To describe the position of an orbiting body you need 6 numbers.  There are different ways to do this:
You can give the position $(x,y,z)$ and velocity $(\dot x, \dot y, \dot z)$.  At a given time $t_0$, and then use Newtons laws to work out the position of the planet at any time in the future.
You can give the orbital parameters:

Eccentricity (the shape of the ellipse)
semi-major axis (the *size of the ellipse)
(inclination, Longitude of ascending node, argument of periapsis) The 3d position of the ellipse.  These depend on the particular choice of reference frame.  For exoplanets it is common to take the reference plane to be the the plane through the star and perpendicular to the line of sight from Earth. This means that many exoplanets have an inclination of about 90 degrees.
Mean anomaly at epoch (The position of the planet on the ellipse at time $t_0$)

And then use Keplers laws to work out the position of the planet at any time in the future.
So we can track them theoretically.
However determining these six values from observations is not easy, and in many cases we don't know all six at all well.  If we can detect a transit we know that the inclination is close to 90, and we might be able to combind transit observations with radial motion observations to get some limits on the other values.  Nevertheless, these six values can't be simply measured from the data.
For most exoplanets they can't be directly imaged (they are too close to the glare of the star, and to dim to be seen) There are a few exoplanets that can be tracked They are usually very large, warm planets that are far from their star.
cf. Transformation of Orbit Elements, State and Coordinates of Satellites in Two-Body Motion


First, Terminology:  Age is how old something is.  Relative age is how old something is when compared to another (older vs younger).  Absolute age is putting a number on that age (I am XX years old).  Absolute model age is an absolute age that is based on a model, but where the age has not been measured directly (based on this cat's tooth, its absolute model age is roughly 3 years old).
Second, the Lunar Crater Chronology, Relative:  A generally known quantity (with some caveats that I'm not going to get into in this answer) is that impactors have been striking surfaces in the solar system since its formation.  So, if a surface has more craters than another surface, on the same planetary body, then it can be assumed to be older than the surface with fewer craters.  So by measuring the spatial density of impact craters on multiple surfaces, you can tell a relative age relationship of those surfaces.
Third, the Lunar Crater Chronology, Absolute Model:  If we are able to tie a specific age, like 3.7 billion years, to a surface that has a specific crater spatial density, like 0.002 craters ≥1 km per square kilometer, then we have an anchor point.  And, to within a factor of ~2 or so, we can say that any surface on that planetary body with 0.002 craters ≥1 km per square kilometer is going to be 3.7 billion years old.  If we are able to anchor several surfaces, with different crater spatial densities, with different radiometric ages, then we can plot them on a graph (crater spatial density on the vertical, age on the horizontal) and fit some function to them.   We can then use that function, or model - which has been calibrated based on absolute ages - to give us the absolute model ages of any lunar surface by measuring their crater spatial densities.  We've done this with data from the Apollo and Luna sample returns.  As the other answerer posted, I did a recent review and re-kajiggering of that function.
Fourth, Dating Individual Features:  That lunar crater chronology can be used to date individual features even if they don't have 1 km craters on them.  We measure smaller craters, and then use another model of how many small craters there are relative to large craters, to figure out how many there WOULD be on that surface if 1 km craters existed on it.  Or, if it's a big or older surface, we have those 1 km craters there and can get the age via that lunar crater chronology function.
Fifth, Rock Abundance:  The DIVINER instrument on the Lunar Reconnaissance Orbiter takes temperature measurements at different times of day and from those can model the rock abundance on the surface.  Basically, a boulder holds onto its heat longer than sand, so the boulder is going to be warmer at midnight than the not-boulders.  What the DIVINER team found, maybe a decade ago, is that young (<1 billion years old) lunar craters have high thermal inertias - large rock abundances - surrounding them.  Rebecca Ghent (et al.) used a few craters that had absolute ages (from Apollo) and absolute model ages (from the chronology curve based on smaller superposed craters) and demonstrated there was a pretty good linear relationship, meaning that for craters <1 billion years old, we now have another method to date those individual features via the DIVINER rock abundance measurements, which, honestly, is easier than counting craters (or at least takes less time).
Sixth, Spectral Slope:  A recent paper demonstrated that young craters also show rays in the far-ultraviolet light, and that removal of those far-UV rays follows a fairly parametrizable relationship, so they were able to get independent ages of young, large craters, too.  Optical maturity studies is not new, but this is the first remember seeing it used in UV for this exact thing.


The first Google-hits suggest that Segue 2 is the smallest spiral galaxy, but that is actually incorrect since it is the smallest spherical galaxy yet discovered. It has a radius of roughly 34 parsec or 110 light years. The original publication on that topic dates back to 2013, but it was again in the news in April 2020.
Then, there are lecture notes on a course Astronomy 626 Galaxies which shows the following diagram:

Also, the lecture notes claim:

Typical values for the scale length [of spiral glaxies] are: $1 {\rm kpc} < h_R < 10 {\rm kpc}$

I agree that this does not answer your question fully, but I got the impression that the lower bound should be around $1 {\rm kpc}$.
References

https://www.nasa.gov/press/2014/september/hubble-helps-find-smallest-known-galaxy-containing-a-supermassive-black-hole
https://www.treehugger.com/just-how-big-is-the-smallest-galaxy-in-the-universe-4862842
http://www.sci-news.com/astronomy/article01147-segue-2-dwarf-galaxy.html
Evan N. Kirby et. al: Segue 2: The least massive galaxy
https://physics.stackexchange.com/questions/148418/


Why planets and stars are always round/oval in size?

The gravitational is radial only. That means if you tried a different shape then you would have regions, like hills, where a change in the angle would experience a much less attraction towards the center. This region would, over time, be attracted to a different angle with a greater force of attraction. Or spherical. This results in a spherical-like shape.


