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21,422 | When a space gas gets pulled together a star is formed. On the other hand, when a massive star dies, it collapses to a black hole. You would think that the initial mass of the gas would be bigger than the star that had existed for billion years and lost mass in the process. So what stopped the space gas from forming a black hole in the first place? | Essentially gas does, it just happens to form a star first. Mass is not the only factor in creating a black hole. You also need for this mass to reach a high density. In the process of doing this, a star usually forms. The energy producing processes in the star interior produce a pressure that balances the gravitation attraction. This prevents a star from reaching a critical density needed for black hole formation. As these energy producing processes run out of useable fuel the star will eventually collapse onto itself creating a black hole. So you can't just take a large volume of gas and create a black hole. Other physical processes occur. | {
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21,843 | To the extent of my understanding, Olber’s paradox states that if the universe was static and homogeneous, we should see a star at every point in the night sky and therefore the night sky should be equally as bright as day. However, since the night sky is dark and non-uniform, it can be said that the universe is not static and not homogeneous. However, if this was already known, what exactly is the “paradox”? Why isn’t it called Olber’s Observation or something else? | Olber's Paradox was created at a time before the idea of a finite universe was accepted. ( It was thought of in the 1600's ). In order to resolve Olber's Paradox, you have to introduce the idea that either the universe had a beginning or it is of finite size. (Note: the solution does not require an expanding universe). So, at the time, it was a paradox. Pretty much all astronomers considered the universe to be static and infinite. Therefore, the fact that their observations didn't fit with what they expected made it a paradox. | {
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22,609 | A possible answer for this is that, light emitted from the galaxies travelled a billion miles all the way to earth, where the hubble space telescope picked up this light through its sensors, and was able to construct an image of the galaxy but if this is true, and galaxies are billions of miles away, shouldn't the light particles emitted from the galaxies be scattered all over the place? after all they have been travelling from millions of years, and have probably collided with asteroids and other foreign objects. What were the chances that about 95% of the photons actually reached earth, giving us a very detailed image. Consider the andromeda galaxy which has a distance of 1.492 × 10^19 mi from earth. If light emitted from the galaxy travels in all directions, then how is it that we can still map out the entire galaxy, evident from the photo below? Shouldn't like half of the galaxy be missing since photons could have hit other objects, and "never have reached earth"? | There are two reasons that often — but not always — light from galaxies millions and even billions of lightyears away make it through the Universe and down to us: Particle number and particle size First, the intergalactic medium (IGM) is extremely dilute. The number density of particles out there is of the order $n\sim10^{-7}\,\mathrm{cm}^{-3}$, or roughly 26 orders of magnitude lower that the air at sea level! That means that if you consider a tube from Andromeda to the Milky Way with cross-sectional area of $1\,\mathrm{cm}^{2}$, it will contain roughly one microgram of matter (thanks to Rob Jeffries for catching a factor $10^6$ error). Second, even if a photon comes close to an atom, it will only be absorbed if its energy matches closely some transition in the atom. Since most of the atoms are ionized (and thus should be called plasma instead, but in astronomy the distinction if often not made), there are no electrons to absorb the photon. The photons are more likely to interact with the free electrons via Thomson scattering, but the Thomson cross section is immensely small $(\sim10^{-24}\,\mathrm{cm}^{2})$, so even if you consider the CMB photons — which have traveled through the Universe almost since the Big Bang — only around 5% of them have interacted with electrons on their way. In other words: The amount of transmitted light depends on two factors: 1) The amount of matter along the line of sight, and 2) that matter's ability to absorb the light. In the IGM, both are tremendously small. When the light enters the inter stellar medium (ISM) inside our galaxy, it may encounter denser clouds with atoms that are able to absorb the light. But usually (although not always) "dense" is still very dilute compared to Earth's atmosphere. Mathematical expression In general, if a beam of light traverses a region of particles, each with a cross section $\sigma$ (measured e.g. in cm$^2$), passing $N$ particles per area of the beam (measured e.g. in cm$^{-2}$), then the opacity of the medium is given by the optical depth $\tau$, defined by
$$
\tau \equiv N \, \sigma.
$$
The transmitted fraction $f$ of photons is then
$$
f = e^{-\tau}.
$$
In general $\sigma$ depends on the wavelength, and thus part of the spectrum may pass unhindered, while another part may be completely absorbed. The figure below (from here ) shows the spectrum of a quasar lying at a distance of 22 billion lightyears, i.e. $10\,000$ times farther away than Andromeda. You see that there are several thin absorption lines (caused by intervening hydrogen clouds whose densities are a factor of 10-100 higher than the IGM), but still most of the light makes it down to us. Because the light we see from this quasar was emitted so long ago, the Universe was considerably smaller at that time, and thus the density was larger. Nonetheless, only a small fraction is absorbed. The farther away the light is emitted, the longer ago it was, which means smaller Universe, and higher density, and thus the more light is absorbed. If you consider this quasar (from here ) which lies 27 billion lightyears away, you see that much more light is absorbed in part of the spectrum. Still, however, much light make it through to us. The reason that it is only the short wavelengths that are absorbed is quite interesting — but that's another story. | {
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22,651 | Does our galaxy moves through space? Or does it stay in a single location? If it does move, what causes it to move? | Does the Milky Way move through space? Yes it does. I'm very fascinated with space, although I don't have a degree or any formal education, I'm still very in love with everything about it and want to learn constantly. Good man Mike. One thing I ask myself is if our galaxy moves through space? It does. When we look at the Cosmic Microwave Background Radiation we see a "dipole anisotropy" due to the motion of the Earth relative to it: Image courtesy of William H. Kinney's Cosmology, inflation, and the physics of nothing See Wikipedia for more: "From the CMB data it is seen that the Local Group (the galaxy group that includes the Milky Way galaxy) appears to be moving at 627±22 km/s relative to the reference frame of the CMB (also called the CMB rest frame, or the frame of reference in which there is no motion through the CMB) in the direction of galactic longitude l = 276°±3°, b = 30°±3°.[82][83] This motion results in an anisotropy of the data (CMB appearing slightly warmer in the direction of movement than in the opposite direction).[84]" 627 km/s is quite fast. See this article , which says it's 1.3 million miles an hour. The speed of light is just under 300,000 km/s or 670 million miles per hour, so the Milky Way is moving through the Universe at circa 0.2% of the speed of light. Also see the CMBR physics answer by ghoppe which talks about the CMBR reference frame, which is in effect the reference frame of the universe. Or does it stay in a single location? If it does move, what causes it to move? I'm afraid I don't know why it's moving. Perhaps it's because the Universe is full of things moving in fairly random directions. Like a gas. Hopefully the question makes sense, if not I can elaborate. It certainly makes sense to me! Edit 13/09/2017 : as Dave points out in the comments, there are other motions, including the motion of the solar system around the galaxy, which is circa 514,000 mph. (See the Wikipedia Galactic Year article). And the motion of galaxies isn't neat and tidy either. | {
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23,027 | The previous generation of stars famously are the origin of all the heavier elements (up until iron?) in the solar system. So a big portion of the solar system mass actually is made up of Carbon, Silicon, Iron and the like because of that. But in the center, and only in the center, there is a star with presumably almost no heavy elements inside. How can that be? Am I wrong about the actual mass concentrations or is there really an imbalance, i.e. is the element distribution really lighter towards the center of the solar system? I would presume that the previous generation of stars just ended in a more or less uniform cloud of debris, from which the solar system formed. But if so, why aren’t there star systems where the star has a very different composition and is kind of a spluttering, dirty fusion machine (metaphorically, I mean)? | The solar system contains very little of elements heavier than Helium - less than 2% by mass. This is reflected in the chemical abundances measured in the photosphere of the Sun. i.e. The Sun does contain heavier elements. Your question is the wrong way around; it is not that the heavier elements have not sunk into the middle, it is that the vast majority of hydrogen and helium that was in the same place as the planets when they formed, did not end up as part of the planets. In fact, even this is only partially true. The mass of planetary material in the solar system is also dominated by the hydrogen and helium in the gas giants. So the conundrum is only why the smaller planets don't have a similar composition to the Sun. The answer to that is temperature and gravity. A small, hot planet just doesn't have the gravity to retain fast moving hydrogen and helium atoms, unless they are trapped in some compound (like water!). Thus, the small planets close to the Sun are depleted of light elements. | {
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23,058 | How does the ISM resist gravity? That's the only force acting on it, and all other particles seem to collect together to form stars. What makes the ISM so special among other particles? | It is not true that the particles in the interstellar medium (ISM) are only acted upon by gravity. For instance, In many cases a significant part of the ISM is ionized, in which case it interacts with magnetic field which permeates the gas and may in some cases be quite strong. In the vicinity of massive and hence luminous stars, radiation pressure may exert a strong force on the ISM. They also emit copious amounts of cosmic rays (i.e. relativistic particles) that transfer momentum to the surrounding gas. Supernova explosions create hot bubbles that expand and sweep through the ISM, resulting in shock waves and galactic outflows. In most cases, however, what may prevent a gas cloud from collapsing is simply its temperature.
Despite all the above processes, and despite gravity being the weakest force, gas clouds do sometimes collapse to form stars. The criterion for doing so is that the gas is dense enough, and that its internal pressure (or thermal energy) is weak enough. This is described by Jeans instability , which formulates the criterion for a cloud of gas to collapse through equating pressure forces, or thermal energy, to gravity. One way to express this is the Jeans mass $M_J$ ( Jeans 1902 ) which is the critical mass of a cloud where thermal energy is exactly balanced by gravitational forces: $$
M_J = \rho \left( \frac{\pi k_B T}{4 \mu m_\mathrm{u} G \rho} \right)^{3/2} \\
\propto \frac{T^{3/2}}{\rho^{1/2}}.
$$ Here, $k_B$ , $G$ , and $m_u$ are Boltzmannn's constant, the gravitational constant, and the atomic mass unit, while $T$ , $\mu$ , and $\rho$ are the temperature, the mean molecular mass, and the density of the gas. In the second line of the equation it is emphasized that $M_J$ increases with temperature, and de creases with density. In other words, if the gas is too hot, or too dilute, the total mass needed to collapse must be higher. In general, gas will not collapse to form stars if the temperature is above some $10^4\,\mathrm{K}$ . If the temperature is higher, the particles simply move too fast. Since various processes may easily heat the ISM to millions of degrees, the gas has to cool before it can collapse. On way to do this is by cooling radiation: Fast-moving atoms collide (either with each other or, more often, with electrons). Some of the kinetic energy of the atoms is spent exciting their electrons to higher levels. When the atoms de-excite, photons are emitted which can leave the system. The net result is that thermal energy is removed from the cloud, until at some point it has cooled enough to collapse. | {
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23,161 | I'm fairly certain people here will have heard about it, already, but apparently, two supernova leftovers clashed some 130 million years ago and some billion billion kilometres away ... What I haven't heard yet, however, is why we should care. I mean sure, it's an interesting phenomenon and measuring it can't have been easy. But now that we've heard it ... what changes? I'll admit it, I don't know particularly much about astronomy, but I'm curious: What's the significance of having achieved this? Why does it matter whether or not we know? | Reasons why this is important: It is the first simultaneous detection of a gravitational wave and electromagnetic signal, and the strongest GW signal yet in terms of signal to noise ( Abbott et al. 2017a ). It spectacularly corroborates the reality of the GW detection technology and analysis. The progenitor has been unambiguously located in a (relatively) nearby galaxy ( Soares-Santos et al. 2017 ), allowing a host of other telescopes to obtain detailed measurements. It shows that GWs travel at the speed of light, a further verification of Einstein's General Relativity ( Abbott et al. 2017b ). It shows that most of the very heavy elements such as gold, platinum, osmium etc. are plausibly produced by merging neutron stars and constrains the rate of such mergers in the local universe (e.g. Chornock et al. 2017 ; Tanvir et al. 2017 ). It shows that short gamma ray bursts — some of the most energetic explosions in the universe — can be caused by neutron star mergers (e.g. Savchenko et al. 2017 ; Goldstein et al. 2017 ). It is the closest detected short gamma ray burst (with a known distance). That the progenitor has also been characterised allows a closer investigation of the interesting physics underlying the ejection and jet mechanisms thought to be responsible for the gamma rays and later X-ray and radio emission (e.g. Margutti et al. 2017 ; Alexander et al. 2017 ). It provides observational constraints on how matter behaves at extremely high densities, testing our understanding of fundamental physics to its limits — for example, the details of the gravitational wave signal moments before merger are diagnostic of the interior conditions of neutron stars at densities of $\sim 10^{18}$ kg/m$^3$ ( Hinderer et al. 2010 ; Postnikov et al. 2010 ). It provides an independent way of measuring the expansion of the universe. Merging binary gravitational wave sources are known as "standard sirens", because the distance to the GW source pops straight out of the analysis and can be compared with the redshift of the identified host galaxy ( Abbott et al. 2017c ). The result agrees with measurements made using the cosmic microwave background and the distance-redshift relation calibrated by other means, verifying our estimation of distances, at least in the local universe. Finally, this event will turn out to be important because it was lucky ; in the sense that the source was detected well-inside the sensitivity horizon of LIGO ( Abbott et al. 2017a ). The detection itself, was not unexpected given the rates predicted based on studying the neutron star binary systems in our own Galaxy (e.g. Kim et al. 2015 ), but the fact that it was so close —
within the closest 5% of the sensitive survey volume where it could have been detected — is fortunate. In the end, if someone thinks none of the above is interesting or important, then nothing I can write will convince them otherwise. The vast majority of people I speak to are curious and fascinated to find out about our cosmic origins and how the universe works. | {
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23,171 | I understood that everything was made of Atoms . An atom is the smallest constituent unit of ordinary matter that has
the properties of a chemical element. Every solid, liquid, gas, and
plasma is composed of neutral or ionized atoms However with Neutron Stars ; the basic models for these objects imply that neutron stars
are composed almost entirely of neutrons, which are subatomic
particles Does this mean neutrons can exist outside of an atom, and that a Neutron star isn’t comprised of an element that is recognisable on the periodic table? | Yes neutrons can exist outside the atom (or nucleus). In free space a neutron will beta decay into a proton, and electron and an anti-neutrino on a timescale of 10 minutes. However, in the dense interiors of a neutron star, the electrons form a degenerate gas, with all possible energy levels filled up to something called the Fermi energy . Once the Fermi energy of the electrons exceeds the maximum energy of any possible beta-decay electron, then beta decay is blocked and free neutrons become stable. This is what happens inside a neutron star and you end up with mostly neutrons with a small fraction perhaps a few per cent electrons and protons. In the outer parts of the neutron star, the protons and neutrons can still arrange themselves into nuclei (but not atoms), but these nuclei are extremely neutron-rich (they would not normally exist in nature) and are only stabilised against beta decay by the process I described above. The very outer envelope may consist of completely ionised iron-peak element nuclei and there may be an ultrathin (few cm) layer of recognisable ionised hydrogen, helium and carbon (e.g. Wynn & Heinke 2009 ). Once the density reaches about $3 \times 10^{16}$ kg/m$^3$ it becomes more favourable for the neutrons and protons to organise themselves into "macro-nuclei" - long strings and sheets of nuclear material, known colloquially as nuclear pasta . At higher densities still, the pasta dissolves into a soup of mostly neutrons with about 1 per cent protons and electrons. The diagram below (from Watanabe et al. 2012 ) shows roughly how these layers are arranged. It should be stressed that this is based on theoretical modelling, with the theory becoming less certain the further into the neutron star you go. Testing these ideas involves nuclear and particle experiments, observations of pulsars, of neutron star cooling, of X-ray bursts, mass and radius estimates in binary systems, pulsar glitches, etc., etc. None of the details have been observationally confirmed beyond dispute, but the basic picture below fits what we know. In particular, the the crust and the n,p,e fluid regions are well understood in theory. The details of the nuclear pasta phases are still the subject of a lot of theoretical work, as are the details of superfluidity in the interior, and what happens in the very central regions (solid neutron core, extra hadronic phases, boson condensation, quark matter) is still theoretically difficult and observationally untested except perhaps to say that the softest equations of state have been ruled out by the existence of $2M_{\odot}$ neutron stars. | {
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24,064 | As stated in several sources, it's supposed that in every galaxy there is a black hole in the middle. My question is, why do these black holes in the middle of galaxies not suck up all the surrounding matter in the galaxy? | You shouldn't think of black holes as "sucking things in". Black holes interact with matter through gravity, just the same as any other object. Think of our Solar System. All the planets orbit around the sun because it has a lot of mass. Since the planets have some lateral motion (they're not moving directly towards or away from the sun), they circle around it. This is known as conservation of angular momentum . When talking about gravity, all that matters is the mass of the objects involved. It doesn't really matter what kind of object it is*. If you were to replace the sun with a black hole that had the same mass as our sun, the planets would continue on the exact same orbits as before. Now, the black holes at the centers of most spiral galaxies do accumulate mass. Some of these black holes have accretion disks around them. These are swirling disks of gas and dust that is slowly falling into the black hole. These gas and dust particles lose their angular momentum through interactions with gas and dust nearby and by radiating energy as heat. Some of these black holes have very large accretion disks, and can generate huge amounts of electromagnetic radiation. These are known as active galactic nuclei . So, long story short, black holes don't "suck". They just interact with things gravitationally. Stars, gas, and other matter in the galaxy has angular momentum, so it stays in orbit around the center of the galaxy. It doesn't just fall straight in. This is the same reason the Earth orbits around the Sun. *Disclaimer: When you talk about things like tidal forces, you do need to take into account the size of the objects. But for orbital mechanics, we don't need to worry about it because the distances between the objects are generally much larger than the objects themselves. | {
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24,590 | XKCD 1944 claims that there is "more gold in the sun than water in the oceans". Is this really true? | The mass of the sun is 1.989 × 10 30 kg. Abundance in the Sun of the elements gives a percentage 1 × 10 -7 % for gold * , so that leaves you with a mass of 1.989 × 10 21 kg of gold. HowStuffWorks states that there is 1.26 × 10 21 kg water on Earth, of which 98% is in the oceans, i.e. 1.235 × 10 21 kg. This would mean the XKCD statement is true: there is 1.6 times as much gold in the sun as there is water in the oceans. * They cite WolframAlpha as their source. Executing SolarAbundance "Gold" there confirms this (mass) percentage. | {
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24,642 | If I had a directional photon-emitting source and placed it inside a black hole pointing upward and out towards the visible universe, I assume the photons traveling at the speed of light would slow and reverse direction back into the center. So if I took the same source and placed it outside of the black hole pointing inward towards the black hole center, can I assume that an emitted photon would travel towards the center faster than the speed of light it is already traveling at? | It doesn't work like that. An observer at the light source (and indeed any observers anywhere else) will always see light travelling (in vacuum) at the speed of light locally. There is also a major problem with your thought experiment. It is not possible for you to have a stationary light source within the event horizon of a black hole. It, and everything else in its vicinity, must be moving inwards. This is as inexorable and unavoidable as is the passage of time for an observer outside the event horizon. In my opinion, the best "visual" way of thinking about the situation inside the event horizon is to imagine your photons of light like salmon trying to swimming upstream, whilst you are on a boat flowing with the stream and releasing the salmon into the water. You will always see the salmon swimming at some speed with respect to your boat. Unfortunately if the stream flows fast enough then the salmon will make no progress and you will both be swept over a waterfall (the singularity) a little further downstream. Likewise, your common-sense fails with the situation of firing light towards a black hole. Light is always measured to have a speed of $c$ locally . It is following through with the consequences of this principle that leads to all the weird behaviour that black holes exhibit. | {
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24,810 | Wikipedia says the Chicxulub impactor is thought to have been a 10-15 km diameter object. Would it have been visible to a (human * ) naked eye before impact? And if so, would it have appeared like a star that grew brighter and brighter each night? * I know, there were no humans at the time. | The answer is yes; for a few nights prior to the impact (assuming they had eyes with a similar sensitivity to our own and could look up!). It could be a bit longer than this if the body was larger than 10 km (it goes up roughly in proportion to the impactor's radius) and could be much longer if the object was a cometary body or had a very high albedo. Details: Impacting solar system objects would have relative closing speeds from around 11 to 72 km/s . We could take the optimal case that the asteroid approaches whilst fully lit by the Sun (which probably precludes the minimum and maximum speed in the range quoted above) and then scale from another similar body - say the asteroid Vesta . This has a diameter of around $a=520$ km, gets as close as $d=1.14$ au from the Earth and has a maximum brightness of about $m=5.2$ apparent magnitude (and is hence just visible to the naked eye) and an observed flux $f = f_0 10^{-0.4m}$ , where $f_0$ is a zeropoint for the magnitude scale. Thus the flux $f_n$ received by a near-Earth asteroid of diameter $a_n$ , at a distance $d_n$ from Earth (in au) and with the same reflectivity would be $$ f_n = f\left(\frac{a_n}{a}\right)^2 \left(\frac{1+d}{1+d_n}\right)^2
\left(\frac{d}{d_n}\right)^2$$ The magnitude of the dinosaur killer would then be $$m_n = m -2.5\log (f/f_n)$$ To be an at all conspicuous naked eye object, $f_n \geq f$ . If we assume the dinosaur-killer had $a_n=10$ km, then $$ d_n^2(1+d_n)^2 \leq 0.0022$$ An approximate solution is obtained by assuming $d_n \ll 1$ and thus we find $ d_n \leq 0.047$ au or 7 million km. Moving at say 30 km/s, then it gets closer by 2.6 million km per day, thus hitting the Earth about 3 days after becoming a naked eye object. Obviously this would be longer for a slower approach speed or for a larger or more reflective asteroid. But shorter for a smaller, faster asteroid or if the asteroid approached from a direction not fully illuminated by the Sun or had a smaller albedo than Vesta. An interesting trade-off to think about is if what is determined by measurement is the kinetic energy of the impactor. This would be proportional to the product of its mass and the square of its speed. The mass will be proportional to $a^3$ . Thus if we fix the kinetic energy and allow the radius to be bigger, then it becomes visible from further away (roughly proportional to $a$ ) and will also be moving slower. i.e. A more massive, but slower impactor will likely be visible for a longer period of time and vice-versa. Another possibility is that the object is of a cometary origin with an icy composition. If that were so then it could be much brighter as a result of sublimation, outgassing and having a bright cometary nucleus and tail. The answer would still be yes, but the visibility period could be weeks (comets are rather unpredictably bright). It thus seems to me that there is a plausible range of parameters and trajectories where a dinosaur-killing asteroid could be observed and then observed to grow brighter over a few nights, but probably not much longer than that unless it was a comet. | {
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25,503 | Early this morning going out on the balcony, I looked up on a star chart app to verify it was Jupiter I was seeing. Then I noticed the alignment of Mars, Saturn, and Pluto on the app. Never being able to identify it before, I stared at where Pluto should be and I'm pretty sure I saw it. My only question is - since it's said that planets shine and stars twinkle, it did seem that Pluto was flicking a bit. Is this normal? Something to do with the relatively low luminosity and greater length of space? | Pluto is something like magnitude 14. The limit for the human vision is somewhere between magnitude 6 (widely accepted) and 8-ish (highly trained observers with perfect vision in ideal conditions using special techniques - and it's a bit controversial anyway). There's zero chance that was Pluto. It was definitely a fixed star. | {
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25,835 | Because light can only travel so fast, all of the light we see in the sky was emitted at a previous moment in time. So if for example we see a supernova or some other great stellar event, by the time we see it, it maybe long over. That made me kind of curious, what is the most ancient light we can see from earth? The universe is supposedly ~13+ billion years old, but we are probably not at the very edge of the known universe so all the light we see is probably less than 13 billion years old. So what is the oldest light we can see? and as an optional follow-up question how do we know the age of that light? I guess the light itself may not actually be literally 'old', but its probably obvious what I'm asking here, put another way: what's the longest distance that now earth visible light emitted has traveled to reach the earth? Though that reformation of the question gets kind of tangled with lensing effects. | The oldest light in the universe is the cosmic microwave background . Roughly 380,000 years after the Big Bang, protons and electrons "recombined" 1 into hydrogen atoms. Before this, any photons scattered off the free electrons in the plasma filling space, and the universe was essentially opaque to light. Once recombination occurred, however, photons were able to "decouple" from the electrons and move through space unimpeded. This relic radiation is still observable today; it has been redshifted and cooled. We can detect light from very distant objects, and we have. It makes more sense to talk about distance in terms of redshift ; the larger the redshift, the farther away an object is. There are a number of extremely high-redshift objects, some of which have had their measurements confirmed, and others of which have not. Candidates include MACS0647-JD (redshift $z=10.7^{+0.6}_{-0.4}$ ), a very distant galaxy. UDFj-39546284 (redshift from $z=10$ to $z=12$ ), another galaxy or protogalaxy (although its distance is uncertain, and its exact nature is disputed). GN-z11 (redshift $z=11.09^{+0.08}_{-0.12}$ ), a very luminous distant galaxy. All of these objects would have formed some hundreds of millions of years after the Big Bang, however, so the light we see from them is much "younger" than that of the cosmic microwave background. 1 I've never liked the usage in this context, as this was the first time they combined; the "re" is kind of misleading. | {
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26,444 | What's at the Center of the Milky Way? In this article it is said that a supermassive black hole lies in the center of milky way galaxy. At its center, surrounded by 200-400 billion stars and undetectable to the human eye and by direct measurements, lies a supermassive black hole called Sagittarius A*, or Sgr A* for short.
