Is a branch on HR diagram a line formed by stars in the same evolutionary phase, or the evolutionary track of an individual star? Here is a HR diagram from Wikipedia, in the well-known form:

Here is another one from the Stellar evolution article:

I know that when stars leave the main sequence, they go to the giant branch through the subgiant branch. I am a bit confused of what we mean when we say "branch". Looking at the first image, we might think that the subgiant branch is the approximate line of stars formed by different subgiant stars on the HR diagram. While looking at the second one, we might think that the expression branch means the evolutionary track of an individual star.
Which one is it?

The answer is: both, but for different contexts.

The HR diagram is used to display the properties of observed stars. A point on the diagram represents a star with a certain effective temperature and luminosity, which can be interpreted as being indicative of the stage of development of the star. Equivalently, the diagram can be expressed in terms of absolute magnitude and stellar classification/type. The diagram was originally constructed in 1911 by E. Hertzsprung and H. N. Russell independently, though we predominantly use Russell's version nowadays.

As one simulates the theoretical evolution of a star, the star traces an evolutionary "track" on the HR diagram, which is a line that runs across the diagram in various directions representing the evolution of the star through its various development stages, as in the second figure in the OP. This is useful for displaying the effect of various parameters on the complicated evolution of stars, see here for example.

These two efforts are related historically: as better telescopes were made, there were more and more stars observed, and more and more points were added the HR diagrams, but eventually we needed computers to handle the data. Along that route, the development of theories of stellar evolution were/are intimately tied to the development of observational astronomy, as experiment and theory always are, and the HR diagram has been a powerful tool for advancing theories of stellar astrophysics.

Looking at the first image, we might think that the subgiant branch is the approximate line of stars formed by different subgiant stars on the HR diagram. While looking at the second one, we might think that the expression branch means the evolutionary track of an individual star.

So you can think of the stars in the first image of the OP as snapshots of each of the stars at a point in their lifetimes. The stars that group in the sub-giant branch (so called for low and intermediate mass stars, but is called the Hertzsprung gap for high mass stars because it's so short lived) are in the part of their evolution where hydrogen burning in their core has ceased and hydrogen shell burning has begun. The evolutionary track in the second image in the OP does indeed show the evolution of a single star through its stages of development.


Is there permanent dry ice on Mars? Are there any places featuring permanent dry ice (carbon dioxide ice) on Mars?
Is there dry ice at the poles?

Yes.
The Southern polar ice cap has a covering of $\mathrm{CO_2}$ about 8 m thick that doesn't completely disappear in summer. It remains in "pits" of up to about 1km in diameter.  The  thinner dry ice layer in the North sublimes completely into the atmosphere in summer. [source]


Leaving the Milky Way If we had a hypothetical spacecraft, would it be possible to exit our galaxy, wait for some time and enter back into another section? Will this method be more efficient that a more traditional flight path?

Not really, for the same reason that you cannot travel west by jumping up in the air and let Earth rotate underneath you, such that you land a little farther to the west.

The reason is that standing on Earth's surface, you already have a velocity toward the east which matches exactly the speed of the surface. Thus, in the reference frame of Earth, you simply jump up and down.

In the same manner, if you use your mad hypothetical spacecraft to fly "up", i.e. away from the Galactic plane, you already have a velocity of $\sim$250 km s$^{-1}$ in the direction of the rotation of the plane, so that in the reference frame of the plane, you simply fly straight up.

Apart from this, in order to leave the Galactic plane, you need to fly roughly 500 lightyears. This will take a long time. You should stay home.


Is our Universe a hologram? I watched a documentary where I heard some very strange theories that are difficult for me to comprehend. It is even difficult for me to explain what I heard. "Objects that fall into a black hole leave a copy of themselves outside the black hole." "We are not living in reality but in a hologram ." Does any of this make any sense to anyone ? O.K. so I went to check the title of this documentary because how can anyone explain it without knowing what I saw. "Fingerprints of God -the Dark Universe" this is the title. I do admit that I did sleep through a good part of it because it was almost 2:00 a.m. Also if anyone wants to watch it then look for it on youtube because that is where I saw it.

I can't give a good answer but I did find the documentary and out of curiosity, watched some of it, skipped forward and found the hologram section you're talking about, it's at the end about 1:18:40 in.  (I'm not going to post a link to the video cause I'm not sure it's correctly copyrighted by those who are showing it)

The video itself isn't bad, but it's basically a science channel level program, which is written to appeal to viewers and explanations are often over-simplified.   This video in particular seemed to try to cover so much ground that it explains very little.   But on the positive side, they have Brian Greene and Leonard Susskind discussing the hologram idea and those guys are both real physicists who've talked about this before.  I don't think it's a main stream idea, just an idea that a few people like.

Here's a magazine article on the subject and an excerpt from Brian Greene's book.

and Here's Leonard Susskind giving a talk on the subject at Stanford, not at a class, more of a lecture cause he mentions that not everyone in attendance might be a physicist, but I've since watched his lecture and he makes a few of these points more clear, at least to me.

And Here's a stack exchange question and answer on a similar subject.    There's also no shortage of articles that can be found on the Holographic principal or the "universe is a hologram", though I think the first is a better set of words to search and the 2nd might lead you to some sub-par articles.

Philip Gibbs - inactive's answer (also in the link above) on what the Holographic Principal implies is better than anything I could say about it, so I'd start there.

As a rule, it's a good idea to post links and if the video is 2 hours long, the specific time as well.   YouTube links can also be removed so they're not ideal.

Edit to answer questions.

Had to think about this one a bit, and sometimes trying to answer "why" is opening a can of worms, but . . . here I go.

  Why do they think that it is our Universe that is the Holograph?

I'm not a fan of the theory myself and I'd guess that not all physicists agree with it either.   One big problem with the theory is that it can't be proven, which really makes it more of a guess based off a mathematical model and not a good theory.  One definition of a good theory is one that can be tested.   If it can't be tested true or false, it's really more of an idea.

As for why Brian Greene likes the theory, you might need to ask him or search his interviews, but maybe he likes it so can sell books - I don't know.  Maybe he likes the idea of the Universe being a Hologram.   He's obviously a pretty smart guy.  You don't get a PHD in physics and get recognition and invited to ted talks and all that by being dumb, but smart guys are wrong all the time when it comes to new theories.  There's no evidence that this theory is true.

Now, I should point out that the Holographic principal as it applies to black holes and the information paradox, all that stuff is good and valid theoretical physics rooted in ideas we understand such as entropy and time dilation and red-shift.   The Holographic principal seems a very reasonably deduced feature of black holes that also, cleverly answers the information paradox, but whether it can be extrapolated to apply to the entire universe is a whole different ball game and to me, it's basically a guess.  Not much more.

  Why couldn't our Universe be reality and a Holograph exists of it
  somewhere else?

Well, I could say "you're right", but the truth is, you're just flipping around the question.  The more relevant questions are whether the theory is true at all, and as I said above (and Susskind said in his video), whether it can be tested.   Susskind's video is really quite good though he starts slowly, but if you're really interested in this it's worth watching.

  What makes them think that this is the Holograph instead of the
  reality?

One thing about Physics to keep in mind.  It's often surprised us.  Physics has given us answers that physicists didn't expect many many times, perhaps more than any other field of study.  Physics is probably more likely to yank the rug of reality out from under us than any other field and with that in mind, it's not a subject to study if you want to feel secure in what you believe and know.    In the days after Newton, Physicists were very confident in what they knew and felt that nothing very new was out there, though the constant speed of light was a puzzle they'd not expected.

Then Einstein came along and pretty much threw everybody's understanding out the window when he started to figure things out and not long after that, the quantum physicists came up with ideas that made Einstein's head spin and to which he directly disagreed.   After that the big bang was unexpected, background radiation was unexpected.  Dark matter was unexpected (except for maybe Fritz Zwicky) and dark energy was quite the surprise when it was discovered.

Being regularly surprised, of-course, doesn't make the hologram theory any more true or false, cause nobody knows, but good physicists these days try to keep an open mind to mathematically valid new ideas.

my 2 cents as a hobbyist.  Hope that wasn't too long and off the point, and please, take what I say with a grain of salt.   I'm not an expert.


What is the meaning of "bright unresolved sources" In many articles talking about telescopes, they talk often of "bright, unresolved sources".

"bright" usually mean that the magnitude should be less than a certain threshold, (or that the signal to noise ratio should bigger than a given value) but what I don't catch the meaning of "unresolved".

For example :

"unresolved sources" in https://docushare.lsst.org/docushare/dsweb/Get/LPM-17/LSSTSRD_v5%202%203.pdf

"bright unresolved point sources" in the introduction of this paper : https://docushare.lsstcorp.org/docushare/dsweb/Get/Publication-86/

Any idea ?

"Unresolved" here means a source of light that appears as a point, not as a disc or a cloud.

For example, the planets can all be resolved to discs by even quite moderate telescopes. Galaxies also can be resolved, they don't appear as point sources.

However, stars, quasars, and most asteroids cannot be resolved by amateur equipment. In the context of the LSST paper, resolved sources are "galaxies" and unresolved sources are "stars and quasars".

  The photometric measurements for resolved sources (galaxies) have to include several standard magnitudes, such as Petrosian magnitudes, as well as appropriate model magnitudes.


How do we know that the mass ratio of hydrogen and helium in the sun is approximately 3:1? From what I learnt, we can find what elements are in the sun by analysing spectrums since each elements has same emission spectrum, and when you look at the sun's spectrum, you can see absorption spectrum missing some lines which is the elements.

But how do we know how much are there? Are there other ways to find out how much of each elements contributes to the sun?

And how can this become an evidence of Big Bang?

Are these even correct?

(Can you explain it easily)

You're asking two questions here but they are related, so that may be OK.

The mass ratio of the Sun is probably not 3:1 though, as the Sun has been turning hydrogen into Helium for nearly 5 billion years.   The interior of the Sun has a considerably higher helium ratio as a result, but the outer edges of the sun - which is what we see and read from spectroscopy, is close to the 3:1 mass ratio.

If this is accurate the Sun overall is closer to about 2.6:1.