Do objects look larger the further away they are, beyond z=1?

Yes this is correct for the standard model. I believe the English language explanation would be that the angular size of a galaxy does not depend on its current distance to us, but its distance when the light we currently see from it was emitted (here using "distance" synonymous with "proper distance"). Though I also believe there are other complicating factors such as the spatial geometry of the Universe.

Look at the first diagram in Figure 1 in this paper, notice how the size of the past light cone in terms of proper distance firstly increases as we go back in time, before reaching a maximum and decreasing to zero at t = 0.


I need help getting mars to look like a circle

I have made sure that the laser is straight by 1. spinning it in the
  eyepiece holder and the laser stayed in the same spot 2. zooming in
  and out and again the laser stayed in the same place. the laser seems
  to be centered so I put it in the middle of the doughnut on the mirror
  and the laser bounces back into the middle of the colimating lens

Very good. You're doing all the right things. Kudos for taking care of your telescope.

Now, about Mars looking like a fuzzball.

The angular size of Mars right now is 6.5 arcsec. The maximum theoretical resolving power of your scope is 0.4 arcsec. In ideal conditions (assuming everything was perfect, which it never is), you'd get something like 16 "pixels" diameter for the image of Mars in your scope. That, I'm sure you realize, is a very low resolution. No wonder it looks like a fuzzball.

Mars is a tricky target. The best times to observe it is around oppositions, which happen once every two years (plus a few months). Mars is nice and big a few weeks before and after opposition, and otherwise it's pretty small. The last opposition was in July 2018, when it measured 24 arcsec. The next one is in October 2020, and it will measure 22 arcsec.

http://www.nakedeyeplanets.com/mars-oppositions.htm

Stick to a few weeks before and after opposition. In the rest of the time, Mars is just too small.

Another source of problems is seeing - or atmospheric turbulence. When seeing is bad, all the high resolution targets (like all planets and the Moon) look bad. Nothing you can do about it. When seeing is good, your scope can work closer to its theoretical resolving power.

https://en.wikipedia.org/wiki/Astronomical_seeing

To predict seeing, go on the Clear Dark Sky site, choose a location closest to where you live, and check the forecast they give you for that location.

https://www.cleardarksky.com/csk/

The row called Seeing is a seeing forecast. When it's dark blue, seeing is good. When it's white or light blue, seeing is bad. No point in looking at Mars when seeing is bad.

Finally, make sure collimation is actually good. You seem to be doing all the good stuff with the laser collimator. That's great. Now, when seeing is good, point the scope at the North Star, defocus a bit (move the focuser in or out of perfect focus a tiny bit) and look at the star image. It looks like a shooting target, a bunch of concentric rings. If those rings are perfectly round and concentric, your collimation is probably good.

http://www.astrophoto.fr/collim.html

This is called a star test, and it's the most accurate test of collimation. It's also the most difficult. So use your laser collimator on a daily basis, which is much easier, and maybe do the star test once or twice, just to keep your laser honest. Once you're convinced your laser technique works well, there's no need for the star test.

Oh, one more thing. Make sure your scope is at thermal equilibrium with the environment. If you're storing your scope indoors, it's probably way hotter than the cold night air. That will create internal turbulence within the scope. It's basically like bad seeing, except it's all inside your scope.

https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/

https://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/

No need to install a fan on the primary (although that would definitely work if you're willing to make the extra effort). Just take the scope outside 1 hour before you begin using it. It will "breathe out" all the heat and it will be close to equilibrium.

Good luck and clear sky!


Globular cluster star density as a function of distance from the center

The usual thing is a King model.

There are indeed free parameters. These are the central density, the "core radius" and a tidal truncation radius.

The background and rationale for these models is given in the link. They provide a pretty good representation of the surface density of globular clusters (or indeed open clusters). They require a numerical scheme to "deproject" from the plane of the sky to 3D.

If you find the nitty-gritty of Abel integrals too tricky in the deprojection, then you could always approximate with a Plummer model. This is analytically and computationally easier to deal with, but lacks a bit of physical realism. Central density and a characteristic radius are free parameters here.


Is there a name for this type of orbit of a natural satellite?

No.  Such an orbit would be co-incidental and dynamically uninteresting.

For a planet of given mass and rotation period, there is a class of orbits, whose period is equal to the rotation period of the planet. If the orbit is prograde this gives an interesting effect: the satellite would remain in the same position in the sky (or if it had an elliptical orbit, it would return to the same position each day)

If the orbit is retrograde, the orbit would not have any particularly interesting features. While it is possible for tidal locking to produce a synchronous orbit, there is no mechanism to stabilise a retrograde orbit like this, and tidal effects will slowly disrupt it.


The constellations are artifacts of visual pattern recognition, many dating back to antiquity.
As the stars are distributed unevenly, naturally some constellations are larger than others.
Ptolemy catalogued 48 of them around 150 CE.
The other 40 were added between 1596 and 1763, either to cover the far southern sky or to fill gaps between the older figures.
Ian Ridpath gives a comprehensive history in Star Tales.

The modern constellation boundaries were standardized in the 1920s by an IAU committee led by Eugène Delporte.
Their goals were to assign each square degree of sky unambiguously to one constellation, to be consistent with scientific literature already published, and to preserve the traditional figures where possible.
Equal size wasn't on the agenda.
The boundaries align with the equatorial grid as of 1875:

The zodiac of 12 equal parts was invented by the Babylonians around 400 BCE and later refined by Ptolemy.
Each sign corresponds to a box 30 degrees wide and about 20 degrees high.
It's important to distinguish between a constellation (a visual group of bright stars) and a sign (a uniform box along the ecliptic).
For example, Virgo the sign spans ecliptic longitudes 150$^\circ$ to 180$^\circ$, whereas Virgo the constellation is somewhat wider and farther east.
In Ptolemy's time the signs and constellations were not far apart:

Ptolemy defined the tropical zodiac in reference to the equinoxes and solstices; for example, the boundary between the signs of Gemini and Cancer lies at the June solstice, wherever that may be among the stars.
Consequently, precession makes the signs drift westward about 1.4$^\circ$ per century relative to the constellations.
In the present day, the zodiac is roughly one sign ahead of its constellations:


What is a possible software for simulating binary star systems?

