize in physics. In light of that, I think we can confidently state that general relativity predicts black holes, but only because we know not only that black hole solutions exist, but also that physically realizable configurations of matter can collapse into black holes (or, at the very least, to objects observationally indistinguishable from black holes, which of course may evaporate in stupendously long but finite timeframes due to Hawking radiation [5]) and that the solutions are stable under small perturbations, i.e., perfect symmetry is not a prerequisite. [1] https://arxiv.org/abs/physics/9905030 [2] https://journals.aps.org/pr/abstract/10.1103/PhysRev.56.455 [3] https://journals.aps.org/pr/abstract/10.1103/PhysRev.108.1063 [4] https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.14.57 [5] https://www.nature.com/articles/248030a0 | {
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43,381 | Around 15 million years after the Big Bang, the ambient temperatures was about $24^\circ {\rm C}$ , which is in a range where water could be liquid. Could liquid blobs of water be existent then? PS: I am not talking about water on the surface of any solid planet. | Let's interpret your question to be about whether the conditions would permit blobs of water to remain liquid, whether or not water existed yet. And the answer is No, because the pressure was by then far too low. Basically, space was already a vacuum, just not as hard a vacuum as intergalactic space is now. It is appealing to imagine an era when the universe was simultaneously dense enough and cool enough for liquid water (and thus perhaps humans) to exist. But alas it is not so. At the time of the creation of the cosmic microwave background, around 370 thousand years after the Big Bang, the temperature was around 3,000K, but the pressure was around $10^{-17}$ atmospheres (see the Wikipedia article Chronology of the Universe , and search for "Recombination"). | {
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43,425 | As @ProfRob stated in his excellent answer regarding the ejection of the Solar System's fifth gas giant , It is for similar reasons that, even though the Sun was probably born in a cluster of $\sim 10^4$ stars, none of those siblings have been firmly identified. So why is it so difficult to find such stellar siblings? NB: Maybe it's because open clusters tend to disassociate quickly and the stars would be scattered everywhere. But wouldn't stars with similar spectra, age, metallicity, etc. be found easily? | Here are the problems/issues: Most stars are born in clusters/associations but a cursory investigation of cluster demographics with age reveals that the vast majority of clusters do not survive to old age. The majority either are never gravitationally bound to begin with or become unbound in the first 10 Myr. The Sun was likely born in a cluster of $10^3-10^4$ stars ( Adams 2010 ). If it dispersed more than 4 billion years ago, then its members would have had time to spread all around the galaxy by now. That is because although the members would be kinematically coherent to begin with, the velocity distribution of stars in the Galactic disc becomes "heated" as stars scatter in the potential of giant molecular clouds and pass through spiral arms ( Wielen 1977 ). They attain a significant velocity dispersion which means they could now occupy positions over a broad swathe of the Galactic disc and be displaced above/below the disc by $\sim 200$ pc. If we argue that the Sun's siblings could be in an annulus, with width $\pm 1$ kpc and thickness 400 pc, then if the solar cluster was $10^4$ stars, then the nearest sibling is expected to be at a distance of 160 pc. Another way of looking at this: The density of stars in the Galactic disc is around 0.1 per cubic parsec compared to the expected density of solar siblings of $2.4\times 10^{-7}$ per cubic parsec.
Thus only about 1 in 400,000 of the local stars might be a solar sibling. How could they be found?
Cluster members would have a similar age and a similar chemical composition. We know the age of the Sun precisely and accurately. We don't have that information for any other star. Age estimates of field stars are imprecise, model-dependent and are of indeterminate accuracy. In the best cases - solar type stars and a little more massive - asteroseismology and the HR diagram position might give an age to about $\pm 0.5$ Gyr. But you need good data to do that and we don't have asteroseismology for the nearest 400,000 solar-type stars (in order to find ONE solar sibling). Even if we did, it would only narrow the candidates down by a factor of 10. What about chemical abundances? There are now big spectroscopic surveys that have observed large-ish numbers of stars. Again, the most robust data comes for stars like the Sun, so that a differential abundance analysis can be done. The more numerous cooler stars have spectra which are more difficult to deal with and there may be systematic abundance errors when comparing with the Sun. Giant stars are bright, with narrow spectral lines, which is good, but their abundances may have been modified during their evolution. Thus we are limited to looking at solar-type field stars. No spectroscopic surveys (yet) have detailed spectra of 400,000 potential solar siblings. Chemical abundance may also not be that discriminatory. In the solar neighbourhood, stars have an abundance dispersion of about a factor of two , centered quite close (a bit below) the solar metallicity. Good quality spectroscopic analysis can give the metallicity to 10%, so enough to resolve the distribution, but it would only whittle down the candidates by factors of a few. Very high quality spectra could look at the detailed abundance mixture, but is only available perhaps for a few thousand stars. The majority also have an abundance mixture that is not very different to the Sun. Thus whilst the evidence is that the detailed abundance dispersion in a cluster is much smaller than the dispersion in field stars ( Paulson et al. 2003 ), it does not look like the Sun is that unusual or has any unique chemical "markers" ( Bensby et al. 2014 ). Some hope may be offered by the detailed mix of s- and r- process elements; the former may be age sensitive (e.g. Jofre et al. 2020 ), while the latter might reveal some peculiarity associated with contamination by nearby supernovae in the Sun's natal environment. At the moment, although candidate solar twins can be found, it is unclear by how much that narrows down the one in 400,000 figure, especially when you consider that abundance peculiarities may also be imprinted by later accretion events or some process associated with planet formation (e.g. Melendez et al. 2012 ) or that the abundance dispersion in a cluster may not be exactly zero (e.g. Liu et al. 2016 ). In summary, the space density of solar siblings is likely to be very low compared with unassociated field stars. The properties of those siblings are not that unusual compared with typical field stars and we lack precise enough measurements of age and chemical composition to have anything more than a list of candidate solar twins at the moment. It's like looking for a needle in a huge pile of other needles. | {
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43,454 | Cosmic background radiation emitted when the Universe was very young still exists. But my wifi signal seems to disappear a short distance from my apartment. Why? | Electromagnetic radiation will continue to travel until it is absorbed. Some of your wifi signal is escaping to space where it may continue traveling for a very long time. However, the strength of your wifi signal will degrade with distance according to the inverse square law . So if you double the distance between your device and the wifi transmitter, your signal will be reduced by 3/4. When you move far enough away, eventually your signal will degrade enough that your device will no longer be able to decode the signal. So the radiation doesn't "disappear" but instead, your device will just be able to pick up less of the energy. Of course, if you separate the device and transmitter by a concrete wall, some of the signal will actually be "disappearing" as it is absorbed by the wall. The CMB (Cosmic Microwave Background) on the other hand, was not emitted from a point source, and is not subject to the inverse square law. Instead, as wikipedia points out, the CMB is very nearly uniform in all directions, but the tiny residual
variations show a very specific pattern, the same as that expected of
a fairly uniformly distributed hot gas that has expanded to the
current size of the universe. So one can imagine that wherever we are in the universe, we can point an antenna in any direction and get about the same amount of signal (excepting Doppler shift) from the CMB. Of course, if we point our antenna towards existing matter, we will expect some or all of the CMB to have been absorbed in that direction. In many cases, astronomers can tell a lot about the composition of gas in absorption based on the narrow frequencies of the CMB that was absorbed. Back in the early 2000s, I spent quite a bit of time looking for cosmic formaldehyde in absorption lines of the CMB near the center of our galaxy, as an indicator for early star formation. | {
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43,471 | There's a lot of concern in the Astronomy community over the deployment of Starlink satellites. For a good discussion, see the related question How will Starlink affect observational astronomy? But why is there so much concern over this problem, given that there are numerous space telescopes ? Presumably a lot more will be launched in the upcoming years thanks to satellite launches becoming cheaper. Aren't space-based observations superior in the first place thanks to the lack of an atmosphere? I understand this sucks for amateur astronomers but is it also a big problem for professional researchers? | It's a problem because there are still lots and lots and lots of ground-based telescopes. Ground-based telescopes are still (by far) the biggest optical telescopes, and the cost of space telescopes is prohibitive for many research projects. It will be a long time before a telescope anywhere close in size to the VLT can be launched. Most space telescopes are specialist devices, observing in a particular part of the spectrum that is blocked by the atmosphere (so there are infra-red, ultraviolet and X-ray telescopes in space) Or doing a specific task (looking for exoplanets, or mapping the positions of stars) Space is getting cheaper, but it will be a long time before it is as cost-effective as ground telescopes in the optical range. And so it will be a long time before all professional telescopes are in space. So it is a problem for professional astronomers. | {
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43,736 | I was just wondering why black hole's gravitational forces are so powerful. I know it's usually explained by Einstein's relativity which states that when an object becomes infinitely dense (a compact mass) it can exert such a force of gravity and warp spacetime. But I also learned about Newton's Law of Gravity equation F = $GM/r^2$ . Considering this equation, if the radius of an object becomes super small, then it can technically have immense gravity. So, can the gravitational pull of a black hole be explained by Newton's Law of Gravity or am I missing something? Thanks. | No you can't and the behaviour of bodies with mass and of light is completely different near a compact, massive object if you use Newtonian physics rather than General Relativity. In no particular order; features that GR predicts (and which in some cases have now been observationally confirmed) but which Newtonian physics cannot: An event horizon. In Newtonian physics there is a misleading numerical coincidence that the escape velocity reaches the speed of light at the Schwarzschild radius. But in Newtonian physics you could still escape by applying a constant thrust. GR predicts that no escape is possible in any circumstances. Further; this numerical coincidence only applies to light travelling radially. In Newtonian physics the escape "speed" is independent of which direction you fire a body, but in GR light cannot escape from (just above) the Schwarzschild unless it if fired radially outwards. For other directions, the radius at which light can escape is larger. GR predicts an innermost stable circular orbit. A stable circular orbit is possible at any radius in Newtonian physics. In GR a particle with some angular momentum and lots of kinetic energy will end up falling into the black hole. In Newtonian physics it will scatter to infinity. Newtonian physics predicts no precession of a two-body elliptical orbit. GR predicts orbital precession. Newtonian physics predicts that light travelling close to a massive body has a trajectory that is curved by about half the amount predicted by GR. Even stranger effects are predicted close to the black hole including that light can orbit at 1.5 times the Schwarzschild radius. The GR approach to gravity is fundamentally and philosophically different to Newtonian gravity. For Newton, gravity is a universal force. In GR, gravity is not a force at all. Freefalling bodies are said to be "inertial". They accelerate, not because a force acts upon them, but because spacetime is curved by the presence of mass (and energy). In most cases, where Newtonian gravitational fields are weak, the consequences of this difference are small (but measureable - e.g. the orbit precession of Mercury or gravitational time dilation in GPS clocks), but near large, compact masses, like black holes and neutron stars, the differences become stark and unavoidable. | {
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43,739 | I saw this answer and read this sentence regarding the distortion of our galaxy when viewed above the galactic plane: Once you got to a point where the entire [sic] galaxy was within your vision, the entire galaxy would look distorted, since our galaxy is 100,000 light years [sic] across, the light from the edges of the Galaxy would be thousands of years old, and would not match up with how the spiral should appear. I am thinking that this could potentially be true, but I am not sure why or how distorted it would be. Does this fact hold, and for what distances will it no longer be distinguishable? Note: I have thought that it will be well-blended because the distance does not spike, it increases gradually, but I need verification. | The light travel time of 100,000 years is quite small compared to the time it takes the Milky Way's spiral arms to complete an appreciable fraction of one rotation. The arms have a pattern angular speed of $\Omega_{\text{p}}=28.2\pm2.1\text{ km}\text{ s}^{-1}\text{ kpc}^{-1}$ ( Dias et al. 2019 ), so they should complete one full rotation on the order of $\tau=2\pi/\Omega_{\text{p}}\approx218\text{ Myr}$ . An observer a distance $h$ above the center of the galaxy would observe light coming from a radius $R$ to have to travel an additional time $$\Delta t=\frac{1}{c}(\sqrt{h^2+R^2}-h)=\frac{R}{c}\left(\sqrt{\left(\frac{h}{R}\right)^2+1}-\frac{h}{R}\right)$$ which is a decreasing function of $h/R$ . Its maximum is at $h=0$ , where it would be on the order of 100,000 years for stars on the outer edge of the disk. Clearly, $\Delta t\ll\tau$ , so there would not be an appreciable distortion, per se, because of the additional distance. While it's true that the objects you'd see would indeed have moved some amount by the time the light reached you, it wouldn't be significant. | {
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43,810 | Modern telescopes go to great lengths to have perfectly shaped parabolic mirrors. My question is, why go to the trouble of having a perfect mirror? Why not take a mirror roughly the right shape, and then correct for the distortion using computers? | correct for the distortion An imperfect mirror does not produce a distorted image - it produces a blurry image. With light-field sensors and phase imaging, one could possibly correct for the blur, but it is much more challenging problem than normal lens distortion correction. Distortion refers to a systematic change in how shapes are projected in an image. It results from a lens or mirror with good, accurate geometry that just does not produce a rectilinear projection. Random imperfections in a mirror do not cause distortion. Every point in the surface of a mirror contributes to every pixel in the result image. If a single part of the mirror is at slightly wrong angle, it does not cause a distortion in one point of the image. Instead, it projects the same image at a slightly different alignment on the same sensor. (1) In the case of a starfield, this would cause ghost images of very dim stars to appear next to the real stars. Repeat this for a thousand imperfections, and the result is just blurry dots. Deconvolution is a process that can be used to remove blurriness, but noise and other uncertainties limit its effectiveness. (1) This may be a bit unintuitive if you think about funhouse mirrors where the image is distorted. Those work differently because they act the part of a planar mirror, where indeed each part of image is reflected by a single part of a mirror. But planar mirrors cannot form an image by themselves, instead the lens in your eye is the critical component of the image accuracy. | {
