Did we discover 10 or 12 new moons of Jupiter? I saw multiple news sites reporting that a team discovered 12 new Jupiter's moons:

c|net - Twelve new Jupiter moons found, including one reckless one
Discover - Jupiter’s Got Twelve New Moons — One is a Bit of a Problem Child

While some other news sites claim that they found only 10 new moons:

nature - Jupiter has 10 more moons we didn't know about — and they're weird
EarthSky - Astronomers discover 10 new moons for Jupiter

Why does the number of moons differ in the different articles? Did the team now found 10 or 12 moons?

Per the Carnegie Science article that Magic Octopus Urn linked from NASA in the comments, a Carnegie Science team led by Scott S. Sheppard noticed something new in spring of 2017 (though some observations occurred as early as 2016). It took a year to confirm the discovery of the new moons.

Ten of the moons orbit in the outer swarm of moons, which is one of the ways to divide the groups of moons. Nine of these follow the pattern of the other moons which orbit Jupiter in retrograde (the opposite direction of the planet's rotation). One they call "oddball" because it orbits prograde (the same direction as the planet's rotation) and has a more inclined orbit than the inner prograde moons.

Two of the moons orbit in the inner group, which is what brings the total number of discoveries up to twelve, as referenced in the first set of articles in your post. Like the other inner moons, these orbit prograde.

One of the retrograde moons and the "oddball" were first noticed in 2016. One of the inner group and the rest of the retrograde moons were discovered in 2017. The second inner group moon was discovered in 2018, then most were announced together on July 17, 2018. (Dates collected from Wikipedia.)

The retrograde moon discovered in 2016 and one of the retrograde moons discovered in 2017 were announced in 2017, this means only ten moons were announced on July 17, 2018, which is why the Nature article refers to ten new moons. The EarthSky article talks some about the ten and the twelve, and it does clarify both groups (divided by inner/outer and year of announcement).


Is C/2012 S1 (aka ISON) the fastest comet on record to-date? This source may be nebulous (no pun intended!) for some. It writes to say

  Starting at around 95,000mph at the start of this month, Comet ISON will eventually reach an astonishing 845,000mph on November 28th when it catapults round the Sun at perihelion

I'm no astronomer; the only comets I even know the names of are

Shoemaker Levy
Swift Tuttle
Wild
Halley Bopp

To the best of my knowledge (feel free to poke me in the rib if i'm wrong!) none of the above list achieved a 3-digit velocity at perihelion. In contrast, going by the figure quoted above, S1 is expected to achieve a mind-boggling 375 km/s!

Is C/2012 S1 the fastest comet on record to-date?

A comet that travels to the Sun from the Oort cloud will be travelling roughly at its escape velocity at perihelion.

The fastest comets will be those that travel from the Oort cloud and plunge directly into the Sun.

If you want the comet to emerge from around the Sun, then its perihelion distance must be no lower than $\sim 7 \times 10^{8}$ m. At that point it will have approximately the escape velocity at that radius $v_{esc} = (2GM_{\odot}/r)^{1/2} = 617$ km/s or 1.39 million miles per hour.

Therefore I expect that lots of comets have attained something like or exceeding this speed before. Comet Lovejoy for example achieved a perihelion distance of $8.36\times10^{8}$ m and was travelling at 536 km/s (1.2 million miles per hour, and in fairly good agreement with the formula I quoted above - the aphelion was not infinity, but $\sim 157$ au, which largely accounts for the difference).


Why Do Planets Revolve Faster When they Are Closer to Their Parent Star? According to Kepler's second law: A planet sweeps out equal areas in equal times (A simple definition). This means that if the orbit is somewhat eccentric, the planet will move faster when on the closest approach to it's parent star, and slowest when at aphelion. So why does this happen, is the curvature of space-time greater closer to the star?

Velocity is a form of kinetic energy, while height within a gravity well is a form of potential energy. For an orbiting body, conservation of energy will keep the total energy constant.

So as a planet moves away from the parent star, it loses velocity and gains potential energy. As it moves closer, it trades the potential energy back for velocity. The point of lowest potential has the highest velocity and the point of highest potential has the lowest velocity.


Which stars did the Sun form with? The Sun formed 4.5 billion years ago, in a molecular cloud. I assume that there were other stars in the vicinity (as is common in molecular clouds). Which stars are they? Where are they now? Are they similar to the Sun?

Most stars form in clusters, so it is very likely that the Sun was part of a star cluster when it formed.

But in On the Dynamics of Open Clusters, the relaxation time of a cluster is calculated to be in the order of $\tau=4\times10^7 \textrm{yr}$. During that time, about one hundredth of the stars will escape from the cluster (i.e. reach escape velocity). The dissipation time of a star cluster is therefore in the order of a few billion years. These figures were calculated for an idealised star cluster containing stars of equal mass. For more realistic clusters the dissipation time may be much shorter. Especially dwarf stars seem to dissipate faster.

It is, therefore, likely that after 4.5 billion years, the Sun has escaped from its star cluster. It would be very difficult to identify the stars with which it was formed, unless you can trace back the position of the Sun and other stars before dissipation, which is impracticable.

As far as I am aware, no sibling stars have been positively identified. The sibling stars should be similar in composition (element abundances), although they do not have to be solar twins as the masses of these stars may vary widely.

EDIT: @adrianmcmenamin mentions an article in Sky & Telescope, where a sibling has been proposed. This refers to an article on arXiv by Ramirex et al. in which the element abundances have been analysed of stars that were earlier identified as possible siblings using galactic dynamics (i.e. tracing back the movement of these stars). Of the 30 stars identified as possible solar siblings using dynamical methods, only 2 stars had similar element abundances, which in itself says something about the accuracy of these dynamical methods...


Where can I find a database of galactic spectra? I was wondering where spectra of galaxies at various redshifts are to be found. I'm looking for ones that can be used to find the recessional velocity of the galaxy and eventually the Hubble Constant. Here's one I found:

The Sloan Digital Sky Survey Data Release 15 contains over 4 million spectra of both galactic and extra-galactic origin from the multi-fiber spectrographs. Of these spectra, 0.7 million came from the original spectrographs during the SDSS-I/II Legacy Survey and the remainder from the upgraded spectrographs as part of the BOSS survey during SDSS-III (see SDSS surveys details pages). These data went through an automatic pipeline that determined the redshift automatically (where the data quality was good enough) and the details of the process are given here.

The spectroscopy data are available through the links to the SkyServer Quicklook which lets you visualize and download the spectra and through the SkyServer Spectroscopic Query Form which lets you search for galaxies. (Try setting the 'Parameters to return' to 'typical' under the 'Spectroscopy' section' and setting the 'Classification=GALAXY' under the 'Spectroscopy constraints' section to only return objects that have been classified as galaxies).


In astronomical interferometry, what values do the points in the uv-plane have? As I understand it, the image of an interferometer is the inverse fourier transform of the information in the uv plane. For each baseline (vector between any two telescopes in the array), representing a certain wavelength along its orientation, there is a point in the uv plane.

(image from lecture notes from nrado.edu)
But what value does each of these data points have? I'd expect them to contain amplitude and phase information, but which exactly?
How do I get these values from the signals of my telescopes? Say, I already computed the phase shifts between the signals, what values do I put in the uv plane?

I was struggeling to understand 2080's question, so I looked up the given reference which might be worth re-quoting:

With this additional information in hand, the question (initially) seemed to be mainly about the  interpretation of 2D discrete Fourier trafo, using the terms of astronomy.
If that interpretation was correct, I suggest e.g. http://bigwww.epfl.ch/demo/ip/demos/FFT/ which offers an intutive and interactive approach to Fourier transformation in general.
Edit: The question has been edited, and the main issue seems to be How do I get these values $V(u,v)$ from the signals $T(x,y)$ of my telescopes? I write "seems" since I am not 100% certain that $T(x,y)$ are indeed the signals of the telescope.
What I know from Math behind a uv plot in interferometry? is that the units used in the u-v-plane are wavelength, e.g. in meters.
References

Lecture by Jason Hessels on Radio Astronomy: The Techniques of Radio Interferometry I: Basics overs an introduction I found quite accessible.


What to expect to see I have telescope with a diameter of 60 mm and a focal length of 900 mm. I have a 2x Barlow lens and eyepieces with focal lengths of 10 mm. Can I see any details of any galaxies? Andromeda for an example? What galaxy is the best for watching with this small telescope? I have f15.
I can view one galaxy but I don't know what exactly. It looks really small with no details.

What do you suggest I try to find in the sky? I viewed Saturn and I was delighted.

Summary: Small telescopes will not show any visual detail in a galaxy. The website What Can You See With Different Telescopes is a good reference. (Thanks to JamesK for referencing this website in a comment to Is the Celestron astromaster 130 EQ Model able to see nebulas and galaxies?) Keep in mind that observing an object visually is much different than photos you may have seen. A camera can take a long exposure to "build up" an image of a faint galaxy. The eye does not behave this way.

Most galaxies are faint when viewed visually. Therefore, details can be difficult to see. (In some cases, it is more a matter of "barely detecting" some detail compared to being able to easily "see" some detail.) For some amateur astronomers, being able to locate an object and see it is a victory. Being able to see some detail in a galaxy is a dream (for a larger telescope).

Here are some of the factors that will affect how much detail you can see in a galaxy:

How large is the telescope? The main purpose of a telescope is to collect light. The more light that is collected, the brighter the object will appear. A 60 mm telescope (2.4 inch, the size of the objective) is small and will not gather much light. Therefore, only the brightest portion of a galaxy will be visible. Fainter details will not be visible. (Perhaps the dust lane in M64 or M104 will be visible if the sky is really dark.)
How dark is the sky? Since galaxies are faint, seeing them well requires a dark sky. Can you see clouds at night because they reflect light? If the answer is no, you can only "see" the clouds because you notice that "stars are missing", then you have a dark sky. If the outline of the clouds are easily visible due to nearby light pollution, then seeing details in "faint fuzzies" will be impossible. What is the magnitude (the brightness) of the faintest star that can be seen? Both of those questions will help to judge how dark or bright the night sky is at your location.
Experience. Experienced observers can see more detail than inexperienced observers. Part is the difference is knowing to take time to study an object instead of viewing it for 5 seconds and saying "oh well".
Does the galaxy have any detail to be seen? (This factor should be obvious but is a good reminder.) Some galaxies are featureless blobs, such as elliptical galaxies. Other galaxies that theoretically have detail may be seen from an angle that makes the detail hard to see, such as a spiral galaxy seen from edge on. Spiral arms can be faint and may be indistinguishable even though the galaxy can be seen due to the combined light of the nucleus and arms.


