Requesting book references for a non-expert person with math background or just a non-expert person I've just finished my masters degree (to be exact previous week ^_^). And I'm completing my collection of applications of my studies. During previous year I started reading about Robotics and Coding theory (but in free times as a hobby). I knew that my field of study has application in Astrophysics but I didn't have enough time to go after this topic as well. But my main question; these days I felt into mood of having a general knowledge in Astronomy etc. everything related to stars and space.

As I'm not seeking the application of my field of study at astrophysics (if I were it was better to put this question at math stack instead of hear) and certainly almost all members here are not familiar with pure mathematics, I will be pleased if someone here introduce one or several nice texts or books which satisfy one of the two following set of conditions:
1- Some elementary sources which people who doesn't have any background can read them. The notes be exciting and the reader wants to continue it until its end.
2- Sources which also use mathematics (specially if the level of the math won't be only some calculus) but the dominant text be about stars and space.

Assume you are suggesting a text to a person as his first thing he will read in this topic. And if you think a series or a number of texts are suitable, please give an order to which should be read first and then which for the next and so on.
(Also please note that I don't like to read guesses! So texts contains we guessed x should be because of y or we think the world is ... are not interesting to me)

Thank you for your attention.

I personally found Stephen Hawking's "A Breif History of Time" to be very interesting and informative.  It is a little bit advanced, but is put forward in plain English.  Here is a link to it on Amazon, and you can "look inside" to see part of the book and see what you think.

http://www.amazon.com/gp/product/0553380168/ref=pd_lpo_sbs_dp_ss_1?pf_rd_p=1535523722&pf_rd_s=lpo-top-stripe-1&pf_rd_t=201&pf_rd_i=055371449X&pf_rd_m=ATVPDKIKX0DER&pf_rd_r=1Z0JYCKHWXEFPK3TS6TM

After reading your comments / clarifications, you might like "A Field Guide to Stars and Planets."  It appears to be a good introduction into astronomy.

http://www.amazon.com/Field-Guide-Planets-Peterson-Guides/dp/0395934311/ref=sr_1_1?s=books&ie=UTF8&qid=1411293771&sr=1-1&keywords=stars+and+planets

As far as covering astronomy from a more mathematical perspective, you might like "A Student's Guide to the Mathematics of Astronomy."

http://www.amazon.com/Students-Guide-Mathematics-Astronomy/dp/1107610214/ref=sr_1_1?s=books&ie=UTF8&qid=1411294172&sr=1-1&keywords=math+of+astronomy

You can take a look inside each of these books to see what you think.


Metals and dust locked into planets I wonder what is the contribution of metals and dust locked into planets, with respect to the amount in the interstellar medium (ISM). In other words, when we measure the metallicity and the dust mass of a galaxy we are sensitive to metals and dust in the various phases of the ISM, and thus the observed value is a lower limit. Is the "correction" a second-order variation?

The amount of metals locked up in planets is completely negligible compared to that of the interstellar medium (ISM).

First, in a typical stellar system, the mass fraction of planets is less than 1%, and of this by far the most mass will be in gaseous planets, rather than rocky planets. For instance, in the Solar System, the mass fraction of all planets is 0.1%, while rocky planets comprises only a fraction of $6\times10^{-6}$.

Second, typically only a small fraction of the gas in a galaxy is locked up in stars. For small galaxies — which are the most abundant — as little as ~5% of the gas is in stars (e.g. Schombert et al. 2001), although for large galaxies like the Milky Way, which are more efficient at converting gas into stars, the ISM-to-stellar gas fraction can be of order unity (e.g. Morokuma-Matsui & Baba 2015).

That is, the fraction of metals locked up in planets is of the order $10^{-5}\text{–}10^{-7}\!$, depending somewhat on the type of galaxy considered.

On the other hand, the fraction of metals locked up in dust is considerable, with typical values of roughly 1/3 of the metals in dust. Thus, if you want to know the metallicity of a galaxy by looking at emission or absorption lines from metals, you must either account for this factor (hoping that it's a universal factor which it may or may not be, see e.g. Zafar & Watson 2013), or measure metals that tend not to deplete to dust grains, e.g. zinc or sulfur.


What is the largest hydrogen-burning star? I am wondering what is the largest known core hydrogen-burning star?  A look at the list of largest known stars on Wikipedia seems to indicate VV Cephei B (at the bottom of the list), but I would like to know for sure if it is the largest known.  In addition to knowing which star it is, I would also like to know its temperature, size, and expected lifetime.

I am also curious to know if the largest known core hydrogen-burning star is similar to what astrophysicists theorize is the largest possible core hydrogen-burning star (given current metallicity conditions in the universe; I know star-formation timescales have a metallicity dependence) and the expected temperature, size, and lifetime of such an object.

I assume by largest, you mean largest radius.

Well it won't be VV Cep B since this is merely a B-type main sequence star.

O-type main sequence stars are known and these have both larger masses and larger radii on the main sequence (when they are burning hydrogen in their cores).

A selection of the most massive objects can be found in the R136 star forming region in the Large Magellanic Clouds. If you look at this list (though I recommend having a look at the primary literature), you will see that O3V stars are listed. Such objects are also present in our Galaxy, for instance in the supercluster NGC 3603 (Crowther & Dessart 1998).

Such stars have masses of maybe $100 M_{\odot}$, luminosities of $2\times 10^{6} L_{\odot}$ and temperatures of 50,000 K. Using Stefan's law, we can deduce radii of $\sim 20 R_{\odot}$.

There are suggestions that even more massive main sequence stars have existed in R136 and NGC 3603 (see Crowther et al. 2010), which are now seen as evolved Wolf-Rayet objects, possibly up to $300 M_{\odot}$ on the main sequence (though this is a model-dependent extrapolation), and these would have had radii $>20 R_{\odot}$.

In the very early universe, population III main sequence stars without metals could have been much more massive and larger.


Why does Kepler's "Big Picture" of comet 67P look so strange? And what is the significance? The observation of comet 67P by the Kepler space telescope has been reported in various science news outlets recently. For example, the image below appeared in the recent NASA news article NASA's Kepler Gets the 'Big Picture' of Comet 67P and the same or similar images can be seen elsewhere. The article says:

From Sept. 7 through Sept. 20, the Kepler spacecraft, operating in its K2 mission, fixed its gaze on comet 67P. From the distant vantage point of Kepler, the spacecraft could observe the comet's core and tail. The long-range global view of Kepler complements the close-in view of the Rosetta spacecraft, providing context for the high-resolution investigation Rosetta performed as it descended closer and closer to the comet.
During the two-week period of study, Kepler took a picture of the comet every 30 minutes. The animation shows a period of 29.5 hours of observation from Sept. 17 through Sept. 18. The comet is seen passing through Kepler's field of view from top right to bottom left, as outlined by the diagonal strip. The white dots represent stars and other regions in space studied during K2's tenth observing campaign.

I don't really understand this image. It is a GIF assembled from several exposures, but in each frame, individual stars appear as tight clusters of dots for some reason, and the comet moves within a bright, narrow band that crosses the entire image in all frames. Why does this looks so strange?
Also, what is the importance of Kepler deviating from it's observations to concentrate on 67P during this period? What is the scientific importance of this series of observations?

above: Image credit: The Open University/C. Snodgrass and SETI Institute/E. Ryan From here.

One thing to keep in mind is that the Kepler instrument is not a telescope like Hubble.  It is a photometer and though it uses CCDs to look at the sky, it doesn't return a picture in the usual sense.  The way it operates is that you only look at the pixels around the object you're interested in because otherwise you'd never get all those pixels transmitted to the ground because there are so many of them.
To observe the comet, they looked at the track it would make across the Kepler field of view.  They chose only the pixels you'd need to look at the comet and its tail, and this turned out to be a 15-pixel-wide swath across the detector.  That band across the image you see is that swath.  You'll notice that most of the picture is black, which are all the pixels that they didn't look at.
The stars in the scene where the image isn't back are the stars that Kepler was also looking at.  However, I'm not intimately familiar with raw Kepler data so I can't tell you why the stars are in clusters of four.  Because all the stars are lined up in the same direction, I would surmise that this is because Kepler was tracking the comet and the stars were moving in the background.  From the way they were going to take the measurements, each 36 hour observation was made up of three observations, so I am at a loss as to why there are four dots per star when this would suggest there would be three.
The answer to the science question can be found from the observation proposal.  To summarize:

Kepler photometry can provide the following key measurements at a time
when no other platform can observe the comet:

Total brightness of the comet. This relates directly to the amount of dust being released and therefore gives a critical ‘total’ activity
measurement, which can be compared with the instantaneous (but local)
measurements from various instruments on Rosetta.
Variability of the comet. There are two sources of variability; changing amounts of dust in the coma, and the underlying light-curve
of the nucleus (its non-spherical shape reflects more or less sunlight
as it rotates).

The measurements from Kepler will extend the total activity curve to
almost the end of the Rosetta mission, meaning that data in the last
months of Rosetta will not rely on extrapolation of the earlier
Earth-based observations. As the final months will be some of the most
exciting – as Rosetta spirals down and makes local measurements from
incredibly close to the nucleus – this context information is very
important in this phase.


Star formation: When a nebula collapses, is it only the gasses that form the star? A nebula is made of dust (elements heavier than He and H) and gas (H and He).

Stars are made when nebulae's collapse and hydrogen begins to fuse. 99% of nebula material goes into making the star. The left-over 0.01 of debris creates planets.

Does 99% of nebula material that makes the star only consist of gas (He/H) and the 0.01% is the dust which makes the planets?

Question Clarification:

Are the constituents of the 99% of mass that makes the star and the 0.01% mass that makes the planets the same?

I.e.

Does both the 99% and 0.01% compose of mainly He/H and a small portion of metals (meaning the star contains heavier elements like Si and Fe etc before fusion?)
OR
is 99% of the nebula composed of He/H, which collapses to form the star, and 0.01% consists of heavier elements that make the planets?