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43,829 | What I mean is an event unfolding that is viewable by naked eye or telescopes, and doesn't take comparing days of footage to see moving pixels. Apart from satellites, is there something that moves or happens so quickly that humans could sense it within a reasonable time? The whole sky is like a still painting, I'd love to see parts of it being "animated". Things on my mind are: Crescent moon occulting a star (revealing or hiding it) Solar and Lunar eclipses Or more exotic things like: Stars exploding (catching that as it happens, is it even possible?) Other planets' moons moving a reasonabl/detectable amount in 30 minutes perhaps Slow pulsar blinking (they are still super fast afaik) Star changing its brightness and not blinking due to haze/turbulence | If it moves or flashes it isn't astronomy, it is meteorology or technology. There are only a few exceptions to this: Meteors are an atmospheric phenomenon, and a meteor will appear to move rapidly across the sky. But because they "come from space" and occur well above the clouds they are often considered to be part of astronomy. As you note, eclipses and occultations happen quickly enough for the changes to be visible. The moons of Jupiter do move notably during the night, and when one moves into or out of shadow, it can appear or disappear over the course of a couple of minutes: easily noticeable. A supernova could brighten quickly enough for the variation to be visible over a night's observation: You wouldn't see it suddenly appear, but over the course of a night it could appear brighter at the end of the night than at the beginning. Similarly, Algol will fade from magnitude 2 to 3.5 over a few hours. It isn't quick enough to notice the change in the moment, but it is quite clear if you are observing over a few hours. Pulsars are too dim to be seen with normal equipment even the bright nearby crab pulsar has a magnitude of 16.5 (and at one flash in 33 milliseconds, it is too quick to see by eye) GRB 080319B was an exceptional object. It was a gamma-ray burst. It gave a flash of gamma rays that lasted a little over a minute. If you had happened to know exactly where to look you could, (marginally with the naked eye, but easily with binoculars) have seen the optical counterpart to the gamma-ray burst. This was created by the formation as a massive early star collapsed to a black hole producing a jet of energy that happened to point our way. The sun is changing and on a small scale it does change on the scale of minutes, but it is hard to see any movement at that scale with basic equipment. If you can get your hands on a Hydrogen-alpha solar telescope you will, be able to see prominances that change over a period of hours. Rarely an asteroid will pass so close so as to be visible. Apopsis will have a close approach in 2029 and will appear as a slowly moving star. Jupiter is active at radio frequencies. You can tune in to Jupiter with the right equipment and hear it changing at rates from a few fractions of a second to a few seconds, see e.g. radiosky.com > Jupiter Central These are exceptions. In general, it is rare for anything so big and powerful that it can be seen over a distance of many light-years to be able to change fast enough that those changes can be noticed by our eyes. | {
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43,907 | Does the term "Hydrogen burning" mean the same as "Hydrogen fusing" in astronomy? If not, then what is the product of "Hydrogen burning"? Assume the product of "Hydrogen fusing" is helium. | In stellar astrophysics, "burning" means nuclear fusion, not chemical combustion. So a star burns hydrogen to helium. (Incidentally, normal chemical burning of hydrogen in air produces water). This might seem to be confusing terminology, but it's not an issue in practice because the regions in stars where fusion takes place (the core, and shells surrounding the core, in big old stars) are far too hot for molecules to exist, so chemical processes simply cannot occur there. Hydrogen fusion in a star core is not a single nuclear reaction. There are two (known) sets of reactions: the proton–proton chain , and the catalytic CNO (carbon-nitrogen-oxygen) cycle . From Wikipedia: The proton–proton chain, also commonly referred to as the p-p chain, 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, whereas the CNO cycle, the other known reaction, is suggested by theoretical models to dominate in stars with masses greater than about 1.3 times that of the Sun. Hydrogen burning is the main way that young stars produce heat and light. (Older stars can burn heavier elements, if they have sufficient mass). But hydrogen burning is also important because it's the main way that neutrons are produced: the burning of heavier elements generally does not change the number of neutrons. (Neutrons are also created when a heavy star collapses into a neutron star ). In a Sun-like star, hydrogen burning is a very slow process. It takes a lot of heat and pressure to overcome the electrical repulsion between two protons and fuse them into a diproton. However, the diproton is extremely unstable, with a half-life of much less than a nanosecond. Usually, a diproton just splits up into two protons, but sometimes one of its protons transforms into a neutron, converting the diproton to a deuteron (a deuterium nucleus). The probability of diproton to deuterium conversion is very small, in the order of $10^{-26}$ . Various p-p chain reactions convert deuterons to helium nuclei. The slowness of deuterium production is a good thing: it means our Sun will continue to burn hydrogen for billions of years. OTOH, it also means plain hydrogen will never be a practical fuel for an artificial fusion reactor. The density in the Sun's core is ~150 g/cm³, which is much greater than any terrestrial material (eg, the density of lead is only 11.3 g/cm³). So a cubic metre of solar core matter has a mass around 150 metric tons. But the rate of hydrogen fusion is so slow that it only produces around 276.5 watts of power, which is similar to the heat production of a cubic metre of compost, as mentioned on Wikipedia . Of course, a 1 m³ compost heap won't keep pumping out heat for billions of years. ;) Wikipedia has a good collection of articles on the various stellar nuclear fusion processes, including: The triple-alpha process , which burns helium to carbon. The carbon-burning process , which converts carbon to oxygen, neon, sodium, and magnesium. The neon-burning process The oxygen-burning process The alpha ladder The silicon-burning process | {
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43,959 | What I have always thought / known is that: Days are based on the period of rotation of the earth. Months are based on the moon. Year is the time taken by the earth to make one revolution around the sun. But if we take a look at the lunation it takes on average 29 days, 12 hours, 44 minutes, and 28 seconds. Why is it that the months then in the Gregorian calendar have 28-29 / 30 / 31 days? | Ancient Egyptians and Mesopotamians came up with “administrative” calendars of 30 days, that were easier to calculate than “real” lunar months of sometimes 29, sometimes 30 days. At the end of the 12 months, they added five days of religious observances, sometimes positive, sometimes negative. When Julius Caesar conquered Egypt, he knew about the very regular Egyptian calendar and appointed his mathematicians and astronomers (if we can really call them that, as Roman science wasn’t very developed) to discuss with their Egyptian peers and devise a better calendar for the Roman Republic (reminder that Caesar was not the Emperor; the first one was Augustus, his [adoptive] son, and not right away). Eventually, it was decided in Rome that, in order to follow with the seasons, the already-existing Republican Roman calendar was to be modified. It contained 10 months of 29 and 30 days plus a “winter” of 59 days, which became divided in January and February. As Julius Caesar was the one who brought this change, he was “honored” by having a month renamed from Quintilis (“Fifth”) to Julius (our July)—and that month was given 31 days, as well as a few others. Also, the beginning of the year was shifted (at this point or later; it’s not clear) to January, making the “numbered months” off with their new rank—e.g. “ Octo ber” is not eighth (“octo”) anymore. Caesar’s reform was not well applied, and eventually, Augustus had to enforce the new calendar, and he was also “honored” with a 31-day month, right after Julius’, so that’s how Sextilis (“Sixth”; and it had 30 days) became Augustus (with 31 days). The rest is basically adjusting here and there to make it a total of 365 days at the end of the year, someone sometime having the “brilliant” idea of alternating 30- and 31-day months (a little like nature “alternates” between 29- and 30-day lunar months). Side note… In some languages, the word for Moon and the word for month is the same—it’s already quite similar in English! For example, Romanian has “Luna” for both, and Korean has ”dal.” | {
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44,049 | The Sun loses about 5.5 million tonnes of mass every second, does this mean Sun's volume is also going down? If so can we tell by how much every year? E:
"by how much" I mean can we tell if sun's radius is going down too | Is the sun shrinking currently? It's the other way around: The Sun is slowly growing hotter and thus is slowly expanding. The Sun accumulates ever more helium in its core as it ages. This growing amount of fusion ash results in the core getting hotter, which in turn results in the Sun expanding in volume. This completely overwhelms what one would expect from looking only at the mass loss due to fusion and solar winds. The solar mass loss rate currently is on the order of $10^{-13}$ solar masses per year. At that rate, it would take on the order of ten trillion years for the Sun to consume or expel all of its mass. That's a much longer time span than the Sun's much faster aging process. The Sun has five to seven billion years or so left on the main sequence. Shortly before the time the Sun leaves the main sequence, it will be about twice as luminous and considerably larger than it is now. Shortly after the the Sun leaves the main sequence, it will be much more luminous and even larger than it was just before it left the main sequence. | {
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44,109 | I read in trigonometry class that 1 minute is equal to 1/60 degrees. So, 'minute' is an angular unit. But also 'arcminutes' are used to measure seperation between celestial objects and also equals to 1/60 degrees.
Are they any similar? If not, then what's the difference? | This can get a bit confusing, because "arcminute" and "minute" are both sometimes used in celestial coordinate systems but mean two different things. An arcminute is 1/60th of a degree, and an arcsecond is 1/60th of an arcminute. That's simple enough, and when talking about small angular distances, it's often much handier to refer to something as being, say, 140 arcseconds across, rather than 0.0389 degrees. So you're likely to see angular sizes or scales quoted in degrees, arcminutes and arcseconds. If you're trying to state the position of an object on the sky, things get a little more complicated, thanks to the commonly-used equatorial coordinate system , which states an object's position on the celestial sphere in terms of its declination and its right ascension. The declination of an object is usually given in degrees, arcminutes and arcseconds. Its right ascension, on the other hand, is usually given in hours, minutes and seconds. Here, one "hour" corresponds to 1/24th of a circle, or 15 degrees. One minute is then 1/60th of an hour, and one second is 1/60th of a minute. So as units of angular separation in this context, an arcminute is different from a minute, and an arcsecond is different from a second. | {
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44,184 | Not sure if that is possible as I couldn't find an answer about it. Are there places in the Universe where there are no gravitational forces? | In two dimensions I think I can infer in a lame, unconvincing and rigorless numerical way that there are likely to be zeros in gravity from a random distribution of objects there can be points of zero gravity. I create a space with 20 randomly distributed point sources, calculate and plot the force field on a 2000 x 2000 grid then choose the smallest grid point and through a minimization routine find a point with arbitrarily small scalar force. I've done everything on a log10 scale, the max, min values are of order +8 and -1 but I can easily find log10(force_magnitude) around -14 by specifying that tolerance in the minimization routine. I can't prove this extends to 3 dimensions nor arbitrarily large space and numbers, but I have a hunch this can be addressed mathematically, so I have just asked in Math SE: What is the relative density and dimensionality of zeros in inverse square force fields to density of sources in (at least) 1, 2 and 3 dimensions? Here are six cases for flavor: There be zeros here! import numpy as np
from scipy.optimize import minimize
import matplotlib.pyplot as plt
# https://astronomy.stackexchange.com/questions/44184/are-there-places-in-the-universe-without-gravity
def mini_me(xy, positions):
return np.log10(get_force_magnitudes(positions, xy))
def get_force_magnitudes(positions, XY):
r = positions[:, None, None, :] - XY
forces = r * (((r**2).sum(axis=-1))[..., None]**-1.5) # all vectors
force_field = forces.sum(axis=0) # vector field
return np.sqrt(force_field**2).sum(axis=-1)
N = 20
positions = np.random.random((N, 2))
x = np.linspace(0, 1, 2000)
X, Y = np.meshgrid(x, x, indexing='xy')
XY = np.stack((X, Y), axis=2)
force_magnitude = get_force_magnitudes(positions, XY)
indices = np.unravel_index(np.argmax(-force_magnitude), force_magnitude.shape) # find the smallest one on the grid
xy0 = XY[indices] # starting point for minimization
result = minimize(mini_me, xy0, args=(positions, ),
method='Nelder-Mead', tol=1E-08)
if True:
fix, ax = plt.subplots(1, 1)
extent = 2*[0, 1]
thing = ax.imshow(np.log10(force_magnitude), origin='lower',
extent=extent, vmax=2)
x, y = positions.T
ax.plot(x, y, '.r')
x, y = result.x
label = str(round(result.fun, 2))
ax.plot([x], [y], 'o', color='none', markeredgecolor='red',
markersize=14, markeredgewidth=2)
ax.text(x+0.02, y, label, color='red', fontsize=14)
plt.colorbar(thing, ax=ax)
plt.title('log10 normalized scalar force')
plt.show() | {
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44,194 | I am an undergraduate in Physics, and I have an opportunity to work with a some of my seniors to learn Computational Fluid Dynamics (CFD) and Magnetohydrodynamics (MHD). I would like to choose my first individual project in order to learn about how these techniques are used in Astrophysics. What are a few basic, illustrative examples of the use of CFD and/or MHD that I might look into and consider working on as my first project? What are illustrative examples of how these techniques are currently used in astrophysics research so I can have a better idea how I might use these skills in the future, and if possible a reference where I can read futher? | In two dimensions I think I can infer in a lame, unconvincing and rigorless numerical way that there are likely to be zeros in gravity from a random distribution of objects there can be points of zero gravity. I create a space with 20 randomly distributed point sources, calculate and plot the force field on a 2000 x 2000 grid then choose the smallest grid point and through a minimization routine find a point with arbitrarily small scalar force. I've done everything on a log10 scale, the max, min values are of order +8 and -1 but I can easily find log10(force_magnitude) around -14 by specifying that tolerance in the minimization routine. I can't prove this extends to 3 dimensions nor arbitrarily large space and numbers, but I have a hunch this can be addressed mathematically, so I have just asked in Math SE: What is the relative density and dimensionality of zeros in inverse square force fields to density of sources in (at least) 1, 2 and 3 dimensions? Here are six cases for flavor: There be zeros here! import numpy as np
from scipy.optimize import minimize
import matplotlib.pyplot as plt
# https://astronomy.stackexchange.com/questions/44184/are-there-places-in-the-universe-without-gravity
def mini_me(xy, positions):
return np.log10(get_force_magnitudes(positions, xy))
def get_force_magnitudes(positions, XY):
r = positions[:, None, None, :] - XY
forces = r * (((r**2).sum(axis=-1))[..., None]**-1.5) # all vectors
force_field = forces.sum(axis=0) # vector field
return np.sqrt(force_field**2).sum(axis=-1)
N = 20
positions = np.random.random((N, 2))
x = np.linspace(0, 1, 2000)
X, Y = np.meshgrid(x, x, indexing='xy')
XY = np.stack((X, Y), axis=2)
force_magnitude = get_force_magnitudes(positions, XY)
indices = np.unravel_index(np.argmax(-force_magnitude), force_magnitude.shape) # find the smallest one on the grid
xy0 = XY[indices] # starting point for minimization
result = minimize(mini_me, xy0, args=(positions, ),
method='Nelder-Mead', tol=1E-08)
if True:
fix, ax = plt.subplots(1, 1)
extent = 2*[0, 1]
thing = ax.imshow(np.log10(force_magnitude), origin='lower',
extent=extent, vmax=2)
x, y = positions.T
ax.plot(x, y, '.r')
x, y = result.x
label = str(round(result.fun, 2))
ax.plot([x], [y], 'o', color='none', markeredgecolor='red',
markersize=14, markeredgewidth=2)
ax.text(x+0.02, y, label, color='red', fontsize=14)
plt.colorbar(thing, ax=ax)
plt.title('log10 normalized scalar force')
plt.show() | {
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44,270 | Thinking about gravitational waves and the fact that they propagate at the speed of light, I was wondering if it isn't suspicious - the speed of light I mean. Does it perhaps point to something fundamental about the spacetime? Is there maybe some connection between EMR and spacetime itself? Or am I seeing things in the tea leaves? | It is very suspicious! It points to the fact that the speed of light isn't just some random speed that light happens to travel at, but is a fundamental property of the universe. In fact, any massless particle will move at the speed of light. This is a consequence of relativity. Energy, mass and momentum( $p$ ) are related by $$E^2 = m^2c^4 +p^2c^2$$ for a particle moving at velocity $v$ less than $c$ , $$p = mv\sqrt{\left(\frac{1}{1-(v/c)^2}\right)}$$ if a massless particle ( $m=0$ ) is moving at velocity less than $c$ , then it would have zero momentum and zero energy. Such a particle could never be detected (since to be detected a particle has to transfer some of its energy and momentum to the detector). Is it possible for a fundamentally undetectable object to exist? That is a matter for philosophers. For the sake of developing models of reality, such particles don't exist. | {