Speed of light in a black hole If I had a directional photon-emitting source and placed it inside a black hole pointing upward and out towards the visible universe, I assume the photons traveling at the speed of light would slow and reverse direction back into the center.

So if I took the same source and placed it outside of the black hole pointing inward towards the black hole center, can I assume that an emitted photon would travel towards the center faster than the speed of light it is already traveling at?

It doesn't work like that. An observer at the light source (and indeed any observers anywhere else) will always see light travelling (in vacuum) at the speed of light locally.

There is also a major problem with your thought experiment. It is not possible for you to have a stationary light source within the event horizon of a black hole. It, and everything else in its vicinity, must be moving inwards. This is as inexorable and unavoidable as is the passage of time for an observer outside the event horizon.

In my opinion, the best "visual" way of thinking about the situation inside the event horizon is to imagine your photons of light like salmon trying to swimming upstream, whilst you are on a boat flowing with the stream and releasing the salmon into the water. You will always see the salmon swimming at some speed with respect to your boat. Unfortunately if the stream flows fast enough then the salmon will make no progress and you will both be swept over a waterfall (the singularity) a little further downstream.

Likewise, your common-sense fails with the situation of firing light towards a black hole. Light is always measured to have a speed of $c$ locally. It is following through with the consequences of this principle that leads to all the weird behaviour that black holes exhibit.


Is it possible to figure out your location in the Milky Way if you are suddenly wormholed to a random, distant location in the galaxy? If you were flying along in your spacecraft somewhere in our solar system and then a large, stable wormhole suddenly opened and you flew into it and were whisked away to some distant location in the galaxy, would you be able to figure out where you were?

Obviously, a large stable wormhole is exceedingly unlikely, and then there's the matter of the power of your onboard computer and telescope(s). But all that aside, I'm wondering if it would be even theoretically possible to figure out where you were, since the constellations would probably be unrecognizable and you wouldn't have an understandable flight path.

Assuming you had access to the relevant astronomical catalogues and data, then yes it would.

Looking at constellations would of course be hopeless and most stellar catalogues only contain star positions for a relatively small volume in our Galaxy. However, there are classes of object inside and outside the Galaxy that would do the job.

A displacement in our own Galaxy would produce a notable displacement in the positions of local group galaxies, like Andromeda and M33, with respect to distant quasars that define the ICRS coordinate reference frame.

Providing one had the means (i.e. telescopes) to make the necessary observations, then it should be relatively straightforward to reconstruct your position in the Galaxy from the new apparent positions of recognisable local group galaxies.

Further possibilities are offered by globular clusters and pulsars.

The colour-magnitude diagrams and chemical compositions of globular clusters are all subtley different, which should enable their identification. They are also spread in a roughly spherical distribution around the Galaxy.

Radio pulsars have unique pulse shapes and timing properties that make them readily identifiable. A possible/probable problem is that many pulsars would be rendered invisible by the change in viewing angle.

In both these cases, a map of the revised spatial locations with respect to the ICRS should enable you to work out where you are compared with the position of the Earth now. I am unsure what precision could be achieved, but then we are unsure of our present location with respect to distance to the Galactic centre by a few percent.


Was there ever a Jupiter transit or Saturn transit? I've watched the Venus transit 2004 with a small telescope, a relatively rare event, considering most people can only observe two in their life time.

If E.T. were sitting in a sunny place on a moon of the outer planets,
sipping on a liquid nitrogen drink, would there ever be (has there ever been) a transit of an inner gas giant?

I guess if E.T. were in Sputnik Planum, there would be no transits due to Pluto's orbital plane being inclined too much wrt. the Ecliptic.

Wikipedia actually has many pages on this subject. Transits of Jupiter, viewed from Saturn are rare. The calculations are done from the centre of the planet, rather than a moon, but the principle is clear: Transits of Jupiter are rare.

The next, non-grazing transit of Jupiter visible from Saturn is in 7541. From Uranus in 2714, and from Neptune in 2188.

But probably the most photogenic transit will be the transit of Saturn from Uranus in 2669.

Saturn will also transit from Neptune in 2061, the only transit which we could live to see. Uranus will not transit from Neptune until 38172, and transits from Pluto or other KBO are even more rare, due to the high orbital inclinations.


Tips for a beginer who wants to learn more about astronomy I have completely no idea about astronomy, but I am starting growing interest in it as I have been always wondering on what has been discovered out there and since I am studying CS, maybe I will learn more than I think as CS and astronomy as relative to some point. How should I start? Can you suggest me a good website for example or something else?

(I know this is an opinion - based question but still I need to make this question here).

To me it sounds that you just want to expand your astronomical knowledge and are not actually looking for a career advice.

I would suggest to just start with reading popular science books.
https://en.wikipedia.org/wiki/Cosmic_Evolution_(book) ,
https://en.wikipedia.org/wiki/A_Brief_History_of_Time for example.


Are there standard algorithms and procedures for creating unique sky maps based on latitude/longitude/date/time? I am trying to build an astronomy app that will use the user's latitude and longitude, along with the current date and time, to create a 3D view of the celestial sphere as seen based on that data.  However, the information I've found so far has been very general and less-than-practical, and has included virtually nothing about how to take date and time into account.

Does anybody know of a more concrete source of information on sky mapping in this way?

If you want to use the user's co-ordinates (longitude and latitude) to calculate the topocentric co-ordinates of an object to a not very high accuracy then you should be able to find a few software libraries (e.g. libnova) that do just that. Many use the algorithms from Jean Meeus' book Astronomical Algorithms. You can use catalogues such as the Bright Star Catalogue to find the equatorial co-ordinates of a star and then use the algorithms in this book to convert to topocentric co-ordinates (also called 'apparent place of a star').
Of course if you only want to calculate the 3D positions of stars, then it would be useless to use topocentric co-ordinates as the distances to the stars are so great that the error introduced by using geocentric (centre of the earth) instead of topocentric (position on earth) is very very small compared to the distance to the stars.

If you want to calculate the position of solar system object (planets, asteroids and comets) than you would need to be able to calculate the position of these objects to a very high precision. Meeus' book and software libraries derived from his book contain these algorithms. For planets you could use VSOP, but many software libraries include these. Meeus' book also includes algorithms to calculate the positions of the Moon and Pluto and the moons of the major planets. For asteroids and comets you can get the data needed for the algorithms from the IAU Minor Planet Center.

For high accuracy the algorithms provided in these libraries will not suffice and you may need to use professional software libraries such as NOVAS or SOFA. But then you are in milliarcsecond accuracy.


Explanation for the first spinning neutron star detected in the Andromeda galaxy? A new class of Neutron stars has been found in the Andromeda galaxy (M31), This is the first time a Spinning Neutron star has ever been detected by astronomers in M31. The article states-

“pulsars” can be found in stellar couples, with the neutron star cannibalizing its neighbor. This can lead to the neutron star spinning faster and to pulses of high-energy X-rays from hot gas being funneled down magnetic fields onto the neutron star.

“It could be what we call a ‘peculiar low-mass X-ray binary pulsar’ — in which the companion star is less massive than our Sun — or alternatively an intermediate-mass binary system with a companion of about two solar masses,” said Paolo Esposito of INAF-Istituto di Astrofisica Spaziale e Fisica Cosmica, Milan, Italy.

While the precise nature of the system remains unclear, the data imply that it is unusual and exotic.
http://www.astronomy.com/news/2016/04/andromedas-first-spinning-neutron-star-has-been-found

Question-
  Could the companion of this Spinning Neutron Star not be a star at all but a Black Hole? This might explain the unusual spin and that all neutron stars do not consume other binary companions stars but rather a Black Hole is in the process of consuming the neutron stars.

The Black Hole can devour all of the outer energy of any active star leaving only the dense core of heavy metals that will eventually cool off entirely creating the rouge brown dwarf.
  Supernovae never derived from dying neutron stars but rather from Binary star Collisions.

In the article it notes that the partner has less mass than the sun, or possibly up to about twice the mass of the sun (quite a range of uncertainty there!) It's not clear, from the article how they made these calculations, measurement of doppler shifts to get orbital velocity I suppose.

The mass suggested there is not especially small, there are pulsar binary systems in which the secondary is one hundredth the mass of the sun, and could be considered to be planets.

However it does rule out a black hole companion, which would have a mass of 3 solar masses at the least, (and more likely more than 5). There is no need to hypothesis a black hole to explain the spin.

The last few sentences in your question don't make any sense to me.


Does New Moon Affect Stargazing After MoonSet? Referencing the following data for December 11th 2017.

https://www.timeanddate.com/moon/@4799274?month=12&year=2017

To get a dark sky for stargazing does it matter if the moon is at 37.8% so long as it is below the horizon when I'm observing?

Since the moon will be set at 1:32PM on the 11th and not rise again until 1:59AM on the 12th I would assume the moon would not affect viewing between those times.

Furthermore is there a guideline as to how close to moonrise the moon will start to have an affect?

For most people the Moon does not significantly impact on stargazing, even when it is full, unless they are trying to view something in a direction close to the Moon.

For people doing astrophotography, the Moon has a bigger impact since the sky is noticeably brighter and results in fainter objects not being able to be photographed.

And for astronomical observatories, some operations are impossible at worst and severely restricted at best since the telescopes are usually larger than "home" varieties and cameras much more sensitive but most noticeably a few days either side of the full moon.

So, depending on what you are actually doing, a ~38% Moon will not be a problem and if you are looking more than about 30° away from the Moon you will not even notice it.