MESA

It is designed specifically for stellar evolution. There are many modules that you can play around with, including binary interaction.


Which database of celestial objects do stargazing apps use?

The Hipparcos is probably all that's really needed.  A lot of them add in portions of the Tycho catalog as well.  But even the Hipparcos catalog has more stars than what most armatures will need, and it's generally impractical memory wise to use the whole thing.  To really go crazy, a torrent for the NOMAD catalog is available.  All of those list only individual stars.  For things like galaxies and nebula the NCG catalog should do, but there are also the Index Catalogue, and Hershel Catalogs.  The most popular catalog would be the Messier, which has no official distribution sources (that I know of), but databases are available if you search this one for example.

If you don't care about the original sources, Stellarium has their star catalogs on their wiki.


What does the surface of Mercury look like?

The MESSENGER probe was able to take many true-color pictures of Mercury. A full list can be found on JPL's Photojournal. It is clear that Mercury is light grey in color.

(source: nasa.gov)

(source: nasa.gov)

(source: nasa.gov)
In terms of the actual surface, Mercury is very similar to the Moon. It's surface is speckled with craters, with some smaller craters inside them, as can be seen in some of the images above. The smoothness varies - note the inside of the larger crater in the first picture. It's quite smooth, save the smaller crater inside it.
See also this pdf.


High quality, detailed images of planets on web

The best NASA images of planets are not from the Hubble Space Telescope; they are from the spacecraft that flew by or went into orbit around them. For Uranus and Neptune, this means Voyager 2; for Pluto, this means New Horizons. For Jupiter, this means primarily Voyagers 1 and 2, and especially Galileo and Juno; for Saturn, Voyagers 1 and 2 and (especially) Cassini.

If you google for "[planet-name] planet gallery", you should get some pretty good images that way; also try adding the name of a relevant spacecraft (if you're stuck, check the Wikipedia page for each planet to learn which spacecraft visited them). A few NASA web pages:

https://photojournal.jpl.nasa.gov/mission/Viking
https://solarsystem.nasa.gov/planets/saturn/galleries
https://solarsystem.nasa.gov/planets/neptune/galleries
https://solarsystem.nasa.gov/planets/dwarf-planets/pluto/galleries

and so forth. (Note that Uranus, even when seen by Voyager 2, is pretty featureless and boring, at least at visible wavelengths.)

You might also try searching at the Astronomy Picture of the Day site. (Not all of those are necessarily public domain, thoug.)


Why is JD 2451545.0 = January 1, 2000, Noon instead of JD 2451558.0?

This Wikipedia article states (correctly) that

  "The Julian Day Number (JDN) is the integer assigned to a whole solar
  day in the Julian day count starting from noon Greenwich Mean Time,
  with Julian day number 0 assigned to the day starting at noon on
  January 1, 4713 BC, proleptic Julian calendar (November 24, 4714 BC,
  in the proleptic Gregorian calendar)"

The Julian year consists of 365.25 days while a Gregorian year consists of 365.2425 days.

Since the introduction of the Gregorian calendar, the difference between Gregorian and Julian calendar dates has increased by three days every four centuries (all date ranges are inclusive).  In addition, 10 days were added to the Julian calendar to obtain the Gregorian calendar at the time of the calendar reform. The current difference between the Gregorian and Julian calendar dates is 13 days: 14 September 2016 Gregorian date is 1 September 2016 (Julian date).

This should explain the extra 13 days in your calculations.


The answer to whether a normal star can contain a core of degenerate neutrons is a yes. Thorne & Zytkow 1977 produced numerically models where a neutron star becomes embedded in the center of a massive giant or supergiant star, surrounded by a large gaseous envelope. In this scenario, the main source of energy becomes not nuclear fusion but instead gravitational contraction from matter flowing from the inner envelope onto the outer core. The energy production ratios are
$$L_{\text{nuc}}/L\approx0.04,\quad L_{\text{grav}}/L=0.96$$
for stars with $M_{\text{tot}}\leq10M_{\odot}$, as above this, convective envelopes form. The models, regardless of mass, predict some degree of shell burning, with hydrogen-, helium-, and carbon- burning layers outside the core. In the inner regions, temperatures are orders of magnitude higher than that required for the CNO cycle, which sets in at around 15 million Kelvin, actually (not 20 million K) and dominates over the p-p chain at about 17 million K.

Thorne & Zytkow did find that for $M_{\text{tot}}<2M_{\odot}$, the envelopes were unstable against radial adiabatic pulsations, implying that it's likely not possible to extend the analysis to the case of the Sun - at that mass, the object is quite vulnerable to instabilities. That said, I am fairly confident that we would be able to tell if there was a degenerate object in the solar core; unlike the red giant or supergiant case, there is no large envelope to hide it, and the mass of the neutron star would at least be comparable to or (much more likely) exceed the mass of the Sun itself.

Some points on the stability of neutron degenerate matter: The problem is actually with small quantities, not large quantities. Small amounts are incapable of remaining bound by their own gravity; the pressures involved are simply too high, and the solar core is nowhere near high-pressure enough to maintain stability. Optimistically, you'd need somewhere in the $0.1M_{\odot}\text{-}0.2M_{\odot}$ range at minimum for stability, though I would be surprised if a mass of degenerate matter this low was produced naturally - a neutron star of a more typical mass would likely have to lose mass somehow.