The Milky Way has the shape of a spiral and rotates around its center, with long curling arms surrounding a slightly bulging disk. It's on one of these arms close to the center that the sun and Earth are located. Scientists estimate that the galactic center and Sgr A* are around 25,000 to 28,000 light-years away from us. The entire galaxy is around 100,000 light-years across. We revolve around the center every 250 million years.Presumbably we rotate beacuse of the BH. When the black hole dies in our galaxy will we be thrown out of the revolving orbit? The shape of the galaxy is expected to change right?It will be some irregular shape not spherical? | Presumbably we rotate beacuse of the BH. No. The galaxy is being held in one piece due to its own total gravity. The black hole is only a small fraction of that. Basically, the BH doesn't matter. When the black hole dies in our galaxy The BH will probably be the last thing left of our galaxy at the end. And even then it will take some incredibly long time for it to evaporate. BH evaporation for very large BHs is basically the slowest process you could imagine. It will be some irregular shape not spherical? The galaxy is not spherical. Its shape is rather more like a round disk (with some irregularities and some features like arms, etc). | {
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26,610 | This question is a follow-up to Do bigger telescopes equal better results? How much bigger does a ground-based mirror have to be to match what a space-based one can do? I guess I'm asking primarily for visible light, but I'm interested in general too. I guess on the ground, you're safe from micrometeorites, so it will probably last longer. At what point does it become cheaper to build a telescope on the moon or something? | It's cheaper. (1) With adaptive optics you can get 0.1 arc second resolution on the ground (admittedly only on a mountain top with particularly good air flow, but still!). This eliminates one of the major advantages of space until you get above several meters mirror diameter. (2) Rocket fairings are the shrouds which protect payloads during the supersonic atmospherics speeds reached during launch. A 5 meter fairing is about the largest that can be flown, which limits the size of the one-piece mirrors which can be launched. (The Dreaded Webb Telescope's mirror is in pieces which will assemble themselves in space -- a very scary and very expensive piece of design.) (3) Servicing a telescope on the top of Mauna Kea or in the high Chilean Andes is a difficult and expensive process. Servicing a telescope in orbit makes that look like small change. (Cost comparable to the cost of building a new giant scope on Earth.) And in-orbit servicing can't even be done with current technology except in low earth orbit. (4) While high resolution is one frontier in astronomy, going deep is another, and going deep requires big mirrors. A 30 meter mirror on Earth gathers much more light than a 5 meter mirror in space. The giant terrestrial telescopes simply do a better job of being light buckets for spectroscopy than anything we can yet put into space. The bottom line is that with the development of adaptive optics, space-based telescopes of currently buildable and launchable size lost their main advantage over ground-based telescopes. And since they're 10x to 100x the cost, they are simply not worth building for many purposes. Space based telescopes still hold a significant edge in parts of the spectrum blocked by the atmosphere such as UV and IR (Webb), and for certain tasks involving long-term high accuracy photometry (Kepler) and astrometry (Gaia). But for general purpose use, the balance seems firmly on the ground's side for large telescopes. This will change if space flight becomes cheaper -- the SpaceX BFR, for example, with its 9 meter fairing and dramatically lower launch costs, offers great hope for space telescopes. | {
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26,687 | Sorry for the absolute begginer question here, but is our sun a part of some globular cluster?
It is something related to Virgo supercluster? | No the sun is not part of a cluster. There are several types of clusters that we see in the sky. The most familiar is the "open cluster", like the Pleiades. These are a group of stars that formed together and have remained close. As the stars drift apart they can become part of a "moving group", a collection of stars that don't appear to be a cluster, but since they share the same age and direction of motion we can tell they used to be a cluster. Many nearby stars are part of the Ursa Major moving group, but the sun is not one. It just happens to be in the same part of the Milky way. The sun was probably part of a cluster shortly after it formed (4.6 billion years ago) but that cluster has long ago broken up. We don't (yet) know of any other stars that seem to have come from the same cluster. Globular clusters, like M13 and Omega Centauri, are larger and have many more stars tightly packed together. They are all rather distant, and the brightest look like slightly fuzzy stars (in fact Omega Centauri was originally thought to be a star). Of course a galaxy is a group of 100 billion stars. We don't normally think of galaxies as a star cluster, because they are so much bigger and the stars in them don't form at the same time. Then there are clusters of galaxies, the Virgo cluster is a cluster of galaxies, and the local group of galaxies is on the edge of this cluster. But the Virgo cluster is not a star cluster. | {
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26,780 | The light from the moon is light being reflected from the sun. The sun, in space, is white. But on Earth, when the light is filtered through an atmosphere, the light appears yellow. So then, why is moonlight white through the atmosphere? | The light from the moon is light being reflected from the sun. This is at least one reason you should not expect the Moon to have the same color. Sunlight hitting an e.g. blue object would appear predominantly blue and similar. So the color we might expect to see from the Moon is going to be adjusted by the color properties of the Moon's surface. As it happens lunar regolith is essentially gray in color (i.e. neutral), so it more or less reflects the spectrum it receives. The sun, in space, is white No it is not. There are a couple of reasons. The Sun is an exceptionally bright object (for us) and will saturate our color vision so that it seems like it's white when we look at it directly. The Sun is actually a yellow star . It's spectrum peaks in the (visual) blue-green range. There are stars that have a red or blue color. White is a tricky one. White is a perceived color. There is no "white" wavelength in the EM spectrum. White is a balance between the three different color sensors your human vision uses. The Sun itself is not white because the color combination "white" we're balanced for is based not on the light from the Sun in the vacuum of space, but on the light that reaches the surface of the Earth and also what is reflected from the Earth (which is part of the ambient lighting we normally experience). why is moonlight white through the atmosphere? It seemed a nice clear sky last night around my neck of the woods and the Moon looked yellow to me. This will vary with your location and atmospheric conditions, but also with your personal perception of color. | {
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26,955 | Do any of the planets have a Roche limit that is strong enough to be felt by an astronaut whilst in orbit? | Roche limit happens where the gravity of the object, trying to pull the object together, becomes smaller than the tidal force (trying to pull the object apart). But the astronaut is bound by not gravity, rather by the electromagnetic interaction between his/her atoms. The own gravity of the astronaut is negligible, compared to the electromagnetic interaction. However, the tidal force affecting an astronaut, should require a little calculation. We can derive the formula of the gravitational acceleration around a point-like body ( $F=\frac{GM}{r^2}$ ), we get $$\frac{dF}{dr}=\frac{2GM}{r^3}$$ (We can ignore the sign on obvious reasons.) Here $G$ is the gravitational constant, $M$ is the mass of the body, and $r$ is the distance. Substituting the values of the Sun, we get $\frac{2\cdot 6.67 \cdot 10^{-11} \cdot 2 \cdot 10^{30}}{(7\cdot 10^8)^3} \approx 7.78\cdot 10^{-7} \frac{\mathrm{m/s^2}}{\mathrm{m}} \approx \underline{\underline{8 \cdot 10^{-8} \frac{g}{\mathrm{m}}}}$ . More clearly, if we are orbiting the Sun just above its surface, a roughly 2m long astronaut feel that his head and foot are pulled apart by around $1.6\cdot 10^{-7}g$ weight. In the case of a $70\:\mathrm{kg}$ astronaut, it is around the weight of $0.0112$ gram on the Earth. The astronaut wouldn't feel it, but not very sensitive sensors could already measure it. This calculation sometimes used $\mathrm{g}$ for "gram", as unit of mass, and $g$ as the (non-standard) unit of acceleration. | {
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27,180 | As I see here , the Sun belongs to the Population I group of stars, which is the 3rd generation of the stars in our universe. 1st generation stars are Population III, 2nd generation are Population II, and 3rd generation are Population I. When the 1st generation (Population III) of stars died, that means most of the hydrogen was burned to helium. Stars die when there is no hydrogen left. Later, the 2nd generation of stars (Population II) appeared and they fuse another portion of hydrogen into heavier elements. If 1st and 2nd star generations burned hydrogen to helium and more heavier elements, then shouldn't like 90% of all universe hydrogen already be converted to helium and something else? If yes, then there should not be enough hydrogen to make the Sun. UPDATE 1 Thanks for all your answers. They are very useful. Now a new subquestion appeared.
When the star dies, like our Sun, it sends out external layers and core becomes white/other dwarf. In this case, new star can be formed only from the hydrogen from the external layer. The questions what is the percentage of initial star hydrogen after burning it to helium goes from this external layer to outer space? | Most of the galaxy's gas is not incorporated into stars and remains as gas and dust. This is not really my area of expertise, but papers such as Evans et al. 2008 and Matthews et al. 2018 seem to suggest that in the Giant Molecular Clouds where most stars in the Milky Way Galaxy form, the star formation efficiency is about 3-6%. So the vast majority of the gas (94-97%) is not made into stars. In very dense environments such as globular clusters, which were formed much earlier in the Milky Way's history, the star formation efficiency get as high as approx. 30%. The canonical quoted rate for "regular" spiral galaxies like the Milky Way is about 1 solar mass of new stars are made per year, which is very low summed across the whole galaxy. Stars also give off a fair amount of their outer, hydrogen rich outer layers during the later red giant phases when the stellar wind is stronger and the atmosphere expands a huge amount (radius of the Sun during the red giant phase will be about what the Earth's orbit is now). Also in the end state when the white dwarf is formed, it's only the core and inner layers that form the white dwarf. The typical white dwarf mass is about 0.6 times the mass of the Sun ( S. Kepler et al. 2006 ) and so there will be a fair amount of unfused hydrogen-rich outer atmosphere left over after the star dies. For higher mass stars, even more of the mass goes into the (ejected at high speed) envelope than goes into the remaining neutron star. These high mass stars are much rarer though; most of the Milky Way's stars are faint, cool M dwarfs. | {
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27,432 | The title basically says it all. As seen from the Earth, is it possible for the Moon to eclipse Venus (or any other planet) or are the orbits inclined such that this never happens? If such an eclipse is possible, is it a frequent or infrequent event? How would I find out when the next one occurs? | As @Donald.McLean said in comments, the answer is yes, the Moon can and does occult the other planets in the Solar System. When something apparently big (like the Moon) passes in front of something apparently small (another planet) it's called an occultation . (I say apparently because from our perspective the Moon appears larger than the planets.) The planets lie* in the ecliptic plane whereas the Moon's orbit is inclined at about 5° to the ecliptic plane, so the Moon crosses the ecliptic plane twice each orbit, so it's not as frequent as it would be if the Moon's orbit were in the ecliptic plane. Occultations of the planets by the Moon can be found here . There is also software ( Occult ) that you can use to generate your own predictions. * Although the planets are said to lie in the ecliptic plane, this is a generalisation - they actually have orbits that are within a few degrees of the ecliptic "plane". | {
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27,627 | As a general question and more in specific regarding the image above: Which part of the milky way, planets, stars, clusters are included in the image? How can I know which entities are included in different pictures? | I wholeheartedly recommend astrometry.net for this sort of thing. It Just Works(tm); running on your image produced this output with absolutely no hints or guidance from me: For avoidance of doubt, I have no association with astrometry.net. | {
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27,633 | I was wondering if any planes had mounted telescopes with the intent to observe the stars. I understand that the atmosphere itself can warp and hinder incoming light and even completely obscure views on cloudy days. Would it be possible to have a telescope which could account for the speed of the airplane and remove motion blur? Would it even be worth it, seeing as the size of the lens would also be limited? Could we see anything interesting that a terrestrial scope of the same size would miss? Hobbes has mentioned SOFIA as a great example of this being done. Does anyone have more information on the benefits of a plane mounted telescope over one on the ground for noninfrared observation? | This has been done. SOFIA is an infrared observatory built into a Boeing 747 SP: SOFIA takes advantage of the fact that some infrared bands are visible at atltitude, these are attenuated by water in the atmosphere so they're less visible on the ground. There have been infrared observatories before SOFIA: The first use of an aircraft for performing infrared observations was in 1965 when Gerard P. Kuiper used the NASA Convair 990 to study Venus. Three years later, Frank Low used the Ames Learjet for observations of Jupiter and nebulae.[20] In 1969, planning began for mounting a 910 mm (36 in) telescope on an airborne platform. The goal was to perform astronomy from the stratosphere, where there was a much lower optical depth from water-vapor-absorbed infrared radiation. This project, named the Kuiper Airborne Observatory, was dedicated on May 21, 1975. The telescope was instrumental in numerous scientific studies, including the discovery of the ring system around the planet Uranus.[21] The proposal for a larger aircraft-mounted telescope was officially presented in 1984 and called for a Boeing 747 to carry a three-meter telescope. The preliminary system concept was published in 1987 in a Red Book. It was agreed that Germany would contribute 20% of the total cost and provide the telescope. Other airborne astronomy is more incidental. Around solar eclipses, you'll often see some astronomy flights. Some of these are tourism, others perform science. These are usually passenger aircraft temporarily modified (scopes installed that look through the existing windows). These take advantage of the fact you can lengthen the eclipse by flying along its path, and you can reach eclipses in places otherwise inaccessible. | {
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27,977 | If I connect an eight foot Yagi or other comparable sized antenna to my oscilloscope and point the antenna at a bright star will I see a voltage on my oscilloscope? I am not interested in turning the voltage into an image just wondering if I would see a voltage increase when it is on a bright star. I’d like to know your thoughts before I take the time to build the antenna. I’m thinking about in the 25cm range. I’ve heard that’s an active area. My oscilloscope will read down to about 20 millivolts. | Stars are too dim for amateur radio equipment. There are two possible radio sources that you can detect: the sun and Jupiter. Jupiter is particularly interesting as interactions between Io and its magnetic field produce beams of radio waves that sweep past earth every 10 hours. These are detectable in the amateur range, at about 20 MHz. Nasa make a kit for detecting these radio signals, or it is possible to use a ham antenna , but of course it must be cut for the frequency of operation. The Nasa kit uses a phased dipole antenna which must be set up in a field or similar as the antenna is about 7m long. Stars are not very good radio sources. Supernovae remnants such as Cassiopeia A or the Crab nebula are much brighter at radio wavelengths. Most supernovae are too distant to be powerful radio sources; radio supernovae are rare . A local supernova would be a radio source but we haven't observed a supernova in the milky way for several hundred years. | {
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28,181 | I have no doubt that we have been to the moon. This question has nothing to do with a moon landing hoax. But, there are two quotes from two different astronauts regarding the size of the earth as viewed from the moon that are puzzling to me. Both quotes talk about how small the Earth looked. Shouldn’t the Earth look very large when viewed from the moon It suddenly struck me that that tiny pea, pretty and blue was the Earth. I put up my thumb, shut one eye and my thumb blotted out the planet Earth. I didn't feel like a giant. I felt very, very small. — Neil Armstong As we got further and further away it [the Earth] diminished in size. Finally it shrank to the size of a marble, the most beautiful you can imagine. That beautiful, warm living object looked so fragile, so delicate that if you touched it with a finger it would crumble and fall apart. Seeing this has to change a man. — James Irwin I know the term large is subjective, but still, the comments seem off. Please let me know what I’m missing. | The Earth is 4 times the diameter of the Moon. The Earth viewed from the Moon will therefore appear to have 4 times the angular diameter of the Moon viewed from the Earth. The Moon is easily obscured by a thumb at arm's length (by a factor of 3-4). Now bring your thumb closer (because you can't fully extend your arm in a bulky space suit) and put on the biggest pair of ski gloves you can find. It is not a great stretch of the imagination to think your thumb, in a spacesuit, would easily obscure something 4 times the size of the Moon. | {
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28,677 | Would that person go blind? Neutrino detectors and the abundance of Neutrinos would detect the upcoming visible show about 3 hours before any visible signs, so there would be time to point certain telescopes that could handle the brightness towards it. I'm curious if an individual with a telescope pointed in that direction would have an unpleasant surprise. Would the scientific community be wise to not announce the massive stellar explosion until after it's visible to avoid potential negative effects from over-eager amateur astronomers. I realize this is a kind of silly question and it might depend too much on the telescope, but I'm curious. | No, it would not be a problem. Supernovae are not at all like flashbulbs – they brighten over a period of many days and dim again even more slowly. Here are a number of different light curves taken from Wikipedia: The rise is fast on an astronomical scale – several orders of magnitude over a period of roughly ten days – but very slow on a human scale. An amateur looking at it would not notice any significant change in brightness, but if the same person came back a few hours later or the next night, the change would be very evident. As far as we can tell, the reason is that the light at peak brightness is caused by emissions from material blown off by the explosion. For example, in Type 1a SNe, most of the light is from the radioactive decay of the huge mass of ejected nickel-56 (half life 6 days). The Wikipedia article on supernovae is quite good and covers this all in more detail. | {
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28,681 | I heard about the radial speed method to discover exoplanets a few years ago, however, these questions kept confusing me. If we know a planet star of mess $m$ is orbiting a star by interpreting the radial speed of the central mass, how can we tell wether this is a even more massive object whose orbiting plane has a larger inclination? Further more, even if we know the angle of our sight and the axis of orbit, can we determine the direction of orbit? Thanks in advance. | No, it would not be a problem. Supernovae are not at all like flashbulbs – they brighten over a period of many days and dim again even more slowly. Here are a number of different light curves taken from Wikipedia: The rise is fast on an astronomical scale – several orders of magnitude over a period of roughly ten days – but very slow on a human scale. An amateur looking at it would not notice any significant change in brightness, but if the same person came back a few hours later or the next night, the change would be very evident. As far as we can tell, the reason is that the light at peak brightness is caused by emissions from material blown off by the explosion. For example, in Type 1a SNe, most of the light is from the radioactive decay of the huge mass of ejected nickel-56 (half life 6 days). The Wikipedia article on supernovae is quite good and covers this all in more detail. | {
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28,969 | I just got the news that the New Horizons space probe has passed by some remote planet on the edge of the solar system. I was surprised that the guy from NASA says that it might take 24 months from us to get the photo of that planet. The solar system is not that big, right? It is slow because the signal transmission is slow, right? But why is the transmission so slow? | New Horizons has just passed the Kuiper Belt Object (KBO) 2014 MU69 also known as Ultima Thule. KBOs form a belt of asteroids (the Kuiper Belt) from Neptune's orbit outwards and of which Pluto is the largest member of the Belt. During the encounter with Ultima Thule, all of the 7 instruments on New Horizons were gathering data (although not all at the same time) and the total data collected is expected to be about 50 gigabits of data (compared to 55 gigabits of data taken during the Pluto encounter in 2015). Since New Horizons is about another billion miles further out than Pluto was and 3 more years have elapsed, there is less power for the (tiny) transmitter and the signals are much weaker. The bit rate is about 1000 bits per second and so the 50 gigabits to transmit this will take 50e9 bits / 1000 bits per second = 50,000,000 seconds or about 579 days. Converting (roughly) to months by dividing by 365.25 and multiplying by 12 shows that it will indeed take about 19-20 months to transmit everything back. The first image at about 300 meters per pixel resolution and so about 100 pixels across the 30 km KBO, should be received on 2019 Jan 1. A second higher resolution image with about 300 pixels across the KBO is expected to be downloaded by 2019 Jan 2. There will be a press conference on 2019 Jan 2 when these images are due to be released and shown. (more details on what to expect when at Emily Lakdawalla's Planetary Society blog entry ) After the initial data download, they expect to perform some analysis to see which images have the best data with 2014 MU69 in the frame. Given the uncertainty in the position of 2014 MU69 and the high speed of the encounter, they had to shoot strips of images and not all will contain the target. These data will be prioritized in the downlink so they arrive on the ground first and can be analyzed first. As mentioned by @luis-g there is also the Solar conjunction which will cause a 5 day period (according to PI Alan Stern in the 2019 Jan 3 press briefing) when reception of the data won't be possible. We would expect this to re-occur in January 2020 but these approx. 10 days don't make a large difference to the time taken which is dominated by the weakness of the received signal after the 15W transmission travels the ~4 billion miles and falls off due to the inverse square law, the corresponding low bitrate allowed by the need to have the transmitted data decodeable and the amount of data to transfer. | {
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29,122 | Light doesn’t accelerate in a gravitational field, which things with mass would do, because light has a universally constant velocity. Why is that exception? | Another way to answer this question is to apply the Equivalence Principle, which Einstein called his "happiest thought" (so you know it has to be good). The equivalence principle says that if you are in an enclosed box in the presence of what Newton would call a gravitational field, then everything that happens in that box must be the same as if the box was not in a gravitational field, but accelerating upward instead. So when you release a ball, you can imagine the ball is accelerated downward by gravity, or you can imagine everything but the ball is accelerated upward, and the ball is simply being left behind (which, ironically, checks better with the stresses you can easily detect on every object around you that are not present on the ball, including the feeling you are receiving from your bottom right now). Given that rule, it is easy to see how light would be affected by gravity-- simply imagine shining a laser horizontally. In the "left behind" reference frame, we see what would happen-- the beam would start from a sequentially higher and higher point, and that raising effect is accelerating. So given the finite speed of light, the shape of the beam would appear to curve downward, and the beam would not strike the point on the wall of the box directly opposite the laser. Therefore, this must also be what is perceived from inside the box-- the beam does not strike the point directly across from the laser (as that point is getting higher then the point across from it where the light was emitted), and its path appears to curve downward. Ergo, light "falls." Indeed, this is the crucial simplification of the Equivalence Principle-- you never need to know what the substance is, all substances "fall the same" because it's nothing happening to the substance, it is just the consequences of being "left behind" by whatever actually does have forces on it and is actually accelerating. Incidentally, it is interesting to note that even in Newtonian gravity, massless objects would "fall the same" as those with mass, but to see it requires taking a limit. Simply drop a ball in a vacuum, then a lower mass ball, then a lower still mass. All objects fall the same under Newtonian gravity. So simply proceed to the limit of zero mass, you will not see any difference along the path of that limit. Nevertheless, Newtonian gravity doesn't get the answer quite right for the trajectory of light in gravity, because Newtonian physics doesn't treat the speed of light correctly. | {
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29,621 | As far as I understand, a moon is an object in permanent orbit around a planet, dwarf-planet, asteroid, etc. If there was another object permanently orbiting this moon, would that be a moon-moon? Is there any natural object of such kind that we know of? If no, why are such objects so unlikely? Edit: this is question is different from Do moons have moons? because I explicitly ask for a reason. Also, the terminology is included here. | No. One explanation for the equatorial ridge on Iapetus involves a now-lost subsatellite , so perhaps such objects existed in our Solar System's past. Subsatellites tend to get rapidly removed by tidal forces causing changes to the orbit. Depending on the parameters of the system, the orbit may decay until the subsatellite collides with the moon or breaks up at its Roche limit. Alternatively it may migrate outwards until the subsatellite becomes unbound: at this point it would become a moon of the planet, though likely on an unstable orbit that would lead to scattering or a collision. The orbital evolution tends to be faster for more massive subsatellites. See Kollmeier & Raymond (2018) for details, who note that any subsatellites orbiting the candidate Neptune-sized exomoon in the Kepler-1625 system would need to be smaller than Ceres to survive. | {
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29,986 | What is the official degree to one decimal point please, of the earth’s rotation in one single day. Can it be confirmed that it is exactly 360.0 degrees using official data? Thank you in advance. | First, we need to decide which definition of "day" to employ. There are several types of days: Apparent solar day : the time between two successive culminations of the Sun (apparent Noon) from an fixed Earth-based observer; Mean solar day : a more uniform, averaged solar day without seasonal variations; Stellar/Sidereal day : the time needed for the Earth to rotate once relative to the stars; SI day : a unit of time containing exactly 86,400 SI seconds defined by caesium atoms. Since we generally refer to the traditional day/night cycle when we say "day", this means a form of solar time. To get a more averaged value, let's use the mean solar day. The current formula linking the Earth Rotation Angle (ERA) to the modern approximation of mean solar time, UT1 (basically the Earth's clock following the mean day/night cycle), is by definition : $$ERA = 2π(0.7790572732640 + 1.00273781191135448 T_u) \text{ radians}$$ Where $Tu$ is the Julian UT1 Date - 2451545.0 So according to this formula, a (UT1) day is 1.00273781191135448 Earth rotations, which multiplied by 360° is about 360.98561° . However, the Earth's rotation and revolution are not constant, and are always changing at somewhat unpredictable rates, so the angle is not perfect, but the changes are very slow. So this is more a modern approximation rather than an exact value. Rounded to one decimal place, this gives you 361.0° , a figure that will likely remain true for at least several millenia. If you want to know the amount of Earth rotation for every SI day, you're in luck: it is possible to consult reports of the Earth's orientation (rotation and polar motion) thanks to the IERS. Values are tabulated for each 0h UTC every day in the IERS publications, allowing to derive the angle that Earth has rotated every 86,400 SI seconds, allowing to scientifically monitor variations in Earth's rotation compared to a very constant unit of time realized by atomic clocks. Nowadays, the Earth's rotation is measured and reported thanks to radio telescopes and VLBI observing distant objects in the universe. Official reports of its orientation, the Earth Orientation Parameters are published on the IERS Website . As an example, here is a graph showing the amount of Earth rotation every 86,400 seconds constructed with IERS data of the last year: As we can see, there are several seasonal, periodic and unpredictable variations in Earth's rotation. | {