Scientists can measure the ratio by observing spectral lines.  Stars make this easy because of how hot they are, but that information is limited to their surface.   We can't see the core of the sun.  Still more than half of the elements have been observed in the solar surface.

See on solar spectroscopy

Or a related question from Physics Stack

The 25% helium by mass is confirmed both by spectroscopy and by quantum physics models based on the modeled expansion as the early Universe goes from hot to cold.   Helium is more stable than hydrogen, but, it requires an intermediate step in it's formation because deuterium has to form first.  Also, though I think this plays a smaller role, free neutrons decay and become protons which become Hydrogen.

More massive elements need hotter temperatures to form but the early universe went from hot (Quark soup), to less hot (formation of hadrons), to static - 92% Hydrogen, 8% helium by element or 75-25 by mass with tiny trace amounts of Lithium.   It stayed that way until the first stars and the formation of heavier elements.

That's a very simplified version.  You can visit the stack exchange answer below for more details or this link here.

Both spectroscopy and quantum physics agree on the ratio, which is always a plus when the same answer is reached two different ways.   There is a lingering lithium ratio problem, but even so, the 75%-25% ratio is pretty good.


Was the radiant of the Perseid meteor shower ever near Vega thousands of years ago? The ancient Chinese story of the Weaver Girl (Altair) and the Cowherd (Vega) tell of starcrossed lovers who are only allowed to meet once a year on the 7th day of the 7th (Chinese) lunar month, roughly around early August, when magpies form a bridge across the silver river (the Milky Way.) This is celebrated in the Qixi festival in China (August 9th in 2016.)

Because of the timing, I'm suspicious that the magpies represent the Perseid meteor shower (August 12th in 2016.)  But I just got my first planisphere, and Altair and Vega are pretty far away from Perseus (although Perseus is still on the Milky Way.)  The Wiki article and others that I've looked into on Weaver Girl and Cowherd does not connect the magpies to any meteor shower.

But the Weaver Girl and Cowherd story is pretty old - up to 2600 years according to Wikipedia.  I don't think the Perseids have always been in Perseus due to the precession of the equinox, but 2600 years ago maybe the "Perseid" shower was closer to the southern triangle, or is that too far away to be explained only by precession of the equinox?

I'm very new to astronomy and I'm not Chinese, but I love this story and I'm looking for more information.

Thanks!

Okay, this turned out to be quite a long answer describing a few arguments I've com up with. The short answer is: maybe, maybe not, but I think it's possible. There are arguments to be made for either side of this discussion, and I don't think you you can ever be absolutely sure unless you find an expert on both astronomy and Chinese folk tales (which I don't claim to be), or invent a time machine to go back 2600 years and check for yourself.

The most obvious argument to be made in favor of the hypothesis that these magpies are the Perseid meteor shower, is that folk tales referencing stars are very likely to be based on astronomical occurrences during that period (such as a meteor shower, perhaps).

The (in my mind) most obvious argument against your hypothesis is what you suggest yourself: Perseus is quite far away from Vega and Altair. However, Perseus is the radiant of the shower, which means that all perseids will appear to come out of Perseus. Because of this, it is perfectly possible for a lot of perseids to appear between Altair and Vega, parallel to the Milky Way. It is, however, impossible for them to go from Vega to Altair, or vice versa, since they wouldn't come out of Perseus anymore.

You suggest that this could be explained by the precession of the equinox, but a quick lookup on Stellarium (which is a great (and free) tool if you're into astronomy) tells me that even 2600 years ago Perseus was quite far away from Vega and Altair (actually, that's quite obvious, since the precession of the equinox causes stars to move relative to the extension of the north pole, but not relative to each other, as far as I know).

Another argument to be made against this hypothesis is that the perseids are by far not spectacular enough to be seen as a bridge over the Milky Way, even when they all appear in the required spot. During the yearly peak of the perseids, you are expected to see about 1-2 meteors per minute, which can be noticeable at best. This argument is then countered by the fact that the ZHR (how much meteor you see per hour) is not the same every year, but can peak on years when Earth passes through a fresh dust cloud left behind by the comet that causes the shower (which, in this case, is the comet Swift-Tuttle, if you're interested). Therefore, it is perfectly possible that the perseid meteor shower was extremely active 2600 years ago, causing the creation of this story. Such peaks are somewhat rare, but not too rare for this hypothesis to actually make sense (in fact, we will be seeing such a peak in 2028).

The fact that the dates of the festival and the peak of the perseids are a few days apart and not exactly the same is not quite an argument against your hypothesis, since it is perfectly possible for the date of the festival to have moved slightly due to differences between the (old) Chinese and Western calendars. I'm not an expert on calendars, however, so I can't say that for sure.

Given all these uncertainties, your hypothesis (along with what you mentioned yourself in your own comments) seems plausible, apart from one thing: we don't know whether Swift-Tuttle was already active 2600 years ago. If it wasn't, the perseid meteor shower obviously did not exist (note that comets don't exist forever: some gravitational variation causes them to exit the Oort cloud and enter the solar system). We don't know when exactly Swift-Tuttle started being a comet.

That is only a weak argument, however, and when it comes to astromical hypotheses, the most convincing argument is usually a principle which is referred to as Ockham's razor: if given multiple explanations, choose the one that requires the least amount of assumptions (or the most plausible ones).

Personally, since the date of creation (2600 years ago) is very vague (the uncertainty on this must be at least a hundred years), I think it is perfectly possible to assume that the perseids peaked somewhere during that period, causing the creation of this story.


Orientation of local star map This image depicts our solar neighborhood
Source
The plane defined by the concentric circles, does it have a relationship to any galactic reference points or the ecliptic?

By inspection of a star charting app, I'd say the circles lie in the celestial equator, ie declination equal 0 degrees.  Stars above the circles are to the north and vice versa.


How bright can white dwarf stars glow as they accrete matter? I realize this is kind of general as it would depend on the size of the white dwarf and the rate of accretion. The general idea I got thinking about is what would happen if a white dwarf star - lets say 0.8 solar mass, well below Chandrasekhar was to accrete a bunch of hydrogen rather quickly.  I thought it might be get very hot and bright, perhaps similar to a red giant for a relatively brief period of time.
Mostly White dwarfs add mass by siphoning from the outer layer from a near by co-orbiting star, but I imagine, that mostly happens quite slowly, not fast enough to make the white dwarf shine like a main sequence star.
Are there studies on how quickly mass can be added in this way and how bright it makes the white dwarf?  Does Hydrogen quickly fuse on the surface of a white dwarf or does it need to build up to a certain thickness first, perhaps creating a mini flash? or does the fusion happen rather quickly.
A very rough calculation, the escape velocity of a white dwarf, which would be similar to the terminal velocity of matter (mostly hydrogen) falling into it is about 5,500,000 meters/second for a 1 solar mass white dwarf.  Source.
The velocity needed for hydrogen fusion is about 20,000,000 meters per second.  Source (under the what makes fusion hard section).   So, white dwarf's aren't massive enough to create fusion upon impact, but the high density and high heat of impact seems like it could create fusion relatively quickly as matter builds up around them, perhaps creating something that perhaps glows like a star and expands rather rapidly - assuming there's a sustained burn and not a kind of very quick explosion of sorts.   I think it could glow very hot and perhaps expand rather large, but I'm just kind of guessing.
If a white dwarf was to crash into, say, a brown dwarf or very low mass star, hydrogen rich, say about .1 solar mass?   Such a brown dwarf would be much larger than the white dwarf but also quite a bit lighter, probably light enough to get penetrated completely though the impact would also likely create fusion and it's difficult for me to imagine what would happen with so violent a crash.   It seems to me that the density of the white dwarf, if it could retain most of the matter from the brown dwarf would be sufficient that you'd be left with a fast burning light weight fusion star, with a very dense core.   This might be an impossible question, but how quickly would a .8 solar mass white dwarf consume the hydrogen from a .1 solar mass brown dwarf, should this rare kind of collision happen.   Would it actually form a star-like object for a period of time, say a million years or so, or would the enormous energy of impact basically blow it apart no new star?  Perhaps a near miss and an inside the Roche limit approach would be better than a direct impact for new star formation.
Thanks.

There might be a problem with your calculation, because by the time a white dwarf made of helium gets up to 0.5 solar masses, not only are the surface layers hot enough to fuse hydrogen, they are actually hot enough to fuse helium!  That's what will happen to the Sun when it ceases to be a red giant.  You are talking about carbon white dwarfs, but they are similar to helium white dwarfs.  So hydrogen fusion at the surface of a white dwarf is a common result-- it is what causes a phenomenon called a "classical nova."  So you are on the right track-- but what you are talking about is a well-known area of study.


Around what apparent magnitude can the naked eye observe an object during full moon For a very rough guideline using healthy/corrected eyes adjusted to the dark, around how bright should an object be to expect it to be visible?

It may shave off 1 or 2 magnitudes, but it depends on many factors: nature of the object (star, nebula, galaxy), altitude (higher altitude has less light scatter), transparency, etc. There's no One Single Answer To Rule Them All.

Light pollution from artificial sources has a greater impact in most cases.


What are the apparent visual magnitudes of Spica A and Spica B? What would be the (theoretical) apparent visual magnitudes of Spica A and Spica B (since they are too close together to be resolved telescopically)?  The Spica System is 77 parsecs away, and the combined apparent visual magnitude of the two stars is 0.97 at maximum (presumably when the stars are not eclipsing each other).  Their orbital parameters and respective luminosities, temperatures, and radii are given on Wikipedia: https://en.wikipedia.org/wiki/Spica

The Wikipedia article says that Spica A has 8 times the luminosity of Spica B (presumably that's at maximum.)  Magnitude is -2.5*log10(I/I0). So if the combined magnitude is 0.97, the magnitude of component A is ~1.1 and the magnitude of component B is ~3.3.


Could the Moon sit on the Earth? Let say that the Earth's Moon was knocked out of orbit and landed on the Earth with some damage but not world ending. Would Earth's atmosphere envelop to include the Moon? Would the moon sink into the Earth? What would happen?