Which eyepiece should I use to view planets with a Slokey Skyways 50080 telescope?

General guidance seems to be to use a magnification of  30 to 50 times the diameter in inches of your telescope's aperture. An aperture of 80mm is just over 3", so we are looking at magnifications of 90x to 150x.
Your telescope has a focal length of 500mm without the Barlow lens, and 1500mm with the Barlow lens. Since magnification is simply the focal length of the telescope divided by the focal length of the eyepiece, we can see that using the 6mm eyepiece without the Barlow lens will give you a magnification of 83x, and using the 10mm lens with the Barlow will give you a magnification of 150x.
Using a Barlow lens will increase the magnification, but it will also reduce the image quality and also the effective aperture, so the image will not be as bright and contrasty. I would suggest forgetting about the Barlow lens (at least initially), and just work your way up from the 25mm eyepiece to the 6mm. There is no point in using the 25mm eyepiece with the Barlow, since that will only give you 60x magnification, and trying to use the 6mm eyepiece with the Barlow is likely to prove frustrating, if not unworkable.
If you enjoy using your telescope, maybe treat yourself to a 3mm eyepiece, so you can get 167x magnification without the Barlow, but read up on telescope lenses (especially "eye relief"), first. Don't just buy the first lens that you find on eBay.
How to See Mars in 2020
Which lens and magnification is best to see planets?


In astronomical interferometry, what values do the points in the uv-plane have?

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.


Is there a spectrum of polarization? Does polarization vary depending on wavelength, from one and the same source?

Yes. Radiation may be polarized when it is scattered, and different media scatter different wavelengths differently.
For instance, neutral hydrogen scatters Lyman $\alpha$ radiation (i.e. photons coming from the $2$$\rightarrow$$1$ transition of hydrogen). The degree of polarization is coupled to the "phase function" which determines in which direction it is scattered. If a Ly$\alpha$ photon is close to the line center (which is at 1215.67 Ångström), it is scattered isotropically (i.e. equal probability of being scattered in any given direction), and will not be polarized, but if it is more than a few Doppler widths from the line center, it is scattered anisotropically, and will then be polarized. Maximum polarization occurs at 90º scattering, where $3/7 \simeq 43\%$ of the light will be polarized.
If the photons are even farther from the line center, they won't be scattered at all (because they're no longer in resonance with the $2$$\rightarrow$$1$ transition), and the radiation is hence unpolarized.
Example of polarization at a particular wavelength
The figure below, taken from Hayes et al. (2011), shows a so-called "Lyman $\alpha$ blob", which is a source of Lyman $\alpha$ that is quite a lot larger than a galaxy. There was some debate whether the radiation is created outside a central galaxy from gas accreting onto the galaxy and cooling — a signature of galaxy formation — or whether it is centrally produced (from stars) radiation that escapes the galaxy and subsequently scatters on the surrounding gas. If the latter scenario is true, you would expect the Lyman $\alpha$ to be polarized to some degree in sort of a ring around the center, because this light would be scattered at roughly 90º toward you. In the figure, the blue cross marks the center, and the red lines' direction mark the direction of polarization and their length mark the degree of polarization (if it were fully polarized, they would have the length of the red bar in the upper right corner).
Indeed it seems that the light is polarized around a central source.


What measurements helped us in understanding 'Oumuamua's location and speed?

Like most asteroids and comets, 1I/'Oumuamua's trajectory was determined entirely by measuring its position in optical images over several days.
The earliest data came from automated, ground-based surveys such as Pan-STARRS and the Catalina Sky Survey,
then targeted follow-ups from various other asteroid observers.
The object's evident extrasolar origin motivated additional observations with more sensitive instruments including the Hubble Space Telescope.
'Oumuamua did not pass near enough to the Earth for a radar observation.
The Minor Planet Center collects those observations here.
In the table of orbital elements, the 0.4 arcsecond residual is the root-mean-square difference between positions computed using those elements and positions actually observed.
In the observation table, the magnitude column indicates which photometric filter was used, e.g. G or R for greenish or reddish bands of visible light.
Pan-STARRS filter w passes a wide band including most of the visible band and some near infrared (Tonry et al. 2012 table 4).
JPL's estimate of 'Oumuamua's orbital elements uses the same observations and includes uncertainties.
They estimate about 1000 km error in perihelion distance (q), 23 seconds error in perihelion time (tp), and 1 arcsecond error in inclination (i) and ascending node longitude (node).
If you follow the Ephemeris link from there, and specify
Ephemeris Type: VECTORS
Table Settings: quantities code=2x
then HORIZONS gives Cartesian position and velocity uncertainties based on dynamical simulation.
These are larger now than when it was observed, and still larger a few months before discovery.


On the consistency of different well-polished astronomy software

If you set the date to 2018-04-06, Stellarium shows the Moon and planets in positions matching the example image.
Any good planetarium software should produce a similar result.

Most likely the vendor cut and pasted two screenshots (note the seam) for April 2018, overlaid "January 1973," and hoped customers would not check.
Perhaps you could ask them to send you an image for approval before printing the ring.


the most accurate system that can show the positions of planets at my birthday

The most accurate system is the ephemeris service from Jet Propulsion Laboratory (JPL): http://ssd.jpl.nasa.gov/horizons.cgi


What is the role of the random catalogue in calculating the two-point correlation function?

The point of the two-point correlation function (pun not intended) is to describe how clustered the galaxies in the universe are. Astronomers want to know if they're all bunched up in tight bundles with huge voids, or maybe they're more or less uniformly distributed.
The two-point correlation function is a way of measuring the distribution of these galaxies compared to a random distribution. In other words, if your distribution of galaxies is not uniformly random (which it isn't for our universe - our universe has a distinct non-random structure of clusters and voids) then your correlation function will indicate the degree to which it isn't random. Let's take a look at the last equation listed in your link.