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30,067 | Google News feed shows me the following. What does the term "Super Worm Equinox Moon" mean and has it ever been used before this 2019 clickbait instance? | All those adjectives being smooshed together signify an uncommon event. That's why you've never seen them together like that before. All 3 conditions have to hold true: It's a supermoon, which means the full moon coincides with the moons perigee or nearest approach. That can make it appear up to 30% brighter than one at apogee (farthest away). These happen about every 13 months and it doesn't have to be exact so usually, we get 2 of them in a row, like we did this year. February had a supermoon and so did March. It's a worm moon, which means it is occurring in the month of March (see @astrosnapper's answer for a better explanation of that). It's during an equinox, basically the first day of spring (or autumn). If any one of those isn't happening then it can't be called a Super Worm Equinox Moon . Apparently, the term supermoon (all one word, by the way) is a relatively recent thing. I tried to view it on Google N Gram viewer, but... It is a particularly bright full moon and it does deserve to have its own terminology, IMO. Update: The rarity of the event is certainly relative. I saw a tweet from National Geographic that we also had a super worm equinox moon 19 years ago. | {
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30,074 | When astronauts are floating about in a spaceship or space station, they nearly always move very slowly. After doing a bit of research I can't see why being in zero gravity would restrict movement to such a degree. It's almost as if there's resistance to their movement, like they're moving through water. | It's more for safety than anything else. Space is a very dangerous place for so many reasons. And making mistakes can very easily cause death. Being weightless does not mean you lose mass, so momentum is just as difficult as ever. But whereas on the ground you can easily use friction to stop, in space if you try to stop against the floor you will just move off it. You can only stop by holding something, or pressing against something close to perpendicular to your movement. As an example, imagine you jumped with all your force from one wall in the ISS. You will notice as you approach the other end that you are travelling at speed, head first, with no safe way to stop. Even reaching out to a handhold on a side wall will whip you round and into that wall, possibly injuring yourself or damaging instruments on the wall. Look at any video from the ISS to see how carefully they move. Similarly, outside the ISS, you want to do everything slowly so you don't damage your suit, miss a handhold or otherwise cause death. | {
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30,313 | At the press conference this morning, the Event Horizon Telescope team didn't say much about Sagittarius A*, which was the target many of us have been waiting for. Is there any explanation anywhere for this omission? | There was a mention of Sagittarius A* during the Q+A portion of the press conference; the team indicated that they hope to produce an image sometime in the future (although they were careful to make no promises, and they're not assuming they'll be successful). That said, I'm not wholly surprised that we ended up seeing M87, rather than Sgr A*, for a couple reasons which the team mentions in their first paper : As Glorfindel said, Sgr A*'s event horizon is much smaller, meaning matter orbiting the black hole has a shorter orbital period. This contributes to variability on the timescale of minutes. The observations of M87 took place over the course of a week - roughly the timescale over which that target varies, meaning the source should not change significantly over that time. Second - and this is the reason I've seen cited more often - Sgr A* lies in the center of our galaxy, and so thick clouds of gas and dust lie between it and us. That results in scattering, which is a problem. There are ways to mitigate this, of course, and the team has spent a long time on this, but it's simpler to just look at the black hole that doesn't have that problem in the first place. That's why M87's black hole is an attractive target. Neither of these are impossible hurdles to overcome, but they're certainly very real difficulties that can't be ignored. | {
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30,339 | There are closer galaxies than Messier 87 for sure, even ours! It sparked my curiosity that they went with one 53 million light years away. Is there a reason for this? | I was surprised too when I first heard they were trying to image M87's black hole. The short answer is because it's really, really big. It is 1500 times bigger (diameter) than our Sagittarius A*, and 2100 times farther away. This makes its apparent size about 70% of that of Sgr A*, which they are also attempting to image. A cursory search of wikipedia's List of Largest black holes shows that there's no other black holes with a combination of size and closeness greater than these two. A couple of other candidates are not too far off. Andromeda's black hole is 50x the size of ours, and at 100x the distance, it would appear half the size of Sgr A*. The Sombrero galaxy is 380 times farther way than Sgr A*, and has a black hole estimated to be 1 billion solar masses, which is 232 times Sr A*, resulting in an angular diameter about 60% of Sgr A*. There appear to be many other considerations to which black holes were chosen, as explained in this similar question. At a guess these would include how obscured each black hole is with foreground dust/stars etc, how active (and therefore bright) the nuclei are, and their inclination w.r.t earth affecting which observatories could observe them at which times. Edit: I've found another plausible candidate. NGC_1600 is 200 M light years away with a central black hole estimated to be 17 billion solar masses heavy. This would put it at about 40% the apparent diameter of Sgr A*. Comparison of the apparent size of the largest nearby black holes And of course obligatory XKCD to remind us how small these objects really appear. | {
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30,343 | In the recently released photo of a black hole shown above, which was created by using data from EHT, why is the lower region brighter than the one above? Is it because of the rotation of the accretion disk? Also what is the orientation of the accretion disk? Are we looking at it head on? | No, you aren't seeing the shape of the accretion disk. Although its plane is almost that of the picture it is far larger and fainter than the ring that is seen. The reason for this asymmetry is almost entirely due to Doppler beaming and boosting of radiation arising in matter travelling at relativistic speeds very close to the black hole. This in turn is almost entirely controlled by the orientation of the black hole spin . The black hole sweeps up material and magnetic fields almost irrespective of the orientation of any accretion disk. The pictures below from the fifth event horizon telescope paper makes things clear. The black arrow indicates the direction of black hole spin. The blue arrow indicates the initial rotation of the accretion flow. The jet of M87 is more or less East-West (projected onto the page), but the right hand side is pointing towards the Earth. It is assumed that the spin vector of the black hole is aligned (or anti-aligned) with this. The two left hand plots show agreement with the observations. What they have in common is that the black hole spin vector is mostly into the page (anti-aligned with the jet). Gas is forced to rotate in the same way and results in projected relativistic motion towards us south of the black hole and away from us north of the black hole. Doppler boosting and beaming does the rest. As the paper says: the location of the peak flux in the ring is controlled by the black hole spin: it always lies roughly 90 degrees counterclockwise from the projection of the spin vector on the sky. | {
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30,579 | As the title says, what happens when a gravitational wave approaches a black hole? I would presume that something interesting happens because of the way spacetime works near black holes but I have no knowledge to back it up. | No, gravitational waves cannot pass through a black hole. A gravitational wave follows a path through spacetime called a null geodesic. This is the same path that would be followed by a light ray travelling in the same direction, and gravitational waves are affected by black holes in the same way that light rays are. So for example gravitational waves can be refracted by gravitational lenses just as light waves are. And just like light waves, if a gravitational wave crosses the event horizon surrounding a black hole it is then doomed to travel inwards to the singularity and can never escape. There is one caveat to this. When we talk about a gravitational wave we generally mean a ripple in spacetime that is relatively small. Specifically it is small enough that the energy of the gravitational wave does not significantly affect the spacetime curvature. So when we calculate the trajectory of a gravitational wave near a black hole we take the black hole geometry as fixed, i.e. unaffected by the wave, and we compute the trajectory of the wave in this fixed background. This is exactly the same approach as we use for calculating the trajectories of light rays. Since light rays carry energy and momentum then, at least in principle, they have their own gravitational fields. But for both the light rays and gravitational waves likely to exist in the universe the energy carried is too small to make a significant contribution to the spacetime curvature. When you say in your question: I would presume that something interesting happens because of the way spacetime works near black holes I would guess you are thinking that the gravitational wave could change the geometry near a black hole, but as described above typical gravitational waves don't have enough energy to do this. It would be reasonable to ask what happens if we give the wave enough energy, but the answer turns out to be that it no longer behaves like a simple wave. Gravitational waves exist in a regime called linearised gravity where they obey a wave equation that is basically similar to the wave equation light obeys. If we increase the energy so much that gravity becomes non-linear (as if the case for black holes) then the oscillations in the spacetime curvature no longer obey a wave equation and need to be described by the full Einstein equations. For example it has been suggested, but not proven, that really high energy gravitational (or light) waves could interact with each other to form a bound state called a geon . I confess that I'm unsure how much work has been done studying oscillations in this regime. | {
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31,678 | Why is Polaris, the North Star, always above (or near) the North Pole? If Earth is tilted, Polaris' path should be in winter 23 degrees away from its path in summer, or not? | You are correct that the axis of the Earth's rotation is tilted with respect to the plane of its orbit by 23 degrees. But it is incorrect that the direction that the axis points changes by a large amount (it should be 2*23 degrees) over a 6 month time span. Your assumption : If axis it pointed at Polaris at (1), then it should be pointed at a different star at (2). The axis remains pointed in the same direction throughout the entire year because the laws of physics are that the axis of a spinning object remains pointed in the same direction unless a torque acts on the body to change its orientation. There are torques acting on the Earth (namely the Sun and Moon), but it takes 13000 years to change the direction from "one way to the opposite way", not 6 months. Correct : The orientation of the Earth's spin axis remains pointed at the same star throughout the year (ignoring the slow, 26000 year precession cycle). | {
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31,775 | I'm researching a scene for a sci-fi novel in which the near-future protagonists observe earth through a station-mounted telescope in Mars orbit. My goal is to understand how much detail they reasonably could discern. The Hubble telescope is probably a reasonable comparison for my purposes. I found images Hubble took of Mars during a close approach, but I don't know if those represent the best resolution possible or simply the resolution that was selected or available at the time, or indeed, if the whole planet was imaged at a higher resolution than the photo published in popular media outlets. Can a Hubble-like telescope observe significantly greater detail than displayed in the article below? If so, what might reasonably be resolved? Large cities? Individual buildings? From Vox.com's Hubble can see galaxies impossibly far away. Here’s what happens when it looks at Mars and Saturn. above: Cropped from Source NASA, ESA, and STScI above: Cropped from Source NASA/Hubble | Forget about magnification. People who know telescopes don't think in terms of magnification. What matters is the angular resolution, or the resolving power: the angular size of the smallest details that you could see in an instrument. Rule of thumb: the resolving power of a telescope with a diameter of 10 cm is 1 arcsecond when using visible light. The numbers are inversely proportional. A 20 cm telescope resolves details 0.5 arcsec in size. A 1 meter telescope resolves 0.1 arcsec. Hubble has an aperture (diameter) of 2.4 m, so its resolving power is 0.04 arcsec. The minimum distance between Earth and Mars is about 55 million km and it only happens very rarely. The maximum distance is 400 mil km. The "average" distance is 225 mil km (but actual distance varies all the time). Let's apply the tangent of 0.04 arcsec at 55 mil km: https://www.wolframalpha.com/input/?i=tan(0.04+arcseconds)+*+55000000 It's 10 km. It would only be able to see the major geographic features. To see buildings (down to the scale of 10 m), it would need a 1000x increase in resolution. That means an aperture of 2.4 km. None of the classic telescope designs can provide that. It would have to be some kind of interferometric design - a large, flat field where several mirrors are placed several km apart and are coupled optically to function as a single huge mirror (well, sort of - this is more of an intuitive explanation). It would be similar to the Navy Precision Optical Interferometer near Flagstaff, Arizona. Some of the wide, flat parts of Valles Marineris might provide a good location for the interferometer. Acidalia Planitia would provide even more space for building huge interferometers, and should be a good place to build structures in general - flat to beyond horizon; it's the place where much of the book/movie The Martian set their story. But any big, reasonably flat field would work. All of the above assumes the distance of closest approach between Earth and Mars. In practice, the distance is greater than that, so aperture must increase. You're contemplating an interferometer with a base of dozens of km if you want to distinguish structures such as buildings. Conceivably, the interferometer could be built in orbit, but you must ensure that the distance between mirrors is maintained with extraordinary precision. On the planetary surface, the ground provides the required rigidity. In space you'd have to... I dunno, use space magic. | {
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31,989 | I've been thinking about black holes, specifically during the final moments before two merge. I'm wondering if black holes, or I guess more specifically their event horizons, are always spherical. It seems to me that in the moments before two merge, their respective event horizons will be stretched, somewhat like how the Moon causes our ocean's tides. I have drawn a (poor) diagram of what I think they may look like. Notice how the event horizons are closer to the singularity on the inner side, this is because the gravity from each black hole is in opposition. The event horizons are further from the singularity on the outer side because the gravity from each black hole adds up. | No need to guess. There's solid research done in this field. Even Wikipedia has some info: As two black holes approach each other, a ‘duckbill’ shape protrudes
from each of the two event horizons towards the other one. This
protrusion extends longer and narrower until it meets the protrusion
from the other black hole. At this point in time the event horizon has
a very narrow X-shape at the meeting point. The protrusions are drawn
out into a thin thread. The meeting point expands to a roughly
cylindrical connection called a bridge. https://en.wikipedia.org/wiki/Binary_black_hole#Shape There are research papers with images showing the results of calculations of the shape of the event horizons during merger. Here's an example: The image above is taken from this paper: On Toroidal Horizons in Binary Black Hole Inspirals We examine the structure of the event horizon for numerical
simulations of two black holes that begin in a quasicircular orbit,
inspiral, and finally merge. We find that the spatial cross section of
the merged event horizon has spherical topology (to the limit of our
resolution), despite the expectation that generic binary black hole
mergers in the absence of symmetries should result in an event horizon
that briefly has a toroidal cross section. | {
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32,059 | I understand this is a silly hypothetical but I'm asking for a 7 year old so please bear with me. Imagine an interstellar stray gas giant comes flying through our solar system. If we were not concerned that it would also steal our atmosphere and create tidal forces that destroyed everything... How close would it need to come to us to exert enough gravity to lift people off the ground and pull them into its own orbit? | TL:DR Jupiter isn't dense enough for its gravity gradient over Earth's radius to produce a 1g tidal acceleration, even right at Jupiter's surface. thanks to PeterCordes Jupiter's gravity will pull on the Earth itself, as well as everything on it. It's not like a vacuum cleaner that selectively lifts small and light objects, the gravitational force will scale with the mass of each object; if the Earth is a zillion times more massive than we are, then Jupiter's gravitational force will also be about a zillion times larger. What that means is that Earth will accelerate towards Jupiter, and we will accelerate along with it, and so we won't "feel the tug" anywhere near as strongly as one might suspect. Instead, let's think about the size of the Earth, and the fact that people on the near side will be closer to Jupiter than the center of mass of the Earth, and people on the far side will be farther away. Since people nearer to Jupiter will feel a slightly stronger acceleration than the center of mass of the Earth, they will feel a quite gentle tug. We'll calculate that in a minute. But believe it or not, people on the far side of the Earth, feeling less of a tug than the Earth's center of mass, will believe they are being pulled in the opposite direction! They won't really be pulled away from Jupiter, but they will not accelerate towards Jupiter as fast as the Earth, and so it will feel like they are being repelled. This kind of force is called a tidal force and this is the picture that's often used with the concept: Source Replace "Satellite" with "Jupiter" The acceleration we feel due to gravity is expressed as $$a_G = \frac{GM}{r^2}$$ where $G$ is the gravitational constant and equal to about $6.674 \times 10^{-11}$ m^3/kg s^2 and M is each mass that's pulling on you. If you put in 6378137 meters and the mass of the Earth ( $5.972 \times 10^{+24}$ kg) you get the familiar 9.8 m/s^2. If Jupiter were 114,000,000 meters or 114,000 kilometers away, the Earth would accelerate at 1 g towards it, but people on the close and far side would accelerate very differently. On the close side, being 6,378 kilometers closer, would feel an acceleration 1.2 m/s^2 greater, so they would feel that they weighed 12% less. And people on the far side would also feel about the same amount lighter because they felt less acceleration than the Earth. If Jupiter were so close that it were practically touching the Earth, it still wouldn't pull is off of Earth, assuming that Earth remained intact. But that wouldn't last very long!!! Earth would be accelerating towards Jupiter at about 20.9 m/s^2, and people on the near side would feel acceleration of 24.8 towards Jupiter, but relative to Earth that's only 3.9 m/s^2, so not enough to overcome Earth's gravity of -9.8 m/s^2. On the far side of Earth it's similar; the acceleration towards Jupiter would be 17.8 m/s^2 but minus Earth's acceleration of - 20.9 it's -3.0 m/s^2 away, but that's also not enough to overcome the attraction to Earth of in this case +9.8 m/s^2. When Earth touches Jupiter, we will feel about 40% lighter on the near side and 31% lighter on the far side of Earth, but we would not leave the surface. However, in just minutes we'd be pulled so deep into Jupiter that we would be crushed by Jupiter's internal atmospheric pressure. It would certainly be fun, but it wouldn't last long! | {
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32,262 | I googled it and checked a few Q&A and there's only things about "Earth's rotation". But why can't we feel the revolution? They say we can't feel the rotation because the Earth spins at a constant speed. Okay, I get what happens for the rotation, but isn't it different when it comes to the revolution? At night we feel the sum of speed of green and blue, whereas at day we should feel the sum of speed of green and minus blue, shouldn't we? In other words, shouldn't we feel the changes in velocity by times? | Firstly the speeds are massively different (about 1000 mph (1610 kph) on the equator for Earth's rotation and 70,000 mph (112,654 kph) for the revolution), so the change is not large. Secondly, the green line is far straighter than it appears in your picture (because the orbit is so large) so Earth's motion around the Sun is pretty close to motion at constant velocity, which Einstein tells us cannot change the outcome of any experiment. | {
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32,410 | Hypothetical question based on my understanding that two event horizons that overlap (touch) can't ever separate again: Imagine a 1 billion solar mass black hole (so the event horizon is massive and very gravitationally weak) is travelling at a velocity of 0.9c through empty flat intergalactic space; now imagine an identical 1 billion solar mass black hole travelling at 0.9c but in exactly the opposite direction so the two are heading roughly towards each other. The black holes' paths, once all the space time warping is taken into account, aren't on a direct collision but the outermost edges of the event horizons will just 'clip' each other, ordinarily only overlap for a fraction of a nanosecond as these two bodies are travelling at such incredibly fast velocities and in opposite directions to each other. So firstly, am I right in thinking that if two event horizons overlap they can never 'unlap'? Secondly, what would happen to this incredible amount of momentum of each other the black holes? Would it just get instantly turned into gravitational energy? Bearing in mind when black holes normally merge, it happens very slowly as black holes slowly move closer and closer together over millions of years giving off gravitational energy as that happens, so not in a fraction of a nanosecond as in this case. And thirdly, what would this look like? Would the event horizons remain fairly spherical and the radiated energy just insane or would they stretch and warp into a kind of long thin elastic event horizon as they shoot past each other and then over time slow down and snap back to each other? | If the event horizons ever touch and become one continuous surface, their fate is sealed - the two black holes will merge all the way in. They can never separate again, no matter what. There are several possible ways to explain it, with varying degrees of rigorousness. An intuitive explanation is that escape velocity at the event horizon equals the speed of light. But nothing can move as fast as light, not even a black hole. In order for the two black holes to separate, parts of one would have to "escape" the other, or move faster than light, which is impossible. EDIT : Another intuitive "explanation" (a.k.a. lots of handwaving) - inside the event horizon, all trajectories lead to the center. There is no possible path from any place within the horizon to the outside. Whichever way you turn, you're looking at the center. Whichever way you move, you move towards the center. If the event horizons have merged, for the black holes to split up again, parts of them would have to move "away from center" (or away from one of the centers), which is not possible. All of the above is about as "rigorous" as "explaining" general relativity with steel balls on a rubber sheet. It's just metaphor. More rigorously, see this paper by Stephen Hawking: Black holes in general relativity As time increases, black holes may merge together and new black holes
may be created by further bodies collapsing but a black hole can never
bifurcate. (page 156) EDIT : Event horizons don't really "just clip each other". Perfectly spherical event horizons are a theoretical abstraction (a non-rotating black hole in an otherwise empty universe). In reality, anything near a BH will deform the event horizon, which will "reach out" towards that mass. If it's a small mass, the effect is negligible. But if two black holes get close to each other, the EHs become egg-shaped, as if trying to touch each other. If they're close enough, then eventually a very narrow bridge will form in between, and the EHs will merge. At that moment, the full merger is decreed and will procede with absolute certainty until it's complete. Nothing can stop it. See this answer: Are black holes spherical during merger? what would happen to this incredible amount of momentum of each other
the black holes? The resulting black hole after the merger is going to have a heck of a lot of spin, if the collision is not perfectly frontal. Whatever energy cannot be stuffed into spin, is probably going to be radiated away as gravitational waves (as others have indicated already in comments to your question). | {
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32,510 | We know most of the objects in the Universe have a spherical or elliptical shape. The object which has less mass and gravitational pull orbits around the nearest object with more mass and gravitational pull. For example: Moon orbits around Earth Earth orbits around Sun Sun orbits around Sagittarius A* which is the center of Milky Way. Thus, is the Milky Way orbiting around some object or perhaps Black Hole? I know that the Milky Way is going towards Andromeda as they are attracting each other and they will collide with each other after 3 billion years to 6 billion years. But it is possible that the Milky way is orbiting around some object at the same time? Perhaps both galaxies are present in a group of galaxies which is orbiting around some object. If the Milky Way is not orbiting around some object then is there any proof found by the scientists for that? | The object which has less mass and gravitational pull orbits around the nearest object with more mass and gravitational pull. Actually, both the heavier and the lighter object orbit around their common center of mass. It's just that the heavier object doesn't move much (has a tiny orbit), while the lighter object moves a lot (has a wide orbit). E.g. our Sun actually orbits the center of mass of the whole solar system, but that motion is tiny, it barely budges. In the case of a double star, where both partners have about the same mass, you can clearly see how both are making similar orbits around their common mass center. Sun Orbits Around Sagittarius A* which us center of Milky Way. With galaxies, including ours, it's a little different. There is no super-heavy thing at the center, around which everything else is orbiting. Not even the very large black hole at the center of our galaxy is heavy enough for that. Rather, galaxies are clumps of matter that create a common gravitational field. Stars, and everything else, are trapped in this common field and orbit around the common center of mass. So the question is that is that is Milky Way is orbiting around some object or perhaps Black Hole. Same idea. There is no single point-object nearby massive enough for our galaxy to "orbit" around it. Our galaxy, along with Andromeda, and a handful of other galaxies, are bound together in what is known as the Local Group. Each galaxy is moving within the common gravitational field of the whole group. The Local Group has a diameter of about 10 million light-years. The Local Group is part of a larger structure, the Virgo Supercluster, which is about 100 million light-years in diameter and has at least 100 galaxies. However, the Virgo Supercluster is more "loose" - it is not gravitationally bound together. | {