It can't happen. The moon is quite big and can't get knocked out of orbit easily.

If it were somehow knocked out of orbit and collided with the Earth, the energy released would melt the entire crust. Huge amounts of matter would be knocked into space. After everything had settled down, you would get a lifeless world, still spherical but now slightly larger.

The moon can't get knocked out of orbit.  If it did, It can't "land". If it did "land" it would be crushed by its own weight, and all life would end in one way or another.


What's the white glow around this star? A few days ago, I took a photo of the night sky in the mountains of Georgia with my DSLR. There, I noticed that one single star (in the center of the part of the photo that I added below) looks very different than all the other stars on the photo. Why does it have this white glow?

Astrometry.net has identified your star field as being part of the Andromeda constellation. The diffuse object in the centre of your image is the Andromeda Galaxy (Messier 31). The bright star to the left of it is Mirach (β Andromedae). Given a reasonably dark sky and averted vision, it is possible to see the core of M31 with the naked eye. Since M31 is approximately 2.5 million light years away from us, it is one of the most distant objects that can be seen with the naked eye.

Diametrically opposite Mirach to M31, and equidistant from it, is the Triangulum Galaxy (M33), although it is only just discernible in your photograph (even here, averted vision helps). If you can see M33 with your naked eye, then you have excellent vision and really dark skies. At a distance of 3 million light years, it is also one of the most distant objects that can be seen with the naked eye.

Astrometry.net

Note that only the core of M31 is visible to the naked eye (as it is in your photograph). If M31 was brighter, the arms of the galaxy would be visible and it would appear to be several times larger than the full Moon, as shown in the composite image below.

NASA


Saint-Exupery describes visit to plateau where he easily finds several meteorites. Is this realistic? In "Wind, Sand and Stars" Antoine de Saint-Exupéry describes a visit to a plateau where it is extremely easy to discern meteorites from stones, since there are no stones.

I picked up one and then a second and then a third of these stones, finding them at about the rate of one stone to the acre. And here is where my adventure became magical, for in a striking foreshortening of time that embraced thousands of years, I had become the witness of this miserly rain from the stars. the marvel of marvels was that there on the rounded back of the planet, between this magnetic sheet and those stars, a human consciousness was present in which as in a mirror that rain could be reflected.

Is this realistic? Are there places on Earth where you can find meteorite on 5 minute walk?
edit: Judging by this NASA article looking for meteorites in Antarctica is relatively hard.

This is only a partial answer; meteor falls are pretty common and so their density will be high in flat areas without a lot of flowing water, they're hard to identify without a good eye or at least a sensitive metal detector.
Hopefully a more quantitative source for the areal density of meteor finds can be found. The numbers are there, and if all rocks in the area that Saint-Exupéry describes are light in color, then picking up and checking every dark rock and checking to see if it has indications of burning or melting, this might be possible. Of course if it were this easy to collect them there then it seems others would have gone and collected a lot more. But those people have no incentive to document it since they can (often) be sold for cash apparently.
Note that the meteors shown in the image from the PRI article linked below are not dark but the dearth of dark meteorites found in Antarctica is a different story (1, 2) though there's been a lot of progress since 1986 when the document you link to was published.
According to Arizona State University's Meteorites page Meteorite Locations:

Since there is an estimated one meteorite fall per square kilometer per year, geologically stable desert regions can show significant accumulations of meteorites. Some desert regions have dozens of different meteorites per square kilometer, though they can be difficult to distinguish from normal terrestrial rocks.

PRI's Want to find a meteorite? Antarctica might be the best place to look says:

This winter, Lanza has been a rookie member of the ANSMET (the Antarctic Search for Meteorites) field team. For 40 years, the project has sent teams of scientists to the bottom of the globe to recover meteorites from all over the solar system, including chunks of the moon, comets, even Mars.
On this trip? Lanza and her colleagues recovered a total of 569 meteorites.
“They are just waiting for us to come and find them,” Lanza says. “You can imagine that blue ice ... and then you just look and there's some dark spots on them. And those are rocks and frequently those rocks are from space.”
Meteorites fall to Earth all the time, but Antarctica is a particularly good place to find them.
“We're lucky with Antarctica,” Lanza says. “It's this big white sheet covered in glaciers. And so these meteorites just become embedded in the ice and start flowing with the glacier. And in some places the glacier will run into something — like the mountain range that we were in — the Miller range. It will slow down and then the winds ... will start to remove some of that ice and that acts to concentrate the meteorites in these locations. So we can actually go there and find many more meteorites than you might imagine would fall in a single location.”

Nina Lanza knows space rocks. In her day job as a staff scientist at Los Alamos National Laboratory, she operates the Curiosity Rover’s ChemCam, using a rock-vaporizing laser to analyze the Martian surface.


Could a pulsar in orbit around a black hole feed the black hole with its radiation? Could the black hole consume the radiation? If so, could the pulsar radiation absorbed by the black hole make it grow steadily more massive over the course of billions or trillions of years?

Having the pulsar emiting radiation without decaying, the black hole not decaying via Hawking radiation and also the orbits of both objects not decaying and in a particular dynamic, I would say yes, veeeery slowly.

Thing is the radiation from the pulsar comes from particles that fall in it (dust, etc) and are "reoriented and shot" (roughly speaking) as high-energy radiation near the poles because the magnetic field.
The black hole would attract even more particles than the pulsar so considering both objects in a space with same density of particles, the black hole by itself should suck in more matter/energy compared with matter/energy from the pulsar beam.

Also consider orbital implications. A black hole (unless supermassive at center of galaxies) and pulsars have equivalent masses. Not the same, but not different in orders of magnitude if the black hole is a stellar remnant as the pulsar.
That means that the orbit will be, probably, more like a double-star system, with the black hole moving also around the center of masses (you can thing about it as Pluto-Charon orbit system).
Now. Consider how stars are formed (from a dust cloud, it starts to collapse, gain angular momentum, etc). Chances are that the orbits and the spins of both objects are in a very similar planes. Think that the beams of the pulsar are shot mostly from the poles (with some inclination though).
I think is almost impossible for a naturally formed stars that one becomes pulsar, the other black hole, and also they orbit in a way that the beams of the pulsar are going to feed the black hole. The beams would just scape from the system, forming ~90º relative to the line from the pulsar-black hole.

Last, about the "feeding". Lets consider that all the above is possible, just very unlikely. We have a scenario where the orbit of a pulsar is ~90º compared with its spin, like Sun-Uranus. Then a fraction of the time it feeds the black hole (twice a "year").
Black holes eat stars sometimes. The size of the black hole grows when it eats, and is one of the theories about the mass of supermassive black holes in the center of most galaxies (just eating... the other is direct formation via a sudden collapse of super-massive dust cloud even avoiding a star formation).
I try o point that this feeding via a pulsar beam would be super slow compared with direct feeding of surroinding matter like the supermassive black holes do.

I would say to the black hole: Sir, eat the pulsar :)

And maybe it would become true: considering very long periods of time, orbits decay. The black hole would maybe collide/absorb the pulsar (as recently detected via gravitational waves)
Finally, depending on the period of time we are talking about, we have to consider black hole evaporation throught Hawking ratiation (it would make it smaller).
On the other hand, I considered pulsars powered by matter orbiting/falling in it. there other types, so a rotation-powered one, after trillions of years, could maybe decay in an almost neutron star.
I don't know about the numbers so maybe these effects are almost zero, not sure.

PD. This is my first answer in StackExchange and I'm not a native english speaker, feedback appreciated ;)


who use the GADGET-2 pack for simulation? can I do it with my PC? I want to do hydrodynamic simulatin for universe at vary redshifts. How can I do it?
or how can I find an simulation of universe for use my work?

Gadget-2 is publicly available. Google works well to find it.
According to their website, it runs also on PCs. But don't get your hopes up, a single PC doesn't have enough computing power to simulate a whole universe. You'll need a bigger machine for that.

They have installation instructions on the webpage (see "Users Guide" under "Documentation"). You will possibly need a linux machine to run it, I don't know if the code compiles well under Windows.


Asteroids between Mars and Jupiter Is the cluster of asteroids between Mars and Jupiter two dimensional as seen in every model of solar system shown in books or television? Or is it three dimensional as i think it should be?

Many models shown in books or television show a very populated asteroid belt but in fact the belt is mostly empty.

To answer your question, the inclination of the asteroids vary a lot going from 0° to 40° although most off them are in between 0° and 30°; See The orbital element distributions of real and modelled asteroids. So yes it would be 3 dimensional.


Can you translate a GPS coordinate to a Galactic coordinate? I am a complete newbie to astronomy and all the related.
So if i say something stupid i am sorry.

But i was wondering if it is possible to translate GPS coordinates like the following:

lat: 41.420042122273024
long: 2.1533203149999736

to its corresponding Galactic coordinates?

I am grateful for all insights that you may offer me :)

PS: if this is the wrong exchange to ask this could you refer me to the right place to ask this question.

Sure you can! Galactic coordinates have the same origin as other J2000.0 systems; the solar system barycenter (center of mass). This is very close to the Sun, usually but not always inside the Sun, because the larger planets, especially Jupiter, pull it around a little bit. You can read a little more here for example, and also read @zephyr's excellent answer.

At first you might ask why the origin of the galactic coordinates isn't the center of the galaxy. I'm pretty sure that the answer is that we don't know where that is! We'd have to know the masses and locations of everything, and of course since most of the galaxy's mass is dark matter we're not going to know where the center of mass is any time soon.

However, the XY plane of the galactic coordinates has been chosen for now, based on an estimate of the Galaxy's apparent equator. Since it's a different plane than our solar system's plane, otherwise known as the ecliptic, the coordinates will be different even though the origin is the same.

Since galactic coordinates are centered near the Sun, the distance of our position from the origin will still be about 1 AU (150,000,000 km).