Visibility of human activity on the moon In this video the host of Test Tube Plus states that

you can go out a buy a laser, point it at the moon, and see the retroreflector left by the astronauts.  When you point at the right spot you'll see a reflection, elsewhere not.  He reiterates that this is not just a hypothetical use of 2nd person for effect, but that you the viewer can go out and try this yourself, and anyone paying any attention to moonlanding conspiracy theories ought to try it
tracks and artefacts at the landing site can be seen with amateur telescopes, and again this is approachable to anyone who cares to look

Neither of these seem correct to me.  Could someone with real numbers weigh in on this?

I seem to recall that seeing artefacts is only possible in the latest high-res images taken from lunar orbit, and the telescope that uses the reflector gets back single photons on some trials and is a much more powerful laser than you could buy!  But, a continuous beam may be different? How big of a laser would you need to see a reflection with a viewing instrument that doesn't cost more than the laser?

The video is hilariously wrong.

However, the principle of laser ranging is more or less right, and it does require the reflectors left behind by the astronauts on the Moon. It's just that the physics and technology involved are far beyond just pointing a toy laser at the Moon.

Project APOLLO (Apache Point Observatory Lunar Laser-ranging Operation) is actually doing this.

http://physics.ucsd.edu/~tmurphy/apollo/basics.html

You need a fairly large telescope to start with - both for collimating the outgoing light pulse, and for receiving as much of the reflection as possible. APOLLO uses a 3.5 meter telescope at the Apache Point observatory.

You need a laser that can generate a high-energy light pulse that is very short. The pulse is injected into the telescope's optical train and sent to the Moon. This is not a laser pointer; it's a high power Q-switched laser for research, a device the size of a refrigerator.

On the receiving end, you need a very good detector also plugged into the telescope. Of the many, many photons sent to the Moon in the pulse, only between 1 and 5 photons make it back down to the detector. You need a detector that can tell the extra 1 to 5 photons from the background noise of light coming from the Moon.

Using this technique, the distance from Earth to Moon can be measured with very high precision.

This is the APOLLO system in action:

Regarding observing human artifacts on the Moon with terrestrial telescopes, I wish it was doable, but it's not. Again the YouTube video is wrong.

The biggest telescope accessible to amateurs has an aperture (diameter) in the range of 1 meter or a little bit larger (the aperture of the largest amateur-owned telescopes currently). The resolving power of a telescope (the size of finest details discernible) depends on aperture - if aperture is measured in mm and the resolving power in measured in arcseconds, then the formula is:

resolving power = 100 / aperture

So a 1 meter telescope has a resolving power of 0.1 arcseconds.

The distance from Earth to Moon is 384000 km (3.84 * 10^8 meters). With a resolving power of 0.1 arcseconds, the size of the smallest detail discernible on the Moon is:

detail size = distance_from_Earth_to_Moon * arctan(resolving power)

or

detail size = 3.84 * 10^8 * arctan(0.1 arcsec) = 186 meters

Anything smaller than 186 meters would be blurred into a single dot. There's nothing we've done on the Moon that's as big as that. It's not possible to see traces of human activity with amateur-level telescopes, even with extremely large meter-class dobsonian telescopes. Even with professional telescopes, we just don't have the aperture yet to resolve that kind of detail.

However, satellites in orbit around the Moon, such as the LRO, were able to image traces of the Apollo missions. That's because they are a lot closer to the site.

http://www.nasa.gov/mission_pages/LRO/news/apollo-sites.html


How small a location is possible to identify when any visible star is at its zenith directly above? How small an area on Earth is it possible to identify with any precision when any celestial object (star, planet, comet, nova) (visible with the naked eye alone) is at its zenith directly overhead?
--Would it be possible to identify an area as small as a town, or village, a square mile, or an acre, or even a building?
--How big an earthly location would be required to narrow it down if such a celestial object is directly above?
  Thanks.
PS:  I've tried to narrow the question down to get a simple to understand answer.  I'm not very technical, nor an astronomer.

Lets assume your observer has a good watch, so she knows the exact UTC time. But lets also assume that she has no special kit, and has no special training.

She might be able to estimate the location of "directly up" to an accuracy of about one hand width (at arms length) which corresponds to roughly 5 degrees.  To achieve better than this, she would need some tools for sighting the position of stars.

With an error of +- 5 degrees, the error on the Earth's surface of any calcuation based on her observation is also 5 degrees of the Earth's surface. That is roughly a circle of radius 550km (350 miles).

Such an observation could prove you were in Mexico and not Canada. But it couldn't distinguish Conneticut and Washington DC, much less a square mile.

To get 1 mile accuracy you would need measurements at least 350 times better, This is better than one-sixtieth of a degree. The most skilled observers did achieve this level of accuracy with naked eye equipment, but not from a single observation. It could be done with a telescope designed for measuring the position of stars, provided the operator was sufficiently highly skilled.

If your observer doesn't have a good watch, she is subject to the problem of Longitude: It is relatively easy to find Latitude just from astronomical observations. It is much harder to find the longitude.


Is nuclear pasta or neutron star crust iron stable outside of neutron stars? When neutron stars collide, they crash into and kill each other in a kilonova explosion, which blasts a lot of their material into space. A huge amount of neutronium, neutron star stuff, is suddenly no longer in the high pressure of the star, and instantly expands explosively into normal matter. In other words, thou shalt not have isolated teaspoons of neutronium. But is the same true for the crust, made of a form of highly compressed iron called nuclear pasta, 10 billion times stronger than steel? Is the crust material of neutron stars stable outside of neutron stars?

"made of a form of highly compressed iron called nuclear pasta,"

This isn't the case. The crust of a neutron star consists of (starting at the surface going down).
(Picture from Watanabe & Maruyama 2012)

Iron-peak nuclei plus degenerate electrons.

Increasingly neutron-rich elements with higher atomic numbers (i.e. not iron) plus degenerate electrons.

Very, very neutron rich heavy nuclei (mass numbers of several hundred) plus degenerate electrons and free neutrons.

Nuclear pasta, where the identity of individual nuclei disappears and the nucleons (mainly neutrons) form long chains, or sheets, surrounded by degenerate free neutrons and electrons.

None of these compositions is stable at low pressures. Specifically, you need to have them surrounded by a high density gas of degenerate electrons to suppress the beta decay of neutrons and neutron-rich nuclei. Putting this stuff into a low density environment would be explosive (but somewhat less) just as in the case of "neutronium" (neutron matter inside neutron stars contains protons and electrons too). This is not because of the beta decay, which is a slow process, but because the constituents have a huge amount of kinetic energy (that is what pressure is).


What telescope is this, and how does it work? In the excellent chemistry series of Periodic Videos' episode Amazing piece of metal (speculum) there is an interesting showing and discussion of Isaac Newton's first telescope.

At 02:36 there is a short cut to a view of a very large, modern telescope mirror. I'm confused by the image because it seems to show what looks like a small 45 degree elliptical, diagonal secondary mirror that you might see near the top end of a long focal length Newtonian telescope, except that it seems to be sitting in front of the primary, facing towards the stars.

Can anyone identify this telescope, and help me understand the basics of how this particular optical system is configured?

edit: I was so focused on the small mirror in the center that I neglected to read the caption. This "very large telescope" may in fact be THE Very Large Telescope, one of it's components at least. But because of all the shiny metal an reflections in mirrors, I still don't really understand what I'm looking at as far as the optical system is configured.

https://www.youtube.com/watch?v=4ESW_NTIhBM

This is indeed the VLT, or rather one of the four 8.2m Unit Telescopes (UT in ESO parlance). Looking at the optical layout (found here along with a description of the telescope), the small mirror in the centre of your screenshot is M3, the tertiary mirror. M1 is the primary (in the background), and M2 is out of the frame towards the top- the white supports you see in the foreground hold it above M1.

One advantage of this design is that you can have two large instrument platforms on the vertical tilt axis of the telescope - this way, you don't have to dangle several tons of instrument from the support structure (Nasmyth focus in the image).


Recommendation for learning about stellar astrophysics I would like to know which are the best books to learn about stellar astrophysics at (just) graduate level.

I have a basic formation in general astrophysics but I'm interested in learning about stars, specially their evolution and constitution. I also have a good mathematical formation so it should not be a problem (but I don't care if it focus on the ideas as long as it is a good book and make itself clear).

I found several books at the university library, but I don't know which of them are good (in my situation at least).

Best book I have used for this topic is: An Introduction to Stellar Astrophysics by Le Blanc.

Excellent book, very good introductory start into this topic.


How did the moon's orbit become eccentric? The Moon's orbit is more eccentric, 0.0549, than most planets. I can understand that planets get eccentric by disturbing each other like under the late heavy bombardment. And likewise for multiple moon systems. But shouldn't the debris after the collision with Theia have been circulirized? Is its orbit being changed by the planets, at every opposition with Venus?

First, your claim that the moon is more eccentric isn't entirely true (lets ignore Pluto):

(Source)

Mercury .205
Mars .094
Saturn .057
Moon .055
Jupiter .049
Uranus .046
Earth .017
Neptune .011
Venus .007

So, it's really not abnormal, though it might be unusually eccentric for a non captured moon, but that's may be due to it being as close to the sun as it is.

Also, our solar system might not be exactly normal.  It's fairly common for solar systems to have a gas giant close to the sun (explained here).  And there are some very odd solar systems - same link.


How did the Earth come to be in orbit around the sun? I'm not sure if this post should be in the physics subject forum, but this seems to fit here too. I have been reading a book about chemistry and how the universe came to be with the theory of stars creating most of the elements we know of. I began wondering how the Earth came to be in orbit around the Sun. I understand that we are in a constant free fall towards the Sun, but our motion causes us to "miss" the Sun stopping us from burning up. However, I'm confused how we were put into motion in the first place. My best guess is that we were captured by the Sun's gravity, and were put into orbit, which would mean the Earth was once a meteor, right? If someone could please explain this, it'd be very helpful because I can't seem to find the answer anywhere.

Thank you

P.S. I'm just a 9th grader, so I don't have much knowledge in astrophysics.

The picture was so much cleaner 20 to 25 years ago. I'll present that nice clean picture first. Stars form from the gravitational collapse of huge clouds of interstellar gas. Those gas clouds inevitable have some net non-zero angular momentum. This forces the gas cloud to change shape from being more or less spherical to being disk-shaped. (Why? That's a different question. Ask it.)