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44,339 | From what I understood, the Milky Way (or stars in the Milky Way) doesn't rotate like a collection of points in a disc due to the presence of some invisible matter. In theory, the angular velocities of all the stars should be the same while linear velocities should decrease as the radial distance increases. But in reality, linear velocities stay almost the same as we move farther from the galactic center. But that will decrease the angular velocity yes, right? And I know that stars in a constellation are usually far apart. So this implies they have different angular velocities around the galactic center/nucleus. Then how does the shape of constellations stays the same? If stars are rotating at different angular velocities, will the constellations get distorted over time? | Yes, the shape of the constellation does and will change over time. All the stars have their own peculiar velocities and have some random motion which over time will ruin all the constellations. However, Even though the stars are moving at rapid speeds, to us, in our sky, due to their enormous distance from us, they appear to move extremely slowly and the constellations can stay the way they are for another few hundred/thousand years until we notice any visible change (Naked eyes observations). Those slow relative changes in position give each star in our sky a particular "proper motion"—a change in angular position. The proper motion of most stars is extremely small, measured in milli-arcseconds per year, where an arcsecond is 1/3600 of a degree, and of course milli means a thousandth of that. Hold up your pinky finger at arm's length: the width of your pinky, expressed as an angle, is about one degree, give or take. One degree is 3,600 arcseconds, or 3,600,000 milli-arcseconds. Proper motion , in astronomy, the apparent motion of a star across the celestial sphere at right angles to the observer’s line of sight; any radial motion (toward or away from the Sun) is not included. The fastest star in our current sky is Barnard's Star, which is moving radially at 110km/s towards us, has a proper motion of only 10.3 arc-seconds (which is the fastest proper motion in our sky). This means, in another 350 years, it would only move around 1 pinky finger-width. Here is the source if you want to read more. | {
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44,443 | Scientists think there may be dozens of undiscovered dwarf planets in the Solar system. What I find intriguing is that, we have been able to detect distant galaxies and clusters, locate supermassive black holes...but how is it that dwarf planets in our own solar system are going undetected? | Dwarf planets are observed by the sunlight they reflect. Undiscovered dwarf planets at the outer boundaries of our solar system are simply fainter than the other things mentioned. If they can't be seen then they can't be discovered. It is possible that they are just about bright enough to be seen but even then, in a single picture they are indistinguishable from the billions of other faint specks of light in the sky. Finding faint, moving extremely faint objects in a series of pictures is a challenging observational project. | {
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44,458 | How close to Earth would Ceres have to be to cause tides of the same strength as by the Moon, above the region it orbits? Ceres has 1.3% the Moon's mass, but that doesn't mean it must be at 1.3% the Moon's distance, right? Is it even possible for Ceres or is Ceres too low-mass to cause similar tides? | As is nicely put on the Wikipedia page about tidal forces, the tidal force is given by $$T=Gm\frac{2r}{d^3}$$ where $T$ is the tidal force (see below), $G=6.67\cdot 10^{-11}\rm\,\frac{m^3}{kg s^2}$ is the gravitational constant, $r$ is the radius of the Earth, and $d$ is the distance between the centers of the two objects. This is not force in correct sense (in Newtons), but is given by $\frac{m}{s^2}$ , like acceleration. You said, that the tidal forces must be the same, so: $$T_M=T_C$$ $$GM_M\frac{2r_E}{d_M^3}=GM_C\frac{2r_E}{d_C^3}$$ $$\frac{M_M}{d_M^3}=\frac{M_C}{d_C^3}$$ $$d_C=d_M\left(\frac{M_C}{M_M}\right)^\frac{1}{3}\approx 89000\rm\, km$$ The height of the center above the surface would thus need to be around 82500 km. But note that the tidal forces of Earth on Ceres would be large. Would Earth destroy Ceres? To answer this, we need an equation for Roche limit $$d=R_m\left(2\frac{M_M}{M_m}\right )^\frac{1}{3}$$ where $R_m$ is the radius of secondary, $M_M$ is the mass of primary, and $M_m$ is the mass of secondary. With inserting the data for Earth and Ceres, we get $$d=11175\rm\, km$$ and $11175\rm\, km <89000\rm\, km$ . Thus, Ceres wouldn't be destroyed. But, as @JamesK said, how would the tides look like? By the equation: $$a_c=a_g$$ $$\frac{v^2}{r}=\frac{GM}{r^2}$$ and $$v=\frac{2\pi r}{t_0^2}$$ we get $$t_0=\sqrt{\frac{4\pi^2r^3}{GM}}=\sqrt{\frac{4\pi^2 (8.9\cdot 10^7)^3}{6.67\cdot 10^{-11}\cdot 6\cdot 10^{24}}}\rm\, s=3.052\rm\, d$$ So, Ceres would make one sidereal rotation period in 3.052 days. If Ceres rotates clockwise, then the one synodic period around the Earth is $((1\rm\, d)^{-1}+(3.052\rm\, d)^{-1})^{-1}=0.7532\rm\, d$ . But if Ceres rotates counter-clockwise, then the one synodic period around the Earth is $((1\rm\, d)^{-1}-(3.052\rm\, d)^{-1})^{-1}=1.4873\rm\, d$ . The two graphs for total tidal forces are drawn below (with their Python code): Python code: import matplotlib.pyplot as plt
import numpy as np
PI = 3.14159265358979
x = np.arange(0, 20, 0.01)
yCeres = []
ySun = []
y = []
for i in x:
yCeres.append(np.sin(i * 4 * PI / 1.487) * 1.098e-6)
ySun.append(np.sin(i * 4 * PI / 1) * 5.05e-7)
y.append(yCeres[-1] + ySun[-1])
plt.title('Total tidal force: Sun and Ceres (counter-clockwise)')
plt.xlabel('Time [d]')
plt.ylabel('Tidal force [ms^-2]')
plt.plot(x, y)
plt.show() Python code: import matplotlib.pyplot as plt
import numpy as np
PI = 3.14159265358979
x = np.arange(0, 20, 0.01)
yCeres = []
ySun = []
y = []
for i in x:
yCeres.append(np.sin(i * 4 * PI / 0.7532) * 1.098e-6)
ySun.append(np.sin(i * 4 * PI / 1) * 5.05e-7)
y.append(yCeres[-1] + ySun[-1])
plt.title('Total tidal force: Sun and Ceres (clockwise)')
plt.xlabel('Time [d]')
plt.ylabel('Tidal force [ms^-2]')
plt.plot(x, y)
plt.show() Compare it to the real current tides: Python code: import matplotlib.pyplot as plt
import numpy as np
PI = 3.14159265358979
x = np.arange(0, 20, 0.01)
yMoon = []
ySun = []
y = []
for i in x:
yMoon.append(np.sin(i * 4 * PI / 1.035087719) * 1.098e-6)
ySun.append(np.sin(i * 4 * PI / 1) * 5.05e-7)
y.append(yMoon[-1] + ySun[-1])
plt.title('Total tidal force: Sun and Moon (counter-clockwise)')
plt.xlabel('Time [d]')
plt.ylabel('Tidal force [ms^-2]')
plt.plot(x, y)
plt.show() Which is pretty natural because the tides are at their highest on full and new moon. | {
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44,567 | I'm a high school junior. I assumed there would be an incredibly huge number of atoms in the universe if not infinite. Recently, I've come across a few articles which claim that scientists believe there are 10 83 atoms in the universe. I'm not sure if I should believe in this, to me, this seems like a small number. In average, a human alone has 10 27 atoms, so, it's kind of unconvincing to me that there's only 10 83 atoms in the whole universe. | This is a reasonable estimate for the number of atoms in the observable universe. It might seem like a small number, compared with the number of atoms in a human only as a result of our brain's inability to have an intuition about very large numbers and exponential scales. There is a very very big difference between $10^{27}$ and $10^{83}$ . How big is the difference? Well $10^{83}- 10^{27}= 9.9999999999999999999999999999999999999999999999999999999\times10^{82}$ A human is only a very small part of the universe. Note two things, firstly this is for the observable universe, the Universe may be isotropic, open and unbounded and if so the total number of atoms in the universe is infinite. Secondly most of the estimates of the number of atoms in the universe that I've seen put the value at about $10^{80}$ | {
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44,583 | This is the page I am referring to. It seems counterintuitive to me that the Sun should be on the opposite side of the barycenter's wobble. I realize I am wrong, but I cannot see why I am wrong. Can someone explain why the wobble is away from Jupiter, not towards Jupiter? Here is an unedited screenshot of the NASA animation - it shows the sun on the opposite side of the green line (barycenter) as Jupiter. My logic says, since gravity is in play here, the Sun and Jupiter should be on the same side of the green line. I have edited NASA's image in MS Paint to show what I think should be happening: | The part of your intuition that is correct is that Jupiter pulls the Sun towards it . The problem is that "pulls towards" does not mean "brings closer"! The gravitational force results in an acceleration towards an attracting body, which is not a displacement or even the derivative of displacement, but the second derivative of displacement. Oscillatory or circular motion has the property that the second derivative carries a minus sign . For example, when the Sun is on the right side of the green circle, its acceleration is to the left , because it is changing from rightward motion to leftward motion. Thus, by being on the opposite side from Jupiter, the Sun is continually accelerating towards it. | {
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44,682 | Most animations and drawings of Black Holes that I've seen usually depict some kind of funnel which is the "entrance" to the black hole; let's call this the front. Are there more than one way to enter the Black Hole, such as the back side (180 degrees from the "front" side)? Can a Black Hole have more than one entry point? (It's kind of difficult viewing a ball {the Black Hole} with multiple entry points around the surface, sucking in matter and light) . | What you are talking about is an embedding diagram . These are ways of visualising the curvature of 3D space by projecting it onto a 2D surface. These can be very misleading - for example, the trajectory of something in freefall around a black hole is not simulated by rolling a ball on this surface. Such diagrams make no attempt to represent the full dimensionality of a real black hole. Black holes exist in all three spatial dimensions. If we are talking about a non-spinning black hole, it is absolutely spherically symmetric and looks and behaves the same approaching it from any direction | {
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44,766 | The Grand Tack Hypothesis states that Jupiter first migrate inward, but it was caught up by the faster inward migration of Saturn, and when the two planets reached 3:2 mean-motion resonance they migrated outward together. As a student who only knows high-school physics, I can imagine that to make these massive gas giants to migrate outward, they should receive a tremendous amount of energy (like how we send satellite into the space). The Nature paper that proposes the Grand Tack Hypothesis ( Walsh et.al., 2011) does not seem to explain why they migrate outward, but references another simulation study (Masset & Snellgrove, 2001) , which is unfortunately to hard for me to understand now. Is there an intuitive explanation of why Jupiter and Saturn can migrate outward? What is the energy source? | First let's try to understand why planets migrate inwards . Planets are formed in a protoplanetary disk; a huge disk of gas and dust that accretes on to a newly forming star at the centre. Gravitational interactions between the planets and the gas in the disk play a very important role in planetary formation and evolution. As planets orbits within the disk, they generate 'spiral density waves'. These are simply changes to the density of the gas in the region local to the planet, which move at a different speed compared to the rest of the material in the disk. The consequence is that the wave exterior to the planet exerts a negative torque and acts to slow the planet down. The wave interior to the planet exerts a positive torque and speeds it up. The net effect is that the negative torque wins, which causes the planet to lose angular momentum and migrate towards the star. On the timescales of planetary formation this is a relatively fast process, and typically affects lower mass planets. When you have a planet as large as Jupiter, something else happens. The torques that the planet exerts on the disk are so strong that the repel the gas away from its orbital region entirely, opening up a 'gap' in the disk. The planet now migrates inwards following the natural evolution of the disk ie. the timescale that gas naturally migrates and accretes on to the star. The planet and the gap migrate inwards together as one. This is a much slower than the process than the one described for smaller planets, hence why Saturn was able to catch up with Jupiter. So, why did Jupiter migrate outwards? When Jupiter and Saturn became locked in an orbital resonance, they formed a common gap in the disk. Saturn essentially cleared the region exterior to Jupiter, reducing the torque exerted on Jupiter by the outer part of the disk. But the gas repelled away by Jupiter in the inner region became piled up at the inner edge of the disk, increasing in density and enhancing its own torques back on Jupiter. The net effect is now that the positive torques from the inner disk overcome the negative torques from the outer disk, and the planets migrate outwards. For this to work, you need quite a specific scenario. Modelling suggests that outward migration can only occur when the inner planet is 2-4x larger that the outer planet. For further reading I would highly suggest this recent review by Raymond & Morbidelli (2020) , which gives an excellent description of our best current models of the formation of the Solar System. | {
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44,976 | Why are there not yet any instruments dedicated to registering time dilation caused by passing gravitational waves? Wouldn't it be interesting to augment LIGO/Virgo capturing of space distortion with simultaneous capturing time dilation (both caused by the same passing gravitational wave)? | General relativity predicts that there are only two possible polarizations of gravitational waves, the so-called "tensor" polarizations $+$ and $\times$ . It turns out you can show that the tensor polarizations actually don't lead to time dilation , making any attempted measurement of it pointless. The short answer, then, is that we don't expect to see any time dilation at all! Now, you could argue that such an experiment would still be useful insofar as it could be used to search for alternative polarizations (the "scalar" and "vector" polarizations) which would indicate that a different theory of gravity is warranted. On the other hand, this would be arguably be redundant, because there are other methods with which we can probe alternative polarizations in interferometric data, either by looking at individual sources or the hypothesized stochastic gravitational wave background (at the frequencies LIGO is sensitive to). An individual transient signal would need five $^{\dagger}$ appropriately aligned detectors to fully characterize contributions of alternative polarizations, but the LIGO-Virgo collaboration was able to search for evidence of scalar and vector polarizations in the signal from GW170814 (more here ) and at least found that purely tensor polarizations were strongly favored over purely scalar or purely vector polarizations. KAGRA has begun observations, and LIGO-India should be completed by the middle of the decade, which will help break some of the degeneracies at work. A search of the stochastic background wouldn't require so many detectors because the signal is not coming from any one place in the sky, so it provides another strategy with which to probe alternative polarizations . The O1 observing run turned up no evidence of backgrounds with scalar or vector polarizations; that said, there was also no evidence of any background at all, tensor polarizations included. It's also possible that pulsar timing arrays may be able to shed light on the issue if a stochastic background is detected and there is substantial evidence for tensor polarizations but not alternative polarizations ( Cornish et al. 2017 ), making some of this moot. $^{\dagger}$ A single interferometer's response to a gravitational wave is a sum of terms corresponding to individual polarizations. In more general theories of gravity, there are up to two tensor modes, two vector modes, and two scalar modes, but the class of interferometers LIGO and Virgo belong to can only measure a particular linear combination of the scalar modes, so we deal with five degrees of freedom. Therefore, five detectors are needed to determine how each mode (or combination or modes) contributes to the signal ( Chatziioannou et al. 2021 ). | {
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45,212 | I'm no expert, but I once studied basic from advanced physics and understand gravity action/reaction escape velocity of 11.2 km/s from the earth surface escape velocity changing as the object go far away Karman line for space (100 km) What I want to know, can a natural object, from a natural process, reach the escape velocity 11.2 km/s and get free of earth gravitation? I have in mind some volcano explosions. Tornados seem weak. Or also a Comet impact. Question is not limited in time, maybe it was possible millions of years before, when activities and gravity might have been different. | Yes, it is not only possible, but has almost certainly happened on Earth. The asteroid that killed the dinosaurs is thought to have produced these high velocity fragments on the order of one-thousandth of the mass of the impactor according to Poveda and Cordero [2008] . In fact, they estimate the volume of escape velocity ejecta from the Chicxulub event as quite substantial in that: the number of fragments with sizes larger than 10 cm and 2 cm is
about 4x10^10 and 2x10^12, respectively The authors of the above paper call these "Chicxulubites" and speculate on the possibility of finding them on the Moon and Mars. The larger the impactor radius and velocity, the higher a percentage of its overall mass will reach escape velocity. In an asteroid impact, the fastest moving ejecta comes from near the center of the crater. The ejecta closer to the edge of the crater has much lower velocity: The equation governing the velocity of the ejecta $V_{ej}$ is therefore partly a function of $r$ , the distance from the center of the crater. $$V_{ej} = \frac{2\sqrt{Rg}}{1+\epsilon }\left(\frac{r}{R}\right)^{-\epsilon}$$ Where $\epsilon$ is a material coefficient of 1.8 for hardpack soil, $g$ is the surface gravitational acceleration, $R$ is the total radius of the equator, and $r$ is the radius at which the ejecta was ejected. Notes: I go into more detail in this answer: What percentage of a lunar meteor strike is blown back into space? . Practically, an object near sea level must achieve higher than escape velocity in order to overcome atmospheric drag as it escapes. | {