Below I wrote a little script in Python using the easy-to-use Skyfield python package. At the moment that I've run the program , the coordinates are:

time (JD):   2458099.18846
time (UTC):  (2017, 12, 11, 16, 31, 23.049599826335907)
latitude (degs):    41.42
longitude (degs):   2.15
galactic (km):     [ -1.46347711e+08  -2.89156773e+06  -2.34254700e+07]
barycentric (km):  [  2.69276456e+07   1.33751808e+08   5.79665590e+07]
and just for fun...
galactic latlon (degs):  [-9.0922867698877123, 181.13191434418721]

Here is the Python script:

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Loader, Topos

load    = Loader('~/Documents/SkyData')
planets = load('de421.bsp')
earth, sun = planets['earth'], planets['sun']

ts = load.timescale()
tnow    = ts.now()
tmonth = ts.utc(2017, 12, range(1,31))
topo = Topos(latitude_degrees=41.42, longitude_degrees=2.15)

position = earth + topo

print "time (JD):  ", tnow.tt
print "time (UTC): ", tnow.tt_calendar()
print "latitude (degs):   ", topo.latitude.degrees
print "longitude (degs):  ", topo.longitude.degrees
print "galactic (km):    ", position.at(tnow).galactic_position().km
print "barycentric (km): ", position.at(tnow).position.km
print "and just for fun..."
print "galactic latlon (degs): ", [x.degrees for x in position.at(tnow).galactic_latlon()[:2]]


How to estimate age of asteroid family Erigon I have one problem:

  On picture below we can see Erigone family asteroids in plane semimajor axis - absolute magnitude. Based on V-type as a consequence of Yarkovsky effect, estimate age of this family bearing in mind that members of this family have spectral of C-type. For average value of change of semimajor axis which belongs to this family take $da=5 \times 10^{-4} \text{AU}$ (for object with diameter $D=1 \,\text{km}$ and time $t=1$ million years). Ignore initial size of family.

I really don't know how to solve this problem (I am studying math, not astronomy). Professor won't tell us how to do it, and I tried to find the answer in his textbook and in two more books in my language, but there is no similar problem at all.

Could you point me in the right direction?

I'm not a professional astronomer, but I'm going to take a stab at this. If anyone sees any problems, feel free to correct them.

The Yarkovsky effect basically means that spinning asteroids over time will be pushed into a further orbit (more precisely, it increases the semimajor axis of the orbit) due to uneven heating and cooling.

Apparently to estimate the age of an asteroid family, using only the Yarkovsky effect, we would start out with this equation:

$$0.2H = \log(\Delta a / C)$$

This equation assumes a fixed geometric albedo for the whole family. $H$ here is the absolute magnitude. $\Delta a$ is equal to $a - a_c$ where $a$ is the semimajor axis.

$$C = \sqrt{p_v} (da/dt)_0 T \cos \epsilon$$

$p_v$ is the geometric albedo.  $(da/dt)_0$ is the maximum Yarkovsky drift rate for a body of size $D_0$, where $D_0$ is an arbitrary reference size (we could use your provided $D$ value above). $T$ is the age of the family that we are trying to calculate. $\epsilon$ is the spin axis obliquity.

Note: I've had to remove my example solution because I could not figure out how to correct my errors from accidentally putting $C$ outside the logarithm.

You can check out more details in my source below.

Source:

"Yarkovsky/YORP chronology of asteroid families" - Vokrouhlickýa, Broža, Bottke, Nesvorný, Morbidelli


How much of carbon, sodium, silicon, and magnesium does the Sun have? I've just begun learning of Astronomy and I can't figure out why any stars would begin their life with such elements if nuclear fusion hasn't created them.

Don't all stars begin life as Hydrogen?

I understand that the presence of these elements can change the luminosity of the given star. Is this also correct?

Stars are mostly hydrogen and helium. The presence of small amounts of other elements does change their luminosity, and their spectra. The heavy elements were formed in other stars.

After the big bang there was a short period in which the conditions were right for hydrogen to fuse into helium. The conditions were not suitable for any heavier elements to form, the state of the universe right after this time was about 3/4 hydrogen, 1/4 helium.

After some time clouds of hydrogen and helium began to form (around clumps of dark matter) and inside some of these clumps, smaller clouds collapsed to form the first generation of stars. These were large and they had no elements other than hydrogen and helium. These stars produced in their cores heavier elements like Carbon and Oxygen, and when they had reached the end of their lives, being big stars, they exploded in supernovae, and these elements were returned to the interstellar space, along with other heavier elements that were formed by the supernova.

The next generation of stars formed from a mixture of hydrogen, helium and the heavier elements formed in the first generation. These stars still had less heavy elements than modern stars, but were more like normal stars. Some of these stars are still around, they are known as Population 2 stars. They are found mostly in globular clusters http://hyperphysics.phy-astr.gsu.edu/hbase/Starlog/pop12.html.

After the population 2 stars had lived their lives and died, either in a supernova or in a planetary nebula, elements that had been made in their cores was returned to the intersellar space, futher enriching it with heavy elements. It was from one of these clouds, seeded with elements from stars that had gone before it, that the sun and the solar system formed.

This is why Carl Sagan says We are star stuff.


How many Moon's Topocentric Conjunction occurs per lunar month? It is clear that only one Moon's Geocentric Conjunction occurs per lunar month, what about the Topocentric Conjunction for a given location?

without scale I can imagine many Topocentric Conjunctions as shown in the pictures below:

yellow disk : Sun
gray disk   : Moon
blue disk   : Earth
yellow box  : Observer's Location
red line    : Conjunction's Line

Let's double-check @barrycarter's conclusion:

Ignoring the rest of the solar system and even the Moon's non-negligible mass, the velocity of the Moon's motion around the Earth is on average roughly 1 km/sec per $v=\sqrt{GM_E/a_M}$ (the vis-viva equation).

The Earth is about 12760 km in diameter, so the Moon traverses the Earth's projection in about 3.5 hours.

A topographic point traverses the Earth's projection in 12 hours with a tangential velocity that's never more than 0.47 km/sec.

That means that the moon's projected position always "moves" faster than a topographic location, so there can never be more than one topocentric superior conjunction per lunar month.

Hopefully there will never be any inferior topocentric conjunctions for this Earth at least.


Could rocks from Earth have reached the Kuiper belt, or Neptune at least? If so, how? This answer  (currently edit 3) to How certain are we that we have not sent life to other planets/moons? begins:

First of all, rocks from Earth are probably just about everywhere in the Solar System. One simple example is this rock found on the Moon. A number of pieces of Mars have been found on Earth, and if that has happened, no doubt there are Earth rocks on Mars. If Earth life can survive a vacuum it has probably gotten to everywhere in the Solar System anyways. (emphasis added)

I started to write the following as a comment:

"...rocks from Earth are probably just about everywhere in the Solar System" is quite a statement, and  ...from which point they will travel throughout the solar system. requires energy. What mechanism accelerates a rock from Earth to the outer solar-system? Mars is not really a distant object compared to the size of the solar system. New Horizons is in the Kuiper belt already. Is it even remotely plausible to suggest that Earth rocks can become KBOs?

Question: Can rocks from Earth have reached the Kuiper belt or Neptune at least? If so, how?
I'll even allow a solid particle of proper space-dust to serve as proxy for the rock for the purposes of this question, even sub-micron in size if necessary, but not simply an atom or a molecule. We know from answers to What is the origin of the dust near the sun? that dust is surprisingly (to me at least) mobile in the solar system due in part to its large surface to mass ratio and interaction with static EM fields and photons for example.

(I have tracked down the reference that I made in my comment on the question.)
Presumably earth rocks once blasted into space (by volcanism or meteorite impact) gain or lose energy primarily by orbital interactions with planets or moons. Since these interactions will occasionally give enough energy to escape the solar system entirely, presumably anything short of that (Neptune's orbit, the Kuiper belt) is also possible. The questions is if this will happen often enough to create a reasonable expectation that an Earth rock could have collided with any given solar system body.
Various attempts to identify the odds have been made.  In one, researchers simulated the path of fragments from the Chixculub impact, and found that many may have made it to the outer Solar System within 10 million years.
This is the paper in Astrobiology


If the space is expanding near the speed of light or the speed of light, why Andromeda is approaching or coming near to our galaxy (Milky way)? Space with distance of 1megaparsec is expanding with rate of 70km/s, and if space with a distance of 4296 megaparsec is expanding at rate of 300,720km/s, then why Andromeda is coming to us? Even there has a dark energy that accelerate the space expansion? Why?

The Andromeda galaxy is about 2.5 Mly away, less than 1 MPc, so if the Hubble flow were relevant, it would have Andromeda moving at about 55 km/s away from us. Andromeda is actually approaching at about 300 km/s, or almost 6X that calculated due to the expansion of space. However, Andromeda and the Milky Way are graivitationally bound along with the rest of the galaxies in our local cluster. The individual (so called, peculiar) velocities of these local galaxies are the result of orbital motion within groups and around the local cluster barycenter.

Expansion dominates on larger spatial scales, such as distances between clusters and superclusters of galaxies.


Is there a general term used to identify the high flat features within Noctis Labyrinthus on Mars? At the western end of Valles Marineris, where it mets Noctis Labyrinthus (highlighted section of image)

there are a series of flat geological formations between the canyons.

Is there a unique term for these formations, or even one borrowed from Earth's geology?

Has the individual sections of these formations been named?

The geological term mesa has been used to describe these structures.

  A transitional zone on Mars, known as the fretted terrain, lies between highly cratered highlands and less cratered lowlands. The younger lowland exhibits steep walled mesas and knobs. The mesa and knobs are separated by flat lying lowlands. They are thought to form from ice-facilitated mass wasting processes from ground or atmospheric sources. The mesas and knobs decrease in size with increasing distance from the highland escarpment. The relief of the mesas range from nearly 2 km to 100 m depending on the distance they are from the escarpment.

Mesa is a term used to describe flat topped isolated geological structures that are composed of horizontal layered rocks (usually sedimentary).

If the flat topped structure is not composed of horizontal layered rocks it can be called a plateau or tableland.


Is there a publicly available Gliese Star List Is there a publicly available list of the Gliese catalogue of nearby stars with their constellations? or a Gliese to Hipparcos List ?

The Gliese catalogue can be found at CDS here.

Given an RA and Dec then you can use the fortran program found here to get the constellation.