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32,562 | Sorry for the noob question but I don't seem to be able to find the answer on the internet. I've been looking through some telescopes and binoculars and noticed that shops typically give different specifications for the two groups. For example, for telescopes I'll often see focal length, aperture ratio, or limit value, but I haven't seen any of those for binoculars; for binoculars, on the other hand, they'll mention exit pupil, field of view, or glass material, none of which I've seen given for a telescope. Why is that? Are the two devices so different as to be incomparable? | With a binocular, all its optical components are fixed - the user can't change them. What's important for the user to know is the size of the front lens, which determines the brightness (and in theory sharpness) of the image, the magnification, and the field of view. These are all useful things to know. A telescope has an interchangable component, namely the eyepiece. The choice of eyepiece determines: The magnification (= focal length of telescope divided by focal length of eyepiece) The field of view, which depends on magification and design of the eyepiece The exit pupil size (diameter of objective divided by magnification) and a few other things Hence, based on a particular eyepiece type you can work out the same information as you get with binoculars. | {
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32,640 | When I go to the NASA HORIZONS webpage https://ssd.jpl.nasa.gov/horizons.cgi and ask for the VECTOR coordinates of the Earth on 1 January 2000, it tells me that Earth has the following coordinates in the J2000 epoch: X =-1.756637922977122E-01 Y = 9.659912850526895E-01 Z = 2.020629118443605E-04 These are measured in astronomical units. Quoting from the webpage: Reference epoch: J2000.0 XY-plane: plane of the Earth's orbit at the reference epoch X-axis : out along ascending node of instantaneous plane of the Earth's orbit and the Earth's mean equator at the reference epoch Z-axis : perpendicular to the xy-plane in the directional (+ or -) sense of Earth's north pole at the reference epoch. But these figures don't make sense! The z value should be EXACTLY zero, since we are measuring at 1 January 2000. Also the x should be exactly 1, and the y should be zero. What is going on? | You made the same fundamental mistake that Anton Gromov made in his question on the sister Space Exploration StackExchange network site: You used the solar system barycenter rather than the Sun as the frame origin. Had you used the Sun, the apparent discrepancy would have dropped by almost two orders of magnitude. A much lesser flaw is that you apparently used midnight rather than noon. The J2000.0 epoch is 12 noon Terrestrial Time on 1 January 2000. This doesn't change things by much, but it is nonetheless important. That half day offset has bitten me more than once. The z coordinate of the Earth's position would not have dropped to zero had you made the above corrections (Sun-centered rather than solar system barycenter, and 12 noon rather than midnight). It would not have dropped to zero even if you had used the Earth-Moon barycenter rather than the center of the Earth. One reason it would not have dropped to zero is that the Earth's orbit about the Sun is not an ellipse. The Earth's orbit doesn't even lie on a plane! The key cause of this non-planar motion is that the Moon's orbit about the Earth is inclined by about 5.15° with respect to the ecliptic, making the Earth bob up and down with respect to the ecliptic 1 . Because the Earth's orbit is not truly elliptical or even planar, the ecliptic plane at some epoch is a time-averaged plane centered about the epoch time that makes the Earth's position with respect to the Sun (or the Sun's position with respect to the Earth) average to zero. A lesser reason is that the J2000.0 mean ecliptic and mean equinox frame was defined over 36 years ago. HORIZONS uses DE431, which was released 6 years ago. JPL's ability to model the solar system improved significantly in the intervening three decades. 1 Strictly speaking, it makes the Sun appear to bob up and down with respect to the ecliptic. The concept of the ecliptic plane dates back to the ancient Greeks, and it retains a vestige of this definition to this day. | {
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32,641 | I thought that solar systems and galaxies are disk-shaped because that is the most stable shape under gravitation. Globular clusters are very old, often times older than their host galaxies, so why haven't they flattened out? | In order to result in a disk-like stellar system there are two conditions that need to be satisfied. (a) The initial gas from which the stars form must have a significant ratio of rotational to gravitational energy. (b) The star formation would have to occur slowly enough, that the gas collapses to a disk before star formation is complete. The formation of a disk from a spherically symmetric state requires a reduction (more negative) gravitational potential energy, whilst conserving angular momentum. This can only happen if there is some means to lose energy from the system in the form of dissipative interactions. A pure stellar system is almost collisionless and there is no means to dissipate energy. That means once a spherical system of stars has formed, there is no way for it to become more disk-like. Thus if star formation takes place in a spherical way, then those stars will end up in a spherical system. Only in cases where the gas collapses to a disk before star formation occurs do we end up with disk-like systems. This is true of the disk of our Galaxy (and of the solar system). In the case of globular clusters (and other types of star cluster in the Galaxy) it appears the star formation is rapid enough that the gas cannot have collapsed to a disk before most star formation is complete. NB. Many globular clusters do have measurable rotation. | {
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32,672 | The Kepler space telescope detects planets based on the dip in brightness caused by planets moving past the star. Wouldn’t that mean that there are an unknown amount of planets that have an orbit that wouldn’t be detected because their orbits don’t cross that path between the star and the telescope? | That's right. The inclination of the orbital plane around stars is considered to be random throughout the galaxy, thus the planets we can detect by the transit method is just a tiny fraction of the planets that we should expect in our stellar neighbourhood. The transit method allows for planetary detection only when the line of sight from Earth to the system is contained, or almost contained, in the orbital plane of the planet. This means that only a tiny range of orbital inclinations on each star are good for detection. Why did I say almost? Because there is some range of inclinations that still would yield a transit. This range is not fixed, and it depends on the distance of the planet to its host star. As you can see in this diagram: Planet A is closer to the star and thus creates a wider shadow. If an observer is located in that shadowed region far away it can detect planet A. Planet B instead is farther from the star and thus its shadow is narrower. It is interesting to note, that even if both planets here share the exact same orbital plane there are places from where you would only detect planet A and never detect planet B (see the green arrows). This is the reason we have a bias towards planets orbiting closer to their star. This effect is in fact quite strong: consider our Solar System from an exoplanetary perspective. If you were located in a random star in the sky, what are the chances you would spot an Earth transit? Well, it turns out that it is way more probable to detect a Mercury transit, even if Mercury is the smallest planet, just because of its vicinity to the Sun. A recent paper showed this diagram of the regions of the sky where some alien inhabitants would spot a transit for each of our planets: As you can see Mercury has the wider strip. Also it's interesting to note that due to these differences in the size of the orbits (let's use the semi-major axis, $a$ , as a reference) and due to small differences in orbital inclinations there is no place in the entire sky from which an alien could detect simultaneously more than four of our planets by the transit method. No place in the universe where all the Solar System's planets would be detectable. The detection method also depends on the relative sizes of the star, $R_s$ , and the planet $R_p$ : A larger star has a larger disk (as viewed from Earth) that can be easily photobombed by a planet and a larger planet can photobomb more easily if it is larger. The result is that the probability of detecting a planet increases as we increase both/either $R_p$ and $R_s$ and increases as we decrease the distance to the host star $a$ . The relation is then of this form: $P \sim (R_s+R_p)/a$ This relation imposes several observational biases. We can see exoplanets that are large and closer to their star, but we can't see planets that are small and farther. That is the reason the first detected exoplanets are the so-called hot Jupiters : giant planets much closer to their stars than Mercury is to the Sun. This diagram shows all the exoplanet detections plotted on size vs. orbital distance: As you can see, small planets are only detectable if they have very small orbits around their stars. We have yet to find a planet the size of Earth (quite small) and with a 365 day orbital period (1 AU distance) using the transit method. There is no reason to think that this is representative of the overall population of planets. The black region of the plot is probably filled with dots, but our instruments can't scout that region yet. The Kepler telescope had a camera with a field of view on which it could detect more than half a million stars, but the actual number of stars monitored during the mission was around 150,000 stars (these stars had good signals and were perfect targets for the mission) . For these 150,000 stars Kepler found 2,345 exoplanets distributed in 1,205 stars. So we can say that for each star targeted by Kepler, the average probability of finding some planets there is around $0.8\;\%$ . That should give you an estimate of the occurrence of orbital inclinations that result in transits. The truth is that this number is too small, because Kepler has several more biases. For example, Kepler only confirmed planets after three transits were detected. Since the Kepler mission lasted for four years and four months we can say that in the best case scenario Kepler was able to detect a planet with an orbital period as long as two years and two months, but this is not even the case since for that to happen a transit should have been detected just at the beginning of the mission, halfway, and at the exact end of it, and this coincidence didn't happen. Thus Kepler had no chance to discover any planet with periods longer than two years (enough for Earth, but not enough for our Jupiter for example), even if the orbital inclination matched perfectly for the transit . So you might expect more possible transits than those actually portrayed by the Kepler telescope. In fact, for planets close to their stars, it has been estimated that the probability of a random alignment to allow a transit gets up to $10\;\%$ . For the case of stars as big as our Sun and planets at the same distance as Earth, the probability of this random occurrence drops to $0.47 \;\%$ . So with all the diversity of planets (in terms of sizes and distances to their host star) it is reasonable to expect a $0.8\;\%$ detection rate for Kepler (if we also add the time restriction to observe three transits). A $0.47\;\%$ is an amazing number! It means that for every Earth-like planet we detect by the transit method we should expect another 213 Earth-like planets orbiting other stars that are undetectable by the transit method. This kind of reasoning has been expanded. We have many difficulties to detect them, but if you mathematically model that difficulty and the corresponding biases associated with the known instruments and you assume random configurations, you can see that each discovery yields statistical significance to the amount of possible planets that are really out there. There are so many detections now that we can finally establish with statistical confidence that there are more planets than stars in our galaxy (even if we have probed an infinitesimal fraction of the entire population), even if this was something that could be expected we have now strong evidence for that thanks to Kepler. This means that there could be around a trillion or more plants just in the Milky Way . Now we are also able to establish some statistical constraints on the occurrence of Earth-like planets (orbiting in the habitable zone of their sun-like star) thanks to Kepler. There are probably around 11 billion planets in our galaxy with these specifications . TL;DR There are many more planets than the ones we can detect by the transit method, between 10 and 100 times more depending on the size and orbital period of the planet you are searching for. | {
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32,859 | An answer to the question of How well would the Moon protect the Earth from a Meteor? mentions as a possibility that the Moon could get knocked into the Earth. What is the smallest change to the orbit of the Moon from being impacted by a large meteor that would cause it to eventually impact the Earth (i.e. "circling the drain")? What timeline would that look like (minutes, hours, days, years, etc)? | As several people have said, this is incredibly unlikely. Part of the reason why is that the "circling the drain" effect you describe doesn't really happen for solid objects much less dense than black holes. Orbits are not "precarious" in that way. So, suppose something large enough and fast enough to change its velocity noticeably, but not large enough or fast enough to shatter it, did hit the Moon. The effect would be to shift the Moon from its present almost circular orbit around the Earth, into an elliptical one. Depending on the direction of the impact, it would either get a bit nearer to the Earth than it is now, once per orbit, or a bit further away (it also might swing North and South a bit). What is important though, is that this elliptical track is stable at least for a while. Suppose it gets knocked into an orbit that is 220000 miles from the Earth at its closest and 240000 miles at its furthest, that is where it will stay. It will not "spiral in". Over a long enough period the gravity of the Sun also comes into play and things may shift a bit, but that is a relatively small effect. Now, suppose that the impact was really big, or perhaps there were a long series of impacts (starting to look like enemy action..) so that the innermost point of the ellipse was eventually driven down to within a few thousand miles of the Earth, somehow miraculously not smashing the Moon to fragments in the process. At this distance it starts to matter that the near side of the Moon is closer to Earth than the far side, so that Earth's gravity pulls on it more strongly. If it orbited closer than about 3000km to the surface of the Earth for long (the Roche limit) these forces would eventually pull it to pieces, and Earth would probably have a pretty set of rings for a short time before internal collisions between the bits caused them to rain down on Earth and kill everyone. Finally suppose the impact(s) was(were) so big that they actually put the Moon into an elliptical orbit whose innermost point was so close to Earth that the Earth and Moon touched. This is manifestly impossible without shattering the Moon, but in that case, the Moon would indeed hit the Earth. The time for the impact would be about 1/4 of the Moons current orbital period, which is to say about a week. | {
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32,920 | I have purchased a custom wedding band from a seller that claims the ring will show the constellations visible at the horizon on a specific date at a certain date and time. However, I have fired up Stellarium and set it up to look at a specific example shown in the product page - and I cannot find any set of parameters which looks like it: Jupiter's not in the right place, some constellations are under the horizon (Virgo, Libra, Leo), some others very high in the sky (Andromeda, Perseus)... Although it's the first time I use Stellarium, I don't see how I could make a mistake: Location set to London Bridge (even the preset London to be sure) Date and time set just like in the example (double-checked 5 times) Cylindrical projection (tried all of them though), offset -30% FoV 210° (to have the 360° mapped to the entire screen - I have absolutely no idea why 180° does not work) Elevation lines horizontal (although I tried moving around to align everything like in the pictures for every projection method - in vain) Here is the example: Am I missing something, or is their sky map wrong? If yes, is it possible that two pieces of well-polished software like Stellarium and this one (looks just as good as Stellarium) have different star locations? How come (isn't there a standard star almanach?)? | If you set the date to 2018-04-06, Stellarium shows the Moon and planets in positions matching the example image.
Any good planetarium software should produce a similar result. Most likely the vendor cut and pasted two screenshots (note the seam) for April 2018, overlaid "January 1973," and hoped customers would not check.
Perhaps you could ask them to send you an image for approval before printing the ring. | {
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32,940 | I live in Essex, England (51.7678° N, 0.0878° E). On 25 July 2019 (hottest day ever in the UK, btw), 06:43 BST, I took this smartphone photo of the Sun. Is that white dot just below and left of the Sun the planet Mars? I've checked with my SkySafari app and it looks like it could be Mars, but I'm just surprised that my relatively cheap phone could take a picture of the planet. Although the photo is quite dark, it was a bright day. Thank you. | That is a camera artefact caused by the bright sunlight reflecting within the lens on your phone. It’s more pronounced than on a large camera because of the small lens size. This is a secondary image of the sun, as the brightness of the source allows for the reflection to be still intense enough to be detected. Here is a photo I took with the same artefact. It appears for me as a cyan dot with a magenta outside, but other photos I’ve taken have one with the same colour as the sun. Notice if you zoom in on yours, there’s a halo around the artefact as well. Also notice that the artefact is directly below the sun in my image, whereas in yours it’s slightly to the left. However, this is because in your image the sun is slightly to the right, and hitting the lens at an angle. As uhoh says , you can see that the sun is the same distance from the centre of the image as the lens flare (or very close to): It should be possible to take a photo with the artefact wherever you want, by re-angling the phone - including inside the sun, where it will be drowned out. I actually took my phone into an Apple Store to find out what this was, where they explained the lens flair effect. Unfortunately I don’t have a source for this, as it was in person. Hopefully the above is convincing enough. | {
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33,018 | Straight from my 7 year old to you, exactly what it says on the cover: What is the hottest thing in the universe? To make it Stack Exchange-friendly, I'll add the following caveats: it should be bounded, as in an actual compact object, or class of objects, or part of an object it should be observable it should be an astronomical object, ie a Quark Gluon Plasma created by collisions at the Large Hadron Collider doesn't count. | Energetic neutrinos have been observed from the core of a supernova ( SN 1987A ). The inferred temperature at the "neutrinosphere" is about 4 MeV (equivalent to 50 billion K - ( $5\times 10^{10}$ K, Valentim et al. 2017 ). These are thermal neutrinos with energies that are characteristic of the temperature of the material that produced them. Hence this material is observable and has been observed. The very centre of the proto-neutron star that is responsible for the neutrino emission is likely to be a factor of two or so hotter, but cannot be observed, even with neutrinos, because the "neutrinosphere" is opaque to neutrinos. By the time this "clears", the proto-neutron star is much cooler - its surface would be orders of magnitude cooler. Arguably we could study the very core of a supernova through gravitational waves if one were to explode in our own Galaxy. Whether this counts as "observing" a hot object, I'm not sure. In a similar vein, we have observed "kilonova" that appear to be due to the merger of two neutron stars. The temperatures generated in these events are also likely to be of order 100 billion K ( $10^{11}$ K), but again these temperatures are not observed directly - the gravitational waves and gamma rays produced in these events are caused by "non-thermal" mechanisms. | {
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33,023 | This answer to Farthest distance to a solar system object that's been measured by radar? mentions that Saturn's rings, and the Uncover Travel post Arecibo Observatory, Puerto Rico – The World’s Largest Radio Telescope For Over 50 Years mentions: Among other accomplishments of the Arecibo Observatory are: Direct imaging of an asteroid for the first time in history. Discovery of water ice deposits at Mercury’s poles. Tracking of near-Earth asteroids to monitor impact risks. Mapping the cloud-covered surface of Venus. Radar imaging of the rings of Saturn, revealing new details of the ring structure. First detection of methane lakes on Titan, a moon of Saturn. First detection of an asteroid with a moon. Question: How did Arecibo detect methane lakes on Titan, and image Saturn's rings? These are pretty remarkable feats from Earth for a single radiotelescope. How were they done? Can citations be found and the examples of the image of the rings and methane lake evidence be shown? | Titan "lakes": Published Open Access in Science: Radar Evidence for Liquid Surfaces on Titan Campbell, D. B., Black, G. J., Carter, L. M., and Ostro, S. J., Science 302 , 5644, pp. 431-434, 17 Oct 2003 DOI: 10.1126/science.1088969 This was a really elegant experiment! A continuous, unmodulated, circularly polarized 13 cm wave was broadcast from Arecibo towards the Saturn/Titan system, and the Doppler shift was used to isolate the returned signal from Titan. Most of the surface is rough, so there is signal returned from areas all over Titan's disk, and since the moon rotates, albeit slowly, returned power from the "left" and "right" sides are shifted to higher and lower frequencies. However during some observing times there was a very strong and pronounced reflection with zero Doppler shift with respect to Titan's known radial velocity, and this peak is attributed to specular reflection. Checks on received polarization confirm that while the power from the rough surface is returned in both circular polarization state, the presumed specular component is only in the expected circular polarization state. As pointed out in @Martin Kochanski's thoughtful answer there is no determination from the radar observation that the returned specular reflection comes from methane. This is simply a presumed component of the presumed lakes, based on known information about Titan's chemistry at the time (2003). We observed Titan on 16 nights in November and December 2001 and on 9 nights in November and December 2002, transmitting at 13-cm wavelength with the 305-m Arecibo telescope and receiving the echo with Arecibo. Titan's rotational and orbital periods are 15.9 days, and our 2001 observations were obtained at a uniform 22.6° (∼800 km) interval in longitude. The 9 observations in 2002 did not provide uniform coverage. The latitude of the subearth track was 25.9°S in 2001 and 26.2°S in 2002, its farthest southern excursion. The round-trip light time to the Saturn system during the observations was 2 hours 15 min, and the limited tracking time of the Arecibo telescope meant that signal reception was restricted to ∼30 min per day, corresponding to 0.5° of Titan rotation (20 km of motion of the subearth point). On one night in 2001 and for most of the 2002 observations (as well as others when we were attempting ranging measurements to Titan), the 100-m Green Bank Telescope (GBT) was also used to receive the echo for the full round-trip time. These data have lower signal-to-noise ratios than those obtained with Arecibo receiving the echo, but the longer receive time corresponding to 2.1° of Titan rotation allowed more subearth locations to be studied. Here is some of the Titan data: Fig. 3. The OC radar echo spectrum at 1.0-Hz resolution for the 2002 observation at the subearth longitude of 80°. The normalized cross section for the specular component of the echo and the RMS slope are 0.023 and 0.2°, respectively. Click for full size Fig. 1. Arecibo radar echo spectra from the 2001 data for five subearth longitudes on Titan. Spectra are shown for both the expected (OC) sense of received circular polarization and the cross-polarized (SC) sense. The ordinate is in standard deviations of the noise. The limb-to-limb Doppler-broadened bandwidth for Titan is 325 Hz. Four of the OC spectra show evidence of a specular component at 0 Hz. Saturn's rings "imaged" (delay-Doppler): From Radar imaging of Saturn’s rings Nicholson, P. D. et al., Icarus 177 (2005) 32–62, doi:10.1016/j.icarus.2005.03.023 The "image" below is not a conventional image, since the Arecibo dish has no way to spatially resolve the transverse expanse of Saturn and its rings. It is a "delay-Doppler" image, using 12.6 cm, ~500 kW radar broadcasts transmitted by Arecibo. The round-trip light time was about 135 minutes. Because Arecibo has limited steering away from the zenith (<19.7 degrees) max Saturn was only available to the dish for 166 minutes even under ideal conditions. The vertical axis shows a delay of about +/- 800 milliseconds which demonstrates spatial resolution, but in the radial or depth direction. The horizontal axis is Doppler shift. The +/- 300 kHz shift represents the orbital velocity of particles in the rings. While the Titan specular reflection above was done with a continuous or CW beam, the delay-doppler imaging technique requires a frequency modulation of the beam with a frequency-hopping pattern. By applying a correlation function using the known pattern to the recorded received signals, components with different return times and different Doppler shifts can be extracted, and the results is then hzstogramed, producing the delay-Doppler image below. This is a standard technique and has been used to image other planets and asteroids: See the following items and refereces within: this answer to What causes “North-South ambiguity” when doppler radar imaging a planet surface equator on? Why do radar maps of the surface of Venus have missing slices? What is the physical geometry of this apparent “eclipse” of a tiny moon of Asteroid Florence? Is delay-doppler radar imaging of NEO asteroids possible only if it spins fast enough? Fig. 2. Delay–Doppler images constructed from data obtained in (a) October 1999, (b) November 2000, (c) December 2001, and (d) January 2003.
Both OC and SC polarizations were combined to maximize the signal to
noise ratio. Note the four bright regions in each image where the delay and
Doppler cells are parallel and where the A and B rings appear to cross over
one another. | {
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33,111 | According to Penrose's research, a non-rotating star would end up, after gravitational collapse, as a perfectly spherical black hole. However, every star in the universe has some kind of angular momentum. Why even bother doing that research if that won't ever happen in the universe and does it have any implications for the future of astrophysics? | All models are approximations, we judge a model on how useful it is. Understanding the collapse of a non-rotating star to a black hole gives insight into the nature of gravitational collapse. Much of the physics of collapse does not depend on spin. The formation of an event horizon, for example. Models can be refined, and in this case, considering rotation leads to further insight, and a non-spherically symmetric structure with multiple singular horizons. All models are necessarily simplifications. But the non-rotating model is still useful. | {