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45,248 | Stars move through the sky very slowly, which is not noticeable in a human lifespan. I’m aware of proper motion of Barnard's Star and things of the sort but I’d like to obtain a noticeable record of changes from 1000s of years ago. There are old star maps I’ve seen online but it is hard to even understand what’s going on. Are there any sources that can explain it simply to me or even show side by side example of old/ancient and new/modern? | In practice, you're probably not going to get anything useful from ancient star maps , for several reasons: Very few of them actually survive from more than a few hundred years ago. Maps (and visual diagrams in general) are hard to accurately reproduce if you don't having printing, so a diagram from a document originally produced, say, 2000 years ago may be rather distorted if the oldest copy you have was made only 1000 years ago, and was itself most likely a copy of a copy of a copy... Even a genuine 2000-year-old map probably wouldn't have the accuracy for you to tell whether an apparent shift in the relative positions of stars was due to actual proper motion or just due to sloppiness in making the original map. What you can potentially use are ancient catalogs with numerical coordinates of stars. This is what Edmund Halley did in the early 1700s, when he compared modern measurements of the positions of stars against their coordinates from the catalogs of Hipparchus, Timocharus, and Ptolemy, and argued that Sirius, Aldebaran, and Arcturus were all shifted by various offsets (15 to 31 arc minutes) compared to their positions 1800 years ago or so. (He also pointed out that in Tycho Brahe's catalog, finished in 1598, Sirius was about 2 arc minutes north of where it was now, indicating the stellar motion had been continuous.) See Aitken 1942 for a reproduction of Halley's 1718 publication. The cumulative proper motions of Sirius, Arcturus, and Aldebaran over the last 2000 years amount to something like half to a full diameter of the moon, so if you had an accurate map from 2000 years ago, you could probably notice the difference (though it would be relatively subtle). But, again, I don't think such a map exists. | {
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45,282 | Many experiments (e.g. changes in Mercury's orbit) show that Newton's Law of Gravity is not exact. According to the theory of relativity, what is the exact formula? | Going from Newton's theory to Einstein's theory is not simple. It's not like you can just add a term to Newton's gravity, like $\textbf{F}=-{GmM \over r^3}\textbf{r} + \textbf{f}(\textbf{r})$ and obtain the right formula. To say it with Feynman's words , you can't make imperfections on a perfect thing, you have to make another perfect thing. And the new perfect thing, Einstein's gravity, is definitely more complicated than Newton's. To understand the extent of the difficulty, let me do a simple comparison between the theories. Newton's theory of gravitation The theory is concerned with two kinds of objects. The matter and the gravitational field. To determine the evolution of these two objects there are two equations, that are usually expressed in an inertial reference system. The first equation shows how a particle of matter at position $x^i$ moves under the influence of the gravitational field $\Phi(\textbf{x},t)$ $${d^2x^i(t) \over dt^2} = -{\partial \over \partial x^i} \Phi(\textbf{x},t)$$ The second equation details how to compute the gravitational field $\Phi(\textbf{x}, t)$ at any point in space and time, if you know the density of matter $\rho(\textbf{x},t)$ at every point in space and time. $$\nabla^2\Phi(\textbf{x},t) = -4\pi G\rho(\textbf{x},t)$$ It is important to notice that $\Phi$ and $\rho$ are both scalar functions: they are functions that map a real number to every point of space and time. These two equations are everything that is needed to describe gravity. The first one shows how the matter moves and the second one describes how the gravitational field adjusts itself to the change of matter distribution. The modified gravitational field will then affect the motion of matter in a different way and so on. Einstein's theory of gravity Here the role of gravitational potential $\Phi$ is taken by the metric tensor $g_{\mu\nu}$ which is a collection of 16 real numbers , one for every combination of $\mu$ and $\nu$ , that can take values 0,1,2 and 3. Actually not all 16 numbers are independent, because $g_{\mu\nu} = g_{\nu\mu}$ , so there are only 10 independent numbers. We go from $\Phi$ , that was only 1 real number for every point in spacetime, to $g_{\mu\nu}$ which carries 10 real numbers for every point in spacetime. This is already looking more complicated. A similar argument can be made for $\rho$ . It gets substituted with the stress-energy tensor $T_{\mu\nu}$ , which again carries 10 independent real numbers. The two equations above get also modified and become a lot more complicated. An important consideration is that Einstein's equations are usually written in a totally generic reference system, to just inertial ones. The first , the one that describes the motion of one particle under the influence of gravity becomes $${d^2x^{\alpha} \over ds^2} = -\Gamma_{\mu\nu}^\alpha {dx^{\mu} \over ds}{dx^{\nu} \over ds}$$ where $\Gamma_{\mu\nu}^\alpha$ is a function of $g_{\mu\nu}$ and its first derivatives with respect to the coordinates $x^{\mu}$ . Notice that in this equations the derivatives are not taken with respect to time ( $x^0$ ), but to the parameter $s$ (that can be chosen to be equal to time, but this is often not the smartest choice). The second equation , which describes how the distribution of matter $T_{\mu\nu}$ shapes the metric tensor $g_{\mu\nu}$ is $$G_{\mu\nu} = {8\pi G \over c^4}T_{\mu\nu}$$ where $G_{\mu\nu}$ is a pretty complicated, non-linear function of $g_{\mu\nu}$ , and its first and second derivatives. This last equation is actually a system of 10 coupled non-linear second order partial differential equations in 4 dimensions and it is a nightmare to solve. Just compare it to the corresponding Newtonian equation, which is 1 linear second order partial differential equation in 3 dimensions. These are the exact formulas according to the theory of general relativity. | {
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45,305 | I was in the desert the other day, in an area that was free of light pollution. While it was easy to see the stars above us, I noticed that near the horizon, even when there is no distance light, we can see almost no stars. Is this normal? What is causing this? Is it air pollution that covers the Earth and is denser when looking at the horizon? Is it just the density of the atmosphere itself? | When you look towards the horizon you are looking through a much greater thickness of air. The air does absorb some light. Dense air near surface absorbs more, and if you look towards the horizon you are looking thought a great distance of dense air. It is not "pollution" per se, though atmospheric aerosols and smoke can exacerbate the effect. Water vapour also absorbs light, and of course any haze or mist will absorb more. If there are wildfires, dust or air pollution, the "extinction" of stars will be greater. Smoke can travel very long distances (I've known smoke from wildfires in Spain to have a very noticeable effect in the UK). In deserts, there can be windblown dust that has a similar effect. However, even with pristine air, the extinction of stars near the horizon will still occur. | {
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45,323 | In physics, it always takes some time for a particle to move from rest to some speed. However, photons (light particles) accelerate instantly from zero to c. How? (A visualization would be helpful.) | "Accelerate instantly" would imply that a photon takes many different velocities at the same point in time. In fact, it would imply that a photon takes on every velocity between $0$ and $c$ simultaneously , but that clearly makes no sense at all - a particle cannot have many instantaneous velocities simultaneously. When a photon is created, it is traveling at $c$ . A photon is always traveling at $c$ , there is no such thing as a stationary photon which is then accelerated. As an imperfect analogy, consider a pressure wave like sound traveling through air - the pressure wave itself is the movement of energy, you can't have a sound wave that remains stationary and does not move. Similarly, a stationary photon does not exist - if it exists, the electromagnetic wave (which is the photon ) must be moving at $c$ . As another analogy, consider what happens when you throw a rock into a pond - ripples expand from the point of contact. How fast were the ripples moving before you threw the rock? That's unanswerable, since the ripples did not exist before you threw the rock. The ripples are simply the movement of water, so the ripples require movement - the ripples come into existence already moving. | {
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45,447 | Are there any stars that use any other elements as its fuel besides hydrogen? If not why not? | For nuclear fusion to occur, there must be a certain temperature and pressure. Generally for heavier elements, higher temperatures and pressures are needed, because the nuclei have a greater electric charge, and so more energy is needed to get them close enough together to fuse. In the centre of a main sequence star there is hydrogen and helium. The energy released by the fusion of hydrogen prevents further collapse and so keeps the star from reaching the temperatures and pressures needed for helium fusion. In a star like the Sun, as the amount of helium builds up, an inert core of helium forms, until (and if) it reaches the required temperature when it will start burning -- as the core has become degenerate matter at this point, the burning will increase the temperature without causing the core to expand and lower the pressure, so the core explodes. There can then be a period with shells of helium burning and hydrogen burning. By this point the star has become an "asymptotic red giant branch" star. In stars heavier than the sun, the process can continue further with carbon, oxygen and heavier elements fusing. However the extreme temperatures at which these processes occur mean that they don't last long, and the star will eventually become unstable and blow itself apart. So, in main sequence stars it is not hot enough to fuse other elements. In evolved giant stars it is hot enough, but it is so hot that the process doesn't last long. | {
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45,578 | I'm trying to teach my sons the difference between using azimuth-altitude coordinates and ecliptic coordinates. In the course of setting up some demonstrations, using Stellarium, I set my location to N 0° 0' 0.00" E 0° 0' 0.00", the time to GMT, and the date/time to noon on March 20, 2021 (March equinox). I expected that the sun would be directly overhead at this time (altitude 90°), but it's not: the altitude is about 88° 08'. I examined several days before and after, and the altitude never reaches 90°. The highest altitude I could find on March 20, 2021 was at 12:07:25, 89° 57' 31.2". What's the gap in my thinking here? Why isn't the sun at the zenith at that time? I'm using Stellarium, but my question is conceptual, not about the software (which I assume to be working correctly). | It never is exactly, directly overhead on the vernal equinox at the Greenwich meridian. You know that it happens on different days in different years so it must be happening at different times on those days. March 20 in 2021 was the day the Sun's declination changed from south to north. It happened at about 0937 UTC and its greenwich hour angle was about 322 degrees. The Sun was directly over a spot in Kenya. | {
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46,667 | I saw this question on Quizlet which said: What is the difference between the Milky Way and the Milky Way Galaxy? And the answer was: The Milky Way is a fairly narrow band of faint diffuse light around the celestial sphere. The Milky Way Galaxy is a spiral galaxy of about 100 billion stars. But isn't the term "Milky Way" means the Milky Way Galaxy? Why is there a difference? If there is a difference, could one include graphs that show the difference between the two? | Milky Way vs Milky Way Galaxy I recommend recognizing and honoring the distinction! The two words being interchangeable is a narrow view that only one well versed in Astronomy can have, and doesn't fit the reality of how ordinary people view it, being the circa 10 10 people who have seen the Milky Way but never having been formally taught about galaxies. The Milky Way Anyone can gaze at the sky on a dark night and know what the Milky Way is without any connection or reference to the concept of a galaxy or even that the Earth orbits the Sun. The Milky Way is what anybody who looks up on a dark night sees, regardless of science. It's simply that milky pattern up there. The Milky Way Galaxy The Milky Way Galaxy is what scientists deduce from observations. It is an abstraction, a model, that fits observation. Of course if I had to bet five dollars I'd say that it is real and we're in it, but that's because I'm a scientist and/or believe in science. But no matter what anybody believes, anyone who looks will see that stuff up there. What at least sighted folks can agree on is that The Milky Way is up there, and what scientists and lay people can tend to agree upon is that it is our galaxy. | {
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46,947 | I just watched one of those videos that shows the scale of the universe, zooming out from one thing all the way out to the sphere that is thought to be the universe. However, they also showed other universes outside our own, and they made it seem like a universe is right next to others in space. I have often heard that our universe is expanding, so I just want to ask: What would happen if one universe expanded enough to collide with another, and what would this look like from inside one of the universes? Source: | The Universe is space and time. If there are other universes they can't be "next to" ours, as "next to" is a statement about the relationship between things in space, and the Universe is not in space. It is space. You can't have two universes colliding at a particular time, because time is a property of the universe. Your mental image of universes as spherical bubbles floating in space is wrong. Unfortunately, there isn't a better visual image. The notion of being outside the Universe and looking at it seems to be fundamentally impossible, because "outside" is a property of space, and space only exists inside a universe. The last part of the video does show "bubbles floating in space" but this is not intended to be a representation of the multiverse (if it exists) but a metaphorical symbolisation of it. It is something that is fundamentally unknown and unknowable. It is beyond our ability to visualise in a realistic way, so it must be visualised in an unrealistic way. This is, at least, our best guess. In fact questions of "outside the universe" are unanswerable. If we could know about something outside the universe, then it wouldn't be outside, it would be part of our universe. So ultimately, Mick is right: Nobody knows. | {
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47,094 | I am using the Earth-Sun distance to teach kids the usefulness of trigonometry. This site is helpful. But one of the more engaged asked what the most accurate measurement is. After some research, I found that elapsed time measurements of reflected radar returns seems to be the way we do it now. By measuring the distance to another planet this way and combining the parallax method, we can calculate accurate distance to the sun. But is this THE most accurate way to measure the AU? What is the error margin? If the AU were changing by .0000001 percent annually, could we detect it? | The astronomical unit was defined to be exactly 149 597 870 700 meters in 2012, so it is no longer a measured value. In fact, it never was a measured value. It was instead a computed value, and it was computed rather strangely. Prior to 2012, the astronomical unit was defined as the distance between the center of the Sun at which a tiny particle in an unperturbed circular orbit about the Sun would yield a value of exactly 0.0172020985 for the Gaussian gravitational constant. That value corresponds to the distance at which a tiny particle in an unperturbed circular orbit about the Sun would have an orbital angular velocity of 0.0172020985 radians per solar day. That value corresponds to an orbital period of 365.256898 days. That value, now called a Gaussian year, was based on measurements available to Gauss of the length of a sidereal year. (The currently accepted value of the sidereal year is 365.256363004 days of 86400 seconds each.) This outdated value in the length of a sidereal year was one of the reasons the astronomical unit was given a defined value in 2012. There were other reasons. One is that that definition made the concept of the astronomical unit a bit (more than a bit?) counter-intuitive. With that definition, uncertainty in the computed value of astronomical unit depended on the ability of solar system astronomers to estimate the Sun's gravitational parameter (conceptually, the product of the Newtonian gravitational constant and the Sun's mass; in practice, a quantity that could be estimated directly as a consequence of models used to generate ephemerides) and the ability to measure time. The ability to measure time (currently about one part in $10^{16}$ ) has far outpaced the ability to estimate the Sun's gravitational parameter (currently less than one part in $10^{10}$ ). This means that if the 2012 change had not happened, the uncertainty in the astronomical unit would depend on the ability estimate the Sun's gravitational parameter, about three parts in $10^{11}$ , or about 0.000000003%. That's a lot better than the 0.0000001% precision asked about in the question. | {
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47,226 | Or a star smaller than a planet? Which star and planet would be an example of this? | The answer depends on whether you mean is any planet bigger than any star, or whether the planet and star have to be in the same system and have been discovered/measured, rather than just that they could exist in principle. There are a few known planets with measured radii that are bigger than the lowest mass stars.