While this protoplanetary disk continued to feed mass to the growing protostar, it also set the stage for the formation of planets. The gas cloud was mostly primordial hydrogen and helium, but it also contained heavier elements thanks to stellar fusion and supernovas in the billions of years that preceded the formation of our solar system.

Those heavier elements behave quite differently than do hydrogen and helium. They have chemistry. The planets started as microscopic clumps of mass of these heavier elements, bound together chemically. These microscopic clumps occasionally collided, eventually forming larger clumps of mass. These larger clumps in turn collided with one another, forming even larger clumps of mass. Eventually the clumps became large enough that they interacted gravitationally, making them grow even larger. This process continued, eventually forming protoplanets, and then planetary embryos, and finally planets.

Temperature in the protoplanetary disk were high near the forming protostar but dropped precipitously with increasing distance from the protostar. At some point, volatiles such as water, ammonia, methane, and carbon dioxide become as solid as rock. This is the ice line, aka the snow line or frost line. Asteroids inside of Ceres' orbit tend to be rocky. Asteroids outside of Ceres' orbit tend to be icy.

Planets that form outside the ice line can grow very quickly and then they can grow very, very large. The stuff that comprises the protoplanetary disk orbits the growing protostar at something other than the rate suggested by Kepler's laws thanks to the pressure of all that stuff in the disk. Thanks to the square-cube law, larger objects aren't as subject to that pressure. Those larger objects orbit at a Keplerian rate. Planets that form outside the ice line grow very quickly and then sweep up gas and ice because they are orbiting at a different speed than the immediate surroundings. The result is gas giants such as Jupiter and Saturn and further out, ice giants such as Uranus and Neptune. Planetary growth is a much more difficult process and much slower process inside the ice line. That's why Mercury, Venus, the Earth, and Mars are rocky and so much smaller than Jupiter, Saturn, Uranus, and Neptune.

That's the pretty picture. The not so pretty picture:

Why are Mercury and Mars so much smaller than Venus and the Earth?
Simulations suggest that the rocky planets should all be more or less the same size. That is not the case in our own solar system, let alone elsewhere.
How could Uranus and Neptune have formed?
Simulations cannot recreate Uranus and Neptune at their current distances from the Sun. The material in the protoplanetary disk should have been too sparse at those distances to form large planets.
Much, much worse, what's the deal with all the weird exoplanets scientists have found?
Scientists have found Jupiter-sized objects orbiting very close to their sun, Neptune-sized objects orbiting where the simple model would have only rocky planets forming, and planets in highly inclined (and sometimes retrograde) orbits that don't make sense.

These simulations (which have become to be very good) and the plethora of exoplanets have pushed the theory of how planets form back into the “that’s funny” stage. ("The most exciting phrase to hear in science, the one that heralds new discoveries, is not “Eureka!” but “That’s funny...”", a quote widely attributed to Isaac Asimov.)


If a star were to suddenly lose nearly all of its stored heat, would it be able to return to its normal state? If not, would it be possible to approximate the maximum heat energy a star could lose before the change became irreversible?

If you suddenly sucked all the heat out of a normal star like the Sun, nuclear reactions would cease, the gas pressure would fall to zero. The star would collapse, initially on a freefall timescale (about 1 hour for the Sun).

The released gravitational potential energy would half be radiated away and half would go into re-heating the interior. The temperature would rapidly rise, the collapse would be halted, nuclear reactions would restart, the star would slowly expand again and a thermal and hydrostatic equilibrium should be restored on a Kelvin-Helmholtz timescale (about 10 million years for the Sun).

One way to see this, is to compare the current thermal content of a star with its capacity for generating new heat by nuclear reactions. e.g. approximate the Sun as a uniform ball of ideal gas at an average temperature of a few million degrees. The thermal content is roughly $10^{41}$ J. The Sun generates this much energy in nuclear reactions in only 8 million years - coinciding with the timescale to reestablish an equilibrium. Or you could compare it with the gravitational potential energy of around $2\times 10^{41}$ J. i.e. There is enough gravitational potential energy to reheat the Sun.

However, not everything would be a continuation. The collapse and reheating would be similar to the star repeating its pre main sequence phase. During this phase the star becomes fully convective and so the core material would get completely mixed with the rest of the star again. This would have the effect of rejuvenating the star as a new zero age main sequence star, but with a slightly higher helium abundance.


Is the mechanism of solar flares on red dwarfs and brown dwarfs the same as that on the Sun? The Sun has solar flares that are caused when there is a magnetic reconnection in the Sun's atmosphere, causing a loop of magnetic field to be ejected at high energy, along with a large number of protons, giving a intense burst of radiation.

Other stars - notably some red dwarfs - have much more intense flares than the Sun does. Is the mechanism the same in these stars as in the Sun, and in larger stars? What causes the flares in red dwarfs, and even brown dwarfs? Why do less powerful stars have more powerful flares?

Jupiter also has a magnetic field, and it's magnetosphere is composed of plasmas. Why aren't "Jovian Flares" seen?

Related: How do large solar flares compare to flares on other stars?

Yes, the basic mechanism is thought to be the same on red dwarfs and at least the hotter brown dwarfs, but the details can be different.

As you say, magnetic reconnection in the corona is the starting point. Well, actually it is fluid motions at the magnetic loop footpoints that is the starting point. The B-field and partially ionised plasma are coupled and photospheric motions put (magnetic) potential energy into the B-field structures.

This potential energy can be released suddenly in reconnection events. These can drive coronal mass ejections or accelerate charged particles along the loop field lines.

A flare occurs when a significant amount of energy goes into accelerating charged particles down the field lines towards the loop footpoints. These charge particles emit radio waves and then non-thermal hard X-rays as they impact the thicker chromosphere/photosphere. Their energy is then thermalised, heating the chromosphere and possibly causing hot ($>10^{6}$ K) material to evaporate into the coronal loops. Here it can cool by radiation and conduction before falling back, or perhaps forming cool prominences.

Similar things must be going on in low-mass stars. The X-ray and optical light curves of their flares do resemble what is seen on the Sun as do the relationships between hard- and soft X-rays and the evolution of plasma temperatures. The details may be different because the temperature and density structure of their photospheres, chromospheres and coronae are a bit different to the Sun's, and there are some indications that their coronae can be denser or sometimes that flares occur in much larger structures than are seen on the Sun. "White light" flares are also more common in M-dwarfs.

Why are flares on red dwarfs so powerful? Partly it is contrast - you are comparing the flare emission with something that is intrinsically less luminous. In absolute terms the flares on the Sun and flares on red dwarfs are not hugely different. What is different is that the flare power as a fraction of the bolometric luminosity and the frequency of large flares can be much higher on M-dwarfs.

The underlying reasons are likely to do with the magnetic field strengths and structures on red dwarfs. Magnetic activity is empirically connected to rotation and convection. Magnetic activity is higher on rapidly rotating stars and those with deep convection zones. M-dwarfs have very deep convection zones (or are even fully convective). They also tend to rotate much more rapidly than the Sun, since their spin-down timescales appear to be much longer than G and K dwarfs. They are thus more magnetically active relative to their bolometric luminosities. It appears that active M dwarfs have very strong magnetic fields (like those in sunspots) covering a very large fraction of their surfaces and this along with the convective turbulence where the magnetic loop footpoints are anchored likely leads to their strong flaring activity.

Brown dwarfs are a bit more tricky. The younger, hotter ones probably behave much like low-mass M-dwarfs (in fact they are M-type objects). The magnetic activity on cooler/older L- and T-dwarfs is much more mysterious. I guess a few flares have been seen, but it is not clear this is related to similar mechanisms as in higher mass and hotter stars. These cool brown dwarfs have neutral atmospheres and the magnetic field is not frozen into the plasma like it is in the partially ionised photospheres of hotter stars. This means that the loop footpoints may not be stressed by photospheric motions in the same way. It is not even clear that brown dwarfs generate a magnetic field in the same way as more massive stars, though it is clear that at least some of them do have magnetic fields.


What does the orbit of multiple star systems look like? How can I find simulations of trinary, 4-nary.. etc. star systems?

What do they look like?

For example the Alpha Centauri system has 3 stars and possible planets. What does that orbit look like?

I cannot find any simulation on here, it includes 2 Centauri A and B does not include the third dwarf star Proxima.

We have been observing the system for many years, I would think a lot of simulations would come out of each new star system we find, or are they all very similar to one another.

In the case of Alpha Centauri A&B and Proxima Centauri,   A&B orbit each other at elliptical orbits, getting almost as close as the Sun is to Saturn and about as far as the Sun is to Neptune.   Proxima Centauri is about 1,000 times further out and it's a tiny compared to the inner 2, so for simulation purposes, it's essentially a 2 star system with a far distant 3rd orbiting the inner 2.    The 3rd star is so far and comparatively small that it has essentially zero effect on the simulation.

Source of Photo

Now, in general, most 3 or more star systems operate in similar ways where you can get a 1 or 2 stars operating as the center of mass and then you need to extend a fair bit of distance for the 3rd star.  3 stars all about the same distance and about the same size can be set up on a computer, but it's unlikely to happen in reality cause instabilities tend to grow.

The Sun-Earth-Moon is effectively a stable 3 body system and not subject to 3-body chaos because the Moon is safely within the stable region of the Earth's Hill Sphere and those 3 objects have clear cut hierarchy in terms of gravitational influence.

Pluto, Neptune& the Sun, for example, is probably long term unstable.   See here

This is Wikipedia's 3 body simulation, but this is probably not something you'd see in solar-systems because of the inherent instability.   In such a 3 body orbital system, one of the 3 bodies would likely either be ejected or 2 of the 3 bodies would collide, creating a more stable 2 body system.   It wouldn't have long term stability.

https://en.wikipedia.org/wiki/N-body_problem

You could potentially get a 3 body system where the 2 outer bodies are in Trojan points to each other (L4 or L5 only).   I'm not sure how common that is but I'd think it happens occasionally, provided the central star had significantly more mass than the 2 outer stars.

L4 and L5 Trojans


Visible satellites real time position API I'm trying to obtain the azimuth and elevation of all (or at least the majority) visible artificial satellites from a given latitude and longitude. Is there a quite stable API that does that for free? I found some APIs that track particular satellite event, ISS passes and so on, but nothing about what I'm looking for.