The most commonly-used estimator is from Landy & Szalay (1993)
$$\xi(r) = \frac{1}{RR(r)}\Big[DD(r)\Big(\frac{n_R}{n_D}\Big)^2 - 2DR(r)\Big(\frac{n_R}{n_D}\Big)+RR(r)\Big]$$

The symbols in this equation are defined as follows:

$\xi(r)$ - The two-point correlation function.
$r$ - The separation between two galaxies.
$DD$,$DR$,$RR$ - The counds of pairs of galaxies as a function of separation in the data catalogue, between the data and random catalogue, and in the random catalogue, respectively.
$n_D$, $n_R$ - The mean number density of galaxies in the data and random catalogues, respectively.

If your data catalogue happens to be completely random, then $DD=DR=RR$ and you find that $\xi(r)=0$ (assuming also that $n_D=n_R$), showing your data catalogue is indeed indistinguishable from a random distribution. If, however, your data catalogue is not defined with completely random galaxies, you'll find $\xi(r)\ne0$, and the degree to which it is non-zero indicates how non-random, or organized your system is.


How to read Babylonians' tablet that were used to track Jupiter's movements?

This news story is based on an article from Science The image is also taken from the same Science magazine article. The text here is a procedural text, it describes the process of using a trapezium to make distance-speed-time calculations, for approximation of the motion of Jupiter along the ecliptic.

You should look at the original article in science and the author's website


That's the job of NASA's Deep Space Network, plus sites in other countries that perform similar functions. NASA currently has facilities in California, Spain, and Australia whose primary function is to communicate with devices sent beyond low Earth orbit.

In addition to these Deep Space Network sites, NASA and the US Department of Defense jointly operate the Tracking and Data Relay Satellite System network. This system comprises geosynchronous satellites outfitted with communications equipment, relay stations in White Sands, New Mexico and in Guam, plus a control station in Greenbelt, Maryland. The European Space Agency has developed a similar system that takes advantage of recent developments in long distance laser communications. These satellite-based systems are intended for use by vehicles near the Earth. More remote vehicles such as probes to Mars, Venus, and beyond generally use the Deep Space Network.


Might the SpaceX Roadster intercept any ongoing surveys?

Well I can't give a mathematically based answer. I can only tell that some man-made objects were recorded by sky surveys. E.g. the Rosetta spacecraft was marked as a newly found asteroid by the Catalina Sky survey although the error was corrected quite quickly.
Rosetta made an earth flyby at this time so it was pretty close.

The SpaceX Tesla will probably not come close to earth for quite a time and as such it probably won't show up in any survey.


If nothing travels at the speed of light, except light, how can a black hole also pull light into itself?

Gravity is a force, and it need not have a "speed"1.

A gravitational body sets up a gravitational field around itself. Note that by the time the light wave approaches the body, this field has already been set up. The gravitational force does not need to "reach out" and "catch up to the light" — it's already there.

In fact, it's not even really a force (though we can treat it as one to an approximation). Gravity bends the fabric of spacetime around it, messing with the meaning of a "straight line". From different reference frames, different lines appear straight. To the light wave, the inspiraling path seems "straight" and thus it follows that path. To an external observer, the light wave is not going straight.

1. It does, but that's the speed at which changes to the field propagate, not the speed at which it "catches up" with other things. Changes in the gravitational field propagate at lightspeed.


Can an infinite universe explain the Fermi Paradox?

The true explanation to the Fermi paradox is not known. There is a very long list of proposed explanations, some of which are listed here:

https://en.wikipedia.org/wiki/Fermi_paradox#Hypothetical_explanations_for_the_paradox

What you're saying is literally the first one on the list, also known as the Rare Earth hypothesis.

https://en.wikipedia.org/wiki/Rare_Earth_hypothesis


Redshifts are plotted because usually that is all that can be measured.
Other quantities could be plotted, but since these are inferred from the redshift itself (e.g. density would just mean plotting $(1+z)^3$), or would need to assume particular cosmological models or parameters (e.g. if you plotted look-back time), then it makes sense to stick with plotting the observed data.


Computing the over-density of Lyman-alpha emitters

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.


The compatibility of the Grand Tack hypothesis with the "core-warping impact" theory of Jupiter's diffuse core

I think this question might trigger more an open-ended discussion, than a definitive answer, but let me try my take on it.
Firstly,

Is it more likely that one of these collided with Jupiter itself at the head-on angle needed for the core-warping collision?

In order to collide, a co-orbital configuration is more favourable than a orbit-crossing encounter at high eccentricity (as seen in Liu et al., extended fig. 2), but their 'high angles' don't quantify high eccentricities properly. This is because in their simulations, all planetesimals are initialized at circular orbits, with initial distances of 5-10 mutual Hill-radii (their k parameter).
Why I think they didn't properly compare apples with apples, is because usually an eccentric planet would have more kinetic energy than allowed in their simulations, reducing the collision cross-section.
This would lead to drastically reduced impact rates at high angles, compared to the data they show. Furthermore in the early scattering scenario that you propose, most planetesimals would be of too low mass to trigger the core mixing.
The mass required for the impactors is, I think, the major point why the two scenarios are not compatible. The initial conditions for the impact scenarios are those of 5 densely packed 10 $m_{\rm oplus}$ planets, of which one undergoes runaway gas accretion and becomes Jupiter. Those conditions are explained to arise due to oligarchic growth. For the 5 planets, this means that 50 $m_{\rm \oplus}$ are packed in a region of 5 AU size. This is already 1/3 of a median class 0 disc mass, and it necessitates a 100% high efficiency when translating pebbles to planetesimals to planets.
Seeing those numbers lets me strongly doubt the realism of the initial conditions.
Concerning the other part of your question,

The few million years" timescale is quite vague. Does anyone know any additional detail which might suggest the impact occurred prior to the Grand Tack's timeframe?