Here is a plot from Chabrier et al. (2008) (and plenty more data will have been added since), which shows the basic picture. This is the mass-radius plot for both stars and exoplanets. It turns out that there are some hot Jupiters that have radii about twice that of Jupiter in our Solar System. You can find examples at exoplanets.org , such as HAT P-67b and XO-6b. These planets are bigger than theory suggests for a "cold" exoplanet, probably because of "insolation" (heating by their parent star) - e.g. Enoch et al. (2012) . On the other hand, the smallest stars, those just above the brown dwarf limit of $\sim 0.075M_\odot$ , that are predicted to have radii (at least once they are a billion years old and have reached the main sequence), of about 1.3 times that of Jupiter. At older ages they can become even smaller - about the size of Saturn (black dashed line). In terms of measurements, there are low-mass objects in eclipsing binaries and also a handful of very low-mass stars that have interferometric radii. For example Proxima Cen (the nearest star to the Sun) is reported to have an interferometric radius of $(0.145 \pm 0.011)R_\odot$ (or 1.44 Jupiter radii) by Demory et al. (2009) and so this is clearly smaller than the biggest exoplanets. If one demands that the exoplanet and star are part of the same system, then although they could exist in principle (as per the discussion above), there aren't any examples (yet). The curves in the plot above are not dependent on the type of star a planet orbits. Therefore, in principle, it might be possible for a $>1M_J$ planet to be found orbiting a (only just) smaller $<0.1 M_\odot$ star, even if it receives negligible insolation. In practice, giant exoplanets are rare around low mass stars, so it could be some time before an example is found. However, a close candidate might be GJ3512b which is an exoplanet with $M\sin i = 0.46 M_{\rm Jup}$ (i.e. this is a minimum mass, since the orbital inclination $i<90^{\circ}$ ) that orbits an M5.5V star quite similar to Proxima Cen ( Morales et al. 2019 ). The star has an estimated radius of $(0.139 \pm 0.005) R_\odot$ and the age is thought to be a few billion years. Looking at the curves in the plot then a cold exoplanet with $i \sim 30^{\circ}$ might be comparable in size to the star. Unfortunately, the exoplanet doesn't transit so no radius measurement is available and it is unlikely to be inflated by stellar insolation because it is in a relatively wide orbit around a faint star An interesting suggestion is that a young exoplanet might offer the best chance of being bigger than its host star. This is because the contraction timescale of a giant planet is longer than the pre main sequence contraction timescale of its star. The curves in the plot above for 1 Gyr and 10 Gyr show this effect, but it is even more extreme for ages $0.1$ Gyr. Thus the best chance of finding planets bigger than their host stars is to look at young systems in star forming regions. Some of these may already have been found using direct imaging, though in my opinion these quite high-mass "exoplanets" ( $>5$ Jupiter masses) orbiting at very large distances ( $>100$ au) are more like binary brown dwarfs. | {
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47,244 | Since the speed of light is the speed of causality, the "now" state of the a faraway object for an observer should be the exact point at which its light reaches the observer, for literally all intents and purposes. My question is whether purporting that something "happened a long time ago but its light is only just reaching us" and that there is a more up-to-date state of an object is a scientifically useful viewpoint, or if it serves no other purpose than to deliver that pang of weird existential nostalgia for the layperson when we say "an alien on planet XYZ would look at earth and see dinosaurs roaming around"? | You're right that astronomers don't really care what's going on in a some galaxy right now ; we care about how they evolve through time, and how its light has been altered during its journey (e.g. redshift and extinction). We don't know how a particular star or galaxy has evolved since it emitted the light we observe, except in a statistical sense, and that "statistical sense" has been obtained through observing similar objects at various distances and hence times, and building physical models that explain the observed properties. That is not to say that the time it has taken light to reach us is not a concept used by astronomers, at least in cosmology and extragalactic astronomy. It is called the lookback time , and is cosmologically linked to the distance from us, the redshift of the light reaching us, and of course also the age of the Universe at the time of emission. It is not uncommon to see plots in the literature showing some property of galaxies as a function of lookback time. Because of these relationships, however, astronomers typically use neither distance nor time, but simply use redshift when talking about some distant galaxy or phenomenon. For less distant phenomena, such as stars inside our own galaxy, which are not cosmologically redshifted, we use parsec (or kiloparsec) which is ~3.26 times a lightyear. That said, I think there is in fact a virtue in "pangs of weird existential nostalgia" when trying to enthrall laypeople. Firstly, I think it's easier to relate to the difference between "when humans started cultivating the land" and "when dinosaurs roamed around" than to relate to "12 thousand lightyears" and "a hundred million lightyears", which are both ridiculously far away. Secondly, I think teaching/reminding people that we always look back in time is a fascinating feature of physics. Furthermore, there are some specific experiments involving short distances where the exact delay between emission and observation is important. Examples include predicting the eclipses of Jupiter's moons, as commented by @antlersoft, and communicating with rovers on Mars, as described in @Nuclear Hoagie's answer . | {
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47,335 | Inspired by this question , I'm curious: Which planet in our solar system, when viewed from which other planet, would appear the largest/brightest? You probably can't really see much of Uranus from Neptune as that question asked, but could you clearly see any other planet from another planet? (Assuming ideal conditions of both planets being at their closest possible distance from each other, and maybe just from an orbiting satellite around the viewing planet so as to avoid any atmospheric effects.) And would it appear larger or brighter than just the stars around it? From Earth, all of the planets we can see are near indistinguishable from stars to the untrained eye without a telescope. But if you were closer to them, are there any planets close enough to each other and/or large enough that one could be seen as obviously a planet from the other? I'd bet on Venus from Mercury based on minimal distance, but Jupiter from Mars would be a good contender given how large it is. | Here is a table with planets, their minimum/maximum distances from the Sun, their minimum distance (assuming the planet's aphelion and its neighbour's perihelion coincide, which they don't, so the actual minimum distance is a bit higher), the size of the largest of the two (indicated with a *), and finally the maximum apparent size $s$ , which is given by $s = 2 \tan^{-1}\frac{\text{radius}}{\text{minimum distance}}$ . All data was taken from Wikipedia. Planet Aphelion (AU) Planet Perihelion (AU) Min. distance (AU) Radius (km) Max. size (") Mercury 0.467 Venus* 0.718 0.251 6,052 66 Venus 0.728 Earth* 0.983 0.255 6,371 69 Earth* 1.017 Mars 1.382 0.365 6,371 48 Mars 1.666 Jupiter* 4.950 3.284 69,911 59 Jupiter* 5.459 Saturn 9.041 3.582 69,911 54 Saturn* 10.124 Uranus 18.286 8.162 58,232 20 Uranus* 20.097 Neptune 29.81 9.713 25,362 7 So the Earth as seen from Venus would be the largest, but as Earth has only half the albedo of Venus, Venus as seen from Mercury will probably be brightest (Venus is largest for us if it's in conjunction with the Sun, so 0% illuminated, but for Mercury it happens if it's 100% illuminated). Jupiter from Mars (or Saturn) is definitely smaller. I've seen other users being able to calculate exact (maximum) brightness and gladly defer to them for that information. | {
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47,530 | It's a classic movie cliche. We see the surface of the Moon with the Earth in the distance and a dark shadow of an invading alien spaceship slowly covering the landscape. We see this in the opening sequence of the 1996 movie Independence Day, and I can think of a few other movies which have done this effect. What's really important is that the spaceship moves into visible frame as it passes overhead. So it's not a shadow from another orbital position, and so that also means the Sun would be directly overhead of the observer. I've thought about where the Moon is in its orbit, where the Earth is in its orbit and where the Sun would be. Is it possible for this to actually happen? If you were standing on the Moon, is there a time where the Earth is directly in front of you and the Sun is directly overhead. So that if something passed overtop of you the shadow would be directly on top of you? | The Moon is always almost exactly the same distance from the Sun as the Earth is (+/- 0.25%) so shadows on the Moon work the same way they work on Earth. What scientists have learned watching cartoons We know from cartoons that if a piano is dropped at noon, directly above us from a very high altitude, we have no indication of this (except for the faint, descending falling piano sound that the audience recognizes) until the piano gets to only perhaps a thousand feet above us. Then a small shadow appears on the ground around our feet, slowly increasing in size. (It's amazing how long it takes to fall a few hundred feet at terminal velocity in cartoons!) If we were to look up, we'd see the small black outline of the piano against the disk of the Sun, growing larger, and larger. As the piano approaches, the shadow gets bigger and darker and more piano-shaped, until the final, gratifying appearance of the smashed piano on the ground, from which the coyote slowly emerges with a big bump on their head. We also know from cartoons that an airplane flying even higher than the piano drop point doesn't make a shadow at all. How this works in the real world. The Sun is about 0.5 degrees wide. If you draw a triangle from the center to the edge of the Sun with us at the vertex, that angle is 0.25 degrees. The ratio of the short to long side is roughly 0.005 and long/short is 200. So an eight foot grand piano starts making a noticeable shadow at 8 x 200 = one thousand feet. The central area of a jumbo jet where the widest part of the wings meet the body is a "blob" about 70 feet in diameter (the fuselage and wings alone are skinnier extensions from that blob) so we can expect it to start making a noticeable shadow at about 70 x 200 at 14,000 feet! Yes, if a jumbo jet flew directly between you and the Sun at this altitude you'd see a fuzzy black shadow fly right over you at about 150 meters per second; it would be about 1/5 of a second blink. Okay, and on the Moon? Is it possible for this to actually happen? If you were standing on the Moon, is there a time when the Earth is directly in front of you and the Sun is directly overhead? So that if something passed overtop of you the shadow would be directly on top of you? With no air nor coyotes to fall into their own falling piano traps, we'll have to assume it's a space ship passing over you on the Moon. Below is a typical space ship we might see in space. The diameter of the saucer section is 127 meters. Using our same triangle we can then calculate that if it passes between us and the Sun closer to us than about 25 kilometers, we'll notice a substantial shadow pass over us. But don't blink! If it is orbiting at at 25 km altitude in low lunar orbit it will be moving at about 1700 m/s and that 173 meter diameter dish will pass by in about 1/10 of a second. Source: quora Just for fun (no piano) | {
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47,545 | I wrote an astronomy olympiad yesterday and there was one task I couldn't figure out an answer to: Estimate at what latitudes it is possible to observe the Moon for at least 24 hours. Under what conditions is it possible to observe the phenomenon at border latitudes? Is it possible at all to observe the Moon for such a long time? | Near the Poles is where unusual things occur. Like, in parts of Greenland, which is pretty close to the north pole, the sun never sets during summers, and the sun never rises during winter. The same way, during some parts of summers and winters over there, the full moon never sets and new moon never rises. But it's not only during winter solstice and summer solstice that you can see the moon all day long. In fact, every month, on some days, you can see it all day long. Also, moon's orbital inclination is around only $5^\circ$ . Due to this low orbital inclination, it can be observed 24 hours long only in some places on the earth. But if its orbital inclination was greater, it could have been observed 24 hours long all around the globe. Verifying with Timeanddate's night sky map You can also verify if this is right or not by using Timeanddate's interactive night sky map. The link is here: https://www.timeanddate.com/astronomy/night/ When you go there, it will first show the map for your location. You can type the city you want to find the map of in the text box on the right. For example, I typed in Nuuk, which is Greenland's capital. Image Credit: Night sky map / Timeandate Then, I went to the night sky map of Nuuk and chose the date as 5th December 2021 (new moon day). As you can see in the image below, on the top left, it doesn't show the rise and set time. This is because it's almost winter solstice and it's a new moon day. Therefore, it doesn't rise at all. Night sky map of 5th December 2021, Nuuk, Greenland. Image Credit: Night sky map / Timeandate I then changed the date to 19th December 2021 (full moon day). As you can see in the image below, it doesn't show rise and set time because it's up all day! It shows the best time to see it too. This day, you could observe the moon all day long. Night sky map of 19th December 2021, Nuuk, Greenland. Image Credit: Night sky map / Timeandate Also, each month, there are days when the moon is up all day long, there are days when it rises and sets, and there are days when it's below the horizon the whole day. To conclude, around the latitudes $60^\circ$ to $90^\circ$ , few days every month, the moon can be observed 24 hours long . | {