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33,292 | Say I'm standing up straight, and I draw a straight line from my core through the top of my head (perpendicular to the ground). What is the probability that that line intersects with a star? EDIT: I'm not trying to exclude any stars. This should include stars which we've observed and stars which we haven't yet observed but can predict due to other things we've determined (like the overall star density of the universe). Also it should include all stars regardless of naked eye magnitude limit. | Summary There's a 1 in 500 billion chance you're standing under a star outside the Milky Way, a 1 in 3.3 billion chance you're standing under a Milky Way star, and a 1 in 184 thousand chance you're standing under the Sun right now. Big, fat, stinking, Warning! I did my best to keep my math straight, but this is all stuff I just came up with. I make no guarantees it's completely accurate, but the numbers seem to pass the sanity check so I think we're good. Caveat the First : The numbers for stars other than the Sun are based on data with a great deal of uncertainty, such as the number of stars in the universe and the average size of a star. The numbers above could easily be off by a factor of 10 in either direction, and are merely intended to give a rough idea of how empty space is. Caveat the Second : The numbers for the Sun and the Milky Way are based on the assumption that you are standing (or floating) at a random point on Earth. Anyone outside the tropics will never have the Sun over their head. People in the northern hemisphere are more likely to have Milky Way stars over their head, with the best odds being people near 36.8° N, because at that latitude straight up passes through the galactic center once a day. 26 Note : You can mostly ignore everything in this answer and just look up the solid angle of the Sun to get the same result. All the other stars are really far away and very spread out. The difference in solid angle subtended is five thousandths of a percent more when we add the rest of the universe to the Sun. Background Let's try to get a somewhat realistic, hard number. To do that, we'll need some assumptions. As pointed out in Michael Walsby's answer 1 , if the universe is infinite (and homogeneous 2 ), there is only an infinitesimal chance of there not being a star overhead, which normal math treats as exactly zero chance. So let's presume the universe is finite. Presumptions Specifically, let's presume the universe only consists of the observable universe. (Look up the expansion of the universe 3 for further information.) Further, let's presume the contents of the observable universe are measured at their current (presumed) positions, not the position they appear to be. (If we see light from a star from 400 million years after the universe began, we would measure it as being about 13.5 billion light years away, but we calculate that it's likely closer to 45 billion light years away due to expansion.) We'll take the number of stars in the observable universe to be $10^{24}$ . A 2013 estimate 4 was $10^{21}$ , a 2014 estimate 5 was $10^{23}$ , and a 2017 estimate 6 was $10^{24}$ , with each article expecting the estimate to increase as we get better telescopes over time. So we'll take the highest value and use it. We'll take the size of the observable universe 7 to be $8.8\cdot10^{26} \text{m (diameter)}$ , giving a surface area 8 of $2.433\cdot10^{54} \text{m}^2$ 9 , and a volume 10 of $3.568\cdot10^{80} \text{m}^3$ 11 . We'll take the average size of a star to be the size of the Sun, $1.4\cdot10^{9}\text{m (diameter)}$ 12 . (I can't find any sources for average star size, just that the Sun is an average star.) Model From here, we're going to cheat a bit. Realistically, we should model each galaxy separately. But we're just going to pretend the entire universe is perfectly uniform (this is true enough as we get farther away from Earth in the grand scheme of the cosmos). Further, we're going to start counting far enough out to ignore the Milky Way and Sun entirely, then add them back in later with different calculations. Given the above presumptions, we can easily calculate the stellar density of the observable universe to be $\delta = \frac{10^{24}\text{stars}}{3.568\cdot10^{80} m^3} = 2.803\cdot10^{-57} \frac{\text{stars}}{\text{m}^3}$ 13 . Next, we need to calculate the solid angle 14 subtended by a star. The solid angle of a sphere is given by $\Omega=2\pi\left(1-\frac{\sqrt{d^2-r^2}}{d}\right)\text{ sr}$ 15 , where $\Omega$ is the solid angle in steradians 16 (sr), $d$ is the distance to the sphere and $r$ is the radius of the sphere. Using $D$ as the diameter, that converts to $\Omega=2\pi\left(1-\frac{\sqrt{d^2-\left(\frac{D}{2}\right)^2}}{d}\right)\text{ sr}$ . Given the average diameter presumed above ( $1.4\cdot10^{9}\text{m}$ ), this gives an average solid angle of $\Omega=2\pi\left(1-\frac{\sqrt{d^2-4.9\cdot10^{17}\text{m}^2}}{d}\right)\text{ sr}$ 17 . At this point, we could set up a proper integral, but my calculus is rather rusty, and not very sharp to begin with. So I'm going to approximate the answer using a series of concentric shells, each having a thickness of $10^{22}\text{m}$ (about a million light years). We'll put our first shell $10^{22}\text{m}$ away, then work our way out from there. We'll calculate the total solid angle of each shell, then add all the shells together to get the solid angle subtended by the entire observable universe. The last problem to fix here is that of overlap. Some stars in the farther shells will overlap stars in the nearby shells, causing us to overestimate the total coverage. So we'll calculate the probability of any given star overlapping and modify the result from there. We'll ignore any overlap within a given shell, modeling as if every star in a shell is at a fixed distance, evenly distributed throughout the shell. Probability of Overlap For a given star to overlap closer stars, it needs to be at a position already covered by the closer stars. For our purposes, we'll treat overlaps as binary: either the star is totally overlapped, or not overlapped at all. The probability will be given by the amount of solid angle already subtended by previous shells divided by the total solid angle in the sky ( $4\pi\text{ sr}$ ). Let's call the probability a given star, $i$ , is overlapped $P_i$ , the solid angle subtended by that star $\Omega_i$ , and the number of stars $n$ . The amount of non-overlapped solid angle subtended by a given shell, $k$ , is then $\Omega_{kT}=(1-P_1)\Omega_1+(1-P_2)\Omega_2+\ldots+(1-P_n)\Omega_n\frac{\text{ sr}}{star}$ . Since we've said stars in a shell don't overlap each other, $P_i$ is the same for all $i$ in a given shell, allowing us to simplify the above equation to $\Omega_{kT}=(1-P_k)(\Omega_1+\Omega_2+\ldots+\Omega_n)\frac{\text{ sr}}{star}$ , where $P_k$ is the probability of overlap for shell $k$ . Since we're treating all the stars as having the same, average size, this simplifies even further to $\Omega_{kT}=(1-P_k)\Omega_k n\frac{\text{ sr}}{star}$ , where $\Omega_k$ is the solid angle of a star in shell $k$ . Calculating Solid Angle The number of stars in a shell is given by the volume of the shell times the stellar density of said shell. For far away shells, we can treat the volume of the shell as being its surface area times its thickness. $V_\text{shell}=4\pi d^2 t$ , where $d$ is the distance to the shell and $t$ is its thickness. Using $\delta$ as the stellar density, the number of stars is simply $n=\delta V_\text{shell}=\delta 4\pi d^2 t$ . From here, we can use the calculation for solid angle of a shell (from Probability of Overlap , above) to get $\Omega_{kT}=(1-P_k)\Omega_k \delta 4\pi d^2 t\frac{\text{ sr}}{star}$ . Note that $P_k$ is given by the partial sum of solid angle for all previous shells divided by total solid angle. And $\Omega_k$ is given by $\Omega_k=2\pi\left(1-\frac{\sqrt{d_k^2-4.9\cdot10^{17}\text{m}^2}}{d_k}\right)\frac{\text{ sr}}{star}$ (from Model , above). This gives us $\Omega_{kT}=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{d_k^2-4.9\cdot10^{17}\text{m}^2}}{d_k}\right) \delta 4\pi d^2 t\text{ sr}$ . Given that each shell is $10^{22}\text{m}$ away, we can substitute $d_k$ with $k 10^{22}\text{m}$ . Likewise, $t$ can be substituted with $10^{22}\text{m}$ . And we already calculated $\delta=2.803\cdot10^{-57} \frac{\text{stars}}{\text{m}^3}$ (from Model , above). This gives us $\Omega_{kT}=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{(k 10^{22}\text{m})^2-4.9\cdot10^{17}\text{m}^2}}{k 10^{22}\text{m}}\right) 2.803\cdot10^{-57}\frac{\text{stars}}{\text{m}^3} 4\pi (k 10^{22}\text{m})^2 10^{22}\text{m}\frac{\text{ sr}}{star}$ $=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\left(1-\frac{\sqrt{k^2 10^{44}-4.9\cdot10^{17}}}{k 10^{22}}\right) 2.803\cdot10^{-57} 8\pi^2 k^2 10^{66}\text{ sr}$ $=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\,2.213\cdot 10^{11}\,k^2\left(1-\frac{\sqrt{k^2 10^{44}-4.9\cdot10^{17}}}{k 10^{22}}\right)\text{ sr}$ From here, we can just plug the numbers into a calculation program. $\Omega_{T}=\sum_{k=1}^{k_\text{max}} \Omega_{kT}$ Where $k_\text{max}$ is just the radius of the observable universe divided by the thickness of a given shell. Thus $k_\text{max}$$=\frac{4.4\cdot 10^{26} \text{m}}{10^{22} \text{m}}$$=4.4\cdot 10^4$$=44000$ $\Omega_{T}=\sum_{k=1}^{44000} \Omega_{kT}$ Results Because of the large numbers involved, it's difficult to just run this in a program. I resorted to writing a custom C++ program using the ttmath library 18 for large numbers. The result was $2.386\cdot 10^{-11}\text{ sr}$ , or $1.898\cdot 10^{-12}$ of the entire sky. Conversely, there's about a 1 in 500 billion chance you're standing under a star right now. Note that we ignored the Milky Way and the Sun for this. The C++ program can be found at PasteBin 25 . You'll have to get ttmath working properly. I added some instructions to the top of the C++ code to get you started if you care to make it work. It's not elegant or anything, just enough to function. The Sun WolframAlpha helpfully informed me the Sun has a solid angle of about $6.8\cdot 10^{-5}\text{ sr}$ , or about 2.8 million times more than all the stars in the universe combined. The solid angle formula above gives the same answer 18 if we provide the Sun's 150 gigameter distance and 0.7 gigameter radius. The Milky Way We could get an approximation for the Milky Way by taking its size and density and doing the same calculations as above, except on a smaller scale. However, the galaxy is very flat, so the odds greatly depend on whether you happen to be standing in the galactic plane or not. Also, we're off to one side, so there are far more stars towards the galactic center than away. If we approximate the galaxy as a cylinder with a radius of $5\cdot 10^{20}\text{ m}$ (about 52000 light years) and a height of $2\cdot 10^{16}\text{ m}$ (about 2 light years), we get a volume of $1.571\cdot 10^{58}\text{ m}^3$ 20 . Current estimates of the galaxy's radius are closer to 100000 light years 21 22 , but I'm presuming the vast majority of stars are a lot closer than that. There are estimated to be 100 to 400 billion stars in the Milky Way 21 . Let's pick 200 billion for our purposes. This puts the density of the Milky Way at $\delta = \frac{200\cdot10^{9}\text{stars}}{1.571\cdot 10^{58}\text{ m}^3} = 1.273\cdot10^{-47} \frac{\text{stars}}{\text{m}^3}$ 22 , or about 4.5 billion times denser than the universe at large. This time, we'll take shells $10^{17}\text{ m}$ thick (about 10 light years) and go out from there. But we need to re-organize the math into a spherical form, so we'll presume the galaxy has the same volume, but is a sphere. This gives it a radius of $1.554\cdot 10^{19}\text{ m}$ 24 , or 155.4 shells. We'll round to 155 shells. $\Omega_{T}=\sum_{k=1}^{155} \Omega_{kT}$ Using our formula from above ( Calculating Solid Angle ), we can start substituting numbers. $\Omega_{kT}=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{d_k^2-4.9\cdot10^{17}\text{m}^2}}{d_k}\right) \delta 4\pi d^2 t\frac{\text{sr}}{\text{star}}$ $=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{(k\cdot 10^{17}\text{ m})^2-4.9\cdot10^{17}\text{ m}^2}}{k\cdot 10^{17}\text{ m}}\right) 1.273\cdot10^{-47} \frac{\text{stars}}{\text{m}^3} 4\pi (k\cdot 10^{17}\text{ m})^2 10^{17}\text{ m}\frac{\text{sr}}{\text{star}}$ $=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\left(1-\frac{\sqrt{k^2\cdot 10^{34}\text{ m}^2-4.9\cdot10^{17}\text{ m}^2}}{k 10^{17}\text{ m}}\right) 1.273\cdot 10^{-47} \frac{\text{stars}}{\text{m}^3} 8\pi^2 k^2 10^{51}\text{ m}^3\frac{\text{sr}}{\text{star}}$ $=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\cdot\,1.005\cdot 10^6\,k^2\,\left(1-\frac{\sqrt{k^2\cdot 10^{34}-4.9\cdot10^{17}}}{k 10^{17}}\right)\text{ sr}$ Plugging this into the program gives $3.816\cdot 10^{-9}\text{ sr}$ , which is $3.037\cdot 10^{-10}$ of the total sky. The odds you're standing under a star in the Milky Way are about 1 in 3.3 billion. Solid Angle Totals Solid angle is: Sun, $6.8\cdot 10^{-5}\text{ sr}$ Milky Way, $3.816\cdot 10^{-9}\text{ sr}$ Universe, $2.386\cdot 10^{-11}\text{ sr}$ Total, $6.800384\cdot 10^{-5}\text{ sr}$ (the extra digits are basically meaningless, adding about five thousandths of a percent to the Sun's solid angle) Milky Way plus Universe, $3.840\cdot 10^{-9}\text{ sr}$ (about 0.6% more than just the Milky Way) References 1 Michael Walsby's answer to this question , is there a star over my head? . https://astronomy.stackexchange.com/a/33294/10678 2 A Wikipedia article, Cosmological principle . https://en.wikipedia.org/wiki/Cosmological_principle 3 A Wikipedia article, Expansion of the universe . https://en.wikipedia.org/wiki/Expansion_of_the_universe 4 A UCSB ScienceLine quest, About how many stars are in space? , from 2013. https://scienceline.ucsb.edu/getkey.php?key=3775 5 A Sky and Telescope article, How Many Stars are There in the Universe? , from 2014. https://www.skyandtelescope.com/astronomy-resources/how-many-stars-are-there/ 6 A Space.com article, How Many Stars Are In The Universe? , from 2017. https://www.space.com/26078-how-many-stars-are-there.html 7 A Wikipedia article, Observable universe . https://en.wikipedia.org/wiki/Observable_universe 8 A Wikipedia article, Sphere , section Enclosed volume . https://en.wikipedia.org/wiki/Sphere#Enclosed_volume 9 A WolframAlpha calculation, surface area of a sphere, diameter 8.8*10^26 m . https://www.wolframalpha.com/input/?i=surface+area+of+a+sphere%2C+diameter+8.8*10%5E26+m 10 A Wikipedia article, Sphere , section Surface area . https://en.wikipedia.org/wiki/Sphere#Surface_area 11 A WolframAlpha calculation, volume of a sphere, diameter 8.8*10^26 m . https://www.wolframalpha.com/input/?i=volume+of+a+sphere%2C+diameter+8.8*10%5E26+m 12 A nineplanets.org article, The Sun . https://nineplanets.org/sol.html 13 A WolframAlpha calculation, (10^24 stars) / (3.568⋅10^80 m^3) . https://www.wolframalpha.com/input/?i=%2810%5E24+stars%29+%2F+%283.568%E2%8B%8510%5E80+m%5E3%29 14 A Wikipedia article, Solid angle . https://en.wikipedia.org/wiki/Solid_angle 15 Harish Chandra Rajpoot's answer to a geometry.se question , Calculating Solid angle for a sphere in space . https://math.stackexchange.com/a/1264753/265963 16 A Wikipedia article, Steradian . https://en.wikipedia.org/wiki/Steradian 17 A WolframAlpha calculation, 2*pi*(1-sqrt(d^2-(1.4*10^9 m/2)^2)/d) . https://www.wolframalpha.com/input/?i=2*pi*%281-sqrt%28d%5E2-%281.4*10%5E9+m%2F2%29%5E2%29%2Fd%29 18 Website for ttmath. https://www.ttmath.org/ 19 A WolframAlpha calculation, 2*pi*(1 - sqrt(d^2 - r^2)/d), where d = 150 billion, r=0.7 billion . https://www.wolframalpha.com/input/?i=2*pi*%281+-+sqrt%28d%5E2+-+r%5E2%29%2Fd%29%2C+where+d+%3D+150+billion%2C+r%3D0.7+billion 20 A WolframAlpha calculation, pi * (5*10^20 m)^2 * (2*10^16 m) . https://www.wolframalpha.com/input/?i=pi+*+%285*10%5E20+m%29%5E2+*+%282*10%5E16+m%29 21 A Wikipedia article, Milky Way . https://en.wikipedia.org/wiki/Milky_Way 22 A Space.com article from 2018, It Would Take 200,000 Years at Light Speed to Cross the Milky Way . https://www.space.com/41047-milky-way-galaxy-size-bigger-than-thought.html 23 A WolframAlpha calculation, (200*10^9 stars) / (1.571*10^58 m^3) . https://www.wolframalpha.com/input/?i=(200*10^9+stars)+%2F+(1.571*10^58+m^3) 24 A WolframAlpha calculation, solve for r: (4/3)*pi*r^3 = 1.571*10^58 m^3 . https://www.wolframalpha.com/input/?i=solve+for+r%3A++%284%2F3%29*pi*r%5E3+%3D+1.571*10%5E58+m%5E3 25 My C++ program code on PasteBin . https://pastebin.com/XZTzeRpG 26 A Physics Forums post, Orientation of the Earth, Sun and Solar System in the Milky Way . Specifically, Figure 1 , showing angles of 60.2° for the Sun, and 23.4° less than that for Earth. https://www.physicsforums.com/threads/orientation-of-the-earth-sun-and-solar-system-in-the-milky-way.888643/ | {
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33,359 | How is it possible to know if a black hole is spinning or not? If a planet is spinning, you can see it clearly but you can't really see a black hole. Next thing would be that matter interacts with adjacent matter and we could see in which direction the matter surrounding the BH spins (like if you spin a ball on water, the water around would spin too in the same direction) but matter can't interact from inside the event horizon to the outside, so matter right at the event horizon would just be interacting with gravity (like the BH has no friction). Now gravity. I would think that you could measure the differences in gravity if a large object is not perfectly uniform but I think a BH has the same gravitation pull on all sides. What am I missing here? How can one even detect or determine by observation that a black hole is spinning, or better yet, measure how fast? | The gravitational field of spinning matter, or a spinning black hole, causes matter around it to start spinning. This is called " frame dragging " or "gravitomagnetism", the latter name coming from the fact that it's closely analogous to the magnetic effect of moving electric charges. The existence of gravitomagnetism is tied to the finite speed of gravity, so it doesn't exist in Newtonian gravity where that speed is infinite, but it's present in general relativity, and for black holes it's large enough to be detectable. Also, for purely theoretical reasons we expect that all black holes are spinning because a non-spinning black hole is the same as a spinning black hole with an angular velocity of exactly zero, and there's no reason why a black hole's angular velocity would be exactly zero. On the contrary, because they are so much smaller than the matter that collapses to produce them, even a small, random net angular momentum of the collapsing matter should lead to a rapidly spinning black hole. (The classic analogy for this is an ice skater spinning faster when they pull their arms in.) | {
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33,463 | Would an object shot from earth fall into the sun? If an object is shot at 107,000 km/h via rocket or otherwise, in the opposite direction to our orbit about the sun, it will be traveling at 0 km/h relative to the sun. The moon is not close enough to the object to have a significant force for the purposes of this question. Will this object start accelerating towards the sun or will it somehow fall into another stable orbit? Could it instead get trapped in the L4 Earth-Sun Lagrange point? | Assume that a spacecraft is instantaneously accelerated at the Earth's surface (disregarding the atmosphere for simplicity). We'll consider this from the Sun's reference frame; in other words, the Sun is stationary and the Earth is moving around it. The spacecraft is accelerated to a velocity which is precisely equal and opposite to the orbital velocity of the Earth around the Sun, making it completely stationary in the instant after the acceleration. What happens next? Well, we can consider the forces acting on the spacecraft: The Earth's gravity causes a force in the direction of Earth. The Sun's gravity causes a force in the direction of the Sun. The stationary spacecraft is therefore going to accelerate towards the Earth and towards the Sun. Since the Earth is moving away quickly on its orbital path, the gravitational force is not enough to pull the spacecraft back into an Earth orbit; however, it will nudge the spacecraft into an elliptical orbit. To demonstrate the situation, I have created a small simulation which can be viewed in a desktop browser. Click here to try the simulation. (You can click "View this program" to check the code, and refresh the page to restart the simulation.) The simulation is physically accurate (ignoring the effects of other planets), but the spheres have been enlarged for easy interpretation. The Earth is represented as green, while the Sun is orange and the spacecraft is white. Note that, while the spheres representing the spacecraft and Sun intersect, the distance between the two physical objects is always larger than 3.35 solar radii. This screenshot shows how the spacecraft has been pulled into an elliptical orbit by the Earth: Finally, we could consider a more realistic scenario where the spacecraft is accelerated until it reaches zero velocity (again, in the Sun's reference frame) at a certain distance from the Earth. At the instant it reaches zero velocity, the engine is stopped. In this case, the result is essentially the same: there are still forces exerted by the Earth and the Sun, so an elliptical orbit will result. The further the rocket is from the Earth when it reaches zero velocity, the more elliptical the orbit. If the Earth is so far away that its gravity is negligible, the spacecraft will fall directly towards the Sun. | {
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33,520 | The common theory is, that if the Sun "shut down", we would see the light for eight more minutes (the time that it takes the photons to reach the Earth). However recently I have read that photons need around 100,000 years to reach the Earth, since the reactions are happening at the Sun's core and gamma rays can't leave the Sun without interacting with other particles, unlike neutrinos for example. Is that theory correct? If the Sun's core "shut down", would we still receive photons (light) for another 100,000 years, with only neutrinos disappearing immediately? | If nuclear fusion were to suddenly stop in the centre of the Sun, then the only clear signature we would have of this is the lack of detectable neutrinos received at Earth, starting about 8 minutes after the reactions ceased. The Sun however would continue to shine for tens of millions of years at roughly its current luminosity. The power source is not "stored" photons. The Sun itself would simply resume the slow gravitational contraction that was halted about 4.5 billion years ago when nuclear reaction rates at the centre were able to increase sufficiently to supply the radiative losses from the surface of the Sun. The characteristic (Kelvin-Helmholtz) timescale for the contraction is about $$\tau_{\rm KH} = \frac{GM^2}{RL},$$ which is 30 million years. i.e. The Sun has enough gravitational potential energy to supply its current luminosity for tens of millions of years. While this is happening, the Sun would approximately maintain its current luminosity, but decrease in radius, meaning that its surface temperature would increase. Once the Sun had contracted to a few times the size of Jupiter (so about 30% of its current radius), the contraction would begin to slow, because the electrons in the core become degenerate and the pressure increases with density by more than expected for a perfect gas. The slowing contraction decreases the rate of potential energy release and hence the solar luminosity. The contraction continues at a slow rate until the Sun becomes a hot "hydrogen white dwarf" a few times the size of the Earth, which then cools to a glowing cinder, with no further contraction, over billions of years (see What would the Sun be like if nuclear reactions could not proceed via quantum tunneling? for some more details). Even if you were to not allow the Sun to contract, it would take some time to radiate it's thermal energy. This timescale is approximately $$\tau_{\rm therm} \simeq \frac{3k_B T M}{m_H L},$$ which assumes the Sun is a perfect gas of protons plus electrons, with an average temperature $T$ . If we take $T =10^7$ K and the current solar luminosity, then $\tau_{\rm therm}=$ 40 million years. On the other hand, if your scenario is just that light from the Sun stops being emitted, then of course it goes dark on Earth about 8 minutes later. | {
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33,628 | The surface temperatures of Mars are about -87C to 5C, which is much colder than that of Earth's.
If Mars has 95% carbon dioxide, which is a Greenhouse gas, why is the surface of Mars so cold? Shouldn't it trap heat and render it hot? | Firstly, Mars has a mean distance from the Sun of 1.524 AU, so by the inverse square law the energy it gets from the Sun is about 40% of what the Earth gets. But the main reason that Mars is so cold is that its atmosphere is very thin compared to Earth's (as well as very dry, see below). From Wikipedia Atmosphere of Mars : The atmosphere of Mars is much thinner than Earth's. The surface pressure is only about 610 pascals (0.088 psi) which is less than 1% of the Earth's value. In comparison, the mean surface pressure on Earth is 101,300 pascals. So the atmosphere of Mars is barely more than a vacuum compared to Earth's. So even though the Martian atmosphere is over 95% carbon dioxide, there simply isn't enough of it to trap much heat. Although carbon dioxide is a greenhouse gas, the predominant greenhouse gas on Earth is actually water vapour . However, water is usually cycled in and out of the atmosphere very quickly in response to temperature and pressure changes. Carbon dioxide is a problem because it stays in the atmosphere for a long time, and its presence shifts the equilibrium temperature upwards from that of the plain water cycle. | {
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33,759 | The following image is the observation made by the Dual Frequency Synthetic Aperture Radar (DF-SAR) onboard the Chandrayaan-2 Orbiter: Due to image size restriction, a lower resolution image is used here. In case you wish to see the original image, visit this ISRO Webpage . In the above image, there are three craters marked Young (Fresh), Intermediate and Old, and we are able to see the difference clearly. This ISRO Webpage states that the difference is due to weathering processes. I can't understand how can such processes undergo in our Moon, where there is no appreciable amount of atmosphere. Is that because of Moonquakes? It would be great if you could explain this. | Aside from the excellent points made in James K's answer , there are other ways to date craters. For example, when the rays of one crater overlay those of another, we know that the former is younger than the latter. We can also estimate ages of large craters by counting subsequent craters inside the crater floor. Crater counting is one of the more common ways of estimating the age of lunar surface features. We can also determine ages of some craters by looking at whether it was affected by known ancient lunar processes. Then there is radiometric dating. Large impacts spread materials across the surface of the moon, so the Apollo mission samples contained ejecta from many craters, some of which are identifiable by things like mineral composition unique to that part of the moon. Once you have a good date for one crater, you can use that data to help determine the age of others. For example, if the floor of a crater has X smaller craters in it, and you know how old that crater is, you can then use those counts to estimate ages of other craters with similar counts. With enough varied data sets, you can generate some pretty good functions for estimating absolute ages. For example, have a look at the image of the Aristarchus Plateau below: NASA (image by Lunar Reconnaissance Orbiter) [Public domain] Notice the two craters at the bottom of the plateau. The one on the left is the crater Herodotus, and the one on the right is Aristarchus. Looking at the differences between the two of them, the first thing you notice is that Herodotus is filled in with the same lava that fills the basin around the plateau, but Aristarchus does not have that fill. So right there we know that Herodotus pre-dates the event that filled in Oceanus Procellarum, while Aristarchus came later. This makes Herodotus at least 1.2 billion years old, which is the current best estimate for the last time lava flowed on the moon. Now, we can do better than that with sampling. If we get a sample of the basalt inside Herodotus, we can use radiation dating to determine when it crystallized, and that would give us a very good upper limit of the age of the crater. And in fact we did get such samples from Apollo, and that, plus crater counting, plus looking at other features around it with known ages, plus the amount of degradation in the crater walls and the coloration of the material, we can tell that Herodotus dates to the Imbrium era, or about 3.8 billion years ago. As for Aristarchus crater, you can see lots of evidence that it is young. For one thing, its interior and ejecta are incredibly bright - it's the brightest crater on the moon. We know from radiological dating and crater counting that similarly bright craters are young. The walls of the crater have minimal slumping, and the ejecta around the crater is still well defined. Aristarchus is one of the 'Copernican' craters, so named because they all formed during the Copernican period, which starts 1.2 billion years ago and goes right up to the present. Other large copernican craters include Copernicus, Tycho, and Kepler, among others. Now look at this wider, high contrast image of the Tycho region. You can see Copernicus and Kepler on the right, and Tycho lower center. Note that the crater rays overlap each other, and by careful examination we can tell the order of the impacts. That helps us further refine the relative ages of the craters, and if we can narrow down the age of one with sampling or other techniques, it helps narrow down the ranges of the others. Btw, the order of creation from oldest to newest Copernicus, Kepler, Aristarchus, Tycho. Here's a pretty good reference describing a number of techniques for dating features on the moon: Relative Crater Age Determination This is an iterative process - each time you narrow down the age of a crater, that can be used as data to further refine the ages of others, But it's not a precise science - age estimates of various craters can still vary by hundreds of millions of years. There is still a lot we don't know about how the Moon formed and evolved. | {
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33,844 | I don't have a telescope but I'm interested in seeing events like eclipses and transits. I'll use the atmosphere as my big natural lens. So I'll watch the upcoming mercury transit at the sunset time where the sun looks bigger than usual. Will this make the black dot of mercury bigger or big enough to be obviously seen by the naked eye? I've never seen it before. Event: https://www.timeanddate.com/eclipse/in/egypt/cairo | I'll use the atmosphere as my big natural lens. So I'll watch the upcoming mercury transit at the sunset time where the sun looks bigger than usual. While the Sun and Moon might seem larger at the horizon, their angular size doesn't get larger. It's an optical illusion. Will this make the black dot of mercury bigger or big enough to be obviously seen by the naked eye? Therefore, no it won't, and never ever look directly at the Sun unless you are using appropriate Eclipse-viewing glasses that are specifically designed for this purpose. Sunglasses are not the correct thing to use, they can block some wavelengths more than others, causing your pupils to dilate while still passing some damaging wavelengths. Properly protected human vision - No. At closest approach, mercury is only 11 arcseconds wide, well below what we can resolve with unaided vision. We can still notice stars because they are on a black background, but it doesn't work the other way around. Mercury's sub-resolution black spot would not be noticeable against a bright disk. Pinhole projector (aka camera obscura ) - No. In this Astronomy SE answer I've said that even using a pinhole projector, you won't be able to see Mercury's transit since they provide very low resolution. | {
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33,899 | Today, Ansa.it released an article that states: [...]. In questo suo peregrinare galattico, il Sole ha attraversato anche
i due bracci della Via Lattea Perseo e Centauro. "Sono zone di alta
densità stellare, in corrispondenza delle quali il Sole e le stelle
intorno rallentano e possono anche fermarsi, [...] (my) translation to English: In this galactic wandering, the sun traversed the two spiral arms of the Milky Way, Perseus and Centaurus.