Retrograde motion and Kuiper Belt Objects As seen from Earth, planets such as Mars and Jupiter exhibit retrograde motion when they are near opposition (from Earth).  I am wondering how this effect extends to very distant objects, such as those in the Kuiper Belt.

Since objects in the Kuiper Belt orbit the Sun at a velocity much slower than Earth (or Mars or Jupiter for that matter) because of Kepler's third law I'd imagine that retrograde motion for these objects is more extreme in the sense that these objects are seen to be in retrograde motion for a longer time and over a larger portion of the sky (in an angular sense) than would be seen for any of the planets.

In an extreme case, is it possible that a very distant Kuper Belt Object (KBO) would never be observed in prograde motion?  I imagine that if you were to observe a KBO that was in the direction of Earth's motion then there would be no contribution from the Earth's velocity on the observation and you would be able to see the object in prograde motion, no matter how slow it was travelling. However, such an object would likely be obscured by sunlight, since objects that lie in the direction of Earth's motion lie above the day/night interface on the Earth.  My question here is that, given real-life observational constraints (like sunlight) how far from the Sun would an object have to be to never be seen in prograde motion?  You can be mathematical.

No, this is not possible.
Kuiper belt objects are in orbit around the Sun and their average distance (semi-major axis) will be beyond the orbit of Neptune. This means that their orbital period will be much longer than Earth's (following Kepler's third law). In the most extreme case (further away) the motion of the object will be virtually non observable and the object will appear not to move with respect to the background stars apart from the apparent motion due to parallax.

These objects will still show apparent retrograde and prograde motion because of the movement of the Earth around the Sun. This prograde and retrograde motion can also be observed in nearby stars where it is used to measure the parallax and thereby the distance to the star. See the figure below for the apparent motion of a star as measured by Hipparcos.

  I imagine that if you were to observe a KBO that was in the direction of Earth's motion then there would be no contribution from the Earth's velocity on the observation and you would be able to see the object in prograde motion, no matter how slow it was travelling.

The direction of the Earth's motion is constantly changing (changes 360° in a year) so the time at which a KBO will be in the direction of the motion of the Earth will be very short.


Did Gaia actually generate complete light curves for 212 Cepheids in other galaxies? The recent BBC news item Gaia clocks speedy cosmic expansion says that a recent Gaia measurement of 212 Cepheid variables in other galaxies within the local group yields a Hubble constant of ~73 km/s/Mpc which is in agreement with a previous measurement using the Hubble telescope.

The results are reported in: A test of Gaia Data Release 1 parallaxes: implications for the local distance scale. I have included Figure 1 at the end here. If I understand correctly, the x-axis or "Photometric Parallax" represents photometric data which is compared to a model to yield a distance, which is then expressed as an equivalent parallax.

Gaia is not a "point and shoot" instrument like the Hubble telescope. It has a slow, extremely steady rotation in order to scan at least 1.5 full revolutions - enough for every start in the scan to pass completely across the CCD array through both telescopes.

But this isn't an optimal observing plan for collecting data on a group of Cepheids. Although there are similarities, the light curves of different stars will be distinct. Am I naive to think that it is necessary to build up a nice, complete light curve with dense points in time in order to use the photometry for precision distance calculations? Was this actually done within Gaia's survey observing program, or was there a special observing plan for the Cepheid measurements?

The data also needs to be good enough to calculate the periods of each star with sufficient precision. The curve of absolute magnitude vs period is quite steep at low periods, meaning the error in the calculated distance is strongly affected by an error in the period.

And could someone help me understand - what is a TGAS parallax?

above: Figure 1 from here.

above X2: "Figure 4 and 4a: Cepheid light curves (click on images for larger versions) Light curves for the twelve Cepheid variables in M100 that have been observed with Hubble. The absolute magnitude, M, is determined from the period of the Cepheids. Adapted from Freedman et al. (1994)." from the tutorial: https://astrosociety.org/edu/publications/tnl/57/realastro3.html

Am I naive to think that it is necessary to build up a nice, complete light curve with dense points in time in order to use the photometry for precision distance calculations?

No, the more I read about it, the more difficult it seems to be. In principle, if you sample regularly and often, you should eventually get a light curve, unless the period is the same as or a multiple of your sampling period.

It's not easy to find how often GAIA observes a given patch of sky, but here it says that after one year of observing

  Gaia has made an average of roughly 14 measurements of each star on the sky thus far, [...]

so they seem to have repeat observations a bit more often than once a month. Cepheids have a period of typically 2 to 45 days (says german Wikipedia - didn't find it in the english article). If your sampling is regular, you should be able to get a decent light curve for at least some Cepheids.

Complete guesswork on my part: Apparently, longer period Cepheids are redder at the minimum. Perhaps they use that to distinguish between the actual period and higher harmonics?

  And could someone help me understand - what is a TGAS parallax?

In the introduction it says

  Tycho-Gaia Astrometric Solution (hereafter TGAS; Michalik et al. 2015)

From the abstract of that paper, it seems to describe their method (they practised with mock GAIA data). Seems like they use the previous Tycho catalogue to improve astrometry and parallax measurements from GAIA data (they mention using Hipparcos parallaxes as a consistency check).

So, figure 1 seems to be Cepheid distance vs. Parallax distance.


Why is Sagittarius A* called so? At the center of the Milky Way, we have a supermassive black hole. It's name is "Sagittarius A*". I was wondering that why is it called so? I mean, why "A*" why not just "A". Is there a naming system for super massive black holes?

Thanks!

Check this wiki link out. "The name Sgr A* was coined by Brown in a 1982 paper because the radio source was "exciting", and excited states of atoms are denoted with asterisks."

There is no unified naming system for black holes. They are usually named after their host galaxy. Others are identified by the name of the survey in which they were observed. A few black holes are catalogued by their constellation and the order in which they were discovered

EDIT: The 1982 paper by Brown and Lo can be found here.


What have I seen in the night sky I was out taking some milky way pictures last night, and saw a much quicker shooting star type thing, only with a much more random movement - any ideas what this could be - this is the RAW file from my camera - converted to JPEG for upload

I took the liberty of looking a bit closer at your image and annotate several features.

If you look closely, you see that several of the brighter objects in your image do the exact same random walk. Those are e.e. the objects A, B and C. This already strongly hints to some shaking of the camera, rather than A being a UFO.
If you look closer at your own image (which you always should), you see those lines all over the image from other weaker stars as well.

This automatically begs the question: How long was your exposure? Or was it composed from several single frames? Because it looks like one of those options  for me.

To substantiate this, let's look at some other features: Feature D doesn't show any spots, thus must have been taken while a shake of the camera was happening, thus must have moved quickly, thus this must be a shooting star. Of course a shooting star would appear like this as well on a non-shaking camera.
For contrast, feature E is also linear, but shows spots, which means while this line was putting light into the camera it must have stopped shaking on regular time intervals. This is a bit weird. But because the angle of the line, just as D, doesn't match any of the streaks that we follow in the figure A, this must also be a self-moving object. Now a self moving object of much lower speed than a shooting star (D), can be a satellite or an airplane.
Then there's feature F which clearly is also slow-moving, but does a weird rectangle that I cannot match to the 'master-shake' of feature A. It could mean that it's also self-moving taken together with one of the corners of the master shake. But then I don't understand why F is doing the cornery thing, but E does not. Another possibility in my opinion is, that E and F are both satellites, which faded in and out of brightness at different moments when the picture was taken, thus a different 'shaking' was imprinted on them.

So yeah, summarizing
Your image does some weird things, most of them clearly caused by camera-shaking. I don't understand all the details in your picture for that matter, but that's far from a reliable UFO claim. In particular because essentially EVERY UFO claim is based on shaky/nonsharp images.

P.S.
Also note how the grass in the lower left is shifted from really dark, to slightly dark. There are two central features in the grass that can be brought to roughly, but not exactly match when moving them over each other in photoshop. This indicates that this is the same patch of grass, just being photographed at two different times, after the grass was moving around a bit.
I suggest a more stable mount for your camera, if you're sure you didn't bounce into it by yourself.


How can we infer the mass of SMBH in galaxies that are not active anymore? I know it is possible to infer the mass of a supermassive black hole (SMBH) by many methods, i.e., stellar orbits for out Galaxy, Iron line profile from the accretion disk, and probably other methods (perhaps from the spectrum of the radiation disk itself, that can be related with the central mass, if supposed thermal in origin).
What I don't know is: how can we infer the mass of SMBH in galaxies that are not active anymore?

I can think of two methods.

Both rely on the dynamics of material surrounding the SMBH, which is affected up to a distance of the order of the "sphere of influence". This is the region where the BH dominates the dynamics as compared to the enclosed mass of the galaxy. The sphere of influence is:

$$ R_{\mathrm{infl}} \equiv \frac{G M_{\mathrm{SMBH}}}{\sigma_{\mathrm{bulge}}^2} >> R_{\mathrm{horizon}} \approx R_{\mathrm{Schwartzschild}} \equiv \frac{G M_{\mathrm{SMBH}}}{c^2} $$

While typically $\sigma_{\mathrm{bulge}} \approx 250 km s^{-1}$, it is well known that $c \approx 300000 km s^{-1}$. This means that the influence of a BH can be felt much further away than its event horizon, which is where the accretion takes place. In fact the ratio between the two distances is about 1000000.

Exploiting this fact, astronomers have been using two methods to probe extragalactic*, quiescent SMBH:

The first method is to observe CO lines (radio astronomy) to trace gas circling the BH. The gas does not need to be near the event horizon, which is much smaller than the sphere of influence. In fact CO observations rely on the gas being relatively dense but cold. Essentially the speed at which the CO gas will rotate is a (quadrature) sum of the declining component due to the stellar mass, plus the Keplerian component due to SMBH mass.
The second method is completely analogous, but relies on measuring the unresolved kinematics of the stars surrounding the SMBH. This can be done in various bands, but most authors use visible line absorptions to measure the velocity and velocity dispersion (and other moments) of the stars in the regions surrounding the BH. If the kinematics cannot be explained without including a point mass in the middle, then you are done.

See this work, for a comparison of the two methods.