This should run on an ESP8266 board, so I can't afford to store positions database or a large ephemeris and do to calculations on it.

I've already used the site N2YO.COM and the API. There is a lot of satellites on it and the most important ones.

https://www.n2yo.com/api/


Why didn't the Event Horizon Telescope team mention Sagittarius A*? At the press conference this morning, the Event Horizon Telescope team didn't say much about Sagittarius A*, which was the target many of us have been waiting for.

Is there any explanation anywhere for this omission?

There was a mention of Sagittarius A* during the Q+A portion of the press conference; the team indicated that they hope to produce an image sometime in the future (although they were careful to make no promises, and they're not assuming they'll be successful).

That said, I'm not wholly surprised that we ended up seeing M87, rather than Sgr A*, for a couple reasons which the team mentions in their first paper:

As Glorfindel said, Sgr A*'s event horizon is much smaller, meaning matter orbiting the black hole has a shorter orbital period. This contributes to variability on the timescale of minutes. The observations of M87 took place over the course of a week - roughly the timescale over which that target varies, meaning the source should not change significantly over that time.
Second - and this is the reason I've seen cited more often - Sgr A* lies in the center of our galaxy, and so thick clouds of gas and dust lie between it and us. That results in scattering, which is a problem. There are ways to mitigate this, of course, and the team has spent a long time on this, but it's simpler to just look at the black hole that doesn't have that problem in the first place. That's why M87's black hole is an attractive target.

Neither of these are impossible hurdles to overcome, but they're certainly very real difficulties that can't be ignored.


Have any planetary systems been found around Population II stars? I always understood that Population II stars were unlikely to have planets as they were virtually all H and He with very little else.

Have recent observations confirmed this, or have planets been discovered around Population II stars?

Kapteyn's star has 14% the metallicity of the Sun and is about 11 billion years old and is in the galactic halo. It has two reported planets.
I don't know if 14% is low enough to count as  Population II, but according to the NASA exoplanet archive it is the lowest metallicity star known to have planets.


How does the gravitation of Jupiter or other big planets affect Kepler's third law? Does anyone know a good reference for this?
I once read how Jupiter's gravitation can be treated as a perturbation. I think they expanded Jupiter's gravitation as an infinite sum of one of the special functions (such as Bessel function, Legendre function, Hermite polynomial, I can't remember).
I am not interested in how the motion of a planet at each time looks like, but just the overall effects on parameters such as period, semi-major axis, and eccentricity.

Short Answer: The gravitation of Jupiter and all the other planets makes Kepler’s third law a bit less accurate than it would be if their gravity were zero.  The gravitational interaction between the planets, however, doesn’t have much effect on either the period or semi-major axis of any of the planets, especially in the short term.
Long Answer:  You might be thinking of this pendulum motion expansion used to describe the orbital oscillations (including the big one between Jupiter and Saturn) here https://history.nasa.gov/SP-345/ch8.htm:

Where $\psi$ is the angle from the vertical, $\omega$ is the angular velocity, and $A^2$ is $g/l$, where $g$ is acceleration and $l$ is presumably the length of the pendulum.
Interplanetary resonances mostly result in precession of planet’s orbits, which are period and semi-major axis invariant.  They mostly change just the longitude of the ascending node.

Other orbital changing effects between planets are Milankovitch cycles which are also period and semi-major axis invariant, according to Wikipedia:

Eccentricity varies primarily due to the gravitational pull of Jupiter
and Saturn. However, the semi-major axis of the orbital ellipse
remains unchanged; according to perturbation theory, which computes
the evolution of the orbit, the semi-major axis is invariant. The
orbital period (the length of a sidereal year) is also invariant,
because according to Kepler's third law, it is determined by the
semi-major axis.

The more significant effects that planetary mass have on adding inaccuracy to Kepler’s third law are through their gravitational effects on the sun when treated as individual two-body orbits.
Kepler’s “laws” describe the motion of the planets.  Kepler arrived at these laws by examining observational data from the solar system in the late 1500s and the early 1600s.  On page 48 of Epitome of Copernican Astronomy Kepler asked himself the question:

How is the ratio of the periodic times, which you have assigned to the
mobile bodies related to the aforesaid ratio of the spheres wherein
those bodies are borned?

And then answered:

The ratio of the times is not equal to the ratio of the spheres, but
greater than it, and in the primary planets exactly the ratio of the
3/2 powers.

A bit of translation is necessary here. The periodic times Kepler references are the orbital periods $p$ of the planets, and the spheres are the radii of the spheres which circumscribe the ellipses of the orbits.  These are the same as the semi-major axes $a$ of the orbits.
So Kepler’s third law as stated was that, for every pair of planets 1 and 2, $(a_1/a_2)^{3/2}=p_1/p_2$.  Our typical modern restatement of Kepler’s third law is that the ratio $a^3/p^2$ is the same for every planet.
Kepler thought that the sun was the center of the solar system.  He didn’t realize that when a less massive object is in orbit with a more massive object, both objects actually rotate around their mutual center of gravity (called the barycenter).  As Newton himself said (thanks UserLTK): “......Kepler knew ye Orb to be not circular but oval & guest it to be elliptical......”: from: arxiv.org/ftp/physics/papers/0107/0107009.pdf source of quote: H. W. Turnbull (ed.): The correspondence of Isaac Newton, II, Cambridge, 1960, pp. 436-437.
Kepler’s original values for the ratios were:

Kepler thought that “a more accurate number will be produced, if you take the times more accurately.”   Kepler attributed all the inaccuracies in his data set to measurement error.  However, Newton proved Kepler wasn’t quite correct.
Newton generalized Kepler’s third law as $$\frac{a^3}{p^2}=\frac{G(m_1+m_2)}{4\pi^2}$$, where G is the gravitational constant, $m_1$ is the mass of the sun, and $m_2$ is the mass of the planet, respectively.  This generalization implies that $a^3/p^2$ will be different for each of the planets, since they all have different masses. The largest differences manifest for the planets with the highest masses, but even for Jupiter, these differences are along the third significant digit, consistent with our current best estimates here:

These values correspond well to my calculated predictions bases on Newton's generalization of Kepler's third law:

To repeat my short answer above, the gravitation of Jupiter and all the other planets makes Kepler’s third law a bit less accurate than it would be if their gravity were zero.  The gravitational interaction between the planets, however, doesn’t have much effect on either the period or semi-major axis of any of the planets, especially in the short term.


Are any of the zodiac stars about to go supernova anytime soon? I have read that Betelgeuse, known as Ardra in Hindu astrology, could go supernova. Are any of the zodiac stars or the stars of lunar mansions about to become a black hole or supernova?

I don't really know what you mean by "Zodiac stars". I suppose you mean the stars of the 12 traditional Zodiac constellations.
Alpha Scorpii, known as Antares, or Jyeshtha in Hindu astrology, is a red supergiant. It is in a similar stage of its lifecycle to Betelgeuse. It is expected to explode in a supernova in the next 10000-100000 years (the timing is not very well understood)
Other potential supernova progenitors are more distant and fainter, such as HD 168625 in Saggitarius, which is not visible with the naked eye. There is also the recurrant nova U Scorpii, a candidate for a type Ia supernova (which occurs when a white dwarf accretes a critical amount of matter) Again it is far to faint, even during an outburst, to be a naked eye star (but it would certainly be visible if it exploded in a supernova).


What is the Neupert Effect? This effect is frequently discussed in the context of solar time series. In particular, it a relationship between soft x-ray (thermal emission) and hard x-ray (nonthermal emission) from the Sun.

The Neupert Effect describes the commonly observed phenomenon that hard X-rays occur predominantly during the rise phase of soft X-rays during a solar flare, as illustrated by the graph below:

In the graph, the first peak is that of the hard X-rays, which you can see occurs during the rise of the other line which represents the soft X-rays.

Source:

The Neupert Effect - Solar Physics at Montana State University


Observing cepheid variables in other galaxies If individual stars cannot be seen in other galaxies (excluding satellite galaxies of the Milky Way) then how can light from cepheids other galaxies be observed?

It is correct that we can't observation stars in other galaxies with the unaided eye. But with telescopes, we can see much more, including individual stars in other galaxies. We certainly have plenty of examples including several galaxies that don't orbit the Milky Way (and some outside the Local Group!)
Indeed, it was because we can detect individual cepheids in other galaxies that we can measure the distances to galaxies. Hubble found cepheids in the Andromeda galaxy in 1924. However, if your telescope does not have the light-collecting power and resolving power it would be nigh on impossible to detect a cepheid against the background of a whole galaxy. We build bigger telescopes to get the light grasp and resolution to do such measurements.
(converted from comments)
We can resolve individual cepheids  (we can't resolve them as a "disc", but we can't resolve many stars even in our own galaxy as a disc) We can resolve them as points of light, distinct from its neighbours, and we can measure the brightness of individual stars.


Would an exoplanet without axial tilt have no seasons? Would an Earth analog exoplanet which rotates around a solar analog, but without axial tilt, have no seasons?

Would it be similar to how Los Angeles does not have very noticeable seasons compared to other parts of the Earth?