These are regions of high density of stars, in correspondence of which
the Sun and other surrounding stars slow down and can even stop, [...] The article is complemented with a suggestive video-clip that depicts the Sun orbiting around the galaxy's center and crossing various spiral arms. Given my lack of an education on the topic, I cannot reconcile the existence of an entity called "spiral arm" with the notion that a star can freely cross several "spiral arms" in its movement around the galaxy's center. My intuition is that if every star within the Milky Way could freely cross the "spiral arms" several times during their life-span, then there would be no such thing as a "spiral arm" at all (because this is supposed to be a mass grouping wherein all matter moves --please forgive me the abuse of the word-- "together"). Where am I wrong? Is the above description of the movement of the Sun accurate? In case of an affirmative answer, is it a very special case or is it a defining characteristic of the entire Orion's arm? In the latter case, can other spiral arms cross each other? | What is a spiral arm? The reason that the Sun, in principle (but see below), may cross spiral arms is that galactic spiral arms are not rigid entities consisting of some particular stars; rather they are "waves" with a temporary increase in density. An often-used analogy is the pile-up of cars behind a slow-moving truck: At all times, all cars are moving forward, but for a while, a car behind the truck will be moving slow, until it overtakes and speeds up. Similarly, stars may overtake, or be overtaken by, the spiral arms. Inside a certain distance from the center of the galaxy called the corotation radius ( $R_\mathrm{c}$ ) stars move faster than the arms, while outside they move slower. Since the stars and the interstellar gas follow the rotation of the arms for a while once they're inside, the density of the arms is higher than outside, but only by a factor of a few (e.g. Rix & Rieke 1993 ). When interstellar gas falls into the potential well it is compressed, triggering star formation. Since the most luminous stars burn their fuel fast, they will mostly have died once they leave an arm. Hence, what we see as spiral arms is not so much the extra stars, but mostly due to the light from the youngest stars which are still inside the arms. Since most luminous also means hottest, their light peaks in the bluish region — hence spiral arms appear blue. Origin of the spiral structure At least the most prominent spiral arms (especially grand designs ) are thought to be created by these long-lived, quasi-stationary density waves ( Lin & Shu 1964 ). The reason that the density waves exist in the first place is not well-understood, I think, but may have to do with anisotropic gravitational potentials and/or tidal forces from nearby galaxies (e.g. Semczuk et al. 2017 ). But in fact even small perturbations may spawn gravitational instabilities that propagate as density waves. In computer simulations of galaxy formation, even numerical instabilities may cause this, so the fact that your simulated galaxy has spiral arms doesn't necessarily mean that you got your physics right. When the luminous and hence massive stars die, they explode as supernovae. The feedback from this process, as well as that exerted by the radiation pressure before they die, may help maintaining the density waves, at least in flocculent galaxies ( Mueller & Arnett 1972 ). Perhaps this so-called Stochastic Self-Propagating Star Formation may also initiate the density waves (see discussion in Aschwanden et al. 2018 ). The rotation speed of the material in the galactic disk is roughly constant with distance from the center (this is mainly due to the dark matter halo hosting a galaxy). Hence, stars close to the center complete a revolution faster than those farther away. In contrast, the spiral pattern rotate more like a rigid disk such that, in an intertial frame, the pattern can be described by a constant angular speed $\Omega_\mathrm{p}$ throughout the disk. However, note that spiral arms are transient phenomena; they appear and disappear with lifetimes of the order of (a few) Myr (e.g. Grand et al. 2012 ; 2014 ). Sometimes you also see multiple spiral patterns propagating with different velocities. The Sun in the Milky Way Note: This section first contained errors based on dubious values for the angular speed of the spiral arms, as pointed out by @PeterErwin and @eagle275. In the case of our Sun, we happen to be located very near the corotational radius $R_\mathrm{c}$ ; we sit at a distance of $R_0 = 8.32\,\mathrm{kpc}$ from the center of the galaxy ( Gillessen et al. 2017 ), while $R_\mathrm{c} = 8.51\,\mathrm{kpc}$ ( Dias et al. 2019 ). Using Gaia data, Dias et al. (2019) find a pattern angular speed $\Omega_\mathrm{p} = 28.2 \pm 2.1\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{kpc}^{-1}$ . At the location of the Sun ( $R_0$ ) this implies a pattern speed of $\simeq235\,\mathrm{km}\,\mathrm{s}^{-1}$ . This is only a little bit slower than (and in fact statistically consistent with) the Sun's velocity of $239\pm5\,\mathrm{km}\,\mathrm{s}^{-1}$ ( Planck Collaboration et al. 2018 ). If the spiral arms were "permanent"*, the timescale for the Sun crossing an arm would hence be gigayears. However, because as described above they're quite transient, we might once in a while overtake a spiral arm. I haven't been able to find firm evidence for whether we have or haven't crossed any arms; Gies & Helsel (2005) argue that we have crossed an arm four times within the last 500 Myr, but base this on matching glaciation epochs with passages through spiral arms (and admit that this requires a lower but still acceptable pattern speed). The article you link to I now wrote to Jesse Christiansen (who the linked article quotes) and asked her if she knows whether or not we are moving in and out of spiral arms; she replied within roughly 8 seconds, tagging Karen Masters who chimed in even faster — they both agree that this is an ongoing debate with no conclusive evidence. Anyway, the article seems to have misunderstood the tweet from Jesse Christiansen. In her animation she shows the journey of the Sun, but shows the galaxy itself as being static, which she did on purpose to keep it simple. Hence, you see the Sun traversing the arms unnaturally fast. | {
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33,902 | After my observation today at the Skytree in Tokyo , I am wondering if this building could be used as a giant sundial? As an aside, it would be interesting to consider how the Skytree might be used to display the time. | What is a spiral arm? The reason that the Sun, in principle (but see below), may cross spiral arms is that galactic spiral arms are not rigid entities consisting of some particular stars; rather they are "waves" with a temporary increase in density. An often-used analogy is the pile-up of cars behind a slow-moving truck: At all times, all cars are moving forward, but for a while, a car behind the truck will be moving slow, until it overtakes and speeds up. Similarly, stars may overtake, or be overtaken by, the spiral arms. Inside a certain distance from the center of the galaxy called the corotation radius ( $R_\mathrm{c}$ ) stars move faster than the arms, while outside they move slower. Since the stars and the interstellar gas follow the rotation of the arms for a while once they're inside, the density of the arms is higher than outside, but only by a factor of a few (e.g. Rix & Rieke 1993 ). When interstellar gas falls into the potential well it is compressed, triggering star formation. Since the most luminous stars burn their fuel fast, they will mostly have died once they leave an arm. Hence, what we see as spiral arms is not so much the extra stars, but mostly due to the light from the youngest stars which are still inside the arms. Since most luminous also means hottest, their light peaks in the bluish region — hence spiral arms appear blue. Origin of the spiral structure At least the most prominent spiral arms (especially grand designs ) are thought to be created by these long-lived, quasi-stationary density waves ( Lin & Shu 1964 ). The reason that the density waves exist in the first place is not well-understood, I think, but may have to do with anisotropic gravitational potentials and/or tidal forces from nearby galaxies (e.g. Semczuk et al. 2017 ). But in fact even small perturbations may spawn gravitational instabilities that propagate as density waves. In computer simulations of galaxy formation, even numerical instabilities may cause this, so the fact that your simulated galaxy has spiral arms doesn't necessarily mean that you got your physics right. When the luminous and hence massive stars die, they explode as supernovae. The feedback from this process, as well as that exerted by the radiation pressure before they die, may help maintaining the density waves, at least in flocculent galaxies ( Mueller & Arnett 1972 ). Perhaps this so-called Stochastic Self-Propagating Star Formation may also initiate the density waves (see discussion in Aschwanden et al. 2018 ). The rotation speed of the material in the galactic disk is roughly constant with distance from the center (this is mainly due to the dark matter halo hosting a galaxy). Hence, stars close to the center complete a revolution faster than those farther away. In contrast, the spiral pattern rotate more like a rigid disk such that, in an intertial frame, the pattern can be described by a constant angular speed $\Omega_\mathrm{p}$ throughout the disk. However, note that spiral arms are transient phenomena; they appear and disappear with lifetimes of the order of (a few) Myr (e.g. Grand et al. 2012 ; 2014 ). Sometimes you also see multiple spiral patterns propagating with different velocities. The Sun in the Milky Way Note: This section first contained errors based on dubious values for the angular speed of the spiral arms, as pointed out by @PeterErwin and @eagle275. In the case of our Sun, we happen to be located very near the corotational radius $R_\mathrm{c}$ ; we sit at a distance of $R_0 = 8.32\,\mathrm{kpc}$ from the center of the galaxy ( Gillessen et al. 2017 ), while $R_\mathrm{c} = 8.51\,\mathrm{kpc}$ ( Dias et al. 2019 ). Using Gaia data, Dias et al. (2019) find a pattern angular speed $\Omega_\mathrm{p} = 28.2 \pm 2.1\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{kpc}^{-1}$ . At the location of the Sun ( $R_0$ ) this implies a pattern speed of $\simeq235\,\mathrm{km}\,\mathrm{s}^{-1}$ . This is only a little bit slower than (and in fact statistically consistent with) the Sun's velocity of $239\pm5\,\mathrm{km}\,\mathrm{s}^{-1}$ ( Planck Collaboration et al. 2018 ). If the spiral arms were "permanent"*, the timescale for the Sun crossing an arm would hence be gigayears. However, because as described above they're quite transient, we might once in a while overtake a spiral arm. I haven't been able to find firm evidence for whether we have or haven't crossed any arms; Gies & Helsel (2005) argue that we have crossed an arm four times within the last 500 Myr, but base this on matching glaciation epochs with passages through spiral arms (and admit that this requires a lower but still acceptable pattern speed). The article you link to I now wrote to Jesse Christiansen (who the linked article quotes) and asked her if she knows whether or not we are moving in and out of spiral arms; she replied within roughly 8 seconds, tagging Karen Masters who chimed in even faster — they both agree that this is an ongoing debate with no conclusive evidence. Anyway, the article seems to have misunderstood the tweet from Jesse Christiansen. In her animation she shows the journey of the Sun, but shows the galaxy itself as being static, which she did on purpose to keep it simple. Hence, you see the Sun traversing the arms unnaturally fast. | {
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33,908 | I'm not sure if this is the correct stackexchange site for this question. I've been reading this article on Vice Motherboard There’s Growing Evidence That the Universe Is Connected by Giant Structures which while admittedly simplified for simpler folk like me put me wondering: Is it at all possible that when we look out in to the distant universe in one direction, we are seeing the same galaxies and stars at a different point in time than if we look out in another direction? Could this occur if the distant body is moving laterally at a fast enough velocity relative to our position, or if the shape of the universe itself loops around? Or is this not a proper interpretation of current theories of cosmology? | What is a spiral arm? The reason that the Sun, in principle (but see below), may cross spiral arms is that galactic spiral arms are not rigid entities consisting of some particular stars; rather they are "waves" with a temporary increase in density. An often-used analogy is the pile-up of cars behind a slow-moving truck: At all times, all cars are moving forward, but for a while, a car behind the truck will be moving slow, until it overtakes and speeds up. Similarly, stars may overtake, or be overtaken by, the spiral arms. Inside a certain distance from the center of the galaxy called the corotation radius ( $R_\mathrm{c}$ ) stars move faster than the arms, while outside they move slower. Since the stars and the interstellar gas follow the rotation of the arms for a while once they're inside, the density of the arms is higher than outside, but only by a factor of a few (e.g. Rix & Rieke 1993 ). When interstellar gas falls into the potential well it is compressed, triggering star formation. Since the most luminous stars burn their fuel fast, they will mostly have died once they leave an arm. Hence, what we see as spiral arms is not so much the extra stars, but mostly due to the light from the youngest stars which are still inside the arms. Since most luminous also means hottest, their light peaks in the bluish region — hence spiral arms appear blue. Origin of the spiral structure At least the most prominent spiral arms (especially grand designs ) are thought to be created by these long-lived, quasi-stationary density waves ( Lin & Shu 1964 ). The reason that the density waves exist in the first place is not well-understood, I think, but may have to do with anisotropic gravitational potentials and/or tidal forces from nearby galaxies (e.g. Semczuk et al. 2017 ). But in fact even small perturbations may spawn gravitational instabilities that propagate as density waves. In computer simulations of galaxy formation, even numerical instabilities may cause this, so the fact that your simulated galaxy has spiral arms doesn't necessarily mean that you got your physics right. When the luminous and hence massive stars die, they explode as supernovae. The feedback from this process, as well as that exerted by the radiation pressure before they die, may help maintaining the density waves, at least in flocculent galaxies ( Mueller & Arnett 1972 ). Perhaps this so-called Stochastic Self-Propagating Star Formation may also initiate the density waves (see discussion in Aschwanden et al. 2018 ). The rotation speed of the material in the galactic disk is roughly constant with distance from the center (this is mainly due to the dark matter halo hosting a galaxy). Hence, stars close to the center complete a revolution faster than those farther away. In contrast, the spiral pattern rotate more like a rigid disk such that, in an intertial frame, the pattern can be described by a constant angular speed $\Omega_\mathrm{p}$ throughout the disk. However, note that spiral arms are transient phenomena; they appear and disappear with lifetimes of the order of (a few) Myr (e.g. Grand et al. 2012 ; 2014 ). Sometimes you also see multiple spiral patterns propagating with different velocities. The Sun in the Milky Way Note: This section first contained errors based on dubious values for the angular speed of the spiral arms, as pointed out by @PeterErwin and @eagle275. In the case of our Sun, we happen to be located very near the corotational radius $R_\mathrm{c}$ ; we sit at a distance of $R_0 = 8.32\,\mathrm{kpc}$ from the center of the galaxy ( Gillessen et al. 2017 ), while $R_\mathrm{c} = 8.51\,\mathrm{kpc}$ ( Dias et al. 2019 ). Using Gaia data, Dias et al. (2019) find a pattern angular speed $\Omega_\mathrm{p} = 28.2 \pm 2.1\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{kpc}^{-1}$ . At the location of the Sun ( $R_0$ ) this implies a pattern speed of $\simeq235\,\mathrm{km}\,\mathrm{s}^{-1}$ . This is only a little bit slower than (and in fact statistically consistent with) the Sun's velocity of $239\pm5\,\mathrm{km}\,\mathrm{s}^{-1}$ ( Planck Collaboration et al. 2018 ). If the spiral arms were "permanent"*, the timescale for the Sun crossing an arm would hence be gigayears. However, because as described above they're quite transient, we might once in a while overtake a spiral arm. I haven't been able to find firm evidence for whether we have or haven't crossed any arms; Gies & Helsel (2005) argue that we have crossed an arm four times within the last 500 Myr, but base this on matching glaciation epochs with passages through spiral arms (and admit that this requires a lower but still acceptable pattern speed). The article you link to I now wrote to Jesse Christiansen (who the linked article quotes) and asked her if she knows whether or not we are moving in and out of spiral arms; she replied within roughly 8 seconds, tagging Karen Masters who chimed in even faster — they both agree that this is an ongoing debate with no conclusive evidence. Anyway, the article seems to have misunderstood the tweet from Jesse Christiansen. In her animation she shows the journey of the Sun, but shows the galaxy itself as being static, which she did on purpose to keep it simple. Hence, you see the Sun traversing the arms unnaturally fast. | {
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33,941 | Most star systems are binary, but why is that? Why would new stars form close to others, and not (relatively) evenly spread? And even if they are clustered close together, why are the majority binary? Or are binaries the ones that survive, and systems with 3 or more stars tend to be unstable and quickly collide? If that is so, what makes binary systems stable? Why do the stars not simply spiral in and collide, or break away shortly after creation? | Collapsing gas clouds fragment into multiple cores because the Jeans mass, that determines the minimum mass that becomes gravitationally unstable to collapse, becomes smaller if the cloud is able to contract without heating up too much. i.e. $$M_J \propto T^{3/2} \rho^{-1/2},$$ where $\rho$ is the cloud density. Thus if the cloud density can increase but the temperature stays constant(ish), then the Jeans mass shrinks and the cloud becomes unstable to further fragmentation. Multiple systems with $n>2$ are inherently unstable unless they are hierarchical. i.e. a star in a wide orbit around a close pair can be stable, as could two close binary systems orbiting each other. The condition for stability is roughly that the separation of the wider star must be 5-10 times that of the inner pair ([dependent on mass ratios and eccentricities - Eggleton & Kiseleva 1995 ). In most other cases, what happens is the ejection of other stars from a multiple system, leaving behind a binary system Durisen et al. (2001) . What makes a binary stable? We'll what would make it unstable? I'm not sure why you think they should spiral in towards each other. This can't happen unless there is some dissipative mechanism, like tidal interactions. Gravitational waves are ineffective in all but close binaries involving compact stellar remnants. They can't "break away" because they are gravitationally bound. EDIT: Note that it isn't true that most systems are binaries. The majority of "systems" are in fact single stars. The binary frequency for solar-type stars is about 50% - i.e. as many singles as binaries; but the binary frequency for the much more numerous M-dwarfs is probably around 30% and so single stars outnumber binary systems, though it is not clear whether that is true at birth Duchene & Kraus (2013) . | {
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33,962 | Stars being flung from black holes in the news made me question if something were heading directly for us at the speed of light, would we even know? Update: Since it's impossible for a star to travel at the speed of light, would we see it coming if it were traveling at 99% the speed of light? and if so, how much warning would we have? | Collapsing gas clouds fragment into multiple cores because the Jeans mass, that determines the minimum mass that becomes gravitationally unstable to collapse, becomes smaller if the cloud is able to contract without heating up too much. i.e. $$M_J \propto T^{3/2} \rho^{-1/2},$$ where $\rho$ is the cloud density. Thus if the cloud density can increase but the temperature stays constant(ish), then the Jeans mass shrinks and the cloud becomes unstable to further fragmentation. Multiple systems with $n>2$ are inherently unstable unless they are hierarchical. i.e. a star in a wide orbit around a close pair can be stable, as could two close binary systems orbiting each other. The condition for stability is roughly that the separation of the wider star must be 5-10 times that of the inner pair ([dependent on mass ratios and eccentricities - Eggleton & Kiseleva 1995 ). In most other cases, what happens is the ejection of other stars from a multiple system, leaving behind a binary system Durisen et al. (2001) . What makes a binary stable? We'll what would make it unstable? I'm not sure why you think they should spiral in towards each other. This can't happen unless there is some dissipative mechanism, like tidal interactions. Gravitational waves are ineffective in all but close binaries involving compact stellar remnants. They can't "break away" because they are gravitationally bound. EDIT: Note that it isn't true that most systems are binaries. The majority of "systems" are in fact single stars. The binary frequency for solar-type stars is about 50% - i.e. as many singles as binaries; but the binary frequency for the much more numerous M-dwarfs is probably around 30% and so single stars outnumber binary systems, though it is not clear whether that is true at birth Duchene & Kraus (2013) . | {
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33,985 | I'm having trouble understanding relative angular/tangential speeds at increasing altitudes above Earth's surface. In particular, I find this comparison of tangential velocities on Wikipedia very confusing. According to it, the tangential speed of Earth's surface (465.1 m/s) is different from the tangential speed required to "orbit" at Earth's surface (7.9 km/s). Why are these different values? My understanding of the Earth was always that material on and within the Earth is orbiting the center of the Earth just like satellties do. Time for multiple questions in one post... Is material on Earth's surface not in free fall around Earth's center? How are geostationary orbits even a thing? Seems like the only orbit that could be geostationary would be standing on the Earth's surface. What changes as you orbit further above the Earth's surface? Does your angular speed increase or decrease? Does your tangential speed increase or decrease? Is magma near the center of the Earth not rotating faster than material in the crust, like in an accretion disk? Can two objects be orbiting (circularly) at the same altitude but with different tangential speeds? Thanks in advance! | 1. Is material on Earth's surface not in free fall around Earth's center? No. Material on the Earth's surface -- or inside it -- is not in orbit, and so is not in free fall. You can temporarily put yourself into an orbit (and thus into free fall) by jumping up into the air, or jumping off a higher surface. When you do this, you are briefly in a very eccentric orbit (one which would take you very close to the center of the Earth, if the Earth wasn't a solid body) -- but then you hit the ground and are no longer in orbit. The Earth rotates in the same way that a spinning top rotates; this has nothing to do with orbits. 2. How are geostationary orbits even a thing? Seems like the only orbit that could be geostationary would be standing on the Earth's surface. Again, the surface of the Earth is not orbiting. The Earth rotates as a rigid body, with (as AtmosphericPrisonEscape noted) residual angular momentum left over from its formation, like a spinning top. Because your angular speed in an orbit decreases the further away you are from the Earth, there will be a point where it happens to match the Earth's spin rate. If you arrange the orbit so it is above the equator and in the same direction as the Earth's spin, then you will always be above the point on the equator: a geostationary orbit. 3. What changes as you orbit further above the Earth's surface? Does your angular speed increase or decrease? Does your tangential speed increase or decrease? Both your angular speed and your tangential speed decrease the further away you get. (Your angular speed would decrease even if your tangential speed stayed the same, because the circumference of your orbit increases with altitude; but in fact the tangential speed decreases as well.) 4. Is magma near the center of the Earth not rotating faster than material in the crust, like in an accretion disk? The Earth rotates approximately as a rigid body, so in general, no. The molten outer core (which is not magma) may rotate slightly slower, while the solid inner core might rotate a little faster, but we're talking about $\sim 0.1$ degrees per year differences , and this has nothing to do with orbits. (The Earth is nothing like an accretion disk.) 5. Can two objects be orbiting (circularly) at the same altitude but with different tangential speeds? Ignoring minor deviations due to things like the non-spherical nature of the Earth, mass concentrations in the crust, etc., the orbital speed for a circular orbit is a function of the altitude only. So two objects in circular orbit at the same altitude must have the same tangential speed. (Note that they can have different velocities , because velocity is a vector quantity -- so you can have two object orbiting in different -- even opposite -- directions at the same altitude, at least until they run into each other.) | {
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34,076 | Gaia is an astrometry spacecraft that's currently operating around the Sun-Earth L2 Lagrangian point. Question: why here? Why not the Sun-Neptune L2 Lagrangian point? By orbiting the Sun at a larger distance, it should be able to get more accurate parallax measurements. Only reason I can think of is cost. I'm not familiar with estimating how expensive space probes cost, but Wikipedia says Gaia cost ~\$1 billion and this is comparable to the cost of the Voyager program, which also cost about ~\$1 billion. Of course Gaia's instruments should be more sophisticated than Voyager's, but there were also two Voyager probes, not one. | Well, you thought about the spatial aspect of a parallax measurement, but not about the temporal one. Gaia's intention is to measure 3D positions as well as 3D velocities. For the distance, you need accurate parallactic measurement, which come in with your orbital period. For a typical Gaia-star with several measurement per year, you'll get 5 values of the parallax after 5 years of time, which you then average. If you'd send Gaia towards Neptune (besides the fact that no one has ever sent an orbiter, to say nothing of a L2 mission that far out) that has a period of 168 years, then after 5 years you'd get... 5/168 th of one paralactic measurement. It simply couldn't achieve its science goals if put around the L2 behind Neptune. Also no one on this planet has any experience in putting something into a outer system L2 point. This is different than putting it into Earth's L2, because reaching the L2 around one of the giants has vast and very precise $\Delta v$ requirements. This would be a massive technological leap, and things don't work that way in space. Small, incremental technological steps are required in an anyways unfriendly environment, to make sure everything works properly and no millions of dollars have been wasted. Compare that to Gaia's predecessor, the Hipparcos satellite , which was parked in geostationary orbit. Now you could still say, why not use Jupiter hypothetically anyways. Well, the orbital period there is still 11 years, and Jupiter's L2 still suffers from the intense radiation environment that is provided by Jupiter's magnetosphere. This would lead to rapid degradation of the CCDs used for scanning across the sky. | {
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34,179 | I remember a few years ago hearing that Jupiter was an anomaly in the landscape of exoplanets. Back then, most Jupiter-mass planets discovered were Hot Jupiters, orbiting very close to their host star. In the last few years since I heard this, many new exoplanets have been discovered. Do we know now if there are many more Hot Jupiters and our own Jupiter is a rather rare occurrence, or if this was an observational bias and there are many more "Cold Jupiters"? | It depends on how you define Jupiter analogues. There are several possible factors, including mass, eccentricity and orbital period cutoffs. Given there's no consistent definition, comparison of results between the various papers is difficult. For example, the recent paper by Wittenmyer et al. considers "cool Jupiters" to be planets with masses greater than 0.3 Jupiters with orbital periods longer than 100 days. These planets do seem to be much more common than the hot Jupiters but this category is a lot broader than just "Jupiter analogues". It includes objects like HD 208487 b, a planet which would be located between Mercury and Venus in our Solar System and has a far more eccentric orbit ( e =0.3) than any of our major planets: hardly a Jupiter analogue. Many of the long period planets have high eccentricities. Imposing an eccentricity cutoff would tend to change things a bit. Other considerations might involve imposing upper limits on the mass, or a different lower limit. The paper notes that their conclusions about the rate of occurrence of Jupiter analogues is consistent with previous studies once the different criteria are imposed. | {
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34,219 | This answer to Can / should the Starlink satellites be blackened? mentions Space News' SpaceX working on fix for Starlink satellites so they don’t disrupt astronomy where SpaceX chief operating officer Gwynne Shotwell is quoted as saying the following: Shotwell admitted that nobody in the company anticipated the problem when the satellites were first designed. “ No one thought of this,” she said. “We didn’t think of it. The astronomy community didn’t think of it. ” Question: Did nobody in the Astronomy community think ahead of time that 12,000 new satellites in LEO might be a problem? Did SpaceX sneak this in under the astronomical observer community's nose by starting "small" ( SpaceX's 4,425 satellite constellation - what's the method to the madness? ) and then "drifting" up to 12,000 satellites without anybody realizing it? Or did the community actually have some good idea that this was likely going to be a problem? fyi it may not stop at 12,000: Space News: SpaceX submits paperwork for 30,000 more Starlink satellites WASHINGTON — SpaceX has asked the International Telecommunication Union to arrange spectrum for 30,000 additional Starlink satellites . SpaceX, which is already planning the world’s largest low-Earth-orbit broadband constellation by far, filed paperwork in recent weeks for up to 30,000 additional Starlink satellites on top of the 12,000 already approved by the U.S. Federal Communications Commission. SpaceX could one day operate 42,000 Starlink satellites if it builds and launches 30,000 in addition to the 12,000 for which it already has FCC approval. Credit: SpaceX | “No one thought of this,” she said. “We didn’t think of it. The astronomy community didn’t think of it.” What utter nonsense. SpaceX are either lying or they didn't bother to go looking for any other opinions. Astrophotographers have been complaining about satellites messing up their photos for years. When Starlink was announced astronomers complained even louder. Go to any popular astronomy forum and search for "starlink" and see for yourself #1 #2 . | {
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34,521 | In a different (but somewhat related) field, some baseball stars have been known to have "baseball eyes." That is, an exceptional ability to visually follow the trajectory of a 90+ mph baseball to a degree enjoyed by 0.1% (one in a thousand) of the human race, as discussed in "Moneyball" by Michael Lewis. I would imagine that would also be an advantage for an astromer, all other things being equal. But has "history"shown this to be necessary? That is, have there been a large or at least disproportionate number of famous and successful astronomers that been identified with eyesight in the top 1% (or higher) of the human population? Conversely, have there been any noted astronomers with notably bad eyesight who have made contributions because their theoretical, intuitive, or other abilities were enough to compensate? | Johannes Kepler Wikipedia: "However, childhood smallpox left him with weak vision and crippled