The very first stars to form indeed consisted essentially only of hydrogen and helium. When stars die, they leave behind them more massive atoms, as you say. These heavier elements are too incorporated in newer stars when they form. This results in stars which start out with a lower portion of hydrogen and helium, thus making them somewhat less effective; see for example this question and answer. The Sun, for instance, is a so-called third generation star, which means that it is made up of materials which have already been used at least twice since the Big Bang, and thus naturally contains heavier elements it has not produced itself.


Why won't the Sun set for days at N66.2 which is below the arctic circle? Correct me if I am wrong, but if we count sunsets by the center of the Sun apparently crossing the horizon then the Sun is supposed to set every day at latitudes under the arctic circle. (Yes if you count in the disc then adjust by 0.27 degrees.)
I was playing around with PyEphem (a python library that claims an approximately 1 arcsecond accuracy), and found that according to it the sun stays up for a couple of days even under it ($66.2<90-23.4$). Can someone explain what is going on?
I created this simple plot to illustrate the issue. The left panel shows the whole period while the right panel shows that the problem is not that the data is under sampled. The plot goes from
2021-06-13 00:12:58.085383+00:00 to 2021-06-28 23:54:05.880661+00:00.

And one more example: Kuusamo is a city at 65°58′N 29°11′E and apparently it also has a few days when the sun doesn't set: LINK1 LINK2

From the PyEphem Quick Reference Guide:

Rising and setting are sensitive to atmospheric refraction at the
horizon, and therefore to the observer’s temp and pressure; set the
pressure to zero to turn off refraction.

It seems likely that, if you're using the default settings, the result returned is including atmospheric refraction, giving the results you would expect to see at the location, where the sun is visible even though a direct line to the disc of the sun would put it below the horizon.

Image is from "What is Refraction of Light" by By Konstantin Bikos and Aparna Kher at timeanddate.com


When the asteroid hit the earth 65 mya did the earth's gravity pull change? By how much? When the asteroid hit the earth 65 million years ago, did the earth's gravity pull change? By how much?

The gravity of Earth did not change substantially.

The strength of gravity on Earth is about 9.8, but varies between 9.76 and 9.83 ms-2(due to Earth not being a perfect sphere)

This can be approximated from the mass of the Earth using $$g=GM_e/r_e^2$$

Where $M_e = 5.97219\times 10^{24}kg$ and $r_e= 6371000m$.

The mass of the asteroid was about $10^{16}kg$ That changes the mass from $5.97219\times 10^{24}$ to $5.97219001\times 10^{24}$. It changes the gravity from 9.8 to 9.800000001

In other words. The change in gravity was utterly insignificant. You probably get a greater change in gravity by walking downstairs.


If someone was above the galaxy, how much of it can he see? If someone was far enough above the milky way to see the entire diameter, how much can he actually see with naked eye? I'm aware that on Earth, we only see stars within 1,000 light-years (roughly) by naked eyes. I found myself wondering if someone is more than 1,000 light years away from milky way then we won't be able to see it?

The Andromeda galaxy is not too different from ours, and yet you can see it with your naked eyes at 2.5 million light years distance. You do need to be under a very dark sky, far from cities, and it does look like a mere white smudge, but it is visible.

Now compare that with the fact that the Milky Way itself is quite visible to the naked eye if you're far from cities...

How do I see the milky way?

...and basically interpolate somewhere in between "little white smudge" and "river of stars filling up half the sky" to get an idea of how our galaxy would look like from its vicinity.

It would not be bright anyway, and you would have to adapt your eyes to complete darkness first. But you would definitely see its outline. Whether you would distinguish anything of the shape of its arms or not, is harder to say, and would probably depend a lot on the actual distance.


Can a black hole be torn apart by SMBHs? Suppose two SMBHs (supermassive black holes) are brought close (e.g. during a galactic merger).
Suppose that a "normal" black hole drifts into the space between the SMBHs.
Could the gravitational pull of the SMBHs be sufficient to tear the normal black hole to pieces?

If so, will we be able to examine the fragments, free from being obscured by the black hole's event horizon?

@peterh's answer contains the essential point, but I think lacks enough background. Let me try to go a bit deeper.

First of all, let's be clear that we're talking about the classical Black Holes (BHs) of General Relativity (GR), and not the true BHs of the as-yet unknown theory of quantum gravity (TQG). To answer your question, TQG does not have a role, because your question is about a regime which is far from where TQG effects would be relevant.

It's contrary to intuition, but the vicinity of a SMBH is a less exotic and less stressed environment than the vicinity of an ordinary BH. If you fell toward a solar-mass BH, the tidal forces would have torn you apart well before you got to the event horizon, while in a SMBH, it happens well after you pass the event horizon. Likewise the Hawking radiation (which can be taken as a measure of the stress in spacetime) is greater outside a solar mass BH than it is outside a SMBH. Wikipedia has a decent article on this which, while it contains at least one howler, on the whole seems to be reliable.

Bottom line: the tidal forces which tear things apart are weaker outside a SMBH than outside a smaller BH. But we've already observed several smaller BHs collide using LIGO, and they don't shred each other, but merge. So we can confidently say that a SMBH will not shred a smaller BH.

Another way to understand why a BH does not shred is to think about what's actually there. The answer is nothing. If you could survive the tidal forces (perhaps by falling into a SMBH) and could avoid the matter which generally forms an accretion disk around a BH (maybe by finding one in intergalactic space where there's no matter to pick up) and fall through the event horizon (EH), you'd notice nothing at all. It's just ordinary empty space outside the hole, at the EH, and inside the EH. Locally (i.e., it the cabin of your (now doomed) spaceship), there's nothing to see and nothing to detect, other than the ever-increasing tidal forces which will eventually tear you apart.  The BH is nothing but curved spacetime and there's nothing else there. What's to tear apart?

I'm using words. The mathematics of GR, which is very difficult but is also well-understood, describes exactly what happens when to BHs approach one another and Hawking's Area Theorem does prove that under GR, merger is the inevitable result.

What about the Theory of Quantum Gravity? Obviously we don't know much about it, since it hasn't been discovered yet, but one of the few things we're sure of is that it will give results exactly* the same as GR at "large" distances. And "large" is this case is probably anything bigger than an atom. Even if the limit is "bigger than a bread box", the two BHs are far too large for their collision to have any TQG effects. The TQG effects all happen down at the singularity the GR predicts at the center of the BH -- but whatever happens there is lost to the outside world.


Shouldn't this cause a fire? This website shows a telescope projecting the sun onto a blackboard:  https://astronomyconnect.com/forums/articles/2-three-ways-to-safely-observe-the-sun.21/
Why isn't the board catching fire?   You can easily start a fire on a sunny day by targeting the focal point of a magnifying glass onto something flammable.  Why isn't the telescope in this picture doing the same thing?

Photo by Luis Fernández García

It could start a fire if the screen is at the focal point of the optical system. That is how you light fires with a magnifying glass.
Here, the blackboard is likely away from the focal point, so you can see the shape of the eclipse (and you get a bigger image) without setting things on fire.
Although this is fairly safe, there are a few things to pay attention to:

If you do this, make sure nobody can walk between the telescope and the screen, because if they go near the focal point, they could get very hot.
Doing this will cause your telescope to heat up. If there are any plastic parts, they can melt.
The telescope in the picture seems to have a small opening. Don't do this with a big telescope. You don't need to collect a lot of light.

Not an answer to the question, but an important note: Observing the Sun is the most hazardous thing you can do in astronomy. Make sure you know what you are doing before you try.


Is there any other theory, apart from the Big Bang paradigm, which describes the birth of universe? The Big Bang theory is the approved theory of the "birth" of Universe. Are there any other theories which can explain how our Universe came into existence?

This is not an answer but a comment which is too long to post as such.

There is no such thing as a scientifically approved theory, there are theories consistent with the evidence some of which are simpler in some sense than others, and philosophy of science tells us that we should prefer those that are simpler.

Also I doubt that the big-bang theory is what you think it is, which is simply that at a certain time in the past the Universe was in a hot dense state which then expanded, forming the lightest chemical elements... The first part of this is a simple extrapolation of the observation of the way that galactic Doppler shift increases with distance and the observed cosmic microwave background. The second part is the consequence of the application of well know nuclear properties of matter.

Now there are plenty of more-or-less widely accepted additions to this, mainly to explain the uniformity and fluctuations in the microwave background, and the emergence of first stars and galaxies etc. But all of these are to some extent speculative since we have no generally accepted unification of Quantum Mechanics and General Relativity or require physics that has no predicted consequence (so far) beyond the phenomena it has been proposed in order to explain.

To make this approximate an answer I will point you to the Wikipedia page on Non-Standard Cosmologies


What Would it Look Like Inside Jupiter's Great Red Spot? From what I've read, Jupiter's 'Great Red Spot' is an unbelievably powerful storm. Do astronomers and scientists have any idea of what that might look like if it were possible to be inside/below it? Would it look like a storm on Earth (thunder, lightning, winds, etc.), or would it look completely different?

You would not actually be able to see that much, the sunlight is less than 4% of the strength at Earth's distance to the Sun. That is still enough to see though (Much brighter than Moon light at night in fact). The problem is the transparency of the atmosphere; we can not look in, you can not look out. You would only see a weak brown or red glow below the cloud cover.

However, there is plenty of lightning!
Strong winds and thunder exist too.

Lightning on Jupiter.


What are these three black cap-like pieces? My son received a secondhand telescope for his birthday. It is a Celestron Powerseeker 114AZ.
I found these three black cap-like pieces separate from other parts. I can’t find what they are or what they do in the instruction manual.
What are they and what do they do?

(click to enlarge)

I am currently using a Celestron Astromaster 130 EQ and I also have a similar type of caps.
Starting from left to right, the first two are caps used to cover eyepiece holder in case the lens is not connected with it, and the last one is a cap for the lens. So I think they might have given you one extra cap as a lens holder.