There are several factors that cause seasons.  In approximate order from least to most speculative:

Axial Tilt:  This is by far the significant factor in determining what the season is on Earth.  As you may know, axial tilt affects the seasons because the light is "spread out" when it falls on an angled surface.  Note that tilt causes different seasons in different hemispheres.  An untilted planet could still have seasons for any of these other reasons.
Eccentricity:  The Earth's orbit is very close to circular, so eccentricity has next to no effect on causing its seasons (and in fact, IIRC, currently Earth's eccentricity's tiny effect is actually opposing its seasons).  An exoplanet with a high eccentricity could easily have planet-wide seasons due to varying distance from its sun.
Star Systems:  As a commenter mentioned, another factor might be how many stars a system has.  If such a solar system is stable (which, surprisingly, it might be, although I can't find the article at the moment), you could get seasons depending on where the stars are in their mutual orbits.  A binary star system, for example, would have periods of winter when the stars align with the planets (so one blocks the other) and periods of summer when the stars are side-by-side.
Star Variation:  (From a comment): believe it or not, the sun actually has "seasons" (here) that are on the order of a decade or so.  These don't have much effect on Earth's climate since on the sun they mainly cause magnetic disturbances, but on an exoplanet the star's magnetic disturbances could be relatively larger, which would lead to significant temperature/brightness variations that would translate into seasons.
Internal Energy:  Planets could have varying/cyclic sources of energy within them, just like solar masses.  In our own solar system, Jupiter's seasons are partly due to this.  On a terrestrial planet, variations might come from, for example, volcanic activity.  Negative feedback loops of this sort will exhibit oscillations.  These sorts of changes might not count as "seasons" for you, though, as they occur over larger timescales.
Stored Energy: Earth's seasons are complicated by the oceans, which greatly affect the intensity of the seasons, especially in the Americas.  In particular, there are various cycles that work like subseasons underlying the main ones.  The Pacific Decadal Oscillation, for example, affects seasons years apart.  I don't see why an exoplanet couldn't have similar cycles that, unlike in the case of Earth, are the main factor in determining seasons.  This could be stored gravitational energy too; for example in a tidally-locked exoplanet/moon system.

If you mean a true Earth-like analogue (in the sense of being exactly-like-Earth-in-every-way-except-the-axial-tilt), then no--it would not have noticeable seasons, since none of the above factors really apply.


Is it conceivable that the presence of chlorophyll (or similar) could be distinguishable from other chemicals on an exoplanet? If an Earth-like exoplanet had a sufficiently extensive plant biosphere (i.e large forests, aquatic algal mats etc), is it at all conceivable that scientists could spectroscopically detect the presence of biochemicals, such as chlorophyll, anthocyanins etc?

If possible, how would the spectra of these biochemicals be distinguished from other chemicals?

Take a look at this article: http://www.nature.com/nature/journal/v513/n7519/full/nature13785.html

Every more or less stable compound has its identifying absorption lines, based, well, on what the compound is made of. The question is, what is the minimum amount for a compound in an given atmosphere at a given distance with a certain sun to be still detectable. In deed, this also depends on how "empty" the light's way to us is. But guessing, you want to know, if we propably could detect an exo-planet with vegetation on it - yes, this is possible under circumstances. Most of all, the area covered by the possible organisms plays an important role.


A vertical stick's shadow at solar noon should be straight north/south, right? At true solar noon, the subsolar point should be directly north or south of you, and therefore a vertical stick's shadow should point directly north or south, right?

The Wikipedia page for Sun Path is confusing me.  It claims that at the equinoxes, a vertical stick's shadow will point 23.5° off of true north or south.  It says that you only get straight north/south shadows at solar noon at the solstices.

  On the equator, the sun will be straight overhead and a vertical stick will cast no shadow at solar noon on the equinoxes. On the vernal equinox, north of the subsolar point (on the equator) the vertical stick's shadow will point a little westwards of true north (NNW) reading 336.5° from true north and little eastwards of true south (SSE) reading 156.5° from true north. On the autumnal equinox, north of the subsolar point (on the equator), the shadow will point a little eastwards of true north (NNE) reading 23.5° from true north (and south of the subsolar, the shadow will point a little westward of true south (SSW) reading 203.5° from true north).

  --https://en.wikipedia.org/wiki/Sun_path#Shadow_of_a_vertical_stick_at_solar_noon

Why would this be true?  My understanding is that true solar noon is defined by when the Sun is directly north or south, and the subsolar point is on your longitude.  Mean solar noon is a different story, thanks to the equation of time.

Your understanding is correct, the shadow of a vertical stick on a level plane always points north or south at solar noon, except in the Tropics where sometimes it will not have any shadow due to the Sun being at the zenith, and at the Poles, where solar noon is undefined. The sun can also be below the horizon at solar noon: places inside a polar circle can experience polar night, when the Sun does not rise for at least an entire day. On those days, even at solar noon, the Sun remains below the horizon.

This erroneous passage seems to originate from edits on 13 April 2012. The original text made more sense:

  On the Equator, the sun will be straight overhead and a vertical stick
  will cast no shadow at noon (solar time) on March 21 and September 23,
  the equinox. 23.5 degrees north of the equator on the Tropic of
  Cancer, a vertical stick will cast no shadow on June 21, the summer
  solstice for the northern hemisphere. The rest of the year, the noon
  shadow will point to the North pole. 23.5 degrees south of the equator
  on the Tropic of Capricorn, a vertical stick will cast no shadow on
  December 21, the summer solstice for the southern hemisphere, and the
  rest of the year its noon shadow will point to the South pole. North
  of the Tropic of Cancer, the noon shadow will always point north, and
  conversely, south of the Tropic of Capricorn, the noon shadow will
  always point south.


Without using absolute magnitudes or isochrones, how might we tell a star's age and evolutionary status? Usual methods of estimating stellar ages involve isochrone approximations. It can also help to estimate a star's radius by correlating its absolute magnitude with effective temperature and apparent magnitude. In the absence of these measurements or observations of tell-tale variability, how might you guess a star's age and evolutionary status?

Given a single, high-resolution $(R\gtrsim 50000)$ spectrum as your only data point, how easy is it to accurately infer the age and evolutionary status of a star? For example, how would the spectrum differ between a red dwarf and red giant, both with of $T_{\mathrm{eff}} = 4000~\mathrm{K}$? Or between two red dwarfs of ages $2~\mathrm{Gyr}$ and $8~\mathrm{Gyr}$?

A good answer could describe how surface gravity $(\log g)$ affects spectral lines (and how this relates to stellar mass and radius), what elements we might observe more strongly at different stages of evolution, and some observational results of gyrochronology.

The spectra of a red giant and a red dwarf are completely different, so there isn't really too much to say about this and distinguishing giants and dwarfs is simple. For example, alkali lines are almost non-existent in red giants, but strong in red dwarfs. The theory as to why this happens is to do with the surface gravity and pressure broadening; it is the stuff of a standard graduate/undergraduate course on stellar atmospheres, not an SE answer.

The fact is that a R=50,000 spectrum with decent signal to noise ratio will quite easily
give you the temperature (to 100K), surface gravity (to 0.1 dex) and metallicity (to 0.05 dex), plus a host of other elemental abundances (including Li) to precisions of about 0.1 dex.

What can you do with this:

You can plot the star in the log g vs Teff plane and compare it with theoretical isochrones appropriate for the star's metallicity. This is the best way to estimate the age of a solar-type (or more massive) star, even if you don't have a distance and is the most-used method. How well this works and how unambiguously depends on the star's evolutionary stage. For stars like the Sun, you get an age precision of maybe 2 Gyr. For lower mass stars, well they hardly move whilst on the main sequence in 10Gyr, so you can't estimate the age like this unless you know the object is a pre-main sequence star (see below).

You can look at the Li abundance. Li abundance falls with age for solar-mass stars and below. This would work quite well for sun-like stars from ages of 0.3-2Gyr and for K-type stars from 0.1-0.5 Gyr and for M-dwarfs between 0.02-0.1 Gyr - i.e. in the range from where Li  starts to be depleted in the photosphere until the age where it is all gone. Typical precision might be a factor of two. A high Li abundance in K and M dwarfs usually indicates a pre main sequence status.

Gyrochronology is not much help - that requires a rotation period. However you can use the relationship between rotation rate (measured in your spectrum as projected rotation velocity) and age. Again, the applicability varies with mass, but in the opposite way to Li. M-dwarfs maintain fast rotation for longer than G-dwarfs. Of course you have the problem of uncertain inclination angle.

That brings us to activity-age relations. You can measure the levels of chromospheric magnetic activity in the spectrum. Then combine this with empirical relationships between activity and age (e.g. Mamajek & Hillenbrand 2008). This can give you the age to a factor of two for stars older than a few hundred Myr. Its poorly calibrated for stars less massive than the Sun though. But in general a more active M-dwarf is likely to be younger than a less active M dwarf. It should certainly distinguish between a 2Gyr and 8Gyr M dwarf.

If you measure the line of sight velocity from your spectrum, this can give you at least a probabilistic idea of what stellar population the star belongs to. Higher velocities would tend to indicate an older star. This would work better if you had the proper motion (and preferably the distance too, roll on the Gaia results).

Similarly, in a probabilistic sense, low metallicity stars are older than high metallicity stars. If you were talking about stars as old as 8Gyr, these would be quite likely to have low metallicity.

In summary. If you are talking about G-dwarfs you can ages to precisions of about 20% using log g and Teff from the spectrum. For M dwarfs, unless you are fortunate enough to be looking at a young PMS object with Li, then your precision is going to be a few Gyr at best for an individual object, though combining probabilistic estimates from activity, metallicity and kinematics simultaneously might narrow this a bit.

As an add-on I'll also mention radio-isotope dating. If you can measure the abundances of isotopes of U and Th with long half lives and then make some guess at their initial abundances using other r-process elements as a guide then you get an age estimate - "nucleocosmochronology". Currently, these are very inaccurate - factors of 2 differences for the same star depending on what methods you adopt.

Read Soderblom (2013); Jeffries (2014).

EDIT: Since I wrote this answer, there is at least one more promising method that has emerged. It turns out that the abundance of certain s-process elements (e.g. barium, yttrium) are enriched quite slowly during the lifetime of the Galaxy (by the winds of dying asymptotic giant branch stars), and more slowly than enrichment with iron and much more slowly than alpha elements like Mg and Si. Thus a measurement of the relative fractions of these elements, like [Y/Mg] can give the age to precisions of a billion years or so (e.g. Tucci Maia et al. 2016; Jofre et al. 2020). This method is probably the best for solar-type stars older than a Gyr, but remains unexplored/uncalibrated for lower mass stars.


Plausible reason for the Hot Neptune gap This NASA article talks about how there's a dearth of hot neptune exoplanets that have been discovered so far.

https://www.nasa.gov/feature/goddard/2018/in-search-of-missing-worlds-hubble-finds-a-fast-evaporating-exoplanet

The article makes no attempt at postulating why this gap may exist. One possible explanation I have is that the silicate material of a terrestrial planet is difficult to vapourise and the surface gravity of a hot jupiter is too great to drive off hydrogen to any great extent, whereas a Hot Neptune lacks both the solid surface and the gravity to maintain its size.