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47,550 | Assuming Ton 618, the largest ultra massive black hole wasn’t the result of feasting on near by matter but was hypothetically once a single star body. How large in terms of size would TON 618 have been if it was a single star? TON 618 is about 11 times larger then the solar system so how much bigger would it have been before collapsing? Again, hypothetically and how long would it have lived for before collapsing into the ultra massive black hole? I’m making a game, but I would like to get that figure somewhat right for the shock value to the player to read the ungodly size of such a star! | Near the Poles is where unusual things occur. Like, in parts of Greenland, which is pretty close to the north pole, the sun never sets during summers, and the sun never rises during winter. The same way, during some parts of summers and winters over there, the full moon never sets and new moon never rises. But it's not only during winter solstice and summer solstice that you can see the moon all day long. In fact, every month, on some days, you can see it all day long. Also, moon's orbital inclination is around only $5^\circ$ . Due to this low orbital inclination, it can be observed 24 hours long only in some places on the earth. But if its orbital inclination was greater, it could have been observed 24 hours long all around the globe. Verifying with Timeanddate's night sky map You can also verify if this is right or not by using Timeanddate's interactive night sky map. The link is here: https://www.timeanddate.com/astronomy/night/ When you go there, it will first show the map for your location. You can type the city you want to find the map of in the text box on the right. For example, I typed in Nuuk, which is Greenland's capital. Image Credit: Night sky map / Timeandate Then, I went to the night sky map of Nuuk and chose the date as 5th December 2021 (new moon day). As you can see in the image below, on the top left, it doesn't show the rise and set time. This is because it's almost winter solstice and it's a new moon day. Therefore, it doesn't rise at all. Night sky map of 5th December 2021, Nuuk, Greenland. Image Credit: Night sky map / Timeandate I then changed the date to 19th December 2021 (full moon day). As you can see in the image below, it doesn't show rise and set time because it's up all day! It shows the best time to see it too. This day, you could observe the moon all day long. Night sky map of 19th December 2021, Nuuk, Greenland. Image Credit: Night sky map / Timeandate Also, each month, there are days when the moon is up all day long, there are days when it rises and sets, and there are days when it's below the horizon the whole day. To conclude, around the latitudes $60^\circ$ to $90^\circ$ , few days every month, the moon can be observed 24 hours long . | {
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47,661 | According to its Wikipedia page : Rotation period: synchronous Eccentricity: 0.0041 But also ...extreme geologic activity is the result of tidal heating... How is it possible? It should not be heated, if its rotation is synchronous, no tidal waves should exist in it. | How can Io be tidally heated while it is in tidal lock? It is tidally locked in a mean motion sense of "tidally locked". That Io is in an eccentric orbit rather than a circular orbit means that tidal stresses can and do build up. Lainey et al. claim that the global energy dissipation in a tidally-stressed moon is given by $$\dot E = -\frac{21}2 \frac{k_2}Q \frac{n^5R^5}G e^2$$ where $\dot E$ is the rate at which tidal energy dissipates, $k_2$ is the moon's second order tidal Love number, $Q$ is the moon's tidal quality factor, $n$ is the moon's mean motion, $R$ is the moon's radius, $G$ is the universal gravitational constant, and $e$ is the eccentricity of the moon's orbit. The ratio $k_2/Q$ strongly depends on the makeup of the moon's interior. Compared to a moon with a solid interior, a moon with a partially molten interior will have a slightly higher value of $k_2$ and a significantly lower value of $Q$ . Io's volcanism is a sign of a moon with at least a partially molten interior. The power of five on the mean motion and moon radius means that a large moon that orbits close to its parent planet will be subject to vastly more tidal stress than a small moon that orbits far from the parent planet. Io is a large moon (larger than our Moon) and it orbits fairly close to Jupiter. Finally, even though Io's eccentricity is small, it is not zero. The fact of $e^2$ means that the tidal energy dissipation strongly depends on eccentricity. Those tidal stresses normally would act to circularize Io's orbit about Jupiter, thereby reducing the tidal stresses. However, Io is also in a 1:2:4 orbital resonance with Europa and Ganymede. These interactions tend to increase Io's eccentricity. This has been hypothesized to lead to an interesting hysteresis loop (e.g., Yoder ). Suppose Io's interior is cool and its eccentricity is very low. This makes tidal stresses very low. This reduces the impact of Jupiter's circularization effects on Io's orbit. The resonance effects now begin to dominate, making Io's orbit become more eccentric. Tidal stresses now become significant and Io's interior warms up. At some point, the tidal stresses that lead to circularization dominate over the effects of Europa and Ganymede. Io's orbit circularizes and Io's interior cools. Rinse and repeat. References: Lainey, et al. "Strong tidal dissipation in Io and Jupiter from astrometric observations," Nature 459.7249 (2009): 957-959. Yoder, Charles F. "How tidal heating in Io drives the Galilean orbital resonance locks." Nature 279.5716 (1979): 767-770. | {
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47,728 | If I'm directly in between the Earth and Moon, what distance from the Earth would I have to be so that the Earth and Moon have the same apparent size? How big would the moon appear compared to it's normal size? | When viewing a sphere of radius $r$ at a distance $d$ from the centre of the sphere, you don't see a circle of radius $r$ . The extreme lines of sight are tangents to the sphere, as this diagram illustrates. A tangent to a sphere or circle makes a right angle to the radius at the point of tangency, so we have 4 similar right triangles. Let the radii of the circles be $r_1, r_2$ and the respective distances from their centres to the origin be $d_1, d_2$ , so the (centre to centre) distance between the two circles is $d=d_1+d_2$ . Then $$d_1 = \frac{d\cdot r_1}{r_1+r_2}$$ $$d_2 = \frac{d\cdot r_2}{r_1+r_2}$$ and the angular diameter $\theta$ is given by $$\sin\left(\frac\theta2\right) = \frac{r_1}{d_1} = \frac{r_2}{d_2} = \frac{r_1+r_2}{d_1+d_2}$$ Using values from Wikipedia for the radii of the Earth and Moon, and their mean distance, we get Body Radius Distance Moon 1737.4 82365.8 Earth 6371.0 302033.2 Sum 8108.4 384399.0 with the angular diameter $\theta\approx 2.41734°\approx2°25'$ or $145$ arc-minutes, which is almost $4.6$ times larger than the Moon's mean angular diameter as seen from Earth's surface, which is $31.7'$ , although it ranges from $29.3'$ to $34.1'$ . The Earth is approximately an ellipsoid, and there are several ways to define the radius of the Earth . In this answer, I'm using the arithmetic mean radius. Another option that makes sense in this context (but not mentioned on that page) is to use a geometric mean radius $\approx6367$ km. A circle of that radius has the same area as a cross-section of the the Earth, in a plane containing the poles, perpendicular to the equator. There's another viewing point behind the Moon, where the Moon just eclipses the Earth. Once again, we get similar right triangles. Let $s$ be the distance from the viewing point to the Moon's centre, and once again $d$ is the distance from the Earth to the Moon. We have $$\sin\left(\frac\theta2\right) = \frac{r_1}s = \frac{r_2}{d+s}$$ So $$r_1\cdot d + r_1\cdot s = r_2\cdot s$$ Hence $$s = \frac{d\cdot r_1}{r_2-r_1}$$ Plugging in the values for the Earth and Moon, we get $s=144133.0$ km, and $\theta\approx1.38134°\approx1°23'$ Those two viewpoints are on the diameter of a circle (actually a sphere) of radius $$q=\frac{d_1\cdot d_2}{d_2-d_1}$$ I'll add a derivation for that below. At all points on that circle the Moon and Earth have equal angular size. As in the previous diagrams, the Moon & Earth circles are blue. The radii are approximately in their correct ratio in this diagram, but the distance between them is (of course) radically reduced, which magnifies the angular size. The large purple circle is the circle of radius $q$ , the small pale purple circle makes it a bit easier to see that the angles are equal. Let $M=(-d_1, 0)$ be the centre of the Moon and $E=(d_2, 0)$ be the centre of the Earth, as in the top diagram. We want to find points $P=(x,y)$ such that $$\sin(\theta/2)=r_1/PM=r_2/PE$$ That is, $$d_2^2((x+d_1)^2+y^2)=d_1^2((x-d_2)^2+y^2)$$ $$d_2^2(x^2+2d_1x+d_1^2+y^2)=d_1^2(x^2-2d_2x+d_2^2+y^2)$$ $$d_2^2x^2+2d_1d_2^2x+d_1^2d_2^2+d_2^2y^2=d_1^2x^2-2d_1^2d_2x+d_1^2d_2^2+d_1^2y^2$$ $$(d_2^2-d_1^2)x^2+2d_1d_2(d_1+d_2)x+(d_2^2-d_1^2)y^2=0$$ $$x^2+2\left(\frac{d_1d_2}{d_2-d_1}\right)x+y^2=0$$ $$\left(x+\frac{d_1d_2}{d_2-d_1}\right)^2+y^2=\left(\frac{d_1d_2}{d_2-d_1}\right)^2$$ Let $$q=\frac{d_1d_2}{d_2-d_1}$$ Thus $$(x+q)^2+y^2=q^2$$ which is a circle centred at $(-q, 0)$ with radius $q$ . Note that $$\frac1q = \frac1{d_1}-\frac1{d_2}$$ If $d_1=d_2$ then $q$ goes to infinity, and the circle degenerates to the vertical line $x=0$ , i.e., the Y axis. Using the previous values of $d_1$ & $d_2$ , $q\approx113249.4$ km. I should mention that these calculations assume the the Earth and Moon are perfect spheres, separated by a constant distance. In reality, none of those things are true, so the true angular sizes of the Earth and Moon are a little different to what I've calculated above. As ProfRob says, the Earth isn't a perfect sphere. It's slightly flattened at the poles, with a flattening factor of $f\approx 1/298.25642$ . The Moon is also flattened, but much less than the Earth ( $f\approx 1/830$ ), due to its much slower rate of axial rotation. Also, the orbit of the Moon & Earth about their barycentre is moderately eccentric, with a mean value of $\varepsilon\approx 0.0549$ , and the eccentricity changes depending on the distance to the Sun. Here's a daily plot of the Earth-Moon distance for 2020, produced using Horizons. | {
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47,769 | Can you explain me in simple words why the satellite in this telescope image appears as a streak? The exposure time is 1 second. | Satellites are moving. They are in orbit around the Earth. Satellites in low Earth orbit are moving at about 7000 m/s relative to the ground. You can work out the orbital speed by $$v=\sqrt{\frac{GM}{r}}$$ where $G = 6.673 × 10^{-11}$ and $M=5.97×10^{24}$ and $r$ is the distance from the Earth's centre = altitude + 6370000 metres. (These values are in SI units, so it will give the orbital speed in m/s. To find the speed relative to the observer, you'd need to take into account the motion of the observer due to the rotation of the earth. In the one second that the photograph is being exposed, the satellite moves. Because it moves, it appears as a streak. | {
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48,155 | The Moon doesn't have any significant atmosphere (surface pressure is $3\times10^{-15}$ bar). Can the absence of atmosphere on the Moon be determined with ground-based observations? When was it first determined that the Moon doesn't have an atmosphere, and with what data? | 19th century observers determined that the Moon is (almost totally) airless using 19th century instruments and observing techniques. Roger Joseph Boscovich (1711-1787) is credited with having discovered that the Moon has no atmosphere in 1753. Wilhelm Beer and Johann Heinrich Mädler established the same conclusion in the 1830s : They also produced the first exact map of the Moon, Mappa Selenographica , published in four volumes in 1834–1836. In 1837 a description of the Moon ( Der Mond ) was published. Both were the best descriptions of the Moon for many decades, not superseded until the map of Johann Friedrich Julius Schmidt in the 1870s. Beer and Mädler drew the firm conclusion that the features on the Moon do not change, and there is no atmosphere or water. The Guinness Book of Astronomy Facts & Feats by Patrick Moore (2nd edition, 1983) discusses solar prominences on page 17: Prominences were first described in detail by the Swedish observerVassenius at the total eclipse of 1733, though he believed that they belongd to the Moon rather than the Sun. (They may have been seen earlier - by Stannyan in 1706, from Berne.) It was only after the eclipse of 1842 that astronomers became certain that they were solar rather than lunar. Page 20 discusses the solar corona, visible during eclipses, which was long suspected to be the atmosphere of the Moon. After observing the eclipse of 16 june 1806 from Kinderhook, New York, the Spanish astronomer Don Jose Joaquin de Ferrer pointed out that if the corona were due to a lunar atmopshere, then the height of this atmosphere wuld have to be 50 times greater than that of the Earth, which was clearly unreasonable. However, it was only after careful studies of the eclipses of 1842 and 1851 that the corana and the prominences were were show unmistakably to belong to the Sun rather than the Moon. Exploration of the Universe Brief Edition , George Abell, 1964, 1969, discusses the lack of lunar atmosphere on page 184: Telescopic observations of the moon, as ell as observations from lunar probes, confirm its expected lack of an appreciable atmopshere. On the earth the air scatters the sunlight around into a certain portion of its night side, producing a twilight zone. On the moon there is no evidence such a twilight zone. Furthermore, when the moon occults (passes in front of) a star, the star's light is observed to blink out suddenly, rather than to do dim gradually as it would if had to shine thorugh an atmosphere around the moon. Triton, the large moon of Neptune, has a very thin atmosphere : Nitrogen is the main gas in Triton's atmosphere. The two other known components are methane and carbon monoxide, whose abundances are a few hundredths of a percent of that of the nitrogen. Triton's atmosphere is well structured and global. The atmosphere extends up to 800 kilometers above the surface, where the exobase is located, and had a surface pressure of about 14 microbars as of 1989. This is only 1/70,000th of the surface pressure on Earth. Voyager 2 photographed atmospheric haze over the horizon of Triton. It also photographed streaks of dark atmospheric material being blown by winds in the atmosphere of Triton . Presumably if the lunar atmosphere was as dense as even the ultra thin atmosphere of Triton it would have visible effects. It is easy and simple to calculate and predict the Moon's orbit roughly. But the precise orbital motions of the Moon is much more complicated to calculate and predict. Those complex calculations are known as lunar theory. Peter Andreas Hansen (1795-1874) was a famous astronomer. Hansen worked out a new version of the lunar theory in the late 1850s which predicted all the lunar movements perfectly - for a few years. And one of the features of Hansen's theory was that the Moon was sort of egg shaped (it is, but not that much) and the near side of the moon is sort of like a giant mountain projecting toward the Earth, and thus father from the center of the moon than the far side is. That would mean that the near side would be sticking out above the lunar atmosphere and all the lunar atmosphere would be on the far side of the Moon, and hidden from Earth by the bulk of the Moon. Simon Newcomb (1835-1909) soon refuted that theory of the shape of the Moon . This was mentioned by The Cornhill Magazine in 1877 on page 724] 8 . H.P. Lovecraft (1890-1937) once wrote that when he was 12 (c. 1902) he wrote a story where there was air on the far side of the Moon, even though he knew that Hansen's theory was false. And Lovecraft was probably far from the last person to write a story where there was air and water and life on the far side of the Moon. For example, the Lunar Trilogy by Jerzy Zulawsky (1901-1911) features a far side of the Moon with atmosphere. https://en.wikipedia.org/wiki/The_Lunar_Trilogy | {