How many members of the IAU typically attend votes? One of the talking points of people how think that the definition of a planet should include Pluto is that during the 2006 vote on the definition of a planet, only a small percentage of the members of the IAU actually voted (183 for, 186 against. 13116 total members), thus making the results of the vote not representative of a consensus among the astronomical community.

Where the number of members who voted that motion uncharacteristically low, compared to most votes taken by the IAU? Does this low number of voters indicate that this decision is not representative of what astronomers working on this issue think?

One of the talking points of people how think that the definition of a planet should include Pluto is that during the 2006 vote on the definition of a planet, only a small percentage of the members of the IAU actually voted.

This argument is very disingenuous or rather naive, or both. One the most disingenuous of all is Dr. Alan Stern, the principal investigator of the New Horizons mission to Pluto who said

  "I'm embarassed for astronomy. Less than 5 percent of the world's astronomers voted." Stern called it "absurd" that only 424 astronomers were allowed to vote, out of some 10,000 professional astronomers around the globe. "It's patently clear that Earth's zone is not cleared," Stern told SPACE.com. "Jupiter has 50,000 trojan asteroids," which orbit in lockstep with the planet.

These are blatant lies.

Regarding the small number of voting members who bothered to vote, Stern was high enough up in the technical food chain to know exactly how professional societies work. Across all societies, voting on resolutions almost uniformly occurs on the last day of the society's general assembly. Most of the members of a technical society do not attend that society's general assembly, and of the few who do attend, most do not attend the final day of the proceedings. A few societies have started to transition to electronic voting, but this is very recent, much more recent than 2006.

Regarding "clearing the neighborhood," Stern either knew or should have known exactly what that phrase meant. If he did know, his statements were a blatant lie. If he didn't know, his statement makes him guilty of academic fraud, which is even worse than lying.

Stern was the lead author of a paper presented to the IAU six years before the vote occurred titled "Regarding the criteria for planethood and proposed planetary classification schemes." This paper developed a metric that shows a 5.5 order of magnitude gap between the eight planets and objects such as Pluto and Ceres. Either he knew what was in that paper / presentation (which makes his later statements a blatant lie) or he didn't, in which case he was guilty of a very serious form of academic fraud, that of forcing a subordinate to add the supervisor's name as an author. That fraud is even worse when the supervisor forces the subordinate to make the supervisor the lead author.

Here's a typical schedule for almost every professional organization's large meeting:

A meet & greet session on the evening before the meeting officially starts. This is a great time to have informal chats, to grab some free food, and to pocket the per diem. One  should carpe per diem whenever the opportunity arises.
An opening plenary session with lots of pomp and circumstance, and lots of blah-blah-blah. This a great time to either meet people in the hallways or to play hooky from the conference and tour the boondoggle of a locale where the conference is being held.
A number of small paper sessions, poster sessions, society committee meetings, and plenary sessions for invited speakers and invited panels. These meetings, along with the vendor exhibits, are why people typically go to these large conferences.
A closing session with lots more pomp and circumstance and lots more blah-blah-blah. This is almost universally where proposals / rule changes / resolutions are voted on. This is also almost universally a great time to take one last look at the vendor exhibits, to play hooky and tour the boondoggle city, or to be already flying home.

Most of the members of a professional organization do not attend the organization's general assembly; they are members for other reasons. Of the fraction of members who do attend, most do not attend the opening session, and even fewer attend the closing session. Voting has traditionally been very low. This tradition is changing as some societies now use electronic voting, but this is very recent.

Since pictures are worth a thousand words, here are a couple of pictures from the 2006 IAU General Assembly voting session, one from the rear and one from the front:

Below are a couple of pictures from the 2015 IAU General Assembly, the first from the opening session and the second from the voting during the closing session. Note that the closing session didn't even use the full auditorium.


death of a red dwarf star / minimum mass needed for a white dwarf? OK, first, I know there's a variety of sizes and types of red dwarf stars and the universe is too young for any of them to have reached the end of their main sequence phase yet, so it's all theoretical and/or modeling.
http://en.wikipedia.org/wiki/Red_dwarf
But what is the theoretical size needed for a star to undergo the electron degeneracy process which turns a small star from at least Jupiter sized, usually bigger, into to earth sized super dense object, where, as I understand it, the electrons are squeezed off the nuclei - electron degeneracy.
It would seem to me that a 7.5% solar mass star, which gradually burns hydrogen into helium but doesn't burn helium, might not have the mass to compact into a true white dwarf but might end it's life looking more like a brown dwarf / super Jupiter - well, talking appearance, not really, because super Jupiter & Brown dwarfs are mostly hydrogen while and an end of life red dwarf should be mostly helium - which, in and of itself, might make the difference.
It's just my curiosity whether all red dwarfs turn into white dwarfs at the end of their burning phase or is there a theoretical mass that's needed for that level of shrinkage to occur?
Thanks.

Stars that have a mass lower than about $0.5 M_{\odot}$ will not ignite helium in their cores, in an analogous fashion to the way that stars with $M<8M_{\odot}$ have insufficiently massive cores that never reach high enough temperatures to ignite carbon.

The cause in both cases is the onset of electron degeneracy pressure, which is independent of temperature and allows the core to cool at constant pressure and radius. [A normal gas would contract and become hotter as it loses energy!]

The end result for a $0.5M_{\odot}$ star will be a helium white dwarf with a mass (depending on uncertain details of the mass-loss process) of around $0.2M_{\odot}$. Such things do exist in nature now, but only because they have undergone some kind of mass transfer event in a binary system that has accelerated their evolution. The collapse to a degenerate state would be inevitable even for the lowest mass stars (which would of course then be very low-mass white dwarfs). As an inert core contracts it loses heat and cools - a higher density and lower temperate eventually lead to degenerate conditions that allow the core to cool without losing pressure.

The lowest mass stars ($<0.3 M_{\odot}$) do get there via a slightly different route - they are fully convective, so the "core" doesn't exist really, it is always mixed with the envelope. They do not develop into red giants and thus I guess will suffer much less mass loss.

The remnant would be a white dwarf in either case and is fundamentally different from a brown dwarf both in terms of size and structure, because it would be made of helium rather than (mostly) hydrogen. This should have an effect in two ways. For the same mass, the brown dwarf should end up bigger because the number of mass units per electron is smaller (1 vs 2) and also because the effects of a finite temperature are larger in material with fewer mass units per particle - i.e. its outer, non-degenerate layer would be more "puffed up". NB: The brown dwarfs we see today are Jupiter-sized, but are still cooling. They will get a bit smaller and more degenerate.

A simple size calculation could use the approximation of an ideal, cold, degenerate gas. A bit of simple physics using the virial theorem gives you
$$ \left(\frac{R}{R_{\odot}}\right) \simeq 0.013\left(\frac{\mu_e}{2}\right)^{-5/3} \left(\frac{M}{M_{\odot}}\right)^{-1/3},$$
where $\mu_e$ is the number of atomic mass units per electron.
Putting in appropriate numbers I get $0.32\ R_{Jup}$ for a $0.07M_{\odot}$ Helium white dwarf versus $1.01\ R_{Jup}$ for a $0.07M_{\odot}$ completely degenerate Hydrogen brown dwarf (in practice it would be a bit smaller because it isn't all hydrogen).

However, it would be interesting to see some realistic calculations of what happens to a $0.07M_{\odot}$ brown dwarf versus a $0.08M_{\odot}$ star in a trillion years or so. I will update the answer if I come across such a study.

EDIT: I knew I'd seen something on this. Check out Laughlin et al. (1997), which studies the long-term evolution of very low-mass stars. Low-mass stars do not pass through a red giant phase, remain fully convective and can thus convert almost all their hydrogen into helium over the course of $10^{13}$ years and end up cooling as degenerate He white dwarfs.


How much does the stellar radius change if seen through different filters? I know that the radius of a star should be different depending on the filter being used to do the observation. That is, the bluer the filter, the smaller the measured stellar radius. I know this effect must be pretty small, however, I would like to get more familiar with it. Can anyone point me to some literature or share their own experiences? Would it be a significant difference in Solar radius if we observe the Sun in x-ray vs radio bands?

The "radius" of the Sun is basically where the optical depth to radiation at any particular wavelength is about 1 (or some times 2/3 is used). This is known as the photosphere and is from where the light that we measure comes from.

The "thickness" of the photosphere - i.e. the depth over which the optical depth changes from being negligible to very large is perhaps a few 100 km at optical wavelengths, compared with $7\times 10^{5}$ km for the radius of the Sun. So the exact definition is important at the level of about 1 part in a 1000.

The light coming from the Sun in a "band" is a mixture of light coming from a variety of depths, depending on the exact wavelength. So your question is a bit ill-defined. If we are looking a the wavelength of a prominent absorption line, then the photosphere is higher up in the atmosphere than if we are looking at a  continuum wavelength. The difference is as I say, at most a couple of hundred km.

It's a bit more difficult to define a solar radius in terms of X-ray or radio emission. The Sun is surrounded by a corona that emits radiation at both of these wavelengths, though that plasma is "optically thin" to the radiation. If we were to shine a beam of X-rays or radio waves at the Sun, they would be totally absorbed perhaps about 1000km above where the visible photosphere is defined in something called the chromosphere.

However, the Sun's outer atmosphere (chromosphere and corona) does not have a plane-parallel geometry; it is shaped to some extent by the magnetic field and you can find spicules and prominences that are optically thick that stick up to greater heights in the atmosphere.


Are there other proofs of the expanding universe apart from the redshift? The theory of the expanding universe is so widely accepted, that the redshift is sometimes used as a measure of distance to far away galaxies.

But is it still possible that the redshift is caused by some unknown phenomena and not by galaxies moving away from each other?

Is there any other proof (apart from the redshift) that the universe is indeed expanding and far away galaxies are moving away from us?

Yes, there is direct, non-red-shift evidence of expansion.