An explanation could be provided by irradiation and evaporation of planetary envelopes.

The introduction in the article by Bourrier et al. 2018, which is the source of the NASA press release, explains that as exoplanets migrate inwards, they are increasingly challenged to hang on to their envelopes due to irradiation and heating (insolation) by their star.

The Neptune-sized planets are less capable of retaining their envelopes because of lower densities and lower "surface" gravities. The idea is that many of them get stripped of their outer envelopes and end up as "Cthonian" planets that are of "super-Earth" size.

A hypothesis is that many hot Jupiters have enough mass and density to maintain a large size for billions of years and hence it is only the hot Neptunes that are depleted.

The recent work on GJ 3470b (cited in the NASA press release) supports this hypothesis by finding a very large atmospheric loss rate from a Neptune-sized object at the edge of the "hot Neptune desert".


Do triple or more times reflecting telescopes exist? Refractors only use the length of the telescope once, reflectors twice, catadioptric telescopes like those of the Schmidt-Cassegrain design three times. Have telescopes been built that reflect the incoming light at least once more?

Cool question!
If you allow for flat mirrors, I've seen five myself. Answer(s) to How did I flip some mirrors around in the dark at 3 AM and change the focal length of a 24 inch Boller and Chivens? show four mirrors and there was one more on the floor below to make the beam horizontal. If you count the aluminized reflective grating as a mirror it was six.
 ALL IMAGES: click for larger size
For more info, see Nasmyth and coudé focus

For more examples of three mirrors built into the telescope itself (as opposed to an eyepiece diagonal for viewing, see answer(s) to

How does this telescope work, and what is it for?
What telescope is this, and how does it work?

For more little mirrors that send light to different instruments, see answers to

What is a pupil slicer, and how does it work with anamorphic optics in VLT's ESPRESSO Echelle spectrograph?
Where exactly is the modification that first corrected the spherical aberration in Hubble's primary mirror?

But three or more curved mirrors is interesting
One example is the Hubble Space Telescope using the Wide Field and Planetary Camera 2 or WFPC2.
Basically it is another whole telescope inside the Hubble, re-imaging the primary focal plane with a second purely reflective optical Cassegrain system to avoid chromatic aberration. (There would be no way to correct a fused silica lens system from 115 to 1050 nm.)
From what-when-how Hubble: The Big Telescope

Figure 3.36 WFPC2 optics. Light enters the optical train from the main telescope at left.


If a solar system were surrounded by a cloud of debris, is it possible for a planet's orbit to intersect it? I'm doing research for a story that I'm writing and I'd like some physics advice on whether a hypothetical system of celestial bodies is possible.

Suppose the following:

A system of planets orbiting a star, similar in size to our own.
A cloud of debris (Asteroids, technological refuse, and dust) surrounds the entire system, similar to the description of a Dyson swarm.
The outermost planet orbiting on a path that intersects the cloud during a small portion of it's year, but is within the cloud for the majority of it's year.

Is this configuration possible within our understanding of physics, or would gravity cause the objects of the cloud and the outermost planet to 'snap' to an orbit that does not intersect? Would the planet 'capture' objects within the swarm in it's own orbit?

My story takes place on the inside of a Dyson swarm or bubble, where interplanetary travel has developed to a reasonable degree, but cannot escape the swarm. An area of intrigue might be the outermost planet. If it can be landed upon while it's orbit is within the swarm, people could 'ride' the planet through to the other side if they stayed there for a number of years or went into stasis there.

Any thoughts or opinions are greatly appreciated.

The solar system IS surrounded by a cloud of debris,left over from its formation. This debris is called the Oort Cloud,& consists of thousands of comet-like bodies,sometimes described as dirty snowballs. None of the known major planets passes through them,but a few years ago some astronomers seemed certain that  they had detected signs of a 9th major planet much further out than the others. They predicted it was only a matter of time before this planet would be discovered,but so far it hasn't been. If there is indeed such a planet & part of its orbit passes through the Oort Cloud,it would probably sweep a lane through it eventually. The planet would not be a nice place to live,as the likelihood is that it has a temperature close to that of liquid nitrogen,& is frequently struck by comets & other debris.


What is wrong with my calculations of Venus' orbital period? I am trying to use Kepler's second law to find the duration of Venus's orbit. I am assuming circular orbits (using Earth and Venus, so low eccentricity). Here is my process:

Assuming that the radius of Earth's orbit is 150 million km, then the swept area in one day is $\frac{1}{365.25}\times\pi\times 150^2 \approx 194 \text{ million km}^2$.

Venus must sweep the same area in the same time. Assuming a orbital radius of 108 million km for Venus, and using $A = \frac{\theta}{360}\pi r^2$, we can find the central angle for the swept sector, that is, the angle traveled in one Earth day:

$194 = \frac{\theta}{360}\pi \times108^2 \implies \theta = 1.90 ^{\circ}$ per Earth day.

Hence the orbital period should be $\frac{360}{1.90}\approx 189$ Earth days.

Of course, the orbital period of Venus is $224.7$ Earth days. The difference between 189 and 224.7 appears to be well beyond the error introduced by my assumption of circular orbits.

What am I doing wrong?

I know this is perhaps a circuitous way of doing this calculation. My goal is to write a mathematics exercise that uses the area of sectors in a meaningful way.

Kepler's laws state that a planet sweeps equal areas in equal times as it moves in its elliptical orbit. It doesn't state that different planets will sweep the same area.

The "equal areas" law can be derived from the "conservation of angular momentum". In fact dA/dt = L/(2m)  (where A is the area, L is the angular momentum and m is the (reduced) mass).

Different planets will sweep out different areas. To calculate the period you used Kepler's third Law: $T^2 = k a^3$ (T= orbital period,a = semi-major axis). If, for convenience you take a in AU and T in Earth Years, then the constant $k=1$.

For Venus, a = 0.72. so $T=\sqrt{0.72^3}=0.61$ or about 223 days.

Hyperphysics has a section on Kepler's Laws


How a star's rotation affects a star in the main sequence? I searched and have found that the questions mainly focus on neutron stars, white dwarves and black holes. This was not what I looked for.

Basically, the bigger the star's mass is, the more intense is its fusion reaction and the shorter is its lifespan in the main sequence. Now, imagine that a star spins much faster. There will be less surface gravity at the equator than at the poles. How does a fast rotation affect the star's fusion reaction?

Will the pressure of the star's mass on its core be lower at the equator and therefore reduce nuclear fusion rate?
Will there be a deeper convection because of the Coriolis effect?
What shall we notice in term of longevity in the main sequence, luminosity and emission spectrum of a fast rotating star compared to a slow rotating one of the same initial mass?

This is a well-studied problem. The effect of rotation on the structure of a low-ish mass star (like the Sun) is summarised by Eggenberger (2013).

Such stars are never observed to rotate so fast that the rotation plays any significant role in their hydrostatic equilibrium, however rotation does play a role by causing additional mixing in the star.

This is important for two reasons: (i) it inhibits the gradual diffusion of helium towards the core, this slightly lowers the opacity in the core and raises it in the envelope (compared with a non-rotating star). This results in a slightly higher luminosity and a slightly higher surface temperature. (ii) More importantly, the additional mixing brings additional hydrogen into the core and this increases the lifetime on the main sequence.

The effects on stars of a solar mass are however unlikely to be very significant in practice, because these stars efficiently lose angular momentum via a magnetised wind during their early lives and the effects of rotation are unlikely to be significant even at several times the solar rotation rate.

The effects on more massive stars can be more severe. These can spin at an appreciable fraction of their break-up rate for a large part of their lives and do not lose angular momentum as efficiently as lower mass stars (they do not have magnetised winds). The effects are described in the canonical paper of Meynet & Maeder (2000); they are more pronounced than for lower mass stars and more complicated because of the radiative envelope and uncertainties in the rotation-dependence of the significant mass-loss.

The hydrostatic effects of rotation are expected to be important early in the main sequence and contribute to a slightly lower surface temperature. At later times the dominant effects are caused by changes in the mixing and diffusion near the core and in the envelope as for the lower mass stars, resulting in higher luminosities and hotter temperatures. The main sequence lifetimes can be extended by 30% due to the additional mixing of fresh hydrogen fuel into the core.


What are the dimensions of LIGO Detector evacuated chambers? The LIGO complex is very large, but I assume that there is a very small chamber that actually does the science since they are measuring very small sheers. How big around is the actual evacuated chamber?

The arms are $4\,\mathrm{km}\,\times\, 1.2\,\mathrm{m}$:

From the LIGO webpage:

  The 1.2 m diameter beam tubes were created in 19-20 m-long segments, rolled into a tube with a continuous spiral weld. While a mathematically perfect cylinder will not collapse under pressure, any small imperfection in a real tube would allow it to buckle (a crushed vacuum tube would be catastrophic). To prevent collapse, LIGO's tubes are supported with stiffener rings that provide a significant layer of resistance to buckling under the extreme pressure of the atmosphere. The tubes must withstand these stresses for at least 20 years.


Is this picture of a digital camera being oversaturated by bright light source i.e. the Sun? Is this picture of a digital camera being oversaturated by bright light source i.e. the Sun?

http://farm4.staticflickr.com/3110/3147227521_396e16bcfc.jpg

It's an artifact from the digital sensor and/or the image processing in some cameras.

There have been a number of digital cameras with this issue. E.g. here is a Sony camera where the sun turned black with a specific setting for the picture profile. (In the example video, it's the S-log picture profile at 0:49.)

Sony called it a bug, and released a fix for it.

This artifact only affects a relatively small number of cameras, and it makes sense to think of it as a bug in the camera image processing: Yes, the sensor is oversaturated, but oversaturation is better represented as all white rather than all black.

If you have an affected camera, it's not just the sun that turns black, it's all oversaturated highlights, like this example with car headlights.


Precision of geocentric gravitational constant I am looking for an answer as to why the geocentric gravitational constant, μ, defined as the product of the gravitational constant, G, and the mass of a body (earth in this case), M, can be calculated to a higher degree of precision than either G or M alone.