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48,295 | I always wonder which planet came first in our solar system? Sources suggest that Jupiter might be the first planet in our solar system, but how did our solar system evolve into an 8 planet system? | The planets (probably) formed by the accumulation of planetesimals , and they all formed over roughly the same time span. Inner protoplanets had access to more matter, so they probably were able to accumulate that matter faster. On the other hand, a lot of the matter in the Solar System is volatile, so it's more likely to condense in the colder parts of the protoplanetary disk. Also, gravity favours a "greedy" formation mechanism. A more massive protoplanet will tend to accumulate matter faster than a smaller protoplanet in the same region, due to its stronger gravity. Most of the matter in the Solar System outside the Sun is in Jupiter, so it's reasonable to assume that Jupiter was an early winner in the matter gathering contest. There are good reasons to believe that Jupiter actually formed closer to the proto-Sun, migrated even closer, and then migrated outwards when it had accumulated a large fraction of its mass. From Wikipedia , the grand tack hypothesis proposes that Jupiter formed at 3.5 AU, then migrated inward to 1.5 AU, before reversing course due to capturing Saturn in an orbital resonance, eventually halting near its current orbit at 5.2 AU. It's hard to know the details of the formation of our planetary system. Fortunately, we now have a lot of data about other planetary systems, including some young systems that are still in the early stages of formation, and that data helps us refine our general theories of planetary system formation, as well as theories about the formation of our system. However, a lot of that data is fairly crude, and it's biased because it's a lot easier to detect large exoplanets close to their star than small distant ones. We can use computers to test theories of Solar System formation. You set up a mathematical model with a bunch of planetesimals, dust, and gas orbiting the proto-Sun, and then crunch the numbers to see if it evolves into something resembling the real Solar System. If it does, your model might be correct. If it doesn't, you adjust the parameters and try again. This process consumes a lot of computing power, since you need to make accurate gravitational calculations involving hundreds and thousands of objects. So the people doing this research generally run their models many times, looking for solid trends, and to help them weed out effects that are due to errors caused by the modeling calculations. These studies have found that it's quite likely that a lot of planetesimals were ejected into the outer Solar System, or even out of the Solar System's gravitational well and into interstellar space. And of course a lot got thrown into the Sun. One current theory of late system formation is known as the Nice model , originally published in 2005. However, as that article mentions, more recent research indicates that the Nice model needs some adjustments. A study from 2011 claims that there's a high probability that our system originally had 5 giants, but one of them got thrown out of the system. Young Solar System's Fifth Giant Planet? , by David Nesvorny. Abstract Recent studies of solar system formation suggest that the solar system's giant planets formed and migrated in the protoplanetary disk to reach resonant orbits with all planets inside 15 AU from the Sun. After the gas disk's dispersal, Uranus and Neptune were likely scattered by gas giants, and approached their current orbits while dispersing the transplanetary disk of planetesimals, whose remains survived to this time in the region known as the Kuiper belt. Here we performed N-body integrations of the scattering phase between giant planets in an attempt to determine which initial states are plausible. We found that the dynamical simulations starting with a resonant system of four giant planets have a low success rate in matching the present orbits of giant planets, and various other constraints (e.g., survival of the terrestrial planets). The dynamical evolution is typically too violent, if Jupiter and Saturn start in the 3:2 resonance, and leads to final systems with fewer than four planets. Several initial states stand out in that they show a relatively large likelihood of success in matching the constraints. Some of the statistically best results were obtained when assuming that the solar system initially had five giant planets and one ice giant, with the mass comparable to that of Uranus and Neptune, was ejected to interstellar space by Jupiter. This possibility appears to be conceivable in view of the recent discovery of a large number free-floating planets in interstellar space, which indicates that planet ejection should be common. | {
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48,316 | HD 84406 , is a star approximately 241 light-years away in the constellation of Ursa Major. HD 84406 will be the first star to be imaged by the James Webb Space Telescope in order to test the focus of the telescope. The star is a spectral type G star and has a high proper motion. Why aren't the other more popular stars like Alpha Centauri or Betelgeuse chosen for the first light instead? Which properties of HD 84406 make it beneficial to the telescope mirror testing? Spectral type? Magnitude? Distance? Does the star have any interesting stuff? Or is it just for the reason that the position of the star is the most convenient for JWST to locate during the test process? | The most important selection criterion is that the star should be available for observation for a prolonged time. Because James Webb observes in the infrared, it must hide the Sun (and Earth and the Moon) behind its sunshield, and can hence only observe some 39% of the sky at any given time (source: NASA ). Webb needs around three months for its optical alignment, so we need a star that has just entered its field of view $^\dagger$ . In addition to this, we also don't want a star in a field that is too crowded, and HD 84406 is located in a rather isolated region in the NW part of Ursa Major: Credit: IAU/S&T/Roger Sinnott/Rick Fienberg with my own annotations. The star should be bright, but probably not too bright (like Betelgeuse), since we don't want to burn MIRI off from the beginning. EDIT : After discussing with several colleagues, damaging MIRI permanently is probably not going to be an issue. Bright sources can however damage the detector temporarily, showing a "ghost image" of the star in subsequent exposures. I've done that myself on a 1.5 m Earth-based telescope, and JWST has a ~20 times larger area, and ~10 times higher resolution, meaning several 1000 times more photons per pixel. Anyway, as you see there are some important constraints, but many other stars would fit as well; there are quite a lot to choose from. And the best part: HD 84406 is a magnitude 6.9 star, so while you can't see it with your naked eye, it should be visible in a pair of regular binoculars. $\dagger$ Stars too close to the ecliptic are only available for a more limited time, so a star that is close to the poles is preferred. | {
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48,498 | I read that the black hole at the center of the galaxy has much less mass than the galaxy itself and that it is somehow held together by dark matter. So now I am wondering if it is possible that there are galaxies with a star at the center? | It won't stay in the center for long. Galaxy nuclei are full of stars. Any star passing by will exchange momentum with the central star and will perturb its position. Stars of similar mass will be able to completely eject the central star from its privileged position. It won't live this long. Massive stars (and it has to be massive, see p. 1) tend to live few million years and die, forming (in the general case) a stellar-mass black hole. On the other hand, galaxies live for bilions of years. So we end up with a very young galaxy with a small black hole in the center. More black holes will sink down, promoting the central black hole growth. Massive stars, as well as their remnant black holes, tend to sink towards the center of the clusters of stars (and the galaxy nucleus has any reason to possess the same dynamics). The object being a massive star is not of great importance, because they quickly (on the timescale of the galaxy formation) convert to black holes. In short, the modern consensus is that galaxies do have a black hole in the center. But even if we start with a galaxy that does not have one (we know that galaxy mergers sometimes eject the central black holes), it will grow its brand new central black hole soon. | {
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48,606 | My 9yo daughter is very into space at the moment and asked a question that my physics knowledge (6th form college, 20 years ago) is way too poor to answer. Her space book tells us that as stars age, they get redder. On the next page, it tells us that Hubble knew the Universe was expanding because distant galaxies were red, and thus moving away. She (and I!) would like to know (in layperson's language please ), how did Hubble know they were moving away and not just old? | The "redshift" measurements that Hubble used to determine his law are based on looking at the spectrum of distant galaxies. That is, splitting the light coming from the object in its constituent colors (e.g. by passing it through a prism). If you look at such a spectrum, you will notice that certain lines (colors) are missing. These are so-called absorption lines, and correspond to the physical properties of specific atoms in the object. The location of these lines are determined by physics and are universal; they are the same for all objects. However, if you look at the spectrum of a distant object you find that all these lines are shifted a little bit towards the red. By comparing to the absorption lines of gases on Earth, we can establish how far the lines are shifted, the so-called "redshift". This is very different to the reddening of older stars. As the stars grow older they cool down, which means that the light they emit has more red light (compare to a fire burning bright yellow when it's hot and a darker red when cooling down). This shift affects the overall distribution of the emitted light, but importantly does not affect the absorption lines, which stay in the same place (color). To show this, here's two spectra from BBC revision notes for exams taken at age 16 . Spectrum from the Sun: Spectrum from a distant galaxy: You can see the lines in the latter have shifted towards the red end. | {
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48,706 | Neutron stars are one type of remnant of a giant star's core after its collapse. Neutron stars tend to rotate at very high speed and the mismatch between its axis of rotation and magnetic pole make it a "pulsar". Over time, the rotational energy is lost and may come to a static state, no rotation. Have we observed any static neutron star(s)? | The absence of evidence of spin cannot be evidence for the absence of spin. We lose the ability to measure the spin of single neutron stars when they slow down below the pulsar "death line" at rotation periods above about 5-10 seconds. Neutron stars in binaries are often measured to be fast rotators, either as (millisecond) pulsars or by monitoring the rotation of accretion hot spots. Neutron stars in binaries can also be slowed by accretion torques and these have the slowest measured rotation rates. The vast majority of neutron stars are not pulsars and are not in close binary systems. They are cool, small and relatively dark. The problem in answering your question is that neutron stars also cool down as they spin down and the slow rotators you seek might be members of this effectively invisible population. | {
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48,714 | Assuming it had no accretion disk, could we still detect e.g. distortions of the background star field? | Yes, easily with a telescope, but not with the naked eye. It is a matter of routine to detect the 1.7 arcsecond shifts caused to stellar positions when seen near to the limb of the Sun. The removal of a photosphere while keeping the mass constant would mean that starlight could travel past the Sun with impact parameters all the way down to the minimum possible at 2.6 times the Schwarzschild radius. This produces deviation angles that can in principle be of any size - light can loop around and come back again or complete several orbits before escaping. However, the big distortions happen within a few Schwarzschild radii of the black hole. The photon ring of the Sun, where light can undergo an unstable circular orbit, would only be 15 km in diameter. This subtends an angle of $0.02$ arcseconds at the Earth, several orders of magnitude below what is resolvable with the naked eye. Edit: Having said that, it is always possible that by careful observation, one might see a strange change in position of a naked eye star if the black-hole Sun passed close enough to it whilst it travelled along the ecliptic. The lensing effect of a point mass can be characterised in terms of the Einstein radius , which for the scenario of a very distant star being viewed from the Earth via a black hole Sun, could be written as $$\theta_E = \left(\frac{4GM_\odot}{c^2 \times 1 {\rm au}}\right)^{1/2} = 2\times 10^{-4}\ \ {\rm radians}$$ When a distant star is seen beyond a lens there are in general two images, one inside the Einstein ring and one outside. The angular deviation caused by the lens is $\alpha = \theta - \beta$ , where $\theta$ is the observed angular separation of the lensing object and the star and $\beta$ is the angular separation if there were no lensing effect. The two solutions for $\theta$ are given by $$ \theta_{\pm} = \frac{\beta}{2} \pm \left(\frac{\beta^2}{4} + \theta_E^2\right)^{1/2}\ .$$ If $\beta \gg \theta_E$ then we can approximate $$ \theta_{\pm} = \beta + \frac{\theta_E^2}{\beta}\ \ \ {\rm or} \ \ \ -\frac{\theta_E^2}{\beta}$$ The first solution indicates that the first (and brightest) image will get increasingly close to the unlensed position $$ \alpha \simeq \frac{\theta_E^2}{\beta} = \frac{4\times 10^{-8}}{\beta}\ ,$$ where $\alpha$ and $\beta$ are in radians. If we argue that an angular deviation needs to be about 0.1 degree to see with the naked eye, then $\beta = 4.7$ arcseconds. i.e. The source needs to get within 4.7 arcseconds of the black hole Sun's position for the brighter primary image to be displaced by 0.1 degrees. This can be thought of a in a different way.