The few-million years are coming from a certain amount of cooling and compactification that is required for the impact to successfully mix Jupiter's envelope. Hence, the impact should not have occurred prior to the Grand Tack.
Taken both those things together, the cooling requirement and the mass/compactness of the initial conditions necessary, I don't think both scenarios are compatible.


The zero of Galactic latitude (i.e. the Galactic plane) was defined by a working group of the International Astronomical Union. The observations used to do this were 21 cm radio observations of atomic hydrogen, which are unaffected by dust absorption near the plane. The details of the observations are found in Gum, Kerr & Westerhout (1960). This paper uses least squares fits to the 21cm data assuming that the greatest concentration of neutral hydrogen is found in the Galactic plane itself. Whether this is the dynamical plane was considered by Gum & Pawsey (1960), as part of the same IAU study. They concluded that it was (bar the location of the Sun with respect to the plane), citing the flatness and relative narrowness of the neutral hydrogen and the coincidence of the source Sgr A*, which even then was thought to be at the Galactic centre, with zero latitude.

Dark matter was not well explored or widely considered in the dynamics of individual galaxies in the late 1950s, however it is not unexpected that neutral hydrogen should lie in the Galactic plane, as defined by the symmetry of the Galactic potential. A rotating mass of gas, capable of dissipative interactions, should settle to a plane perpendicular to its angular momentum vector.

The 21cm definition of Galactic coordinates cannot itself define the position of the Galactic plane in 3d space unless the Sun is exactly in the plane. Detailed observations of star counts and the dynamics of stars around the Sun and perpendicular to the plane have revealed this isn't quite the case. The details are given in How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it? ; the Sun is currently around 17 pc above the Galactic plane and heading upwards. This is close enough that it makes very little difference to the apparent flatness and narrowness of the 21cm data.


How many stars would it take to draw a line across the middle of the sky that appeared solid?

This is more of a question of human perception, than astronomy. I was going to answer your question with this: "One, it just needs to be close to Earth." But, I decided it wasn't THAT funny. Anyway, stars are essentially point sources as far as our eyes are concerned. Typical visual resolution is about 0.02° or 0.0003 radians. Assuming from horizon to horizon is 180° (or π radians) that calculates out to roughly 10,000 stars. I'd probably increase that by 50% or 100% to be sure. You do understand that you can't line stars up side by side, I hope. There's no need, its about the angular distance, not absolute distance between them, that matters. They can be light years apart as long as they appear to be within about 1 arcminute of one another, our eyes will see them as a single object - subject to the psychological aspects of keeping color and apparent magnitude roughly uniform as well.


What is a name?
A name is a word, that is reasonably unique, that is used to identify a person or thing. When a child is born there is no word that identifies it, and so its parents have to choose a "name".  Similarly when a new astronomical object, such as an asteroid, is discovered, it has no word that identifies it and so it is assigned first a number and later a name.
But there is already a word that identifies our sun and moon:  "the Sun" and "the Moon". There is no need for a new name because they already have a name.  Notice that English (unlike some other languages) gives us a typographical clue that these are names, because the first letter is capitalised.  The Moon is a moon of the Earth and the Earth is a planet orbiting the Sun.
Earth, Moon and Sun are the correct names in English.


galaxies in a specific radius in different shapes of universe

It seems clear to me that this is a homework question. I believe what the question is after is to look at the equations for distances in the universe as a function of curvature and see how this curvature affects things. Read, for example, sources like this one or this one.

As with most of things in cosmology, there's lots of equations and many ways to write them. A relevant equation would be the following angular diameter distance:

$$D_M = (1+z)^{-1}
\begin{cases}
\frac{1}{\sqrt{\Omega_k}}\sinh\left(\sqrt{\Omega_k}D_c/D_H\right), & k = -1 \\
D_c, & k = 0 \\
\frac{1}{\sqrt{\left|\Omega_k\right|}}\sin\left(\sqrt{\left|\Omega_k\right|}D_c/D_H\right), & k = +1 \\
\end{cases}
$$

In this case, you can see the three forms it takes for a closed universe ($k=-1$), a flat universe ($k=0$), and an open universe ($k=+1$). It should be pretty clear that if you chose some volume with some radius calculated by $D_A$, you'd get different volume sizes and thus different amounts of galaxies in each volume for the three universe types.

By looking at the equations and playing around with them, you should be able to convince yourself which universe type results in the largest volume.


How do we know that 2MASS J0523-1403 is a red dwarf?

The brown dwarf "limit" is about $0.072 M_{\odot}$ at solar metallicity (e.g. Chabrier et al. 2000) and is composition dependent. It gets a little higher in metal-poor gas and a little lower in metal-rich gas. $0.064 \pm 0.012 M_\odot$ (the third significant figure is superfluous) is within one error bar of that limit, which in itself is only a 68% confidence limit, not an absolute limit of what is possible. Therefore this object could be a star or more likely, it is a brown dwarf.
The relationship between spectral type and mass is not well established. It requires researchers to actually measure the mass of objects near the stellar/brown dwarf boundary. A spectral type of L2.5 would a priori indicate that this was probably a brown dwarf, and that is indeed what the mass measurement suggests.
The paper referred to by the wikipedia article you cite, in support of a L2.5V spectral type, is by Dieterich et al. (2013). Those authors say the spectral type is L2.5. The wikipedia authors made up the luminosity class.
To establish whether something really was a brown dwarf I think you would have to detect (or not) neutrino emission from its core. A sort-of proxy is to look for lithium. (Old) objects with lithium must have a mass lower than about $0.06 M_\odot$ and are almost certainly brown dwarfs. Objects that have depleted their lithium, which fuses at lower temperatures than hydrogen, are most likely stars, but could just be brown dwarfs. For example, I imagine that this object would have depleted its lithium, so it doesn't really tell you one way or another.
In the end, it isn't hugely important to know. There isn't much difference between a ball of very slowly contracting gas that is approaching its minimum degenerate radius and a slightly more massive ball of gas that has a little bit of fusion going on and a stable radius that is ever so slightly larger.
Postscript: I just checked and the mass determination comes from Filipazzo et al. (2015). This is not a dynamical mass estimate. They use evolutionary models and their estimate of the bolometric luminosity of the object to estimate the mass, assumimg that the age is somewhere between 500 Myr and 10 Gyr. The error bar here is just the range given by these age estimates (an older object would be higher mass). The error bar could be hugely underestimated. When dynamical masses for low-mass stars/brown dwarfs are measured, they are usually considerably larger (by 30% or more) than obtained from evolutionary models, for poorly understood reasons (e.g. Ireland et al. 2008). So from that point of view, maybe this object is a star (with a larger mass).