*(extragalactic means outside of our own galaxy)


If jupiter was the densest planet on the solar system with low volume, could it be a star without multiplying its mass 80 times? Astronomers or Scientists say that Jupiter's Mass would have to be multiplied 80 times to become a star (M Mass). What if Jupiter was denser, would there be thermonuclear energy because of the density?

To achieve thermonuclear fusion you need high temperatures and high density.  If you were able (by magic) to compress and heat the core of Jupiter then fusion would occur.
But there is no mechanism that can cause Jupiter to increase it's temperature and density like this. In the core of a proto-star, gravity will provide the force needed to increase the pressure and density and the release of gravitational energy will provide the temperature to allow fusion to start.
Moreover, when fusion does start, the release of energy causes the core to stop contracting and prevents the density from continuing to increase.  The release of fusion energy stabalises the collapse of the ball of gas, and allows stars to shine for perhaps billions of years.  If you started fusion in the core of Jupiter (for example by setting off a nuclear bomb) it would cause that part of core to expand, lowering density and ending the fusion.  The stable state for a body the size of Jupiter is to be a planet, not a star.
Increasing density by changing the composition to one with larger nuclei actually makes it harder to start fusion.  Larger nuclei have a larger positive charge. This makes it harder to get them to fuse. So you need much higher temperatures.
So the only way for Jupiter to become a star is to increase it's core temperature and density. But that can't happen except by greatly increasing its size.


If I were to point into the sky, how many galaxies would be in this line? If I point in the sky and draw a line from the tip of my finger to the end of the universe, how many galaxies would this line cross? I had a discussion with my dad and he said due to the universe being so empty the line wouldn't cross any solid matter. I said that the universe is so enormous in size that there would be many galaxies and stars crossing the line.
Is this calculable? If yes which one is correct?

Take a look at the "Deep field" image from Hubble: http://www.spacetelescope.org/science/deep_fields/

You can see in this tiny part of the sky, hundreds of galaxies, but between there is lots of dark sky. Most lines of sight don't end on a galaxy. I'd say dad was right. I would estimate, from the deep field, that you have about a 1/10 chance of pointing at a galaxy.

Even if your finger does point at a galaxy, the distances between stars are vast. Think of two footballs separated by the distance between Europe and China as a model for the distances between stars. A finger pointing at a galaxy would be unlikely to point exactly at one of the stars.

This is probably a Good Thing. Olber's paradox supposed that if each line of sight ended on a star, the night sky would be a uniform blaze of light.


How much would you weigh if Earth stopped orbiting the Sun? I have a curious question.

The reason why astronauts are weightless in space is because they are orbiting Earth. If they suddenly stopped orbiting they would fall directly into Earth. Earth is doing the same around the Sun, so it would fall directly into the Sun if the Earth stopped orbiting the Sun.

If the Earth stopped rotating and orbiting, obviously ignoring the adverse effects of doing this and having Earth hold it's current position in it's orbit, how much would a 150 pound person weigh if they stood on suitable position on Earth.

This question is kinda out there, but it's a thought-provoking question for me and hopefully others.

In a sense, this is one of those "If I break the laws of physics, what do the laws of physics say will happen" kinds of questions.

Ignoring that, a 150 pound person would still weigh about 150 pounds if the Earth miraculously stopped orbiting the Sun. Almost all of a person's weight results from resistance to the Earth's gravity field. The tidal gravitational forces from the Moon and Sun are on the order of 10-4 ounces of force.

Note: I'm answering the question from the perspective of "what would a bathroom scale say". A doctor's scale would say a 150 pound person weighs 150 pounds regardless of whether the person is standing on the Moon, the Earth, or a very massive planet. A doctor's scale measures mass. A bathroom scale measures apparent weight.

So what would a bathroom scale say? If the Earth is at one astronomical unit from the Sun, whether the Earth is orbiting at 30 km/s or falling straight toward the Sun doesn't matter one iota. All that matters is that the Earth is one AU from the Sun. In both cases, that the Sun is 1 AU distant makes a 150 pound person weigh about $1.8 \times 10^-4$ ounces less at noon or at midnight than at sunrise or sunset.

The difference between the orbiting Earth and the in-falling Earth is that these tidal forces are more or less constant in the case of the orbiting Earth. They grow ever larger in the case of the in-falling Earth as it gets closer and closer to the Sun.

These tidal forces become quite large by the time the surface of the Earth just touches the surface of the Sun. Now our 150 pound person would weigh only 73.5 pounds per a bathroom scale if he was standing at that point where the surface of the Earth just touches the Sun. He would also weigh about the same at the point on the Earth furthest from the Sun. He would however weigh considerably more, about 188 pounds, if he happens to be at a point where the Sun is on the horizon.

If, on the other hand, some mysterious, magical force makes the Earth stop orbiting the Sun and makes this change permanent (i.e., the Earth stops moving with respect to the Sun), a 150 pound person closest to the Sun would weigh about 1.4 ounces less, while a 150 pound person furthest from the Sun would weigh about 1.4 ounces more. A 150 pound person who sees the Sun at the horizon would still weigh about 150 pounds in this scenario.


Pictures of large galaxies The Andromeda Galaxy is tens of thousands of light years in diameter. That suggests that the most distant stars in the Andromeda galaxy is up to tens of thousands of years "behind" in their orbit around the galaxy, compared to the closest stars. Have anybody generated pictures of the galaxy, which shows the approximate real position of stars? Is this effect visible when looking at the Andromeda galaxy through an amateur telescope?

M31 is actually something like 141,000 light-years in diameter (there has been some variation in our understanding of its size, but 141K is good compromise).  M31 is 2.54 million light-years from us.  The rotational speed of stars near the edge of the galaxy's disk is in the vicinity of 200 km/sec.

Given all that, in one year a star in the vicinity of the outer edge of the galaxy's disk is going to traverse 6.3 x 109 km.  At 2.54 x 106 light-years away, the apparent movement of such a star in the course of a year is going to be about 1.5 x 10-8 degrees, or 0.000054 arcseconds.

The Keck Telescopes in Hawaii, one of the most powerful telescopes in the world, has an angular resolution of 0.04 arcseconds.

As you can see, the best telescope in the world cannot distinguish any change of position of any star at that distance over the course of a year.  It would take about 740 years before such a star would move far enough for a telescope with the angular resolution of the Keck to distinguish any movement at all.

So, it isn't visible through an amateur telescope either.


The drift of the Moon's equatorial rotation velocity Are there any citation or documentation showing the measured degradation of the Moon's equatorial rotation velocity (thus affecting its moon's axial rotation)?

I am looking for the drift of the Moon's axis rotational speed, presumably over 10,000 years or more.

Since the Moon is tidally locked to the Earth, it rotates with the same orbital period as it orbits the Earth.  Thus, tracking the Moon's rotational velocity is essentially the same thing as measuring the changes in its orbit - if you know one, then you know the other.
The recession of the Moon has been measured using laser ranging (see links in the answer to this question) to be about 3.8 cm / year.  From that rate of change of orbital semimajor axis, you could use Kepler's third law to work out a rate of change of the Moon's orbital period, and thus of its rotational period.


How to align equatorial mount in the Southern Hemisphere? I'm new to astronomy and recently got myself a telescope with an equatorial mount. I've found lots of resources online showing how to align the telescope but they all operate on the assumption that you live in the Northern Hemisphere and align to True North or Polaris.

What is the procedure for those of us who live on the other half of Earth? Pointing my telescope north, I still suffer problems finding and tracking objects in the sky.

I'm going to post answers for both hemispheres, because there are good points from both sets of directions I've found.

For the northern hemisphere:

  The "declination drift method" is the most accurate way to accomplish this. The method is straightforward, but it does require some time and patience.

  First, aim the mount's polar axis roughly at Polaris. Now point the telescope at a star that's somewhat above the celestial equator and as close to south as you can judge by looking opposite Polaris. Put in a high-power eyepiece. If the eyepiece has cross hairs, center the star on them. Otherwise put the star on the north or south edge of the field and defocus it a little. Turn on the clock drive, and ignore any east-west drift.
  If the star drifts south in the eyepiece, the polar axis is pointing too far east.
  If the star drifts north, the polar axis is too far west.
  Shift the polar axis left or right accordingly, until there is no more drift.
  Now aim at a star that's near the celestial equator low in the eastern sky.
  If the star drifts south, the polar axis points too low.
  If the star drifts north, the polar axis points too high.
  Again, shift the polar axis accordingly.
  Now go back and repeat from the beginning, because each adjustment throws the previous one slightly off. When all visible drift is eliminated the telescope is very accurately aligned, and you can take long deep-sky exposures.

Via Sky & Telescope

For the southern hemisphere:

  Part 1

  Direct your telescope to a bright star that's low-ish (about 20 degrees) on the Eastern Horizon, near the Celestial Equator (i.e. 0 degrees DEC). The Orion nebula would be good to look at in mean time (given the right time of the year - as it may be too high or not visible).
  At this point, the east/west rotation of your mount has little effect, leaving you to correct the angle of elevation of the mount.
  Now, if the star drifts NORTH, your polar axis is too low - change the axis to angle further up in the sky (i.e. the latitude knob on your eq mount)
  But if the star drifts SOUTH, your plar axis is too high - change the axis to angle further down toward land (i.e. the latitude knob on your eq mount).
  You can tell if the star is drifting north or south by allowing the star to drift for a little bit - now, move your telescope to "catch up with the star" using the DEC control. If you see it heading north, then it's drifting north and vice-versa!

  Part 2

  Now, point the telescope to a bright star on the Meridian and on the celestial equator i.e. right above your head somewhere. This is so the elevation of the mount (as corrected in the steps above) will have minumal effect, allowing you to accurately correct the east/west rotation of the mount.
  Now, if the star drifts NORTH, your polar axis is too far EAST - so rotate your mount further West (i.e. rotate your mount clockwise while looking down at it).
  Now, if the star drifts SOUTH, your polar axis is too far WEST - so rotate your mount further East (i.e. rotate your mount anti-clockwise while looking down at it).