Let's say you could detect shifts of about 0.1 degree with the naked eye (that's about a fifth of the diameter of the full moon). For small angle deflections $$\Delta \theta \simeq 4 \frac{GM}{bc^2}\ ,$$ where $b$ is the impact parameter (roughly speaking, how close the light gets to the black hole) and $\theta$ is in radians. If we let $\theta = 0.1 \times \pi/180$ radians, then $b=3400$ km. Thus star positions would be significantly deviated if they crossed within a region of angular radius $\sim 3400/1.5\times 10^{8}\times 180/\pi = 0.0013$ degrees or 4.7 arcseconds from the black hole Sun. i.e. The same result. Thus we might expect to see major and perhaps visible deviations in the position of a star if the black hole Sun got within 5-10 arcseconds of it on the sky. It would then be an exercise to see whether any sufficiently bright stars do pass within 5-10 arcseconds of the Sun's path along the ecliptic... | {
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48,726 | What would be the highest possible magnitude $m$ of some star that could be viewed with a naked-eye? I am acquainted with this question , but mine is about the ideal conditions. In order to achieve maximum possible efficiency, I would suggest these points to be considered: Find (or better, construct) a monochromatic star with the peak wavelength at the maximum of the human perception ability (in order to avoid bolometric correction and ensure maximum efficiency). Find the most experienced astronomer possible with extraordinary viewing abilities. Place him out of the atmosphere, therefore in the middle of nowhere (just an astronomer and a star of magnitude $m$ ). Let him use the averted vision technique. Do you suggest any other points that should be considered? What is then the maximum possible magnitude $m$ ? | Yes, easily with a telescope, but not with the naked eye. It is a matter of routine to detect the 1.7 arcsecond shifts caused to stellar positions when seen near to the limb of the Sun. The removal of a photosphere while keeping the mass constant would mean that starlight could travel past the Sun with impact parameters all the way down to the minimum possible at 2.6 times the Schwarzschild radius. This produces deviation angles that can in principle be of any size - light can loop around and come back again or complete several orbits before escaping. However, the big distortions happen within a few Schwarzschild radii of the black hole. The photon ring of the Sun, where light can undergo an unstable circular orbit, would only be 15 km in diameter. This subtends an angle of $0.02$ arcseconds at the Earth, several orders of magnitude below what is resolvable with the naked eye. Edit: Having said that, it is always possible that by careful observation, one might see a strange change in position of a naked eye star if the black-hole Sun passed close enough to it whilst it travelled along the ecliptic. The lensing effect of a point mass can be characterised in terms of the Einstein radius , which for the scenario of a very distant star being viewed from the Earth via a black hole Sun, could be written as $$\theta_E = \left(\frac{4GM_\odot}{c^2 \times 1 {\rm au}}\right)^{1/2} = 2\times 10^{-4}\ \ {\rm radians}$$ When a distant star is seen beyond a lens there are in general two images, one inside the Einstein ring and one outside. The angular deviation caused by the lens is $\alpha = \theta - \beta$ , where $\theta$ is the observed angular separation of the lensing object and the star and $\beta$ is the angular separation if there were no lensing effect. The two solutions for $\theta$ are given by $$ \theta_{\pm} = \frac{\beta}{2} \pm \left(\frac{\beta^2}{4} + \theta_E^2\right)^{1/2}\ .$$ If $\beta \gg \theta_E$ then we can approximate $$ \theta_{\pm} = \beta + \frac{\theta_E^2}{\beta}\ \ \ {\rm or} \ \ \ -\frac{\theta_E^2}{\beta}$$ The first solution indicates that the first (and brightest) image will get increasingly close to the unlensed position $$ \alpha \simeq \frac{\theta_E^2}{\beta} = \frac{4\times 10^{-8}}{\beta}\ ,$$ where $\alpha$ and $\beta$ are in radians. If we argue that an angular deviation needs to be about 0.1 degree to see with the naked eye, then $\beta = 4.7$ arcseconds. i.e. The source needs to get within 4.7 arcseconds of the black hole Sun's position for the brighter primary image to be displaced by 0.1 degrees. This can be thought of a in a different way.
Let's say you could detect shifts of about 0.1 degree with the naked eye (that's about a fifth of the diameter of the full moon). For small angle deflections $$\Delta \theta \simeq 4 \frac{GM}{bc^2}\ ,$$ where $b$ is the impact parameter (roughly speaking, how close the light gets to the black hole) and $\theta$ is in radians. If we let $\theta = 0.1 \times \pi/180$ radians, then $b=3400$ km. Thus star positions would be significantly deviated if they crossed within a region of angular radius $\sim 3400/1.5\times 10^{8}\times 180/\pi = 0.0013$ degrees or 4.7 arcseconds from the black hole Sun. i.e. The same result. Thus we might expect to see major and perhaps visible deviations in the position of a star if the black hole Sun got within 5-10 arcseconds of it on the sky. It would then be an exercise to see whether any sufficiently bright stars do pass within 5-10 arcseconds of the Sun's path along the ecliptic... | {
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48,806 | This is a backyard astronomy question. My middle-school-aged son and I would like to measure the apparent angle between two easily-visible stars more precisely than measuring with our outstretched hands and fingers. Our current technique gives measurements in fingers instead of degrees, and is really inconsistent depending on how we need to twist our arms, and his arms seem to be growing longer! What are some inexpensive ways we can get a better measurement of the angle between two stars? | A tool such as Jacob's staff or a cross-staff can be used. This is essentially two pieces of wood in a cross shape, one of which can slide on the other. By Original: Fantagu Vector: Majo statt Senf - Own work based on: Jakobsstab3.jpg, CC BY-SA 4.0 By aligning the arms of the cross to the two stars you can find the angle between them. A practical variation of this is the Sky Ruler (from BBC Sky at Night magazine) . This is marked in degrees, so you don't need to do any trigonometry, and is made of card so you don't need carpentry skills. | {
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48,833 | The Moon's orbital period is generally given as 27.3 days, and the radius of Geostationary orbit as 26,200 miles. However, using Kepler's Third Law, if I raise 27.3 to the power of 2/3, and multiply the result by 26200, I get a value of a bit over 237,500 miles - about 1000 miles less than the actual value - for the distance from Earth to Moon. Am I making a silly mistake somewhere? | In addition to the issues raised in Ralf Kleberhoff's answer, you need to account for the mass of the Moon. The correct form for Kepler's third law is $$a^3 = G(M_1+M_2) \left(\frac T {2\pi}\right)^2$$ With regard to calculating the Moon's semi major axis given its orbital period and given geostationary orbit data (period of 1 sidereal day and semi major axis of 26199 miles) this becomes $$a_\text{Moon} = a_{GEO}\left(\left(\frac{1 \text{sidereal month}}{1 \text{sidereal day}}\right)^2 \left(1+\frac{M_\text{Moon}}{M_\text{Earth}}\right)\right)^{1/3}$$ An easy to remember value for the Moon/Earth mass ratio is 0.0123. This is accurate to several decimal places. (A more precise value is 0.012300037.) Do this calculation right and you'll get 239066 miles . (The published value is 239071 miles.) Error analysis Not accounting for the mass of the Moon makes your result small by about 0.41%. This is the largest source of error in your calculation. Using one day instead of one sidereal day makes your result small by about 0.18%. Using 27.3 days instead of 27.321661477 days makes your result small by about 0.08%. Using 26200 miles instead of 26199 miles makes your result too large by about 0.0038%, which is negligible. All of the non-negligible error sources make your result a bit smaller than my calculated value of 239066 miles or the published value of 239071 miles. | {
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48,836 | NASA has just released a telescope alignment evaluation image from JWST for the star HD 84406 2MASS J17554042+6551277. It looks like this: Higher resolution at Wikipedia To my untrained eye, the star looks like it would look when viewed from a mirror that has a small smudge at the center. There are multiple lines at 45 degree originating from the star (and also couple of fainter ones at left-center and bottom-center of image). Is this because of the brightness of the star (it is "only" 258 about 2000 light years away)? Or are there some image post-processing steps that have not been applied to the image? Would the star look the same, if, for example, Hubble Space Telescope photographed it? | Any telescope will have diffraction of the light due to the edges between mirror and non-mirror. This sets a fundamental limit to the telescope's resolution, given by the size of its primary mirror. Earth-based telescopes will usually not have this problem, since the atmospheric seeing will dominate, but James Webb is in space, and so is "diffraction-limited". The diffraction pattern depends on the shape of the mirror, as well as on anything that is in front of the mirror, i.e. the secondary mirror and the arms holding it (called "struts" or "spider"). In general, an edge will result in spikes perpendicular to that edge. A round mirror, such as Hubble's, results in concentric rings, but its secondary mirror is mounted on a plus-shaped spider, and hence point sources observed with Hubble will have plus-shaped spikes, as seen below: NGC 6397 globular cluster. Credit: NASA/ESA/H. Richer. James Webb's mirror is made up of 18 hexagonal segments. The edges of the outermost mirrors hence all follow three different directions, which are aligned at 60° to each other. This is the reason for the six brightest spikes. Additionally, Webb's secondary mirror is mounted on three arms. The two lower arms are actually aligned with the hexagonal pattern (on purpose, I presume), so the diffraction caused by those fall on top of four of the six spikes. But the upper boom is vertical in the images, and thus gives the fainter, horizontal spikes. You can see the alignment here on this "selfie" (created using a specialized pupil imaging lens inside of the NIRCam instrument that was designed to take images of the primary mirror segments instead of images of space): Credit: NASA. Below you see the resulting diffraction pattern from various spiders: Diffraction is seen for point sources, which will typically mean stars, because they are bright. The brighter a source is, the brighter the spikes will be. If you exposed for long enough, you'd also see a diffraction pattern from the fainter point sources. Almost all the other sources in this image are galaxies. In principle you also see diffraction from extended sources, but as John Doty comments the pattern will be convolved with the surface brightness of the source, smearing out the spikes. The bright galaxy seen ~edge-on at ~10 o'clock in the JWST image is a very distant one, with a redshift of 0.285 (source: Grant Tremblay's tweet ), placing it at a distance of almost 3.8 billion lightyears. | {
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48,855 | We know that the Moon doesn't have an atmosphere; thus the sky is seen dark. Also, from this question, the angular diameter of the Earth viewed from the Moon is a bit bigger than $2^\omicron$ at the perigee, meaning that it will block the Sun completely (the Sun has an angular diameter of $0.538^\omicron$ ). If a total lunar eclipse occurs at the perigee can the astronauts on the Moon see the Earth? Or do they just see it like other dark parts of the sky? | The Sun eclipsed by the Earth was actually recorded in 2009 by the Japanese lunar satellite Kaguya (athough this was strictly speaking not a total eclipse as such; see explanations below) Here the report (from which the photo sequence was taken) https://global.jaxa.jp/press/2009/02/20090218_kaguya_e.html Here the corresponding video in (near) real time The fact that the ring gets bigger from the top is due to the circumstance that the event happened exactly at Earth rise, so the bottom part is blocked off by the Moon's surface. To avoid any misunderstandings of what can be seen here: as mentioned in the linked referencem this was a penumbral lunar eclipse, i.e. the Sun was only partially eclipsed by the Earth as seen from the Moon. So the diamond ring effect is here not the end of a total eclipse (as we are used to from solar eclipses as seen from Earth), but it is due to the uneclipsed part of the sun just rising over the lunar horizon. In the earlier part of the video it was still blocked by the Moon's surface. Due to the fact that the Moon has no atmosphere this looks then pretty much like a total eclipse up to this point A true total eclipse of the Sun by Earth was however recorded in 1969 by the Apollo 12 astronauts on their way back home from the Moon. See this video, taken with an (apparently hand-held) 16mm movie camera | {
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48,858 | We know that each ellipsoid has 3 diameters named $2a$ , $2b$ , and $2c$ . The Earth and all planets, in general, are ellipsoids (Saturn is the best example because it's the most oblate planet in the solar system). But all we have read and heard are two types of diameters: equatorial diameter, and polar diameter. So where's the third one? I mean, the equatorial diameter itself should differ. One is 12756 km in the case of Earth. But what about the equatorial diameter that is perpendicular to the other equatorial diameter? | It is possible for a rotating body in hydrostatic equilibrium to be a triaxial ellipsoid. This solution was found by Jacobi in the mid 1800s, thus it's known as the Jacobi ellipsoid . An example in the Solar System is Haumea. The dwarf planet Haumea is believed to rotate in just under 4 hours. This rapid rotation causes the dwarf planet to be elongated in appearance. However, this requires a high rotation speed. As Anders Sandberg explains here , If a deformable self-gravitating initially spherical body rotates it will become an ellipsoid. For low rates of rotation this is an oblate spheroid with a circular cross-section, the Maclaurin case. As the rate of rotation becomes higher this state becomes unstable, and it turns into an elongated Jacobi ellipsoid. For even higher angular momentum these become unstable, and the object does break in two. For further details on this solution, please see this page by
Richard Fitzpatrick, courtesy of The University of Texas at Austin. | {
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48,876 | How to get phase-folded data (-0.5,0.5)? For phase-folded value to the interval (0,1), I use: $$ \phi = \frac{H_{JD}-T_0}{P} \mod 1$$ $\phi$ is phase H JD is time T 0 is the reference time P is the period T 0 should correspond to $\phi$ = 0 | It is possible for a rotating body in hydrostatic equilibrium to be a triaxial ellipsoid. This solution was found by Jacobi in the mid 1800s, thus it's known as the Jacobi ellipsoid . An example in the Solar System is Haumea. The dwarf planet Haumea is believed to rotate in just under 4 hours. This rapid rotation causes the dwarf planet to be elongated in appearance. However, this requires a high rotation speed. As Anders Sandberg explains here , If a deformable self-gravitating initially spherical body rotates it will become an ellipsoid. For low rates of rotation this is an oblate spheroid with a circular cross-section, the Maclaurin case. As the rate of rotation becomes higher this state becomes unstable, and it turns into an elongated Jacobi ellipsoid. For even higher angular momentum these become unstable, and the object does break in two. For further details on this solution, please see this page by
Richard Fitzpatrick, courtesy of The University of Texas at Austin. | {
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48,941 | The passage of time is relative depending on whether one is the stationary observer or the object/particle traveling at the speed of light (or close to it). I get this, kind of. But, when we talk about "it takes light X [units of time] to travel Y [units of distance]", where does time dilation come in? In the case of light's journey from the Sun to Earth, if we imagine the photons doing the actual speed-of-light travelling, is it 8 minutes for "them"? Or is it 8 minutes for us to observe their journey?
Apologies in advance if it's a stupid question. | If I interpreted this article correctly, then the answer to my question should be: 8 minutes is what we perceive, whereas for the photon the journey is instantaneous, due to the fact that it travels AT the speed of light, not close to it, at which point "the photon itself experiences none of what we know as time: it simply is emitted and then instantaneously is absorbed, experiencing the entirety of its travels through space in literally no time. Given everything that we know, a photon never ages in any way at all." | {
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48,945 | In the Darian calendar entry on Wikipedia we read (emphasis mine): The Martian year is treated as beginning near the equinox marking
spring in the northern hemisphere of the planet. Mars currently has an
axial inclination similar to that of the Earth, so the Martian seasons
are perceptible, though the greater eccentricity of Mars' orbit about
the Sun compared with that of the Earth means that their significance
is strongly amplified in the southern hemisphere and masked in the
northern hemisphere. While I see that due the high eccentricity of Mars, it makes the seasons to have far-from-equal length ( At Earth this is the case as well - though not significant), but I don't see why the seasons significance should be "strongly amplified" in the southern hemisphere. If the southern hemisphere has a longer (say) Winter, so the northern hemisphere will have a longer Summer. How does this kind of symmetry fail. Edit: I think I have found a possible solution: probably around the time of South hemisphere Summer, Mars is closer to the Sun. Though if it so, the South Summer is shorter in time. | Your solution is correct. Mars has a perihelion that is, coincidentally, quite close to the southern Hemisphere summer solstice. Perihelion is actually about one (Earth) month before the solstice. The last Perihelion was on the third of August 2020 and the last southern summer solstice was on the s econd of September . The intensity of sunlight is therefore significantly stronger during the southern hemisphere summer. This chart, from " Solar Radiation Incident on Mars and the Outer Planets:
Latitudinal, Seasonal, and Atmospheric Effects " shows that during the Northern summer, no part of the planet receives more than 400 cal/cm²/day, and mid-latitudes receive about 350. Whereas at Southern solstice, much of the Southern hemisphere receives more than 500 cal/cm²/day | {
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