Is Jupiter just a super earth with hydrogen atmosphere?

Jupiter probably originated with a dense core, and then accreted a gaseous envelope. To call the core "rocky" could be misnomer, since the planet formed much further out from the Sun where it may have been possible to form ices. The exact location of Jupiter's birth and the composition of its core are not known, but the composition was unlikely to have been the same as Earth. It is also probable (again, theory not fact), that Jupiter's core may have reached multiple Earth masses before accreting the gaseous envelope.


How to calculate the time for the solar disk to pass the horizon and transit's lines

The time it takes the sun to cross the horizon is given at:

Expression for length of sunrise/sunset as function of latitude and day-of-year
How long does a sunrise or sunset take?

Re the time it takes to cross the transit line, this doesn't really answer your question, but, to a good approximation, the time varies between 128 seconds at the equinoxes and 140 seconds at the solstices, regardless of latitude or longitude.

More specifically, the formula is $\frac{128}{\cos (\text{dec})}$ seconds, where $\text{dec}$ is the Sun's declination. You can calculate the Sun's declination itself using the formulas at https://en.wikipedia.org/wiki/Position_of_the_Sun

The calculation here is relatively simple:

The Sun travels $360 \cos (\text{dec})$ degrees in a 24-hour day, where $\text{dec}$ is the Sun's declination
When the Sun is transiting, the motion is perpendicular to the transit line (the Sun's motion is entirely in azimuth, not in altitude)
Therefore, all of the Sun's angular motion translates to motion across the transit line; in contrast, the Sun rises and sets a (non-perpendicular) angle (except at tropical latitudes on the two days where the Sun passes directly overhead), so sunsets and sunrises take longer than $\frac{128}{\cos (\text{dec})}$ seconds
Since the Sun has an angular diameter of 32 minutes or $\frac{8}{15}$ degrees, it takes $\frac{\frac{8}{15}}{360 \cos (\text{dec})}$ of a day for the Sun to cross the transit line
Since a day is 86400 seconds, this works out to $\frac{128}{\cos (\text{dec})}$ seconds.

Caveats and nitpicks:

I assume there are 86400 seconds between successive noons. This is incorrect for two reasons:

The time between noons is not exactly one day. The cumulative difference forms the Equation of Time but the day to day difference is quite small
The Earth's day is slightly longer than 86400 seconds, which is why we need leap seconds

The Sun's angular diameter actually varies based on Earth's distance from the Sun, but 32 minutes is a good approximation
I assume the Sun's altitude doesn't change while it's transiting. While this is a good approximation, the altitude does change slightly
I assume the Sun's declination doesn't change while it's transiting. The change in declination is very small, so this is a reasonable assumption
There are probably other assumptions I made implicitly that I am not noting here.
Refraction is not an issue, since the motion we are discussing is azimuthal and not in altitude.
I used the "one over cosine" form above to make things easier for non-mathematicians. The more compact form would use "secant".
The calculations I did for this problem are disorganized, but available at: https://github.com/barrycarter/bcapps/blob/master/STACK/bc-solar-transit.m


Does the Sun's light travel fast enough to have a straight path to Earth?

Light travels in straight lines in spacetime, but not necessarily straight lines through space, and the same is true for free-falling orbits of test particles like satellites (when thrust-less). However, in relativity what constitutes 'space' means taking some 'slice of now' through spacetime, which is of course depends on the reference frame one chooses to use.

For weak, slowly changing gravitational fields appropriate to the Sun, a conventional choice is the static weak-field metric
$$\mathrm{d}s^2 = -\left(1+2\frac{\Phi}{c^2}\right)c^2\,\mathrm{d}t^2 + \left(1-2\frac{\Phi}{c^2}\right)\underbrace{\left(\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2\right)}_{\mathrm{d}S^2}\text{,}$$
where $\Phi$ is the Newtonian gravitational potential. Thus, we can compare to a hypothetical flat spacetime, with a flat Euclidean space given by $\mathrm{d}S^2$. The trajectory of light follows null geodesics ($\mathrm{d}s = 0$), so the light travels as if traveling through a medium of refractive index
$$n = c\frac{\mathrm{d}t}{\mathrm{d}S} = c\sqrt{\frac{1-2\Phi/c^2}{1+2\Phi/c^2}}\approx 1 - 2\frac{\Phi}{c^2}\text{.}$$
A changing refractive index does make light bend; this is exemplified in the simplest case by Snell's law, although the continuously varying case found here needs the more general Fermat's principle.

You shouldn't take the analogy of a refractive medium too literally. Some differences between an actual medium is that there is no dispersion as the refractive index is independent of frequency, and that there is an additional effect of gravitational redshift not present in refractive media. But the latter is separate issue from the trajectories of light rays, so we can ignore it for our immediate purposes.