From wikipedia

  ...for celestial bodies such as Earth and the Sun, the value of the product GM is known much more accurately than each factor independently. Indeed, the  limited accuracy available for G limits the accuracy of scientific
   determination of such masses in the first place.

Can someone explain this? I can't find a good answer anywhere I look.

Ignoring details such as the oblateness of the Earth, atmospheric drag, third body influences such as the Moon and the Sun, relativity, ..., the period of a satellite of negligible mass (even the International Space Station qualifies as a "satellite of negligible mass") is $T=2\pi\sqrt{\frac {a^3}{\mu_\mathrm{Earth}}}$. Neither Newton's gravitational constant nor the mass of the Earth are involved in this expression. This means that, ignoring those details, calculating $\mu_\mathrm{Earth}$ is merely a matter of calculating a satellite's rotational period and its semimajor axis.

Humanity has lots and lots of artificial satellites in orbit, and the people who model the orbits of those satellites don't ignore those details. A few of those satellites were specially designed to enable the determination of the Earth's non-spherical gravitational field (e.g., GRACE and GOCE), and a few were specially designed to enable extremely precise orbit determination (e.g., LAGEOS). Even with all of those details, the Earth's gravitational parameter is a directly inferable quantity (i.e., knowledge of G is not required). Moreover, the value is known to a very high degree of precision.

The Earth's mass? Not so much. The most precise way to "weigh the Earth" is to divide the high precision Earth's gravitational parameter by the low precision universal gravitational constant G. There's a problem here, which is the notoriously low precision of the gravitational constant when expressed in SI units.


Did we adapt to Earth or Earth to us? I often hear that Earth is a unique planet because it has life. We also see a proper balance between plants and humans and other animals. Why is life not possible on another planet? Often people say because the environment of other planets is so different from the earth and it's hard to find a planet like ours. That seems pretty satisfactory; I can not live on Mars or Saturn.
On the other hand, I find it little misleading. Why do we imagine ourselves on other planets to see if life is possible on other planets? There might be different creatures that are used to different environments. There might be creatures that are adapted to ${\rm H_2SO_4}$.
I have two questions. First is my reasoning true? And if so what makes life rare on other planets or makes the earth so unique?

Your reasoning is at least true in part.
First off.  We have only one example of life.  It is based on carbon chemistry and water solutions. As we only have one example, we don't know if fundamentally different chemistries are possible. We just don't know if the Earth is unique or special, or whether life is commonplace throughout the universe.
If you head over to https://worldbuilding.stackexchange.com/ you will find lots of people speculating of different biologies, including those with a sulphur and sulphuric acid based chemistry
However, given our understanding of chemistry, it does seem likely that most life in the universe will be carbon/water based.  Carbon is the only element that produces complex structures, and is also one of the most common elements.  Oxygen is another very common element, but it is very reactive and so will tend to react with hydrogen to form water.  Water also happens to be one of the very best solvents.  So carbon seems to be the best element to build biological compounds and water is the best solvent to carry those around.
This means that is is a reasonable hypothesis that life will be found most often on planets that have surface temperatures and pressures that allow for liquid water.  That is a pretty narrow range.
We are adapted to Earth.  We have had 4 billion years of evolution to adapt to Earth, even as Earth changed.  For example we have adapted to an oxygen-rich atmosphere.  We could have adapted to Sulphuric acid (at least to dilute sulphuric acid).  Oxygen is as toxic as acid to those bacteria that haven't evolved to tolerate it.
Whether the Earth has adapted to us is more controversial.  This is the Gaia hypothesis of James Lovelock "that living organisms interact with their inorganic surroundings on Earth to form a synergistic and self-regulating, complex system that helps to maintain and perpetuate the conditions for life on the planet." (wikipedia)  However "the Gaia hypothesis lacks unambiguous observational support and has significant theoretical difficulties" ( David Waltham. Lucky Planet).
We don't know if the Earth is unique, nor even if it is special.  Our current ability to detect planets would struggle to detect the Earth, if it orbited another star. So we don't know how planets there are that have liquid water on the surface. We also don't know how many of them harbour life in any form. We don't know if life is possible in alternate chemistries. Even if life turns out to be common, we might still be special in that we have oxygen-breathing, multicellular, complex and intelligent life.
If it turns out that life is rare... we don't know why it would be. Life started pretty soon on Earth.  If it turns out that life is actually rare we would need to explain that. But we don't have enough data to do so yet.


Computing the over-density of Lyman-alpha emitters Let's assume that I have observations of $N$ galaxies in a projected area $A$, within a redshift range $\Delta z = z_{max} - z_{min}$. What is the correct way of computing their number volume density? I guess the volume has to be derived multiplying $A$ by the differential Luminosity Distance $D_{L}$ computed at $z_{max}$ and $z_{min}$, i.e., $D_{L}(z_{max}) - D_{L}(z_{min})$, but I am not sure.
Also, what is the more appropriate way of evaluating the "significance" of any over-density.
For instance, assuming that the observed galaxies are Lyman-$\alpha$ emitters, does it exist any computation of the "standard" number density of Lyman-$\alpha$ emitters, to compare with?

Luminosity function
Your approach is in principle correct, but would give the number density $n$ of Lyman $\alpha$ emitters (LAEs) in a "luminosity volume". Usually, the comoving volume is used, since you can then more easily compare densities at different redshifts $z$. If $n$ changes with $z$, you then know that it's not just because of the expansion of the Universe, but due to some astrophysical process.
Moreover, since $n$ is a strong function of galaxy luminosity $L$, you would do this calculation in bins of luminosity. You then get the so-called luminosity function (LF), $n(L,z)$. LFs are often fitted with the completely phenomenological Schechter function, though there is evidence that the bright end of the LF is less steep than this form would suggest.
Filter shape
Note though that, observationally, your approach is only approximate: If your redshift range $\Delta z$ comes from the width of your (presumably) narrowband filter, then this method is only exact in the case of a perfect "square" filter, i.e. one that transmits 100% in some range, and 0% outside this range.
In reality, filters are described by some non-square transmission function (could be a Gaussian, but really it has to be measured, as it's not a simple functional form).
Hence, galaxies with redshifts falling in the wings of your filter has to be brighter in order to be detected, than galaxies falling in the center. In other words, your LF will be less complete the fainter the galaxies, and your inferred number density will be a function of brightness.
To account for this, you must in principle fold in the exact shape of the filter. In practice, it is not uncommon to neglect this effect though, and just use an "effective" width of the filter, typically the FWHM.
Overdensity
To compute an overdensity requires you to assume a "reference" LF. There isn't one standard LF, as this will depend on whose observations you prefer. But a good choice could be e.g. Wisotski et al. (2018) parametrizes the LAE LF based on deep MUSE exposure of the Hubble (Ultra) Deep Field.


How massive can a cold solid object be? I understand that a neutron star is dense enough that adding more matter will increase the amount of degenerate matter directly, and the limit to its size is about 1.4 solar masses.

But if an object were formed not by an explosion that crushes it down to neutron matter, is it necessarily so?

Given normal atoms of any desired species, and it is piled up carefully so it does not get hot. When it gets around one solar mass, will it necessarily collapse into a neutron star, or can it get more massive?  Will such an object eventually collapse catastrophically or can it have a degenerate matter core and a substantial thickness of less-compressed matter on top and normal matter on top of that?

What is the largest possible mass of a cold solid object?  Can it be several solar masses?

By cold I mean that it’s not a star, puffed up by ongoing consumption of energy. Whatever it's made out of is not being “used up”.

The limit to the neutron star mass is at least $2M_{\odot}$, since at least two have measured masses as large as this (e.g. Demorest et al. 2010).

The answer to your question is the same as the answer to what is the maximum mass of a neutron/quark star, since if you compress matter of any sort, this is ultimately what it will become.

The answer to this question is also unknown and depends on the uncertain equation of state (the relationship between pressure, density and composition) at ultra-high densities, but must be somewhere between the $2M_{\odot}$ I mentioned above and an upper limit of around $3.5M_{\odot}$, which is imposed by General Relativity and an equation of state where the speed of sound equals the speed of light (the hardest possible equation of state).

Whether these highly compressed objects are "solid" is also a matter of debate and research. The conventional picture of an old-ish white dwarf is that it is indeed a crystalline solid. The bulk of the interior of neutron stars of moderate mass is almost certainly a fluid of neutrons along with some protons and electrons. The neutrons may solidify at very high densities and this is thought to be one of the "harder" options for the equation of state.

To answer a comment by @zephyr - cold in this context means that the Fermi energies of degenerate fermion species are very much higher than $kT$. White dwarfs and neutron stars (older than a few seconds) are "cold", despite having interior temperatures, in the case of the latter, of a billion degrees. Making them even colder does not change the size/density of the object but does enable the interior of a white dwarf to crystallise.


Wavelength and radial velocity What is the relation between the wavelength and the radial velocity? I have a data: wavelength and intensity and I would like to plot intensity with respect to radial velocity - similarly as in the picture (source)

EDIT

I have a file of alpha H with 2 columns - wavelength (in angstrom) and intensity (normalized for 1) and I would like to plot intensity with respect to radial velocity. I use in gnuplot:

c = 299792458
lambda_0 = 6562.817*10**(-10)
plot 'file.asc' using (c*($1*10**(-10)-lambda_0)/lambda_0):2

Why I get the wrong range on the x-axis?

$$v_r = c\left( \frac{\lambda - \lambda_0}{\lambda_0}\right),$$
where $\lambda$ is the observed wavelength and $\lambda_0$ is the wavelength at rest. This gives an equivalent velocity with respect to the central wavelength of the line.

In your plot, what is shown is a line profile where the wavelength separation from $\lambda_0$ for that line (or from the measured centre of the line, I can't tell which) has been converted into a velocity.

Note that it makes little sense to plot more than a single line profile in this way.


How to derive the depth in my image (limit Magnitude )? I saw that some survey said that their depth is 27.5 magnitudes.
May I ask that if I have a dataset with magnitude and their errors, for example I have J band j_m, j_cmsig, j_msigcom, j_snr from Sextractor.
How can I derive the 3 sigma magnitude depth in this fields to compare with the other survey.

It's the magnitude where the average signal-to-noise ratio (I assume that's what j_snr represents) is 3.