The past temperature of the Cosmic Microwave Background Radiation (CMBR) has been directly measured and found to be substantially higher than it is today.  Its reduction in temperature over time is direct evidence of expansion.  Here are the details:

According to this paper, the CMBR was measurably hotter in the past (less-technical synopsis here).  The researchers observed absorption lines in a gas cloud located in a distant galaxy and found that the pattern of lines seen could only be explained if the CMBR temperature at time of absorption was between 6 K  and 14 K (it's 3 K now).  This temperature is consistent with the expected temperature for that galaxy's redshift (9 K).  Note that the temperature was measured from the specific pattern of lines seen and not from how much the lines had been red-shifted; this measurement would yield the same temperature even if there were no red-shifting.  Since a hotter temperature implies a higher density, this cooling of the CMBR over time is direct evidence for expansion of the universe.

Additional comments

What is the relationship between red-shift and absorption lines?

Inspired by a conversation with uhoh in the comments:

In my answer I refer to a "pattern" of "absorption lines".  For those not versed in the topic allow me to explain.

When a light shines through a cloud of gas specific frequencies of light get absorbed.  When this light is then shone through a prism the blocked frequencies will appear as black lines in the spectrum (see illustration below).  The exact lines that appear and their positions in the spectrum (the "pattern" of "absorption lines") depends on the elements present in the gas and the gas's environment.  The effect is most clearly seen with a light that emits photons at all frequencies; this kind of light is known as black-body radiation.  Although emitting light at all frequencies, a black-body radiator will emit the most light at a particular wavelength; the location of this peak is referred to as the black-body's temperature.  The CMBR discussed in the question is an example of black-body radiation.

Source: Doppler Shift, Edward L. Wright
(Excellent site BTW, the FAQ is worth a look for more info on red shifts and cosmology in general)

As light is travels through (expanding) space it's wavelength and the wavelengths of the absorption lines stretches at a fixed rate for all frequencies.  Let's say that at time of emission/absorption a spectrum shows lines at wavelengths of 1, 3, and 5 nm1.  After the photons have traveled for a certain amount of time, all of the spectrum's wavelengths will appear to have doubled2.  The line formerly at 1 nm is now seen at 2 nm, the one formerly at 3 nm is now seen at 6 nm, and the one originally at 5 nm is now seen at 10 nm.  Though their absolute frequencies change over time, the ratio of the lines' wavelengths (and frequencies) relative to one-another remains constant.

The precise amount that a given object's spectrum is shifted directly correlates with its distance.  As seen in the diagram above, close objects (like the Sun) show no red-shifting.  As one looks at objects further and further away one sees increasing amounts of red shift3.

In the discussion in the answer above, it's this pattern of relative positions in the lines that is affected by the CMBR temperature at time of absorption and not the degree to which the lines have been shifted.

1 To put it technically, this point is at $z=0$ where $z$ indicates the magnitude of the shift, positive for red shifts (moving away) and negative for blue shifts (approaching).  A more in-depth discussion of this topic (including the precise definition of $z$) can be found here.
2 The wavelength-doubling (frequency-halving) point is at $z=1$
3 It should be noted that since there is some uncertainty in the rate that the Universe is expanding at, red-shifts do not refer to precisely-known distances.  Thus astronomers and cosmologists rarely refer to the distances to distant objects in absolute terms of, say, light-years or parsecs, preferring, rather, to use the amount of red-shift observed (the $z$ mentioned above).

The mechanism behind red shifting isn't that the photons themselves are changing rather it is that the very space that the electromagnetic waves are moving through is expanding.  (Photons are both particles and waves; no, it's not exactly intuitive.)  This constant stretching of space stretches the light's wavelength giving rise both to the effect of red-shifting and the increase in a given photon's red shift over time.

Douglas Hofstadter, CC A-SA 3.0

How does red-shift relate to the CMBR?

In the comments Alchimista asked "Isn't CMBR actually the quintessence of red-shift anyway?"
(I'm assuming you're using the common, and not cosmological, meaning of "quintessence")

Yes, the current CMBR temperature (3 K) is generally agreed to be the result of relatively high-energy photons (3000 K) emitted about 380,000 years after the Big Bang that have had their wavelengths stretched over time by the expansion of the Universe towards the red (i.e. cooler or lower-energy) end of the spectrum.  This expansion was inferred by Hubble et al. from the observation that smaller and dimmer galaxies (as seen from Earth) have a greater shift in their spectra.  The farther the apparent distance, the greater the observed shift.  Using this apparent distance-correlated red shift we can infer that the Universe was smaller in the past and thus denser with a higher temperature for the CMBR.  Based on observed red shifts of distant galaxies, we can then deduce, but not directly measure, what the CMBR temperature was at each distance.

What the authors of the above paper did was make a direct measurement of the CMBR's temperature at a specific time in the past.  The measured temperature is higher than it is today which implies a denser and thus smaller Universe.  The researchers further found that the directly-measured temperature fit neatly with that inferred from the observed red shift of the galaxy being studied.

In a nutshell, the chain of inference is swapped:

For reasoning based on red-shifting:
Increasing red shifts with apparent distance (directly measured) ⇒ Expansion ⇒ Denser Universe in the past ⇒ Higher CMBR temperature in the past.
For a direct measurement of past temperature (as with this paper):
Higher CMBR temperature in the past (directly measured) ⇒ Denser Universe in the past ⇒ Expansion ⇒ Observed red shift.

These two inference chains based on different sets of evidence neatly complement and support one-another.

One thing to note is that the CMBR wasn't created by expansion (at least not directly) rather it's expansion that explains its current temperature and uniformity.  Per the Big Bang theory, the early universe was very dense; so dense and hot that all matter was a plasma of subatomic particles, opaque to photons. At about 380,000 years after the Big Bang the Universe had cooled (through expansion) enough that protons and electrons could combine to form neutral Hydrogen gas (which is transparent). The CMBR is the light that was set free at this time and has been cooling ever since.


How is it possible that the CMB approaches the earth from all directions? I presume that the photons from the CMB approach the earth from all directions, otherwise we couldn't detect them with a picture where it is present everywhere in the universe with a tiny anisotropy.

Now the CMB is a result of the big bang. Another result of the big bang is that in all directions we see galaxies moving away from us, like we are in the center of the universe (which is probably not the case).

But why are all the galaxies (except a few) moving away from us while the CMB is approching us. What is the reason for that difference? Or aren't those photons coming from everywhere?

You are comparing apples and oranges. The CMB is like the light from a distant galaxy, which while the galaxy is receding the light approaches us. CMB is the light emitted by universe at recombination. The matter that emitted the CMB is/was receding from us with a very high red-shift.


Will we be able to see our galaxy if we travel far enough? Will we be able to see our galaxy if we travel far enough? When we leave earth far enough, we'll be able to see Earth from distant, is it same with the galaxy? if we travel far enough, will we see the shape of our galaxy with naked eye?

Do we have to travel really far to see our galaxy like a dot? or just far enough to realize the size and shape of our galaxy?

In theory, yes; in practice probably not ...

i) It's perhaps helpful to take the Andromeda galaxy as a first working example. It is somewhat bigger than our own Milky Way galaxy but not so big as to be a problem.

When we look at Andromeda, it appears to occupies an area about 3 degrees by 1 degree. By comparison, the Moon appear about 0.5 degree across. Andromeda is a good target to see, a little faint but it is just about visible with the naked eye on a good night.

Andromeda is about 2.5 million light years away so, even travelling at the speed of light, it would take 2.5 million years to get far away enough from the Milky Way so we could look back and see our galaxy like we see Andromeda today.

( In Star Trek, the writers seemed broadly to limit themselves to adventures within our galaxy for the same reason: within the realms of sci-fi, perhaps they could justify getting around the Milky Way within a reasonable period of time - not so to travel between the galaxies )

ii) To reduce the apparent size of our galaxy to a dot would require travelling at least 10x further or probably more - depending on how you are observing. 50 million light years, anyone?


Why are gravitational perturbations stronger at larger semi-major axes? Why are mechanisms such as the Kozai-Lidov effect more prominent at at large semi-major axes?

If we had a binary system of a primary and secondary body, with the third perturber as the Sun. Why is it that the secondary feels more perturbations from the gravitational influence of the sun the further away it is from the primary?

"Why is it that the secondary feels more perturbations from the gravitational influence of the sun the further away it is from the primary?"

In brief, it is because the net perturbing acceleration on the secondary is just the (vector) difference between (a) the accelerative attraction towards the perturbing body experienced by the secondary, and
(b) the accelerative attraction towards the perturbing body experienced by the primary.

Thus, the closer the secondary is to the primary, the more nearly equal in size and direction are those two attractions towards the perturber, and the closer to zero is their vector difference.  Another result is that the the more similar are the changes in velocity in size and direction produced by the perturbing accelerations in the primary and the secondary, and the closer to zero is the resulting perturbation in their motions relative to each other.

This has been known for a long time as a consequence of Newton's 6th corollary to the laws of motion: "If bodies, anyhow moved among themselves, are urged in the direction of parallel lines by equal accelerative forces, they will all continue to move among themselves after the same manner as if they had been urged by no such forces."

Because there can be immense variety in the possible trajectories of the primary and secondary apart from the perturbations, any detailed illustrations can speedily develop massive and intricate trigonometrical expressions.

But in all cases, including that of the Kozai-Lidov effect, the scale of the effect, however intricate its form, depends on the size of the net perturbing accelerations.

A highly simplified configuration can at least show by example how a net perturbing force, affecting the relative motion of primary and secondary, is almost directly proportional to the first power of the distance between primary and secondary -- though it also depends of course on further factors due to changes in angular configuration.

The diagram below indicates some highly simplified configurations.

Suppose first that the primary (E) and secondary (M) are for one instant in line with the perturbing body (S), with M at M1 between E and S.  Let s stand for distance ES, and d for distance EM (with d << s).  Suppose also that the mass of E and M are negligibly small relative to the mass of S (though not negligible relative to each other).