hands, limiting his ability in the observational aspects of
astronomy." He made great use of Tycho Brahes great systematic observations in his theoretical work. He did not need exceptional eyesight for his developments in optics and telescopes. | {
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34,775 | Type Ia supernova are used as standard candles . But they also are transient events. This means that to determine the distance of a galaxy using supernovae, you have to wait for one to occur. How often do type Ia supernovae typically occur in a galaxy ? | There is a lot of scope to provide a very detailed answer here. The rate depends on what sort of a galaxy you are considering and when, what its star formation rate is (or was) and what its total stellar mass is. A good reference is the Annual Review of Astrophysics article by Maoz et al. (2019) . This says that for a Sbc galaxy like the Milky Way, the specific rate ( the rate per unit of stellar mass in the galaxy) of type Ia supernovae is $10^{-13}$ per year per solar mass of stars. For a Milky Way stellar mass of $\sim 6\times 10^{10} M_\odot$ , this means a rate of 0.006 per year, or one type Ia SNe every $\sim 200$ years. In the local universe this rate scales as about $M^{-0.5}$ , where $M$ is the galaxy mass (valid over about 5 orders of magnitude for $M>10^7 M_\odot$ ; Brown et al. 2018 ). So the specific rate of type Ia supernovae is larger in smaller galaxies. The rate also decays after a galaxy is formed, or after a burst of star formation as something like $t^{-1}$ . | {
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34,827 | In about 5 billion years the Sun is predicted to become a red giant and have more than 200 times its current size, reaching a radius of about 5 AU when largest. I wonder what spectral class the Sun would have then, I think somewhere from M1III to M5III. Do we know a red giant star that is similar to what the Sun is predicted to become? | Models for the future behaviour of the Sun do vary, mainly as a result of uncertainty of mass loss during the red giant (H shell burning) and asymptotic red giant (H+He shell burning) phases. A highly cited paper by Schroeder & Smith 2008 claims that the Sun will reach its maximum size of about $256 R_{\odot}$ (1.18 au) at the very tip of the red giant branch (and not at the end of the asymptotic giant branch phase, which is suggested by some other models). This maximum size will occur 7.6 billion years in the future (not the 5 billion years of popular literature), when the Sun's mass will be reduced to about $0.7M_{\odot}$ and have a surface temperature of 2600 K (or about 2300 Celsius). This would have a spectral type of M6III, or perhaps even M6I. Do we know of a star similar to this? It depends what you mean by similar, but there are unlikely to be any stars like this in our Galaxy. The reason for this is that star formation began in our Galaxy about 12 billion years ago. But a 1 solar mass star like the Sun requires about 12 billion years (or a bit more) to reach the tip of the red giant branch. Even were this just about possible time-wise and we were to find a red (super)giant star of about $0.7 M_{\odot}$ , it would be highly unlikely that the star would have a similar chemical composition to the Sun. That is because stars born early in the life of our Galaxy would have very low concentrations of metals like iron and nickel that are only produced inside stars and only present in the interstellar medium when a generation of stars have lived and died. So my perhaps pedantic answer, is that there aren't any big red giants we can see now that started life as 1 solar-mass stars and have a similar chemical composition to the Sun. | {
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34,963 | Since we see the Triangulum galaxy M33 from a quite vertical position (contrary to our Milky Way and a bit the Andromeda galaxy) it should be easy to image the black hole in the center of it, shouldn't it? Why did they prefer to first image the black hole in a galaxy which is about 20 times farther and thus harder to photograph the black hole in its center? | M33 does not appear to contain a supermassive black hole: in fact there's no evidence that it contains a central black hole at all. The upper limit on the mass of a central black hole based on the dynamics of the core region is a few thousand solar masses. Merritt et al. (2001) " No Supermassive Black Hole in M33? " derive an upper limit of 3000 solar masses on a central compact object in M33, noting that this is still consistent with the M-σ relationship between the mass of a supermassive black hole and the velocity dispersion in the stellar bulge, using which they obtain a predicted mass of 2600–26300 solar masses. Gebhardt et al. (2001) " M33: A Galaxy with No Supermassive Black Hole " obtain an even smaller upper limit on the mass of a black hole of only 1500 solar masses (their best fit mass is zero, i.e. no central black hole), which they state is significantly lower than the predicted mass from the M-σ relationship. | {
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35,218 | If a star is at a distance of one lightyear, how old are its photons when they reach earth (from the photons’ perspective)? If time dilation is near zero at light speed, can we assume that the light that we see today from a distant star has the same age as when it was emitted? | Photons can't have a perspective. If we have a particle with mass, we can imagine taking a frame of reference in which that particle is at rest. We can then see things "from the particle's perspective". But there is no frame of reference in which a photon is at rest. Photons always move at the speed of light in every frame of reference. If I try to set up a frame of reference which is moving at the speed of light there is a singularity. The universe has no time, and the whole of space is squashed into two dimensions. So in a very real way, a photon doesn't have a perspective. We can only consider time in a frame of reference that includes it. It makes perfect sense to say that the photons are one year old in our frame of reference. And that is the best we can say. | {
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35,254 | If the sun is constantly converting the mass into energy, then will its gravitational field continue decreasing? | If the sun is constantly converting the mass into energy, then will its gravitational field go on decreasing? It's a very interesting question and the answer is yes! The solar constant indicates the mean solar radiation of electromagnetic waves (mostly in visible and near infrared light and I'll answer based on that. While the conversion of mass matter † to energy in the Sun's core now represents a loss of mass proper matter, it turns out that that energy (trapped in the Sun and slowly diffusing towards the surface) will have the same gravitational attraction as the matter it came from until it actually escapes the Sun! There is some prompt mass and energy loss via neutrinos and it's significant, perhaps several hundred keV per neutrino I simply don't know the number yet. I'll ask a separate question about it. I'm guessing that losses due to the stellar wind are small, but I'll update here as soon as the following is answered: How much mass does the Sun lose as light, neutrinos, and solar wind? update: the answer there is that loss via neutrinos is only about 2.3% of the radiative loss, and on average loss via solar wind and coronal mass ejections is about 4E+16 kg/year, or about another 30% relative to the radiative loss described below. The value $I$ is about 1360 Watts per square meter at $R$ = 1 AU which is about 150 million kilometers or 150 billion meters. So the total energy lost per second $P$ is $$P = 4 \pi R^2 I$$ Taking the time derivative of $E = m c^2$ we get $$\frac{dE}{dt} = P = \frac{dm}{dt} c^2$$ so $$ \frac{dm}{dt} = \frac{1}{c^2} \ 4 \pi R^2 I$$ That means that the value of the mass that we use to calculate the Sun's gravitational attraction changes by about 4.3E+09 kilograms per second, or 1.3E+17 kilograms per year. The Sun's current mass is about 2.00E+30 kilograms, so this effect changes by a very tiny fraction per year, about 6.7E-14. Over the age of the Earth of 4.5 billion years, that's 3E-04, or about 0.03% if the Sun's output were constant. It has probably changed over this time of course, so this is just a rough estimate. † Thanks to @S.Melted's answer for clarifying this. added to my answer by @Tosic: The Earth feels no torque from any force during this (the force from radiation is radial), which means its angular momentum is conserved. This means $$R_1 v_1 = R v$$ $$R_1 \sqrt{\frac{GM_1}{R_1}} = R \sqrt{\frac{GM}{R}}$$ $$M_1 R_1 = M R$$ We can see the Earth's orbital radius would change by a factor of 0.03% as well (M1 and M are solar masses). | {
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35,260 | Could Hawking radiation be detected with instruments available today? If not, what sort of instruments would be required to observe it? Would there be any way to confirm through observations the existence of Hawking radiation? I'm primarily asking about techniques that we could reasonably implement with current technology or something that could be implemented in the near future, but if you need to get close to a black hole to be able to observe the radiation and couldn't do it from a spacecraft in Earth or solar orbit, then that would be the answer. | If the sun is constantly converting the mass into energy, then will its gravitational field go on decreasing? It's a very interesting question and the answer is yes! The solar constant indicates the mean solar radiation of electromagnetic waves (mostly in visible and near infrared light and I'll answer based on that. While the conversion of mass matter † to energy in the Sun's core now represents a loss of mass proper matter, it turns out that that energy (trapped in the Sun and slowly diffusing towards the surface) will have the same gravitational attraction as the matter it came from until it actually escapes the Sun! There is some prompt mass and energy loss via neutrinos and it's significant, perhaps several hundred keV per neutrino I simply don't know the number yet. I'll ask a separate question about it. I'm guessing that losses due to the stellar wind are small, but I'll update here as soon as the following is answered: How much mass does the Sun lose as light, neutrinos, and solar wind? update: the answer there is that loss via neutrinos is only about 2.3% of the radiative loss, and on average loss via solar wind and coronal mass ejections is about 4E+16 kg/year, or about another 30% relative to the radiative loss described below. The value $I$ is about 1360 Watts per square meter at $R$ = 1 AU which is about 150 million kilometers or 150 billion meters. So the total energy lost per second $P$ is $$P = 4 \pi R^2 I$$ Taking the time derivative of $E = m c^2$ we get $$\frac{dE}{dt} = P = \frac{dm}{dt} c^2$$ so $$ \frac{dm}{dt} = \frac{1}{c^2} \ 4 \pi R^2 I$$ That means that the value of the mass that we use to calculate the Sun's gravitational attraction changes by about 4.3E+09 kilograms per second, or 1.3E+17 kilograms per year. The Sun's current mass is about 2.00E+30 kilograms, so this effect changes by a very tiny fraction per year, about 6.7E-14. Over the age of the Earth of 4.5 billion years, that's 3E-04, or about 0.03% if the Sun's output were constant. It has probably changed over this time of course, so this is just a rough estimate. † Thanks to @S.Melted's answer for clarifying this. added to my answer by @Tosic: The Earth feels no torque from any force during this (the force from radiation is radial), which means its angular momentum is conserved. This means $$R_1 v_1 = R v$$ $$R_1 \sqrt{\frac{GM_1}{R_1}} = R \sqrt{\frac{GM}{R}}$$ $$M_1 R_1 = M R$$ We can see the Earth's orbital radius would change by a factor of 0.03% as well (M1 and M are solar masses). | {
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35,431 | This might sound like a strange question, but something got me thinking about it recently. The opacity of plasma in stellar interiors can get quite high, making for shorter free-paths for photons. In these conditions I guess that the light you could theoretically gather, supposing you have a pair of indestructuble eyes submerged in the solar interior, would be the one emitted by the plasma in your immediate surroundings, right? So if the opacity is high enough I can imagine places inside a star like the Sun where there is the same ambient illumination as a typical moonless night here on Earth. My questions are: Is this line of reasoning correct? Are these conditions actually possible inside a star? Where exactly inside a star are these conditions possible? The answer ProfRob gave an excellent answer, as always, but in case someone wants a plain language summary we talked about this: If the optical depth of the solar plasma is about a few micrometers this means that, when I'm inside the Sun, I'm observing the plasma only in an extremely small spherical volume (bacterium sized in fact) around my eye and that the rest of the Sun is hidden from me, as if it didn't existed. No photons outside that extremely tiny volume around my eyes will reach them. This is because there's high opacity. But because of opacity we also know that the plasma in that tiny volume must absorb a lot of light, and that energy rises the temperature of the plasma. You might not be seeing a lot of material but this material absorbs so much light from the surroundings that it is itself re-emitting light in the form of thermal radiation. The plasma is in fact a black-body radiator. In thermal equilibrium the absorbed energy (by the opaque plasma) is the same as the emitted one (as black-body radiation). Thus, if we were outside of thermal equilibrium, like during a supernova event, we might get a different result. In any case what you would see if your eyes where inside the stellar plasma is the same as what you would see if you pressed your eyeballs to the surface of a star (assuming that the temperature inside is the same as that of the surface, which is only true for a small depth (as you go deeper the temperature increases and you would see it even brighter). | No, it's not. The radiation field in the interior of the Sun is very close to a blackbody spectrum. If you look in any particular direction the brightness (power per unit area) you see is $\sigma T^4$ , where $\sigma$ is Stefan's constant. Even at any particular wavelength it is always the case that a blackbody of higher temperature is brighter than a blackbody at lower temperature. Given that the interior temperature might be $10^7\ \mathrm K$ , then the surface brightness is $5.7 \times 10^{20}\ \mathrm{W/m^2}$ , compared with the $1400\ \mathrm{W/m^2}$ you would get by looking directly at the Sun ( please don't do this ). Note that most of this power comes out at X-ray wavelengths, but because of the properties of a blackbody, the brightness at visible wavelengths will still be plenty brighter than that of the solar photosphere (see below). A possible source of confusion is this term "opacity". When things are in thermal equilibrium, which the interior of the Sun is, then they emit the same amount of radiation as they absorb. So high opacity also means high emissivity. Details for interest: The opacity, $\kappa$ in the solar interior ranges from 1 cm $^2$ g at the centre to about $10^5$ cm $^2$ g just below the photosphere. To estimate the mean free path of photons we need to multiply this by the density $\rho$ and take the reciprocal: $$ \bar{l} = \frac{1}{\kappa \rho}\ .$$ The density varies from 160 g/cm $^3$ at the centre to about 0.001 g/cm $^3$ just below the photosphere. Thus the mean free path is about 6 micrometres at the center and is actually quite similar just below the photosphere (it peaks at around 2 mm about three quarters of the way out towards the surface). Thus your "view" of the stellar interior is of a foggy sphere with radius of no more than a few times $\bar{l}$ . The fog however is tremendously bright - as outlined above. The brightness at particular wavelengths is proportional to the Planck function $$B_\lambda = \frac{2hc^2}{\lambda^5} \left(\frac{1}{\exp(hc/\lambda k_B T) -1}\right).$$ Thus at $\lambda=500$ nm (visible light), the ratio of brightness for blackbodies at $10^7$ K (solar interior) to 6000 K (solar photosphere) is $4.2\times 10^{4}$ . i.e. Even just considered at visible wavelengths, the interior of the Sun is about 40,000 times brighter than the photosphere. | {
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35,575 | From the LIGO website , black hole mergers have been observed between black holes with a mass up to roughly 50 $M_\odot$ . Are there no black holes with a mass above 100 $M_\odot$ or is this an observational bias? Why haven't we observed any mergers between black holes with a mass in the 100 - 1000 $M_\odot$ range? | It is quite likely there is an astrophysical upper limit to the mass of a black hole that can be produced during the core collapse of a massive star, caused by the pair instability supernova phenomenon. There isn't an observational bias against detecting more massive black holes in the range of 100 to a few hundred $M_{\odot}$ . Details: The frequency of the gravitational waves is twice the orbital frequency of the binary system. The general scenario is that early in their evolution, a mrging binary system will be orbiting at relatively long periods (seconds !), but as gravitational waves take energy out of the orbit, the system becomes more compact, the orbital period gets smaller and the frequency of the emitted gravitational waves goes up. This continues until the black holes get so close together that their event horizons merge. Very roughly, we can derive (from Kepler's third law, not going into detail), using Keplerian orbits $$ f_{\rm max} \sim \left( \frac{GM}{\pi^2 a_{\rm merge}^3} \right)^{1/2}\ ,$$ where $f_{\rm max}$ is the peak frequency at merger (when the gravitational wave signal is also maximised), $a_{\rm merge}$ is the separation of the masses at merger and $M$ is the total mass of both black holes. If we let $a_{\rm merge} \sim 2GM/c^2$ , the sum of the two Schwarzschild radii of the black holes, then $$f_{\rm max} \sim \frac{c^3}{GM} \left( \frac{1}{8\pi^2}\right)^{1/2} \sim 2\times 10^4 \left(\frac{M}{M_{\odot}}\right)^{-1}\ {\rm Hz}$$ Now, LIGO is limited to observing frequencies above about 20 Hz. The sensitivity drops off rapidly below that because of seismic noise and other factors. If the mass of the merging black holes exceeds some critical value then the frequencies of the gravitational waves they produce never get into the sensitivity wndow of LIGO. Using the expression above, we can estimate that this happens only if the total mass exceeds $1000 M_{\odot}$ . Observing the mergers of more massive black holes would require a detector that is sensitive to lower frequencies, probably beyond the surface of the Earth (e.g. LISA ). This calculation is only good to a factor of 2 or so, but we can check it. GW150914 had a total mass of around $65 M_{\odot}$ and merged at frequencies of about 120 Hz. Since $f_{\rm max}$ scales as $M^{-1}$ this suggests 360 solar mass mergers should be just about detectable, but clearly demonstrates that LIGO could detect black holes of 100-200 solar masses. What's more, at a given distance and frequency, the signals from such mergers would be more powerful than for less massive black holes -- something like $h \propto M^{5/3}$ , which means the volume in which the mergers would be visible goes as $M^5$ . Thus more massive black hole binaries would have to be extremely rare in order to have evaded detection. The astrophyscal reason for an upper limit is the pehenomenon of pair instability supernovae (e.g. Farmer et al. 2019 ), which blows the star apart rather than leaving a black hole (or any other kind) of remnant. This likely happens for stars with initial masses of $130+\ M_{\odot}$ , and means that leaving behind black holes with $M > 50M_{\odot}$ is very difficult, with even lower mass limits for stars with a metallicity more similar to the Sun, since they lose more mass in stellar winds during their lives. For initial masses of $250+ M_{\odot}$ it is possible that the pair instability supernova mechanism ceases and direct collapse to a black hole becomes possible. In which case their might be a population of $300+ M_{\odot}$ mergers that are just below LIGO's sensitivity window. New Earth-based gravitational wave detectors like the Einstein Telescope and Cosmic Explorer aim to push their low frequency response down to a few Hz and might be capable of detecting mergers in the 300-1000 $M_{\odot}$ range. This means you could not get a merger pair between about $100 M_{\odot}$ and $300 M_{\odot}$ (unless they themselves were the products of a merger). | {
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35,914 | I understand that solar radiation causes material to vaporize out of a comet into dust but why does the dust then trail behind the comet like a "tail"? Assuming gravity is the only applied force acting on the comet, shouldn't all of the material, including the dust, be travelling at the same speed due to conservation of momentum? What causes the dust to travel slower than the comet's nucleus? In other words, why does the dust form a "tail" and not a "cloud"? | There are two forces that can cause the formation of a tail: the solar wind and radiation pressure. The first misconception in your question is "the dust [travels] slower than the nucleus". The tail is not left trailing behind the comet, it is pushed away from the comet by the sun. When the comet is moving away from the sun, the tail is in front of the comet. Now radiation pressure is small but real. When light shines on something there is a small force. This pushes dust back from the comet in the direction opposite to the sun. The dust is still affected by gravity and a curved dust tail results. The ultraviolet light from the Sun ionises the gas and gives it an electric charge. The solar wind carries magnetic fields and the gas (or more properly plasma ) follows these fields in a straight line back from the sun. So space around the Sun is not empty. There is powerful light and magnetic fields that are strong enough to push the dust and gas released by the comet away from the coma, and form the tail. | {
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36,040 | All of the Galilean moons are large and close enough to Jupiter that they can completely eclipse the sun and allow a solar eclipse to happen. My question is can an observer from Earth, see the moon(s)' shadow(s) on the surface of Jupiter with a telescope or are the shadows too small? | Yes, you can see it (I have seen Io's shadow on Jupiter and we were happy it was visitor night so that we could share the view with guests) A 50cm mirror and 125x magnification allows you to see it when the air is not too disturbed and when you know where and when to look. Likely a somewhat smaller telescope will do, too, as light sensitivity is not too crucial - more the resolution and magnification so that you can still see the tiny black dot of the shadow. The shadow has about the same diameter as the width of some major clouds bands on Jupiter. Stellarium is a great tool to find the right time to look at Jupiter to find these transits. Find a local amateur observatory and they sure will be happy to share this view with you (possibly not before late northern summer due to the rise time of jupiter in 2020) | {
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36,054 | I’ve seen a lot of photos showing Mercury in front of the Sun when it passes by and you can see just how tiny it is in comparison. Here’s a great example: So I’d expect if I was travelling towards Mercury in that photo that as the planet got bigger in my view so would the Sun and eventually I wouldn’t be able to see the edges of the sun because it’s so big at this distance... makes sense that an object gets bigger the closer you get to it! However what confuses me is that if you were on the surface of Mercury... this is the view you’d get where it’s about a 6 times larger than when viewed from Earth. So how is this so small when it’s taking the whole view up when we’re viewing it further away from mercury in the first photo? But we’re closer in the second photo but now the sun is smaller... How is the first photo possible (it’s an actual photo taken by NASA) why isn’t it taking up the whole sky in the second photo when you’re actually on the planet? | From where we stand on Earth, Mercury is pretty small about 13 arcseconds across at most. The sun, by comparison is about 1800 arcseconds across, so if you are to see Mercury as a disc, you need to magnify your image a lot. And that makes the sun appear very very big. It only appears very big because it has been magnified. But if you are on Mercury, you don't need to magnify the image of the sun. The sun on Mercury is about 5000 arcseconds across. Big, but not filling the sky. That is just because it hasn't been magnified | {
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36,171 | Eddington waited for a total solar eclipse to happen to be able to observe gravitational lensing of the stars behind the Sun. And nowadays, amateurs can do the same thing . Of course, the Moon is much less massive than the Sun, so it doesn't bend light as much. But since Eddington's time, our observation capabilities have greatly increased. Is it possible to detect the gravitational lensing of stars that are behind the Moon? | Measuring the gravitational deflection of light by the Moon is just out of reach of current observational techniques. The angular deflection caused by the lensing of a distant background object by a foreground (nearby) object is given by $$\theta \simeq 4 \frac{GM}{Rc^2},$$ where $M$ is the mass of the lensing object and $R$ is the closest projected distance of the ray from the centre of the mass. For the Moon, the maximum deflection would occur when the ray just grazes the limb of the Moon and would equal 26 micro-arcseconds. At present the most accurate instruments for measuring precision positions are VLBI radio observations of point sources, where 10 micro-arcsecond relative positional precision can be "routinely achieved" ( Reid & Honma 2014 ). To use this technique you need bright, radio point sources to be close in position to the Moon. This is certainly possible and has been done many times using the Sun as the lensing object, although the claimed accuracies of a short time-series of measurements on single sources is an order of magnitude lower and so probably wouldn't work for the Moon (e.g. Titov et al. 2018 ). The Gaia astrometry satellite has likely end-of-mission positional precisions of about 5 micro-arcseconds for bright stars, but it is unclear what the precision is for a single scan (which would be needed because the Moon moves!). However, to counterbalance that, one could also average the lensing effect over many stars surrounding the Moon and of course average this for different stars with the Moon observed at different times over the entire Gaia mission. Unfortutunately , although in principle this could be done, because of the way that Gaia scans the sky it is always pointing away from the Moon and so there will be no observations towards the Moon that would be capable of revealing these small deflections (e.g. see here ). | {
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36,191 | When we talk about the reason for the seasons, we usually have to dispel the misconception that seasons are caused by being close and far away in the Earth's elliptical orbit. And usually, we mention that the Earth is actually closest to the sun in January, in the dead of winter (for the Northern hemisphere). But when did astronomers first have the Earth's orbit measured carefully enough that they knew the Earth was slightly closer during January? How was that measurement made? How accurate were the first measurements? They didn't they measure the size of the disk of the sun very, very carefully, did they? Perhaps with a pin-hole camera? That seems like it would be very difficult to do. Additional I guess if we are talking about long enough ago, they would have thought it was the Sun's orbit that brought it closer because either because of an eccentric (the idea that the orbit of an ancient planet had a center that was offset) or because of epicycles bringing the Sun closer on the circle on a circle. I just wonder what sort of observation they might have made. One of the nicest comparisons I've seen can be found here. If it was me with the tools available in ancient times, I'd probably use a rotatable camera obscura , and maybe a cone with markings to place at the center of the image of the sun, to exaggerate the effect of the size differences. Second Additional Based on JdeBP's answer I want to see if I have the correct concept. (I would put this in comments, but comments can't be nicely formatted.) Doing a search for the dates and times of the solstices and equinoxes, and finding the time between those dates and times, I found the lengths of the upcoming seasons. Summer 2020 is 93 days, 15 hours, 47 minutes Fall 2020 is 89 days, 23 hours, 0 minutes Winter 2020 is 88 days, 21 hours, 7 minutes Spring 2021 is 92 days, 17 hours, 54 minutes Summer 2021 is 93 days, 15 hours, 49 minutes If we subtract from 1/4 of an astronomical year, we get about: $$
\begin{matrix}
Spring & +1.4 \: days & & Summer & +2.4 \: days \\
Fall & -1.4 \: days & & Winter & -2.4 \: days
\end{matrix}
$$ From there it seems like with a geocentric model with an eccentric, we could get a good approximation for the date of perihelion. I'll have to think about the details of how to get there, though. | Hipparchus, not Kepler Kepler got the conic sections right, and Newton gave us the mechanics. But the question is about when people knew that the Earth was closer to Sol in one part of the year than others, and Hipparchus knew that , even though he wasn't too hot on the values of the orbital radii. Hipparchus' version of the eccentric model had Sol's (purported) circular orbit around the Earth not centred upon the Earth, but 1/24th of an AU away. Therefore Sol (purportedly) orbited at varying distances from Earth. This was, after all, the whole point of the eccentric model, to explain non-uniform apparent motion through variation in distance. Perigee and apogee were known in the times of Hipparchus and Ptolemy. Hipparchus even worked out when the furthest point (apogee) was. Ptolemy furthermore made an error based upon knowing that his placement of the apogee in Gemini was the same as that of Hipparchus 280 years beforehand, declaring that perigee and apogee were fixed. They of course were not. Hipparchus placed apogee at 5.30° Gemini. Astronomers in the 9th century in Baghdad applied the same calculations to their measurements and placed it at 20.45° Gemini. As for how this was observed, it was not done by measuring the Sun's appearance at all (although Hipparchus did do that). Ptolemy and Hipparchus had a geometric model of a true geocentric circular orbit versus the (purported) eccentric circular orbit of Sol. It incorporated the equinoxes and the solstices. By observing the times of the equinoxes and solstices, the lengths of the periods between them, they were able to determine trigonometrically all of the other orbital parameters, which included placement of perigee and apogee. That points of closest and furthest approach existed was known in the 2nd century BC, as was their angular locations relative to the solstices; they've been in the models from then onwards. That they moved around took about 11 centuries after that to discover. The correct conic sections and the idea of both bodies orbiting around a barycentre came somewhat later, but that wasn't the question . Further reading Hugh Thurston. (1994) " Appendix 2: Calculation of the Eccentric Quotient for the Sun, and the Longitude of its Apogee ". Early Astronomy . New York NY: Springer. DOI 10.1007/978-1-4612-4322-9_1 James Evans (1998). The History and Practice of Ancient Astronomy . Oxford University Press. ISBN 978019509539. Viggo M. Petersen and Olaf Schmidt (June 1968). "The Determination of the Longitude of the Apogee of the Orbit of the Sun according to Hipparchus and Ptolemy". Centaurus . Volume 12. Issue 2. Pages 73–96. DOI 10.1111/j.1600-0498.1968.tb00080.x | {
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36,291 | Why does the Parker Solar Probe slow down as the distance from the Sun increases? Image credit: Wikipedia user Phoenix777, CC BY-SA 4.0 | Why does the “parker solar probe” lose speed as the distance from the sun increases? Because energy and angular momentum are individually conserved quantities in the two body problem. Except for where the Parker Space Probe has close fly-bys with Venus, the gravitationally interactions between the Parker Space Probe and the solar system are very closely modeled as a two body problem (the Sun and the probe), plus very small perturbations from the planets. One way to express conservation of energy in the two body problem is the vis-viva equation, $$v^2 = \mu\left(\frac2r - \frac1a\right)$$ where $\mu = G(M+m)$ is the sum of the central body's standard gravitational parameter and that of the orbiting body, $r$ is the distance between the two bodies, $a$ is the semi-major axis length (a constant), and $v$ is the magnitude of the velocity vector. Note that the mass of the Parker Space Probe is so much less than that of the Sun that one can drop the Parker Space Probe's mass from the expression $\mu = G(M+m)$ , resulting in $\mu = GM_{\text{sun}}$ . Note that the only variable on the right hand side of the vis-viva equation is the radial distance. As radial distance increases the square magnitude of the velocity vector (and thus the magnitude of the velocity vector) decreases. Without mathematics, conservation of energy dictates that the sum of an orbiting body's kinetic energy and gravitational potential energy must remain constant. As the orbiting body moves further from the central body, the orbiting body's potential energy increases, which means its kinetic energy must correspondingly decrease. This in turn means the orbiting body's velocity must decrease. | {