  Part 3

  Repeat these two steps a few times to increase the accuracy of your plar alignment - repeating Part 2 will allow you to get Part 1 more accurate.

Via OZScopes

Edit to provide less-precise-but-more-prompt method

  Polar alignment for southern hemisphere observers is very similar to techniques used for the northern hemisphere. The major difference is the lack of a bright pole star like Polaris near the south celestial pole (SCP) to aid orientation used in several alignment procedures.

  Rough polar alignment using the latitude scale of your scope is identical to the procedure used in the northern hemisphere.

  Either look on a map, use Google Earth, or consult an almanac to find your observing site's latitude. Unlock any latitude adjustment screws on the sides of the mount and turn the latitude adjustment screws until the index on the polar axis reads your latitude. Tighten the adjustment screws if needed to secure the latitude setting. (You may also need to loosen the center pivot bolt by turning the hex nut to allow the equatorial mount head to be tilted.)

  Latitude adjustment of the mount

  Now complete the polar alignment by turning the entire mount (not either axis; both should be clamped tightly) to align the upwards end of the polar axis with south on the horizon.

  Another more accurate rough method is to point to Sigma Octantis. This star is located only one degree from the SCP, the point in the sky around which all the other stars appear to rotate, and where the polar axis of a properly aligned equatorial mount should point.

  First, set up the mount so that the polar axis is pointing south.

  Second, unlock the declination clamp and move the scope in declination so that the tube is parallel to the polar axis. Your declination setting circles should read 90° in this orientation. Clamp the declination lock.

  Note: These last steps involve moving the entire mount. Don't use either the RA or Dec motions to change the position of the tube.

  Move the mount in altitude and azimuth until Sigma Octantis is in your finder's field of view or centered in your finderscope. At magnitude 5.5, it is a fainter star--just slightly brighter than the unaided eye limit--so your finder will definitely be helpful here.

  Sigma Octantis and the SCP

  Tweak the position of the mount by again moving the mount, this time centering Sigma in the eyepiece field of view. Altitude can be adjusted using the latitude adjustment screw or shortening-lengthening tripod legs.

  For greater accuracy, you can point directly at the SCP using an offset from Sigma Octantis. (This can be done after aligning the optical axis of your finderscope with the polar axis.) Use the chart above or these patterns of southern circumpolar stars to find the SCP. Draw an imaginary line toward the SCP through the Southern Cross stars Gamma Crucis and Alpha Crucis (down the long axis of the cross). Draw another imaginary line toward the SCP at a right angle to a line connecting Alpha Centauri and Beta Centauri. The intersection of these two imaginary lines will point you close to the south celestial pole.

Via Celestron


Are there any known neutron star - supergiant binaries? After looking for more information about TZOs (Thorne-Zytkow Object), I came up with a question that I had trouble finding answers on the Internet. Do we know of any future TZOs, supergiants being orbited by/orbiting a neutron star whose orbit is decaying at a rate that will cause the supergiant to merge and form a TZO before it dies in a supernova? If so, what are they? If not, why can't we find them?

One possible formation pathway for TŻOs involves a massive star undergoing mass transfer to a neutron star companion. A common envelope forms, and subsequent friction leads to a merger. High-mass x-ray binaries (HMXBs) are therefore good candidates for progenitor systems. Hutilukejiang et al. 2018 simulated TŻO formation through a number of channels, and found that 3 of 114 known HMXBs in the Milky Way involve a massive supergiant companion capable of filling its Roche lobe and engulfing the neutron star, potentially leading to a TŻO. The three systems listed are  SMC X-1, Cen X-3 and LMC X-4.
Progenitors of TŻOs through other channels may be harder (if not impossible) to find. Random collisions between neutron stars and supergiants in globular clusters are effectively unpredictable on an individual basis. Pulsar kicks from a supernova in a supergiant-supergiant system can also lead to TŻOs but are also difficult - again, if not impossible - to identify, given that kicks are not well understood. I would therefore assume that common envelope evolution is the best bet for identifying progenitors.


Earth Rotation and Orbit Geometry Let Line A be the projection of the axis of Earth's rotation onto the plane of the eliptic.  Let Line B the line connecting the center of the Earth and center of the Sun.

At what times during the year is Line A perpendicular to Line B?

The axis points in pretty much the same direction all the time, though it does wobble slightly, ranging between 22.1 degrees and 24.5 degrees every 41,000 years.   (Source)

Perpendicular between two lines is a 2 dimensional property, so, while the axis is never "straight up" so to speak, there is a 90 degree angle between the axis and the line between the center of the sun and the center of the earth during  the spring and fall equinox.  Those are the only times.


Finding equation for energy density of matter I'm trying to get an equation for the energy density of matter of the universe $\rho(t)$, assuming the number of particles is conserved and rest mass energy is much greater than kinetic energy. $\rho(t)$ should be a function of the scale factor of the universe $a(t)$, mass of particles $m$, and the number density of the particles $n(t_0)$.

I'm not sure how to approach this since $\rho(t)$ is in units of $\frac{Mass}{Length^3}$, $m$ is in units of $\frac{Mass}{Particle}$, $n(t_0)$ is in units of $\frac{Particle}{Length^3}$, and $a(t)$ is in units of $Length$. I guess there could be a factor of $\frac{a(t)}{a(t_0)}$ or $\frac{a(t_0)}{a(t)}$, but I don't know where I'd get that from.

Any help?

EDIT! I forgot that $a(t)$ is dimensionless.

Energy density $\rho \sim nmc^2$ if kinetic energy is unimportant (and you are ignoring dark energy and other forms of energy density).

But the number density of particles varies as $a^{-3}$.

Thus
$$ \frac{\rho}{\rho_0} = a^{-3},$$
where $\rho_0 = n(0)mc^2$ is the present day energy density and the present day scale factor is 1.


How long would a photon take to reach us if it's emitted from a galaxy receeding at c? A galaxy at the boundary of the Hubble sphere is receeding from us at the speed of light, right? If it emits a photon now, how long will it take to reach us?

Actually, this is a perfectly good question if interpreted in a reasonable way, i.e., interpreted as asking how long will light take to get to us if emitted at comoving age 13.8 billion years from a galaxy whose comoving distance is currently increasing from us at rate c (and the Hubble law with H=70 tells us that comoving distance is 4.3 Gpc or 14 billion LY).  That is not an infinite time, as that is not the definition of the Hubble sphere.

The answer to how long it would take, less than infinity, depends on the cosmology model, and is a little hard to calculate because cosmology calculators (such as http://www.astro.ucla.edu/~wright/DlttCalc.html) tend to focus on light reaching us now-- as that's what we see.  For example, that calculator gives that a galaxy that is receding from us at rate c now would have had to emit light 9.25 billion years ago for us to be seeing it now, but that's not what was asked.

As mentioned, it is not easy to calculate the time it would take light emitted at the Hubble sphere now to get to us, but it is not infinite.  The Hubble sphere would only be the edge of what we could ever see if the Hubble constant were constant with time (as in an accelerating universe completely dominated by a cosmological constant), but even though the expansion does seem to be accelerating, it is not yet completely dominated by a cosmological constant (assuming that is what is causing the acceleration).  Hence, we would indeed eventually see the light emitted at the Hubble sphere now, but it might take a very long time, perhaps 50-100 billion years but that's just a guess without doing a real calculation with a full cosmological model.


Nicolas Copernicus discovery As I was reading about the heliocentric model, a question came up: How was Nicolaus Copernicus able to figure out that the sun is at the center of the solar system, and that all planets orbit around it, even though the telescope had not been invented yet? During his time the geocentric model was the standard.

Copernicus did not figure out that the Sun was at the centre of the solar system. He merely proposed that a heliocentric model simplified the calculation of planetary orbits. Epicycles (circles within circles) were still needed in the Copernican model, for instance for the orbit of Mars, but there were fewer of them than in the Ptolemaic model.

The Copernican model received a boost when Kepler, discovering that planetary orbits were elliptical, removed the need for epicycles completely, thus increasing Copernicus' advantage over Ptolemy.

The first evidence that the Copernican model was a true representation of actual reality came with the observation of stellar parallax in 1806. Until then, it was preferred only because it was a simpler description - a much simpler one, after Kepler.


What is meant by the notation $A^{\times{B}}_{\div{C}}$; where A,B,C are real numbers? I have come across a few papers using the notation $A^{\times{B}}_{\div{C}}$; where A,B,C are real number. For example, $3000^{\times{3}}_{\div{4}}$.
An example can be Eqn (4) in Stern & Laor (2012) "Type 1 AGN at low $z$ – II. The relative strength of narrow lines and the nature of intermediate type AGN", which reads as follows:

$$\frac{L_{\rm bol}}{10^{43}\ \rm erg\ s^{-1}} = 4000^{\times 4}_{\div 4} \left( \frac{L_{\rm [O\ III]}}{10^{43}\ \rm erg\ s^{-1}}\right)^{1.39}$$

What is the generic meaning of this notation?

A confirmation of the explanation is given slightly further on where it talks about how the uncertainty is derived:

In equation (4) we use the 0.6 dex scatter of $\log L_{\rm UV}$ around
the $L_{\rm UV}$ versus $L_{\rm [O\ III]}$ relation, as an estimate for the uncertainty
in deriving $L_{\rm bol}$ from $L_{\rm [O\ III]}$.

The term "dex" indicates a base-10 logarithm of a unit, which translates into a factor of $10^{0.6} \approx 4$. So translating the error expressed in the logarithm of the quantity into an error expressed on the quantity itself, this becomes "multiply-or-divide by 4", which here is rendered $4000^{\times 4}_{\div 4}$.
Pela's excellent answer explicitly goes into detail about what this implies for the probability distribution. I usually prefer notation like $10^{x \pm \sigma}$ which sticks to more familiar notation and explicitly implies something vaguely log-normal rather than normal, although perhaps it gets a bit cramped when asymmetric error bars come into play.


Where do the symbols $\lambda$ and $\beta$ come from for celestial latitude and longitude? I'm reading Episodes From the Early History of Astronomy, and the author mentions the system of celestial coordinates based on the ecliptic. In this system, the two coordinates are latitude and longitude, denoted by $\lambda$ and $\beta$. Where does this choice of symbols come from?
$\lambda$ is "l" like in "latitude", I guess. But why $\beta$?