With these approximations, the accelerative attractions of S on M and S on E are respectively $ k/(s-d)^2 $ and $ k/s^2 ,$ and the net perturbing acceleration on M is the difference $ ( k/(s-d)^2 - k/s^2 ) .$

Putting $ s(1-d/s) $ for (s-d), and using the binomial expansion of $ 1/(1-d/s)^2 $ , one sees that the terms in $ k/s^2 $ cancel, leaving the net perturbing force as $ k/s^2 * (2d/s) $ , plus terms in higher powers of d/s, i.e in $ k d^2/s^4 $ and so on.

Where d is very much smaller than s, the higher-power terms in d/s can be neglected, and then the net perturbing acceleration on M at M1 in the chosen example-configuration is closely approximated by  $ +2 k d / s^3 $ , away from E and towards S.

If instead the configuration has M at M2 so that E is in line between M and S, then the net perturbing acceleration on M clearly becomes $ ~ -2 k d / s^3 $ , i.e. away from E and away from S.

If instead M is at M3, with the line EM3 at right-angles to ES, and if also the angle ESM3 can be treated as small enough that its cosine can be approximated as 1 and its sine as d/s , then it is easily found that the net perturbing acceleration on M at M3 is approximately $ k d/s^3 $ towards E, again neglecting higher powers of d/s.

If M is at an intermediate position M4, and D represents the angle ESM4, it can be seen by using a little more trigonometry that the net perturbing acceleration on M at M4, under the assumptions already made, has a component parallel to line ES of approximately $ +2 k d cos D / s^3 $, and a component perpendicular to line ES (acting always towards line ES) of approximately $ k d sin D / s^3 $.

All the components are proportional to the E-M separation d, to the extent that the omitted series of terms in higher powers of d/s can be treated as negligible as has been done here.


How to convert from Cartesian J2000 coordinates to Cartesian Galactic coordinates? I'm using a star database (the HYG database) which has xyz positions and velocities for the stars. This has +X towards the vernal equinox, +Z towards north celestial pole, +Y towards Right Ascension 6 hours and it's based on J2000 and centered on the sun.

I'd like to convert these to a Cartesian Galactic coordinate system centered on the sun, with +X pointing to its center and +Z angled so that it shoots out of the galactic plane on the north side.

This should just be a simple rotation, but after a lot of searching I can't seem to find the rotation to use, nor some way of converting it through a third system.
Does anyone know how to obtain the correct rotation matrix here? (or some other method of rotation)

To transform an equatorial $(x, y, z)$
into a galactic $(x_G, y_G, z_G)$, the
Gaia data release 1 documentation, section 3.1.7, uses

$$
\begin{bmatrix} x_G \\ y_G \\ z_G\end{bmatrix}
= \mathbf{A}'_G
\begin{bmatrix} x \\ y \\ z \end{bmatrix}
$$

where

$$
\mathbf{A}'_G = \begin{bmatrix}
{−0.054876} ~ {−0.873437} ~ {−0.483835} \\
{+0.494109} ~ {−0.444830} ~ {+0.746982} \\
{−0.867666} ~ {−0.198076} ~ {+0.455984}
\end{bmatrix}
$$
is composed of three coordinate frame rotations
$$
\mathbf{A}'_G = \mathbf{R}_Z(-l_\Omega)
~ \mathbf{R}_X(90^\circ - \delta_G)
~ \mathbf{R}_Z(\alpha_G + 90^\circ)
$$
with the north galactic pole at equatorial coordinates
$(\alpha_G = 192.859^\circ, \delta_G = +27.128^\circ)$
and the galactic equator crossing the celestial equator at galactic longitude
$l_\Omega = 32.932^\circ$.

This is equivalent to the transformation defined in the
Hipparcos catalogue, volume 1, section 1.5.3.

Murray 1989 and
Liu 2011
give similar matrices whose elements differ after 6 decimal places, which matters only if you require sub-arcsecond precision.


How much does a linear lightyear of space expand in a year? Now that we know that the universe is expanding, how much does a linear lightyear of space expand currently in a year?

A recent estimation of the Hubble constant is 71.9 (km/s)/Mpc.

This means there is an expansion of

$$\left(71.9\ \mathrm{\frac{km}{s\ Mpc}}\right)\left(\mathrm{\frac{60\ s}{min}}\right)\left(\mathrm{\frac{60\ min}{hr}}\right)\left(\mathrm{\frac{24\ hr}{day}}\right)\left(\mathrm{\frac{365\ day}{yr}}\right) = 2,267,438,400\ \mathrm{\frac{km}{yr\ Mpc}}$$

To apply this expansion over a light year:

$$
\frac{2,267,438,400\ \mathrm{(km/yr)/Mpc}}{3,261,564\ \mathrm{lyr/Mpc}} = \boxed{695.2\ \mathrm{\frac{km/yr}{lyr}}}
$$

which is about $7.3483\times10^{-11}\ \%$ per year or $432\ \mathrm{miles/year}$.

A linear light year of space expands by 695.2 km/yr with a given Hubble constant of 71.9 (km/s)/Mpc.

Edit:
A better measurement released on 2018-02-22 using parallax found the Hubble constant to be 73.45 ± 1.66 (km/s)/Mpc.


What is $\phi '$ in orbital mechanics? For the last week or so, I have been teaching myself orbital mechanics within the context of Braeunig's Rocket and Space Technology.

I noticed a symbol, $\phi '$, and was wondering what context that was used in? I think it is an angle measurement, but I know that it is almost equal to the semi-major axis on earth.

From the box labeled Geodetic Latitude, Geocentric Latitude, and Declination, found about half way down after selecting Orbital Mechanics here.

This appears to be the geocentric latitude ($\psi$), which is the angle between the equatorial plane and the point on the surface of the ellipse. It can be calculated from the geodetic latitude (also known as the geographic latitude) ($\phi$) and the eccentricity of the ellipse ($e$):
$$\psi(\phi)=\tan^{-1}\left(\left(1-e^2\right)\tan\phi\right)$$
If $e=0$, then the two lattitudes are the same, because the focus is at the center of the ellipse.

Image courtesy of Wikipedia user Wmc824 under the Creative Commons Attribution 3.0 Unported license.

It might make more sense to use this latitude to describe the position of an orbiting object, but the vector from the focus to the object's position $(r,\theta)$ is perpendicular to the tangential component of the object's velocity vector, meaning that the geodetic latitude may be better.


How often do comets survive passage by the Sun? I had heard that comet ISON might not survive a close pass by our Sun, and I was curious about the odds of how other comets had fared. So, how often do other comets survive a close pass by our sun? Also, are there any connections between a comet's size or buildup(of material) that makes it more or less likely to not survive?

Key factors:

How close is perihelion? Too close and it may be destroyed on its first pass. We know Halley's Comet, which has a perihelion of about 0.6AU, has been orbiting for over 2000 years, passing the sun every 74-76 years and is still going strong.
How big is it? Every pass loses material, so a bigger comet could last longer.
What is its composition? Darker comets can absorb more energy when close to the sun, which may accelerate outgassing and breakup.

And of course some comets hit large objects - Shoemaker-Levy hit Jupiter and was destroyed.


How are "parallel fields" implemented on the Hubble Space Telescope? The Phys.org news item Hubble is the ultimate multitasker: Discovering asteroids while it's doing other observations mentions the Hubble Space Telescope's use of parallel fields:

A couple years ago, the NASA/ESA Hubble Space Telescope was observing very distant galaxies, some of them billions of light years away. It was part of the Frontier Fields Project (FFP,) an ambitious observing program which aimed to observe six massive galaxy clusters. The FFP pushed Hubble to its limits.
[...]
This image is a couple years old now, and it shows what's called the parallel field for the Abell 370 galaxy cluster, about 4 billion light years away. It's called a parallel field because of the technique used to capture it.
A press release explains parallel fields this way:
"While observing each cluster with one of the cameras on Hubble, the team also used a different camera, pointing in a slightly different direction, to photograph six so-called 'parallel fields." This maximised Hubble's observational efficiency in doing deep space exposures, imaging a myriad of far away galaxies."

Question: How are "parallel fields" implemented on the Hubble Space Telescope? How are two cameras able to take different exposures in different directions at the same time? What optical components points them in different directions?

As the question Instrument aperture sizes on Hubble Telescope shows, the focal plane area is large enough to focus on several instruments at the same time (but with each capturing a different area).
If two objects of interest are separated by a certain angle (the instruments are fixed within the focal plane), the telescope can be rotated so that two instruments are exposed, one to each object.  If desired, the telescope can later be rotated to swap the instruments on each object, so that both objects get a wider spectral coverage, for example.

No optical components point the instruments (cameras/spectrometers) at different objects; rather, the whole telescope orients on one direction in space, and by rotating around the main mirror axis to a particular angle (matching the positions of the fixed instruments) can choose to expose multiple instruments to multiple objects in that Field-Of-View.  All instruments (except the WFC UV/IR sensors) are always exposed to something, but data collection/transmission stresses the systems, so 'parallel' viewing is used selectively.

Only the Wide Field Camera uses a pick-off mirror (steerable for UV or IR sensors). The other instruments are aligned with their patches of the focal plane and used COSTAR lenses for correction until Service Mission 4 removed COSTAR and replaced the actual instruments (which now have built in corrections). The Wide Field Camera was replaced several times and had its own corrective mirrors instead of a lens via COSTAR.


Is there a database for identifying the falling light I saw? I witnessed a light fall from the sky. It was visibly about as bright as Mars, so brighter than meteorites I've seen, was just a point of light with no tail, and traveled much much faster than an aircraft. The horizon was blocked by a house, so I do not have any frame of reference for how far away it hit the ground (it could as well have been 3 miles or 10,000 miles away), but I do know the coordinates for where I was, and the direction is was relative to my position.

Is there some database I can use to search space events to learn what it was that I saw?