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36,294 | Io ( source ): The Moon ( source ): | It's due to the larger relative apparent size of the Sun. When the source of light is a point source the shadow is harder , and when it is extended it is softer . Jupiter is approximately 5 times more distant from the Sun than the Earth, so the Sun is approximately 5 times smaller in the sky. *Source: University of North Carolina CS | {
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36,326 | Its common knowledge that people used to think that the sun is a ball of fire or molten metal, but when did science start to prove otherwise? | I think it's maybe not the case that there was a moment when the astronomy community conclusively rejected the ball-of-fire hypothesis; astronomers simply accumulated more and more evidence against it. If you want to put a rough date on it, you could put your finger somewhere in the middle of the 19th century, as by then, other ideas had taken hold. Back in the classical period, Anaxagoras had proposed that the Sun was a heap of molten metal . I don't know whether this was widely accepted by his contemporaries. The idea of the Sun as a ball of metal or fire certainly persisted for some time, though perhaps largely for lack of any better ideas. We didn't even understand oxygen and combustion until the work of Lavoisier and others in the late 18th century, so detailed calculations were presumably out of the question for a millennium or two after Anaxagoras. I don't know when calculations of how long combustion could sustain the Sun were first done, but it appears to have been not more than several decades after the theory of combustion was developed. Why? Well, we can say that by the middle of the 19th century, the predominant explanation for the Sun's luminosity was not the burning of coal but instead gravitational potential energy. By the 1860s, it was widely known that chemical reactions could only power the Sun for a few thousand years. We also now had a potentially viable alternative: a decade earlier, Hermann von Helmholtz had begun exploring the idea that gravitational contraction of some sort, by what we now call the Kelvin-Helmholtz mechanism , was the source of energy, with gravitational potential energy being transformed into heat $^{\dagger}$ . Around the same time, Lord Kelvin suggested that meteors falling into the Sun provided the necessary energy, a similar mechanism to Helmholtz's. I believe astronomers continued with the contraction hypothesis through the turn of the century - I've seen an article written around 1900 to that effect. However, during the early and mid- 1900s, quantum theory and nuclear physics were being developed, and the work of Eddington, Bethe and others would lay the groundwork for our current understanding of solar energy production. Previous models (including, finally, Kelvin-Helmholtz contraction) were now known to be insufficient because they allowed the Sun to shine for only thousands or millions of years, and geologists had established that Earth itself was much older than this. Fusion, on the other hand, allows the Sun to survive for billions of years - a timescale that matches up well with the age of the Earth. We also knew that hydrogen and helium were the dominant constituents of the Sun and other stars; while Wollaston and Fraunhofer had performed the first solar spectroscopy observations in the early 1800s, the true composition of the Sun was not accepted for more than a century, when Cecilia Payne made a detailed study of spectral lines. $^{\dagger}$ While this does produce heat in various bodies, including T Tauri stars, it is not significant in most stars beyond that stage. | {
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36,613 | I'm trying to explain to my daughter that the Summer Solstice is not a day of the year, but an exact instant in time. My search shows that this will be about 21:43 or 21:44 UTC tonight. To make my point I searched for a more precise time. I figured I could find a time accurate to at least the second, or hopefully several places after the decimal , but I couldn't find anything. Can anyone show me a source for the precise time of the Solstice to at least the second? To the millisecond? The microsecond? | TL;DR: In a couple of weeks it might be possible to say precisely (within a second, perhaps a fraction of a second) when the summer solstice did occur. But until then, sub-minute estimates should be treated as fraudulent. The reason you are seeing different times for when the summer solstice will occur is that different websites use different models of the orbits of the bodies that comprise the solar system and different models of the Earth's orientation in space. It's very important to remember that "all models are wrong, but some are useful" . There are several issues with finding the exact time of when solstices occur. One is that the solstices are extrema of the Sun's declination. Without a model of derivatives, finding extrema (e.g., the solstices) mathematically is a much harder problem than is finding zero crossings (e.g., the equinoxes). More importantly, a very precise prediction would require multiple very precise models. While many models of the orbits of the bodies in the solar system are very, very good, none are good enough to get the timing of the solstices down to the millisecond. Even more importantly, the tilt of the Earth's orientation in space with respect to its orbit about the Sun needs to be modeled. The best model, the IAU 2006/2000A precession-nutation model has over a thousand terms (1365 terms, to be precise). And even then, "all models are wrong." The people at the International Earth Rotation and Reference Systems Service (IERS) know this very well. The IERS is the organization that is responsible for modeling time as measured by the Earth's rotation and for modeling how the Earth is oriented in space. There are some aspects of the Earth's orientation in space that are not quite predictable. The US Naval Observatory, one of the key members of the IERS, publishes daily updates of observed deviations from model predictions. It take a couple of weeks to fully digest measurements made by astronomers worldwide. | {
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36,629 | This question arose when I thought that on Normal Days , at least I have observed bright sun while going outside, going to office, in park etc. for less than a second. It happens almost daily. Not deliberately but that's how things work I guess. Similar example is playing cricket. Many times sun comes in the way of ball in the air and again I have a glimpse of sun. To me, it is very common as I've seen in international cricket matches on TV. On the other hand, in case of the days of Solar Eclipse , on one or two occasions, when the sun wasn't very bright due to clouds, I've deliberately observed the partial eclipse just for a glimpse, just to confirm it's actually cut . And I confirmed it. (How fast it was? It was not like I stood still and observed the sun for less than a second and then closed eyes. It was like this: I started moving my head and sight from earth surface and in a faster way moved my head towards sun and without any break, stopped at safe location on earth again where there was no sun) . And it did literally no harm to me. So my doubt is: Why there's so much theories that you must not stare sun even for less than a second or even a glimpse by mistake? If you see news on eclipse day, you'll see full news about this. But on any other normal day, no such news. Is there something special, something more bright on solar eclipse day? Will it damage eyes? | Glancing at a partial solar eclipse is about as dangerous as glancing at the Sun on any other day. If you look at the Sun a few minutes after sunrise or a few minutes before sunset, when the Sun's altitude is low, the light is filtered through a lot of air, and most of the ultraviolet is scattered, so it's a lot less dangerous than looking at the Sun in the middle of the day. However, during an eclipse, people tend to look at the Sun for more than just a few minutes, and that's not healthy for the eyes. The big danger of looking at a solar eclipse without proper protection occurs when it's a total (or nearly total) eclipse. When the eclipse is near totality, the sky is dark, so your pupils dilate, letting in more light. But the light rays coming from that small sliver of Sun are just as bright as normal, and have the same proportion of UV (ultraviolet) as full sunlight. Those UV rays can cause a lot of damage to the retina very quickly, and that damage can be permanent. | {
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36,632 | Titan is the only body in the Solar System other than Earth to have large bodies of surface liquid. Since Saturn is much more massive than the Moon, I would expect for tides, if they're present to begin with, to be much larger than the ones here on Earth. | Glancing at a partial solar eclipse is about as dangerous as glancing at the Sun on any other day. If you look at the Sun a few minutes after sunrise or a few minutes before sunset, when the Sun's altitude is low, the light is filtered through a lot of air, and most of the ultraviolet is scattered, so it's a lot less dangerous than looking at the Sun in the middle of the day. However, during an eclipse, people tend to look at the Sun for more than just a few minutes, and that's not healthy for the eyes. The big danger of looking at a solar eclipse without proper protection occurs when it's a total (or nearly total) eclipse. When the eclipse is near totality, the sky is dark, so your pupils dilate, letting in more light. But the light rays coming from that small sliver of Sun are just as bright as normal, and have the same proportion of UV (ultraviolet) as full sunlight. Those UV rays can cause a lot of damage to the retina very quickly, and that damage can be permanent. | {
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36,652 | I'm following the steps in this paper on how to calculate geocentric longitude. In section 3.4.2 it says the Geocentric rectangular co-ordinates of Mercury then are: Xg = Xh−X0 = 0.990 A.U.
Yg = Yh−Y0 = −0.259 A.U. Converting these into polar form, we get the geocentric distance and longitude as: rg = √(X2g+Y2g) = 1.024 A.U.
λg = tan−1(Yg/Xg) = 345.3◦. For the life of me I cannot get the inverse tan of (Yg/Xg) to give me 345.3. This is what I'm doing in python: Xg= 0.990
Yg= -0.259
math.degrees(math.atan(Yg/Xg)) = -14.66091789971255 Any help greatly appreciated. | Glancing at a partial solar eclipse is about as dangerous as glancing at the Sun on any other day. If you look at the Sun a few minutes after sunrise or a few minutes before sunset, when the Sun's altitude is low, the light is filtered through a lot of air, and most of the ultraviolet is scattered, so it's a lot less dangerous than looking at the Sun in the middle of the day. However, during an eclipse, people tend to look at the Sun for more than just a few minutes, and that's not healthy for the eyes. The big danger of looking at a solar eclipse without proper protection occurs when it's a total (or nearly total) eclipse. When the eclipse is near totality, the sky is dark, so your pupils dilate, letting in more light. But the light rays coming from that small sliver of Sun are just as bright as normal, and have the same proportion of UV (ultraviolet) as full sunlight. Those UV rays can cause a lot of damage to the retina very quickly, and that damage can be permanent. | {
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36,684 | Since gas giant consist of most gas components, where do we establish their "surface"? My take is basically to take the limit in which all light is opaque. For example, in this photo: The surface, then, will be the limit of the black blackground with the planet. Any other way to formally define the "surface" of a gas giant? | There are two common definitions in use for the surface of gas planets: The 1-bar surface: As pressure increases, the deeper in we go into the gas planet, we will hit a pressure of 1 bar at some altitude. Gas at this altitudes will usually sit deep enough in the gravitational well and be of a near-uniform density and temperature, as to not be influenced by exterior parameters, for example the solar wind. Therefore, the altitude of the 1-bar level will remain essentially constant, for short astronomical times. The $\tau=2/3$ -surface: This is the altitude, from which photons can escape freely into space. This happens at an average optical depth $\tau$ of 2/3. It is essentially what you see in your image as the limit of the black background. For the sun one end of the photosphere is the average $\tau=2/3$ -surface, and for transiting exoplanets this is identical to the measured transit radius at that wavelength. There is no hard relation between those two surfaces, but in general their altitude will not be different by more than a scale height, as at around 0.1-1 bar the gaseous atomic and molecular bands become enormously pressure broadened, which makes the atmosphere quickly opaque at most wavelengths, for the usual gas giant components. | {
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36,780 | Why don't our neighbors have much nitrogen? You would think that, without 'nitrogen-fixing' organisms and such, there might be more..... | Nitrogen, with a molecular mass of 28 atomic mass units, is too light to have remained in Mars's atmosphere. Carbon dioxide, with a molecular mass of 44 amu, could (and does) exist on Mars, but it is rather sparse. Venus's atmosphere appears to contain a small amount of nitrogen when viewed on a percentage basis. "Only" 3.5% of all of the gases in Venus's atmosphere is nitrogen. But that's not a particularly good metric. Venus has so very, very much carbon dioxide in its atmosphere that the CO 2 dwarfs every other constituent. When viewed in terms of mass, Venus has a good deal of nitrogen in its atmosphere, four times more nitrogen than the Earth has in its atmosphere. | {
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36,802 | As the title says, when did we realise with reasonable confidence that our star is not going to be going out in a supernova blaze of glory? I ask because a while ago I read " The Songs of Distant Earth " by Arthur C. Clarke which was from 1986 (but based on an earlier story) and a plot point is that the sun went supernova. I was surprised that this was still considered a possibility in the 1980s but accepted it. However, today I began reading " The Gods Themselves " by Isaac Asimov, written in 1972, and in it a character states that the Sun is too small to go supernova. This confused me as I thought both Clarke and Asimov were both scientifically knowledgeable and informed. I can find plenty of information online about how our Sun will likely end and why it won't go supernova but not any information on how long we have known (and therefore whether Asimov was ahead of his time or Clarke was uninformed in a way that would suprise me)? | I think the definitive work is that of Hoyle & Fowler (1960) . They argued that supernovae were produced by two possible mechanisms - what they called an implosion/explosion or an explosion within degenerate matter. Both of these mechanisms required very high internal temperatures ( $>2\times 10^{9}$ K) and they argued that this could only be achieved for both mechanisms if the star is more massive than the Chandrasekhar limit of $5.8M_{\odot}/\mu_e^2$ , where $\mu_e$ is the number of mass units per electron (1 for hydrogen, 2 for helium, carbon, oxygen, 2.25 for 56Fe etc.); otherwise it could achieve a degenerate state, supported by electron degeneracy pressure, and cool indefinitely without contracting any further. In addition, they successfully identified these two types of supernovae with Type I and Type II supernovae and argued that very high mass stars accounted for Type II supernovae via the implosion/explosion mechanism, whereas the Type I supernovae arose from the ignition of degenerate material in less massive stars, but still larger than the Chandrasekhar mass, which is $>1.2M_{\odot}$ for all plausible compositions. They even then conclude that stars up to $10M_{\odot}$ may lose enough mass to avoid a Type I supernova and become stable white dwarfs. Though it seems not to be explicitly stated in the paper, it is implicit that the Sun could not explode as either type of supernova. As an aside - I reminded myself of the plot of "Songs of Distant Earth" (I read it loooong ago). Clarke's story posits that the "neutrino problem" (the lack of detected solar electron-neutrinos), which was a thing back in the 60s, was actually because something odd was going on in the Sun's core. So I don't think that Clarke was suggesting that under normal stellar evolutonary circumstances the Sun would imminently explode. Indeed it was well known at that time that the Sun should continue on its main sequence life for billions of years. | {
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36,925 | I know Neil Armstrong placed a mirror on the Moon and people shoot lasers there which get reflected, thus measuring the time the light needs they can conclude the current distance of the Moon. But the Moon is periodically moving away from the Earth (when annular eclipses are possible) and getting closer again (when total eclipses become possible). Sometimes we hear of records like "the Moon is as close as it hasn't been for decades". I remember, about one year ago, I saw a very close and bright full Moon, almost outshining most stars. So how is it concluded that the Moon would constantly move away from the Earth? Is it just based on the physical prediction of celestial mechanics, like calculation that a day on Earth would get longer and the Moon's orbit slower? | There are, I think, at least four parts to this argument: the first being the theoretical argument that ties it all together and the remainder being observational evidence for the Moon's orbit increasing in size. 1. The underlying theoretical argument. This, of course, is the idea that tidal braking causes the Earth to slow down in its rotation and the Moon to move further out in its orbit . (Once upon a time this also caused the Moon to slow its rotation, until it became tidally locked, with its rotation period = its orbital period.) This predicts two things: the Moon should be gradually getting further away from the Earth over time; and 2) The Earth should be slowing down in its rotation (days getting longer) over time. There are at least three sets of evidence supporting this. 1. The increasing distance of the Moon, as derived from lunar laser ranging. The key point here is that it's the average distance that's increasing. Since the Moon's orbit is elliptical, the distance varies over the course of one lunar orbit, and there are additional variations introduced by the gravitational influence of the Sun, other planets, the not-perfectly-spherical shapes of the Earth and Moon, etc. What this means is that if you trace the Moon's distance over time, you will see it increasing and decreasing, but you will also see that the average distance is gradually getting larger. (And this is what has been measured with the laser ranging experiments; e.g., this 1994 article by Dickey et al. which found that semi-major axis of the Moon's orbit was increasing by about 3.8 cm/year.) 2. The slowing down of the Earth's rotation. As required by basic physics, if the Moon is getting further out in its orbit, and thus gaining angular momentum, there has to be a compensating loss of angular momentum, and this happens via the Earth gradually spinning more slowly. This has been measured in at least two ways: A. Timing of historical lunar and solar eclipses. There are Chinese records of eclipses going back to roughly 400 BC, and Babylonian records back to almost 800 BC, as well as more recent Greek, Arab, and European records. These can be used to estimate changes in the length of the day, in two ways. First, some records include an approximate time of day, so we can run our calculations backwards and figure when an eclipse visible at, say, Babylon in 200 BC should have occurred. And even when we only know the day (and location) of an eclipse, we can predict where it should have been visible. As explained by this article describing a 2016 study by Stephenson et al. : "The oldest event in the catalog, a total solar eclipse that occurred in 720 B.C.E., was observed by astronomers at a site in Babylon (now modern-day Iraq). But, working backward, today’s astronomers would have predicted that the eclipse should have been seen a quarter of a world away, somewhere in the western Atlantic Ocean. The discrepancy means Earth’s rotation has gradually slowed since the 8th century B.C.E." B. Geological and paleontological measurements of the length of the day. * It turns out there are some geological records which can be used to derive the number of days in a lunar cycle, or the number of days in a year, millions or hundreds of millions of years in the past. One way is via the analysis of "tidal rhythmites ", which record alternating effects of ocean tides. Since tides have cycles of twice per day (high and low tides) due to the Earth's rotation and twice per lunar orbit ( spring and neap tides ), you can use the combination to work out how many days there were in a lunar orbit. There can also be yearly variations, which allows you to work out how many days there were in a year when the tidal rhythmites were formed. Another way is by studying certain fossils, where the growth of part of the organism is recorded in both daily and annual variations: for example, the growth of a shell can vary over the course of a day, but also over the course of a year (e.g., more growth in summer, less in winter). Putting this together allows you to figure out how many days there were in a year when the organism was alive. A fascinating recent example is this careful study of a 70-million-year-old fossil of a rudist (an extinct type of mollusc), which reveals that a year had 372 days back then, so that each day was 23.5 hours long instead of 24 hours. | {
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37,045 | Volcanism on Io is caused by the fact that it is tidally heated . There are four moons that are closer to Jupiter than Io with higher eccentricities, yet they don't seem to have any volcanism at their surface. Why isn't the tidal heating enough to generate volcanism on any other moons of Jupiter? | It’s because they are much smaller than Io. Tidal forces are differential forces, that is, they result from the difference in gravitational pull on one side of a body compared to the other. When an object is small, the difference in distance to the two sides of it is necessarily small as well. According to Wikipedia , Amalthea, the largest of those four innermost moons has a long axis that is only 250 km, and the others are smaller yet. The strength of the tidal heating scales as the body’s radius to the 5th power , quite a strong dependence. The small mass plays a role, too, but the dependence on radius is stronger. For this same reason, small objects on Earth don’t feel tidal stretching from our Moon, yet the Earth as a whole does. On a related note, the inner three of those four satellites are inside Jupiter’s Roche limit , the orbital distance at which a purely self-gravitating body should be pulled apart by tidal forces. But they survive because most small bodies aren’t held together primarily by gravity (just as your own body isn’t) - the internal electromagnetic forces between atoms (which manifests macroscopically as the rigidity of the rock) is the main source of the bodies’ structural strength. | {
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38,348 | Swift–Tuttle Comet has orbital period of 133.28 years. Yet, people say that debris and dust from this comet causes this meteor shower. This does not make any sense. Or do they want to say that from 1992, every year, for 28 years Earth picked up debris left behind so many years ago while passing the August part of orbit? It's logical to assume that all this debris would be exhausted by now... | The comet sheds material each time it get close to the sun. The sun heats this dirty snow ball, the ice evaporates and tears dust and smaller rocks with it. These dust particles then follow a similar orbit around the sun as the comet. But as they are ejected with some velocity and react differently to the solar wind & radiation, their orbit is slightly different from their parent comet.
Different orbit means generally different orbital velocity and, over time, the particles spread out along the comets orbital path. They form a steam of material that constantly dissipates and is replenished each time the comets approaches perihelion. And each time the earth passes through the part of the solar system this stream occupies, we see meteor. Here is a nice interactive animation: https://www.meteorshowers.org/view/Perseids | {
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38,355 | This popped up into my mind just now. The Moon is tidally locked to Earth, and also has a significantly eccentric orbit. This means that its orbital velocity near periapsis is considerably faster than its orbital velocity at apoapsis. Thus, for the Moon to always show one side to Earth, its rotation around its own axis has to slow down and speed up depending on where it is in its orbit, but doesn't that violate the law of conservation of angular momentum? And if its rotation around its axis was constant then it would get out of sync and not be tidally locked, no? | You are right, that would be weird if the Moon speeds up and slows down this way to always show the exact same side to the Earth. That's why it doesn't. At some point in the orbit the Moon's rotation (its phase) lags behind and at some point it is too fast. It is (pretty much) constant and does not adjust to the varying orbital velocity. That's (one reason) why it wobbles. And this wobble is what's called libration . But the Moon is still tidally locked as its rotation period is still equal to its orbital period. | {
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38,357 | I am an analytical chemist with some interest in amateur spectroscopy. Since astronomers use echelle spectrographs to study the high resolution spectrum of the stars, someone suggested to post the query here. An echelle grating produces a spectrum with several overlapping orders. One can then use a cross-disperser, which could be a prism or a grating, which can separate the overlapping orders and then we obtain a very high resolution spectrum. I want to know how a real echelle spectrogram looks like -as if we were looking at the 2D spectrum with our naked eyes? I have a small echelle grating (79 grooves/mm, 75 $^o$ blaze) and tried to visually generate a 2D spectrum with the help of a plastic transmission grating. The light source was an ordinary fluorescent lamp Pictures are posted below. The experiment is too elementary. A horizontal slit made by two blades as an entrance on the shorter rectangle of a shoebox. The light from the slit falls on the reflective echelle. On the larger side of the shoebox, I made a window to take a picture. Picture 1: A vertical slit made in a cardboard box, the long side of the echelle is perpendicular to slit An expected overlapping order spectrum seen. Picture 2: A horizontal slit made in the cardboard box, now the long side of the echelle is parallel to the slit Now, if we hold a plastic transmission grating (Edmund) in our hand, with its rulings perpendicular to the echelle, and view the echelle through that transmission grating, I can photograph the following 2D spectrum. Picture 3: Hold a transmission grating with its rulings crossed with the echelle and view picture no. 2. Unfortunately, this looks like a replica of a spectrum rather than separation of orders. **The main question: How should one hold the transmission grating in order to visually see the separated orders. Picture no. 3, was supposedly, separation of orders but it does not look like it. If this were a true 2D spectrum we should see some increasing distance between Hg green lines. All separated orders look the same resolution. Is there any other simple way to see a true cross dispersed spectrum with eyes using a simple cardboard type box? The key problem seems to be the issue how should we hold the transmission grating in order to see a true 2D spectrum unlike Picture 3.** EDIT After Prof.ELNJ's suggestion in the answer , I wanted to confirm if this is what he meant. The grating has an arrow with a pencil (from ThorLabs) which shows the blaze direction because by viewing in this direction one can see a "rainbow" of colors. So the arrow is towards the viewer, and I hold a transmission grating "crossed" with the echelle ruling. Rough view of what appears to the eye | You are right, that would be weird if the Moon speeds up and slows down this way to always show the exact same side to the Earth. That's why it doesn't. At some point in the orbit the Moon's rotation (its phase) lags behind and at some point it is too fast. It is (pretty much) constant and does not adjust to the varying orbital velocity. That's (one reason) why it wobbles. And this wobble is what's called libration . But the Moon is still tidally locked as its rotation period is still equal to its orbital period. | {
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38,363 | In the Astronomy Picture of the Day from August 8, 2020 (" Crescent Saturn "), the caption states From Earth, Saturn never shows a crescent phase Why is this? We can observe phases on planets such as Venus which have an apparent size a lot smaller than Saturn. Saturn is not tidally locked to the sun ( only Mercury is ). So why does the sunlit face of Saturn always face Earth? | Phases are just different perceived illuminations of an object at different illumination and observing angles. If the observer is, with respect to the object, located in a similar direction as the light source shining on the object then you should expect to see the vast majority of the object illuminated, if the observer is located in the opposite direction you would see the object back-lit, and if you are at right angles observing the object with respect to the direction of the light source you would see the object half-illuminated. Since Mercury and Venus are always inside Earth's orbit and move with different rates around the Sun as Earth does, the Earth (the observer) is able to locate itself at any angle with respect to the light source that shines on the observed planet. This means that you can see any phase of Mercury or Venus (except from a perfect 100% illuminated phase due to the body of the Sun blocking the view). Here you have an example for Venus: Now think about what happens from the point of view of an Earth's observer for the exterior planets. The Earth will never have a chance to see the planet's back-lit side since there's no position in its orbit that would allow for this. As seen from the other planet, the Earth is always close to the Sun, so it can be seen almost exclusively during the day, which means that from the Earth you almost exclusively see illuminated regions of the planet. Mars is the closest exterior planet, so the Earth manages to gain enough elongation to see a bit of the night side, but it is a tiny fraction of the disk as viewed from here. Here you have a picture of the phases of Mars as seen from the Earth: This is even worse for far away planets. Saturn is so far away that from its vantage point, that the Earth is basically always close to the Sun (to the light source). From the Earth, Saturn seems always fully illuminated, by an extremely small margin that allows to see a slim crescent of darkness in perfect conditions. Only with spacecraft like Cassini and Voyager have we been able to see what Saturn looks like from behind. The first time humanity did this was in 1980 with photographs like this one (from Voyager): Before that we had never seen the night side covering more than a percent of the disk. Even the shadow of Saturn cast over the rings is nearly impossible to spot from the Earth. Look at this amateur photograph that shows precisely a bit of that shadow over the rings behind Saturn (bottom-right part): That shadow is almost non-existent from Earth's point of view, and it's all because Earth's orbit is inside Saturn's and Saturn is far away from the Earth. The Earth is always so close to the line connecting the light source (the Sun) and the illuminated object (Saturn), that you shouldn't expect any more phases than "full" from here. Simple geometry. | {
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