German: $\lambda$ (or l) like Länge (longitude) und $\beta$ (or b) like Breite (lattitude). A century ago it used to be a language many science papers were published in and English was far from dominant.
Mind though that it is just symbols of notation and you can define them as you see fit - if you define them. They are just one choice among many possible and one among a few commonly used, depending a bit on context where you come from.


How do we calculate the image circle of a telescope at prime focus? I'm trying to find a suitable CCD and telescope combination. So, I want to make sure that the image circle of the telescope circumscribes the CCD. But I'm unable to find anything that tells me how to calculate the same.

Image Circle will vary with scope design. Just about all scopes will cover the field of a 1.25" eyepiece barrel, and scopes with 2" focusers will usually cover the field of a 2" barrel (There are some exceptions where the field is bigger than 1.25" but doesn't cover the full 2"). Basically, you need to check the spec for the scope - and in any case the size of the focuser usally sets an upper limit. (With smaller SCTs and Maks, the size of the baffle tube can also limit the image circle).

Most small refractors for astrophotography will cover an APS sized DSLR sensor.
If you're looking for something to cover a full frame DSLR sensor, the choice gets more restricted - quite a few of the field flatteners/reducers for smaller refractors are only designed for APS coverage.

If you want to go for an unusually large sensor, then your choice is even more restricted - something like Takahashi's FSQ106ED astrograph has an 88mm diameter image circle in native mode, large enough for some medium format sized sensors, or 44mm with the reducer or extender - but there's a high price tag to go with it.

However, if you're looking at large format astonomical CCDs, they're usually very expensive too. Most of the more affordable astro cameras have smaller sensors, so are less demanding on image circle size.


Why did the big bang not just produce a big black hole? Questions I've often wondered about:

If all the matter and energy were concentrated at a single point at the big bang, why wasn't that a black hole, or why didn't it form one?

If the reason #1 above didn't form a black hole is one of several explanations such as inflation or whatever else, then why didn't all the mass and energy form a big black hole at some finite time after this big bang happened?  For instance, I've (possibly incorrectly) heard that inflation made the universe the size of an orange.  Well why didn't it form a black hole then?  Or once the universe expanded to, say, the size of the moon. Why not then? Just insert whatever reasonable size you want in place of "orange" or "moon."  The question is why didn't a black hole form out of all the matter and energy after the Big Bang?

Your problem essentially arises from trying to apply Schwarzschild black hole logic, the assumptions of which are pretty much maximally violated at the big bang.
The following were true at the big bang, and violate the usual black hole formation logic.
The event occurred everywhere in space, not actually a point. In particular, the energy was uniformly distributed everywhere. The net gravitational potential was therefore near zero, and there was no one point to which everything could collapse. Furthermore, since stuff was everywhere, there was no expanse of vacuum (in a flat spacetime, no less) outside the collapsing region. And furthermore, things were moving rapidly, were in a highly excited state, and were not in thermal equilibrium (until inflation hit, and then things were too diluted and causally disconnected to collapse en masse).
As for what we could describe the universe as at time $t=0$, we have no idea. General relativity has a spacelike singularity there, and it subsequently cannot say anything beyond that. It is expected that general relativity is not a correct description of spacetime in the pre-inflation era, in large part because quantum mechanics introduces significant effects in those conditions, and it is well known that the two theories are incompatible.
Edit:
This question has been asked on the physics SE many times.
https://physics.stackexchange.com/q/20394/55483
https://physics.stackexchange.com/q/3294/55483
https://physics.stackexchange.com/q/26435/55483
Perhaps the answers there will be illuminating.


Declination calculation My task is to calculate current declination and right ascension for some artificial Earth satellite. For example ISS.

I am using latest TLE data for extracting keplerian orbital elements and then calculate current cartesian orbital state vectors in ECI frame.
From the position vector I directly calculate declination and right ascension, using cartesian to spherical coordinates conversion.

Right ascension gets calculated correctly but I am having problems getting the same result for declination as the one displayed by satellite trackers.

Speaking of satellite trackers, I don't understand how is their declination value even correct.

Declination is an angle measured north or south of the celestial equator.

Celestial equator of course is on the same plane as the equator of Earth.

Then how is it possible that, when a satellite (ISS in this example) is almost directly above Earth's equator, satellite trackers calculate declination to be around -29° ?

Shouldn't the declination angle be zero?
My calculations for this "ISS above equator" example did give declination to be close to 0°.

Who is doing the calculations wrong and why?

You are forgetting parallax.

Something which is very distant and lies on the celestial equator will have a declination of 0°, but if it is nearby then its declination will only be 0° to an observer located along Earth's equator.

From your location in Zagreb, something like the ISS, even if it is crossing Earth's equator at a longitude of ~16° will appear to be in the southern sky, and your declination coordinate will likely be south of 0°.


About more than 15 stars moving in straight line through the sky I had seen more than 15 stars are moved in a straight line through the sky just like a train. And the distance between each star was not same, some of them were near and some of them were far. The intensity of the light of each star was also different. I had seen this phenomena on 21st November 2019 at Indian time 5:30 AM in my place. What could that have been?

These will be SpaceX's second batch of 60 Starlink communications satellites. They were launched on 11 November 2019, and are still in roughly the same orbit, although they are slowly dispersing. Enjoy the show. There should be plenty more. SpaceX plans to launch several thousand of these satellites. Unfortunately, this is not good news for astronomers, since they may interfere with ground-based observations.


Life cycle of a gas planet? I am interested in learning more about the (simplified) life cycle of gas planets which are not brown dwarfs (meaning less than 13 Jupiter masses). It obviously starts off with their creation within a protoplanetary disk. For Jupiter, things then happened quickly:

Jupiter formed in a geologic blink. Its rocky core coalesced less than a million years after the beginning of our solar system, scientists reported Monday in the Proceedings of the National Academy of Sciences. Within another 2 million or 3 million years, that core grew to 50 times the mass of Earth.

Source: Washington Post: Jupiter is oldest planet in solar system
But what happens afterwards? How long would a gas planet like Jupiter live (assuming that the central star of the system would not consume it)? What will happen to the radual density distribution over time? On which parameters (mass, radius, initial density, ...) does the evolution depend?
Related

Terminology question: gas giant vs gas planet
Will Jupiter eventually stop shrinking?

The life cycle is just one of gradual cooling and settling. The heavier elements will sink towards the centre; the planet will get slightly more dense - probably reaching about 90% of its current radius. It will essentially become a cold mini-white dwarf made mostly of hydrogen and helium and supported by cold electron degeneracy pressure. The density is probably not high enough for it to crystallise before differentiation of the chemical elements has completed.
The density profile will not change very much from what it is now. If anything the desnsity gradient will become slightly shallower become because a degenerate gas of heavier elements has more electrons per unit mass and so exerts a higher pressure for the same density. Since the thermal conductivity is extremely high then a temperature gradient will gradually be eliminated in the inner regions.
These processes take place on tens of billions of year timescales. Jupiter will be quite similar to the way it is now by the time the Sun becomes a white dwarf.


Have caged molecules ever been observed in space? astrosnapper@'s thorough answer to the question What is precipitable water vapor in millimeter-wave radioastronomy and how is it measured? links to the paper Observing Conditions for Submillimeter Astronomy which discusses spectral windows of atmospheric transmission in the 100 ~ 1500 GHz range. These windows are "delineated" or limited by strong absorption lines of molecular water vapor in the atmosphere.

Figure 2 is shown below. The caption reads

  Fig. 2. Energy levels of the lowest rotational states of water with transitions labeled by the frequency in GHz. The lines at 557, 752, 988, 1113, and 1229 GHz delineate the principal submillimeter windows.

and displays ortho- and para- water separately.

Curious to find out what those were, I went to Martin Chapin's (also here) site Water Structure and Science which discusses the physics of ortho- and para- water, and describes experiments on single water molecules trapped and isolated in C60 fullerene "cages" (Endohedral fullerenes):

ortho-Water and para-Water

Question: Have caged molecules ever been observed in space?

These could be trapped in fullerenes (which I believe have been observed) or otherwise.

Fullerenes have been detected in space, including C60, C60+ and C70 and appear to be responsible for the diffuse interstellar bands (DIBs), which are absorption features caused by the interstellar medium. The question of endohedral fullerenes is an interesting one, it is discussed in the paper "Interstellar fullerene compounds and diffuse interstellar bands" by Omont (2016).

In section 3.1.2, the situation for metallofullerenes M@C60 is discussed:

  Analysis of the optical spectra of metallofullerenes thus
  presents the same difficulties as do C60 or C60-. It is therefore not surprising that even theoretical studies of their
  energy levels and excitation are lacking, incomplete or ambiguous, and we are still waiting for laboratory spectra.
  The accuracies of wavelength determinations are still far
  from being useful in constraining possible associations of
  visible DIB features with metallofullerenes. Estimates of
  line strengths are practically always lacking.

From this it would appear that more lab work is necessary before we are able to identify metallofullerenes from astronomical spectra.

In section 7, Omont describes the situation for hydrogen molecules in fullerenes (H2@C60), with section 7.2 discussing the prospects for finding interstellar H2@C60:

  It seems that the presence of endohedral H2 would not
  change the basic properties of interstellar fullerenes much,
  making it difficult to reveal its presence, except thanks to
  specific spectroscopic features. However, even if it were
  significant, the presence of endohedral H2 should be difficult
  to confirm because the abundance of H2@C60+, for instance,
  could hardly be expected to exceed or even reach 10−9.
  There is thus practically no chance to ever detect its near-IR H2 2 µm absorption. UV spectroscopy is more sensitive
  by orders of magnitude than absorption of weak IR lines.
  It can detect H2 column densities lower than ∼1014 cm−2
  (e.g. Gillmon et al. 2006). But even so, the UV detection
  of H2@C60 seems out of reach.

A quick check of the papers citing Omont (2016) doesn't turn up any detections of endohedral fullerenes, so it looks like the situation remains as described.