This could be a satellite, such as the ISS, which appears as a bright, fast moving star. You can look for times when the ISS is visible on websites like http://www.heavens-above.com/

It might be a meteor (not a meteorite, that is the stone, not the streak in the sky). Meteors travel much faster than satellites. On any given night you will see several meteors each hour. A meteor that is "as bright as Mars" (which varies in brightness, and can reach -3 but is more likely to be between -1 and 0) is a less common event, but not exceptionally rare. Reports of Fireballs (those with magnitude -4 or brighter) are collected by various bodies, such as the uk meteor network


Why do aliens have to be carbon based lifeforms? So correct me if I'm wrong, but when searching for extra terrestrial life, scientists usually search for signs they would expect for life on earth to thrive in. We look for conditions on planets that meet certain criteria e.g close enough to a sun for water, environments for carbon based life forms to exist. I was wondering; why cant an alien be a sulphur based life form that for the sake of argument drinks lead and breathes in chlorine like we do oxygen.

I want to start with a disclaimer that I'm not a chemist and hopefully I don't screw any of the chemistry up. If I do, please let me know.

I think Bill Oertell hit the nail on the head. Just because you can imagine a species which can drink lead doesn't mean it's actually physically possible. Only specific atoms can form the basis of the (extremely) complex organic molecules needed to form life and Carbon happens to be the atom which can do that. The primary reason for this is that Carbon has four valence electrons and its outer shell can hold eight, allowing it to form strong and varied bonds that most other atoms just can't form. As such, Carbon is the most logical element to form the backbone of organic molecules.

Other ___-based Lifeforms

Moving down the periodic table, you can see that Silicon has the same properties (4 valence elections, 8 possible) and you might be inclined to think that you could form some sort of silicon based life. This topic has been widely considered, but you run into a number of problems that make Silicon-based life unfeasible. The primary reason is that Silicon is just much bigger and more massive than Carbon (having a nucleus with 28 nucleons compared to Carbon's 12). This makes it harder for Silicon to form the necessary bonds that Carbon can readily form and you're out of luck right form the get-go. Similar issues occur with other "Carbon group" elements, except they're even further exacerbated by even higher masses.

There are many other "organic" configurations chemists have considered such as boron-based life or sulfur-based life but they all have some sort of problem that makes it hard for it to exist. Some are more possible than others, but Carbon-based life, as we know it, just seems to be the best solution. If you gave someone all the elements in the periodic table and said, make organic life, they'd probably come back with Carbon-based life because its the easiest. You don't have to overcome as many obstacles.

Observations

Aside from the impracticality of non-Carbon based life, you have to consider that everything we've seen in the universe, as far as organic chemistry goes, points to the fact that the Universe is filled with Carbon-based chemistry. Astronomers have been able to look out at comets, interstellar dust clouds, and all sorts of places and time after time, they see complex, Carbon-based, organic molecules. For example:

We've found Polycyclic Aromatic Hydrocarbons (PAHs) all over the solar system and galaxy,
Astronomers have also found things like i-propyl cyanide in meteorites and in interstellar dust $-$ an organic molecule which is a necessary ingredient in amino acids,
Organic molecules have been found (and believed to have formed) on Ceres.

The point I'm driving at here is that the universe seems ready-made to start forming Carbon-based life everywhere. Everywhere we look, we see organic molecules which have Carbon as the backbone of their structure. Nowhere do we see complex silicon based molecules (or something even crazier). This is highly suggestive that Carbon is the way to go if you want to quickly and easily form life.

There are probably a 100 other reasons to favor Carbon over something else, but that delves into more complicated chemistry than I've touched on here and I'm not as knowledgeable about that. I would think this question would get a more in-depth chemistry answer on the chemistry stack exchange.


How was precision astrometry done before digital imaging? Did they just put a ruler on the photographic plate? Quite some precision must have been required to discover Neptune from its influence on Uranus for example.

Prior to digital imaging then photographic plate negatives were analysed with scanning microdensitometers to produce astrometric catalogues. Many of these catalogues are still in use today, they are valuable sources of early epoch positions that enable proper motion measurements.

For more details you could look at the descriptions of the SuperCosmos project http://ssa.roe.ac.uk// (which is based on Schmidt plates) or the UCAC4 catalogue, http://ad.usno.navy.mil/ucac/readme_u4v5, which uses plates to get positions and proper motions for faint stars.

In the good old days, before even these catalogues existed, astronomy groups would have copies of the whole Schmidt and Palomar sky surveys. You would put the relevant plate on a massive, concrete-based, hydraulic X,Y measuring machine, with a binocular microscope. You would measure sets of Standard stars, get a 6-coefficient fit to convert x and y into RA and Dec,then find your objects on the plate, measure x,y, calculate RA and Dec. Then snap a polaroid to use as a finder chart at the telescope.

I was doing this as late as 1995 before the advent of the Digitised Sky Survey.


How can the universe be expanding? I assume expansion implies an increase in the volume of the universe. But if the volume of the universe is increasing, does that not imply the area it is expanding into's volume is being reduced?

In 1929, Hubble published his discovery that the sizes of galaxy redshifts and proportional to galaxy distances. That is, nearby galaxies are less redshifted than further away galaxies. This implies that further away galaxies are travelling away from us faster than nearby galaxies, which implies that we live in an expanding universe. We do not expand into anything, but rather, space is stretching. The Earth is not expanding, nor are you, me, or any other matter in the universe. It is space itself that is stretching.


Solar Noon: meridian crossing time versus time of maximum elevation According to the first paragraph of the Noon Wikipedia page, solar noon is the moment when the Sun crosses the meridian and is at its highest elevation in the sky. The wording there suggests that there is an exact identity between the solar meridian crossing time and the solar maximum elevation time on a given day. However, later in same article in the Solar Noon section it states that,

  Solar noon is the moment when the Sun transits the celestial meridian – roughly the time when it is highest above the horizon on that day.

(Emphasis mine)

The use of the word "roughly" really surprised me. I was strongly under the impression that these times were necessarily equal for sub-polar regions. Am I wrong and missing something, or is the wiki article just poorly worded?

There are a couple of things you have to think about in order to understand the problem at hand. The first is that there are three coordinate systems at play. The first is the celestial coordinate system, which is a coordinate system based upon the latitude and longitude of the Earth (the celestial equator is the Earth's equator projected out into space). See this post.

The second coordinate system is that of the solar system: The plane of the solar system (which contains the sun and planets, and traces out a line in the sky called the ecliptic). Now because of the tilt of the Earth's axis with respect to the normal vector of the plane of the solar system, the celestial equator is not the same as the ecliptic. What's more, is that the relative position of the ecliptic and the celestial equator change throughout the course of the year.

Lastly, there is the altitude/azimuth coordinate system which locked to the observer. The zenith (the point at which all horizons are equidistant from the point you are looking at) is $+90^{\circ}$ in altitude and is a degenerate point in azimuth, and all horizons are $0^{\circ}$ altitude but lay on a unique azimuthal arc stemming from the zenith and intersecting the horizon.

Below is a diagram showing the relationship between the first two coordinate systems, the celestial coordinate system and the coordinate system of the solar system.

Why does all of this matter for the question you've asked? It's because as you look due East and due West, you're looking at $0^{\circ}$ declination (the celestial equator intersects the horizon at exactly due East and due West) whereas the ecliptic intersects the horizon at different points. These points change throughout the year as the relative position of the ecliptic changes with respect to the celestial equator. Now, the amount the sun travels on the ecliptic over a 24 hour period is $0.9856^{\circ}$ ($360^{\circ}/365.25 \text{ days}$; actually I may want to use a sidereal day here but nonetheless it's about a degree).

Due to this, there is potentially a little bit extra in altitude that the sun may travel throughout the course of a day (since the path of the ecliptic is neither aligned with the celestial coordinate system, nor aligned with the local alt/az coordinate system of the observer; unless you're at the north or south poles in which case your local alt/az coordinate system is the same as the celestial coordinate system, however both are still mis-aligned with the coordinate system of the solar system). I think the word "approximate" needs to be there.

All told, I don't personally know how to quantify the extra bit of altitude the sun may gain from the misalignment of the ecliptic with lines of constant altitude and from the sun's motion along the ecliptic, but I would welcome references and calculations from other people. This little bit extra may not be a really small number. In other words, you might want to see how the rate of change of altitude of the sun at that point changes simply due to the Earth's rotation and how it may compare to the rate of change of altitude due to the motion of the sun along the ecliptic.

EDIT:
I also just looked to see what the ecliptic looks like on the sky (as seen from Philadelphia), and how it compares to the local alt/az coordinate system. Below is a picture with all coordinate systems present to try to illustrate the point I was trying to make.

The green line is the meridian, the orange(ish) lines are alt/az coordiantes, the blue line is the celestial equator, and the red line is the path of the ecliptic. Hopefully you can make this out sufficiently well.


Is Diffraction of light observed in gravitational lensing/microlensing? As far as I understand the wave-optics effects,such e.g.diffraction at a caustic, are hardly observable in the gravitational lensing. Are there any observations where the diffraction paterns of the visible light have been seen? If not visible, which wavelength?

I'm not sure how to define the "edges" or obscurations in a case of gravitational lensing.  The image we produce on Earth is the combination of our local telescope lens elements and the "grav lens" a zillion Kessel Runs away from us.  Even if you use the premise that the black hole itself is a central obscuration which should cause a Spot of Arago, I don't think any other diffraction-like effect is observable from such a great distance.  All the wavefronts of  light which passes thru the GravLens region continue to expand, ending up near-parallel from Earth's viewpoint.


Do meteorites streak across the sky in the same direction? Once while camping with my family as a child we all watched a meteor shower in the summer (in the month of August).

All the meteorites streaked across the sky in the same direction.

Obviously, this was the result of the Earth passing thru a group of meteors but I've always wondered.

Do meteorites always streak in the same direction? While I know it's not likely to be exactly the same direction, but generally speaking. Would one say meteorites always travel for example: east to west?

I thought this was a plausible fact since the Earth is travelling around the Sun in a constant direction, and additionally it's always rotating in the same direction.

While meteorites can travel in all directions. Does the Earth's gravity pull them straight down towards the surface resulting in the streak to be always in the same direction.

No, meteorites of the same shower look as coming from the same point in the sky. They are usually named according to the point they seen to come from, e.g. the leonids, or the perseids. The point, they seem to originate is a result of the relative velocity of the meteorites to the earth.

Meteor showers are thought to be remnants of comets, hence the similar orbits of the meteorites within a shower.

Here a more complete list of meteor showers.


How to understand exactly why gravity darkening happens on rotating stars? Phys.org's TESS delivers new insights into an ultrahot world links to KELT-9 b's Asymmetric TESS Transit Caused by Rapid Stellar Rotation and Spin–Orbit Misalignment (readable in arXiv)
The assymetric dip in the light curve comes from a near-polar transit across a rotating, oblate star where the poles are hotter and therefore brighter due to gravity darkening:

KELT-9’s high internal angular momentum ($\nu \sin(i)$ = 111.4 ± 1.3 km/s) flattens it into an oblate spheroid, making the equatorial radius of the star larger than the polar radius. Additionally, the star’s abundant centrifugal force near its equator distorts its hydrostatic equilibrium, causing its effective temperature to vary by nearly a thousand Kelvin over the surface of the star. These two effects of stellar oblateness and varying effective temperature – together commonly referred to as gravity darkening (Barnes 2009) – change the total irradiance on KELT-9 b (Ahlers 2016).

Those links discuss gravity darkening but don't offer a simple explanation.
Wikipedia's Gravity darkening says:

When a star is oblate, it has a larger radius at its equator than it does at its poles. As a result, the poles have a higher surface gravity, and thus higher temperature and brightness.

Question: Why exactly does the increased surface gravity at some places on a given star lead to a higher temperature at those locations? Is it related to the difference in scale heights? The surface brightness relates to the temperature at the photosphere, is the reason simply that a higher pressure therefore higher temperature is needed to support the same density in a higher gravitational field?

Figure 2. (Left) KELT-9 b begins its transit near the star’s hot pole and moves toward the star’s cooler equator. Our transit analysis directly measures the stellar inclination (i★), the planet’s projected alignment (λ), and the orbital inclination (i.e., the impact parameter b). We find that KELT-9 varies in effective temperature by ∼ 800 K between its hot poles and cooler equator. (Right) KELT-9 b’s phase-folded primary transit from TESS. The transit depth steadily decreases throughout the eclipse, indicating that KELT-9 b begins its transit near one of the host star’s hotter poles and moves toward the dimmer stellar equator.

The argument goes something like this.
Hydrostatic equilibrium means that the local pressure gradient is proportional to the local density multiplied by a latitude-dependent local gravity. If the pressure just depends on density and temperature, this means that those quantities will also just depend on latitude and therefore will be constant along an equipotential surface. i.e. pressure, temperature and density are functions of the effective gravitational potential $\phi$.
For stars with radiative outer envelopes, the heat flux is proportional to the temperature gradient, multiplied by some stuff (like inverse opacity) that just depends on density and temperature.
But
$$\nabla T(\phi) = \frac{dT}{d\phi}\nabla \phi = f(\phi)g_{\rm eff}$$
If we now say that at the surface $\sigma T_{\rm eff}^4$ equals the radiative flux, then we recover the Von Zeipel gravity darkening law that $T_{\rm eff}$ is proportional to $g_{\rm eff}^{1/4}$.
The missing step in this argument is to show the $f(\phi)$ is constant. Given that the photosphere is defined as where the optical depth is some fixed value (usually 1 or 2/3), and may be assumed to depend only on temperature and density, then this also lies on an equipotential. But $dT/d\phi$ also only depends on $\phi$ and so must also be constant along an equipotential.
For more detail, although missing the last paragraph above(!), see https://www.astro.umd.edu/~jph/Stellar_Rotation.pdf .
The situation is much more complex for stars with convective envelopes or differential rotation and I think can only be tackled through detailed modelling.


What is the angular diameter of Earth as seen from the Moon? ...today I must have left my brain home. I know it's a simple calculation but I keep failing simplest calculations today.

What is the angular diameter of Earth as seen from the surface of the Moon?

You can calculate the angular diameter of the Earth using the equation:
$$a = \arctan \frac{D}{d}$$
where $a$ is the angular diameter, $D$ is the physical diameter of the Earth, and $d$ is the distance from the Moon to the Earth.

The equatorial radius of the Earth is $r_E = 6378.1 \textrm{km}$, the diameter is therefore $D= 2 \times r_E = 12756.2$.

The mean distance to the Moon is $d=384399\textrm{km}$. This gives a (mean) angular diameter of $a=1.90065$°.
At Perigee the distance is $d=362600\textrm{km}$. This gives an angular diameter of $a=2.01482$°
At Apogee the distance is $d=405400\textrm{km}$. This gives an angular diameter of $a=1.80226$°

These Moon-Earth distances are as seen from the centre of the Moon. To calculate the diameter from the surface of the Moon, you'll have to subtract the position of the observer along the Moon-Earth axis.

If the observer is on the Moon's equator and the Earth is at zero hour angle (i.e. on the local meridian), the distance to the Earth needs to be subtracted by $r_M=1738.14\textrm{km}$. This gives the following values:

The mean distance to the Moon is $d=384399-1738 = 382661\textrm{km}$. This gives a (mean) angular diameter of $a=1.90928$°.
At Perigee the distance is $d=362600-1738 = 360862\textrm{km}$. This gives an angular diameter of $a=2.02452$°
At Apogee the distance is $d=405400-1738 = 403662\textrm{km}$. This gives an angular diameter of $a=1.81001$°

The angular diameter of the Earth from the surface of the Moon is, therefore, between $a=1.80226$° (at apogee and the Earth is near the horizon) and $a=2.02452$° (at perigee and for an observer at the equator and when the Earth is at maximum altitude on the meridian).

Or about 2 degrees.


At the moment of total solar eclipse what are exact distances of Moon & Sun to the Earth By total eclipse I mean the Moon and Sun are seen equal in size, neither of them is bigger.

For example, the Moon's average distance is 384000 km and the Sun's average distance is 149.5 million km. But when this distances occurs, are the Moon and Sun seen equal in size? (from Earth of course)

First off, your definition of "total solar eclipse" is different from that generally used. A total solar eclipse occurs when the moon is directly in front of the sun and is bigger so it blocks the whole of the photosphere. Also note that the sun and moon are both somewhat oblate (not perfect spheres, but flattened) and not exactly the same shape, but for the sake of estimation I'll use the diameter information from wikipedia.

The moon has an average diameter of 3,474 km, and the sun has an average diameter of 1.392684 million km. So the Sun is 400.9 times larger than the moon. When the sun is 149600000 km distant, then if the moon is $149600000/400.9 = 373200$km, the sun and the moon would appear equal in size.

When the moon is 384000km from the centre of the Earth, the distance to the Earth's surface is smallest at the sub-lunar point, about 378000km, but still greater than 373200, so when the Sun and Moon are both at their average distance, an annular eclipse occurs. Annular eclipses are slightly more common than total eclipses.

However if the distance from an observer to the moon is closer than 1/400.9 of the distance to the sun, then a total eclipse is possible, and this allows you to generate infinitely many solutions to the question "How far is the moon during a total solar eclipse"


Look up current positions and velocities of solar system objects I am working on an orbital dynamics code and, for fun, I would like to model the Sun-Earth-Moon system with my code.  I can look up the masses of each object just fine, as well as average distances and relative velocities (which I can also calculate).  However, I would like to model the Sun-Earth-Moon system in great detail.  I would like to know the positions and velocities of each object on any given day (today works just fine) so that I can represent the orbits of these objects with great accuracy, hopefully even reproducing their actual ellipticity.  Where can I find values for:

current Sun-Earth distance
current Earth-Moon distance
current Sun-Moon angular separation
current velocity of Earth relative to Sun (or relative to Sun-Earth(-Moon?) center of mass)
current velocity of Moon relative to Earth

and, maybe even

current velocity of Sun relative to Sun-Earth(-Moon?) center of mass.

Is there any resource that provides these current values?

Is there any resource that provides these current values?

Yes. The JPL HORIZONS on-line solar system data and ephemeris computation service provides these values, and much more.


How would tidal effects behave in this setup: An icy moon orbiting a gas giant, as the gas giant orbits a red dwarf? Let me know if this is more appropriate for worldbuilding.

Essentially, I'd like to know how tidal effects would behave in these circumstances (if it's even stable):

An icy moon orbiting a gas giant, with the gas giant orbiting a red dwarf.

Here is a detail that may manipulate the answer:

I want the moon (if possible) to live in the red dwarf's habitable zone. It's okay if it periodically leaves the habitable zone, so long as it returns.

I just need to know where tidal locking occurs (if at all), and how the moon receives most of it's heat (through internal heating like tidal forces or though external heating via the red dwarf).

I'm not sure exactly what your question is, but Tidal locking is fairly common for moons.   If the Gas giant was in the habitable zone of the red dwarf star, then it would need to be fairly close and as a result the Moon would need to be fairly close to the Gas giant - it should be tidally locked to the gas giant, not the star.  (A moon being tidally locked to a star is probably impossible because the planet will always have the greater tidal effect on the moon).

So your scenario, you have a moon, tidally locked to a gas giant planet, which orbits a star.   Tides don't exist in a perfectly circular tidal lock because the pressure is constant, but there's no such thing as a perfectly circular orbit, so there will always be some tidal squeezing as the moon moves around the planet.

The real question is how eccentric the Moon's orbit is around the gas giant planet, because it's eccentricity will determine it's tidal squeezing and any libration.

As far as periodic habitable zones, that would depend on how eccentric the orbit of the gas giant was around the star.  You could have it moving in and out of the habitable zone, but it would presumably be a short-period orbit, so it would be a matter of days outside, then days inside, perhaps a week or two outside and a week or two inside, unless you went highly eccentric.

All these are general points.  There's lots of variables.

There's no one answer for where the Moon would receive most of it's heat - you'd need to work out the specifics or orbit, distance and solar temperature and then give the Moon a specific orbit and then you could work out an approximation of where more heat comes from.

and this feels like worldbuilding to me.   It's too hypothetical for astronomy.


Can someone explain me this diagram from an article? How should I understand this 59 degree mentioned below, what does it mean? I could not get it. The photo is a diagram from an article dedicated to the invention of the ecliptic. The photo is from my tablet (the title could be seen)

The Earth's equator is at an approximate 23.4 degree angle to the plain of its orbit around the Sun. That means that to an Earth-bound observer the Sun appears to move north and south of the celestial equator (which is merely the projection of the Earth's equator on to the sky) during the course of the year.

In fact this is what gives us the seasons - so in the the Sun is at maximum northern declination (the equivalent of latitude when projected on to the sky) at the time of the Summer Solstice (around 21 June - hence summer in the northern hemisphere) and at maximum southern declination at the time of the winter solstice (around 21 December - hence northern winter).

But this movement also means different rising and setting points - in the Northern hemisphere the Sun's rise is between north and ast in the summer and between south and east in the winter (unless, that is, you are within the Artic circle in which case it will not set/rise at all for at least some period in the summer and winter).

There are a variety of online tools that show you this effect - this one: https://www.timeanddate.com/sun/iraq/baghdad?month=12&year=2017 - shows that in Baghdad the sun rises - in 2017 - at 61 degrees East on 21 June and at 118 degrees East on 21 December - a difference of 58 degrees.


What do the acronyms in "AT 2017 gfo" mean? About 11 hours after the detection of GW170817 by Ligo/Virgo and the quickly succeeding detected GRB170817, a transient afterglow in the optical-nIR frequency appeared which showed many signs of r-process nucleosynthesis. This transient was given the identification AT 2017 gfo.

  "SSS17a, now with the IAU identification of AT 2017 gfo"

My questions are simple. What do the acronyms "AT" and "gfo" mean? And for that matter, what does the acronymn "SSS" stand for? I tried to do a quick search through the IAU website but all I could find were general instructions on how to come up with your own astronomical identifier. I suspect "AT" means astronomical transient, but I can not confirm.

You are correct. The AT is short for Astronomical Transient. The follows the year and an arbitary sequence of letters to uniquely identify the event, which is assigned by the Transient Name Server

This is the official IAU designation for Supernova Candidates, however true supernovae will be re-designated "SN YYYYX". The neutron star merger was not a supernova, so retains its AT designation

This is described at (source) with more details at Transient Name Server (TNS)


How will nearby galaxies some day expand beyond our horizon? (Neil deGrasse Tyson) In his appearance on Television contained in the video The Mystery That Keeps Neil deGrasse Tyson Up At Night, Astronomer andeducator Neil deGrasse Tyson says:

I don’t want to be blamed if you can not get to sleep tonight.
This dark energy, in the future, will render the universe so large, having accelerated so significantly, that all the galaxies of the night sky will have accelerated beyond our horizon.
All the galaxies are the source of our knowledge of the big bang.
Everything we know about the history of the universe comes to us from these galaxies.
If they accelerate beyond our horizon, the next generation of cosmic explorers will only have the stars of the milky way to think about.

What does this mean, in simple terms if possible. How will nearby galaxies some day expand beyond our horizon (if that's what he's saying and if it is correct)?

Assuming that Neil deGrasse Tyson is referring to the cosmic event horizon, he's right. But that doesn't really mean that galaxies "fall out of sight". He's right that they will disappear, but not because they leave the horizon.

Nearby galaxies don't disappear

First of all, there's a small thing to note that I'm sure he just left out to simplify matter a bit:

The Universe expands and drags galaxies along, but for galaxies that are sufficiently close, their mutual gravitational attraction keeps them "locked". Hence, we will always be able to see the galaxies in our Local Group. In the case of Andromeda, we will even end up merging, and I suppose that after a sufficiently long time, we'll merge with the other galaxies in our group, simply because we will keep orbiting around each other, so at some point we'll come so close so as to merge.

Galaxies don't leave the observable Universe…

Second, and more importantly, galaxies do not leave our observable Universe. The observable Universe consists of the region of the Universe from which light has had the time to reach us since Big Bang, and the "boundary" between the observable and the un-observable Universe is called the particle horizon. As time goes, the light from an ever-increasing number of galaxies will have had the time to reach us. This number will never decrease. Once inside our observable Universe, a galaxy will stay inside.

…but fewer and fewer galaxies will be able to send a light signal to us…

However, the distance at which galaxies may emit a light signal today, that we may receive some day in the future, only asymptotically increases to a finite value (of some 17 billion lightyears; see e.g. Davis & Lineweaver 2004), while the galaxies themselve continue to increase their distance. I suppose this is what Neil deGrasse Tyson refers to: Even though the galaxies we see don't vanish from sight, we will never see what's going on in them today.

…and in practice, even the ones we see now will disappear…

Moreover, even the galaxies that we do see will become increasingly redshifted until eventually all light from them will be in the radio regime. Furthermore, photons will arrive at a smaller and smaller rate, until for all practical purposes, they will be invisible to us.

…especially if $w>-1$

The above considerations hold true if the Universe expands, and accelerates, as we think it does. However, there is a possibility that the acceleration itself will increase in magnitude. If the equation of state $w$ of the dark energy thought to be responsible for the acceleration is smaller than $-1$, the distance to the point where the Universe expands faster than the speed of light may decrease indefinitely, such that galaxies will indeed leave the horizon. Eventually, even gravitationally bound galaxies will be separated, stars will be torn out of the galaxies, planets will be pulled from the suns and finally disrupted, culminating in a Big Rip.


Why do particles from a comet that result in meteor showers spread out mostly along the comet's orbit? My understanding of the meteor showers that repeat yearly is that the Earth passes through the orbit of a comet that is nowhere near us, and might not even exist anymore, yet it's orbit is filled with a stream of dust in the same heliocentric orbit by way ahead of or behind their cometary source by tens of millions of kilometers or more.

How do these particles spread out along the comet's orbit so far and yet remain tightly correlated transverse to the orbit? Some meteor showers like the "Unicorn shower" linked below can be quite short, lasting only a fraction of a day!

What makes some meteor showers continue for days, while the "Unicorn shower" can be shorter than one hour?
Meteor shower predicted to be best viewed BEFORE midnight (for reasons other than the moon)?
Why are these objects moving at Vastly Different Speeds along the same orbit?

Left: From What (actually) is Jupiter doing to this year's Perseids meteor shower? Right: From Why would the Perseids meteor rate fall off after maximum faster than the increase before maximum? - click each for full size

The cometary particles need not retain the exact same orbit as the comet in order to stay close together near perihelion, thus where their orbit passes Earth's orbit.

Slight changes in the orbital parameters of the released dust particles can include slighly higher or lower aphelion, and thus change the revolution time and the time of perihelion (Earth) passage enough to spread out the particles. Over many revolutions it will eventually spread out over the whole orbit.

Additionally the orbits by the dust particles are subject to changes due to interaction with the solar wind and solar radiation which will over time change their orbits (yorp effect, Yarkovsky effect, radiation pressure etc).

This behaviour also explains that the meteor showers will be stronger in some years when Earth passes through a part where particles have not yet had time to more equally disperse.

Addendum: The other question on tails How can comets have tails if there's no air resistance in space? is related. Yet the (visible) tails are caused by either (ionized) gas or (small) dust particles (nm ... µm) which scatter the light. While these dust particles certainly also can make meteors, the larger the particles are, the less efficiently radiation pressure works on them, so their orbital evolution is much slower. And comets do release also a lot of bigger particles from µm up to cm (excact size distriubtion subject to disussion and analysis especially as follow-up to Rosetta mission, also depending on comet size and age) which are not easily visible in the visible light when observing it from Earth.


Would a tablespoon of a neutron star remain intact? I've heard people say that a tablespoon of neutron star would weigh over a billion tons. If we ever could take a tablespoon of one would it still remain intact with the same density?

If we take neutron star material and somehow transport it somewhere for examination (say the Earth!), the results would be catastrophic. At say a density of $\sim 10^{17}$ kg/m$^{3}$ the neutrons have a number density of $\sim 6\times 10^{43}$ m$^{-3}$ an internal kinetic energy density of $3 \times 10^{32}$ J/m$^{3}$ (calculated using the relevant equations for an ideal gas of degenerate neutrons at this density). So even in a tablespoonful (say 20ml, which would have a mass of 2 billion tonnes!), there is $6\times10^{27}$ J of kinetic energy (15 times more than the Sun emits in a second, or a few billion atom bombs) and this will be released instantaneously.

The energy is in the form of around $10^{39}$ neutrons travelling at around 0.1-0.2$c$. So roughly speaking it is like half the neutrons (about a billion tonnes) travelling at 0.1$c$ ploughing into the Earth. If I have done my Maths right, that is roughly equivalent to a 50km radius near-earth asteroid hitting the Earth at 30 km/s.

The neutrons in a dense neutron star gas are relatively stable (beta decay is blocked by electron degeneracy). The expansion described above would allow beta decay into protons and electrons, but as this happens on timescales of 10 minutes, it is hardly relevant to the initial destruction. However, you would end up after a few tens of minutes with an expanding cloud of ionised hydrogen a few light minutes across.

The minimum possible size to gravitationally bind neutron star material is thought to be around $0.15 M_{\odot}$ (see here). The equilibrium electron density (there are always some electrons and protons present in neutron star material) for lower masses is too low to block neutron beta-decay.