Light deflection is quantified by an impact parameter $b$, which can be thought of as the perpendicular distance between the Sun and a hypothetical flat-spacetime trajectory, and a scattering angle $\theta$, the angle between that and its actual path in the limit of infinite distance. (Note that the illustration in wikipedia assumes a repulsive rather than attractive potential, but otherwise conceptually it's the same situation.)

Modeling the Sun as having a spherically symmetrical gravitational potential $\Phi = -GM/r$, the leading-order light deflection is,
$$\begin{eqnarray*}
\theta = 4\frac{GM}{c^2b} &\quad&\left(\text{if}\;\frac{GM}{c^2b}\ll 1\right)\text{,}\end{eqnarray*}$$
which is twice what one would get for a particle at light speed under Newtonian mechanics. For light ray just grazing the outer edge of the Sun (or being emitted from there), this angle is about $1.7$ seconds of arc. Completely radial light rays would not be bent in this spherically-symmetric case.


What is meant by the bifurcation of hydrostatic equilibrium shapes in planetary formation?

I think all this means is that up to some critical rotation rate, the equilibrium shape of a rotating, self-gravitating fluid is an oblate spheroid. That is, this shape defines a unique global minimum of energy.

Above this threshold there are two possible equilibrium solutions. Apparently one is an oblate spheroid, whereas the other is a triaxial ellipsoid. These will define local energy minima, but the triaxial ellipsoid is the global minimum.

The gory mathematical details are no doubt contained in Chandrasekhar's textbook on equilibrium ellipsoidal figures. Pages 3-5 of Iurato (2014) appear to summarise the historical development of this bifurcation in the equilibrium solutions.


Candidate major astronomical events circa 600-541 mya?

There was japanese research into nebula winters, to find correlation of weather records with starbusts, dark bok clouds and supernovae:

research reference

A global winter caused by a proto-stellar clouds, called bok globules, is possible too, because there may be 100 times more of them than we can see, they are abot 1ly diameter and 2-50 solar masses, completely dark. Perhaps they would dim the sun by at least 10-40 percent.

You can also find a document that regroups current research for Nebula Winters, corresponding to image 1. the geology evidence is the following:

Bodiselitsch et al. (2005) found iridium anomalies at the base of all cap carbonates both after
the Marinoan and Sturtian glaciations; iridium and other platinum-group elements are typical
proxies for extraterrestrial (including exosolar) materials and are much more abundant than in
the Earth’s upper mantle and crust. These authors interpreted the anomalies to be the product of
the rapid accumulation of extraterrestrial material once trapped in the surface ice and later
transported to the ocean floor via the melting of the ice at the end of the glaciation.
Nevertheless, the excursion of iridium may also reflect the enhanced flux of extraterrestrial
material due to an encounter with a nebula.
Moreover, 244Pu is expected to remain in the Fe/Mn crusts of interest. Fields et al. (2005)
estimated the 244Pu production by a supernova explosion to approximately 0.9 ×10-7
solar
masses. The expected flux of the 244Pu at the Earth is approximately 10-16 kg m-2
yr-1
. A
significant amount (0.1 Bq m-2
) of the accreted 244Pu remains in the sedimentary rocks, even
considering the half-life of 244Pu (80 Myr).
The existence of abundant 100-μm spherical iron balls in pelagic sediments is also possible
direct evidence. For example, Miono et al. (1993) argued that the spherules accumulate in
Paleozoic-Mesozoic bedded cherts. These spherules are generally believed to originate from
25
micro-meteorites. The nebula winter model can be tested by similar but further advanced
investigations, including the analysis of the abundance of the isotopes and platinum family
elements of the minerals separated from spherules in the pelagic sediments, detailing their
exosolar nature. In addition, mineralogical investigations on the ages, the abundances of
platinum family elements, and the isotope ratios of the minerals of the exosolar grains in the
pelagic sediments at that time, will clarify the influence of the galactic environment on the
natural history of the Earth.

The theory is supported by time correlation, which is fairly close to starburts, which occur every 100Ma.

Specific evidence of historical astronomical events require the presence of supernovae elements, i.e. Iridium, in rocks from snowball times.

Other reasons for the Cambrian explosion are the edvent of bilateral animals, 600Ma, the last snowball event from 630Ma, PH and oxygen instability in the ocean, algea blooms. I found this pic which gives an idea of the oxygen concentrations from the cambrian:


Why do people choose 0.2 as the value of linking length in the friends-of-friends algorithm?

The friends-of-friends (FOF) algorithm (Huchra & Geller 1982; Press & Davis 1982, Davis et al. 1985) for finding groups of particles can be used for various numerical problems, not just cosmology, and linking lengths may vary between these problems. The larger the values you use for $b$ (the linking length in terms of the mean interparticle distance), the more particles you will include in your group of particles. Since the density falls off with distance from the center of the group, this means that the average density will be smaller for larger values of $b$.
In cosmological applications of the FOF algorithm, particle groups are called halos, and experience shows that using $b = 0.2$ result in an average enclosed overdensity of close to $\Delta = 180$ times the average "global" density. This value of 180 equals the virial overdensity predicted by the spherical collapse model in the Einstein-De Sitter cosmology (see e.g. Mo et al. 2010). Also, the value of 180, or perhaps 200, is also typically the overdensity used for another popular halo finder, the so-called spherical overdensity algorithm.
That $b=0.2$ results in $\Delta \simeq 180$ is actually only true for an isothermal density profile; the exact value will depend on the actual profile and halo concentration $c$. Surhud et al. (2011) offer an analytical solution to the overdensity as a function of $b$ and $c$.


How massive does a moon have to be to have a stable magnetic field?

Scientists (and science fiction writers) have speculated about the possbilities of life under the surfaces or on the surfaces of large moons in our solar system or large exomoons of large exoplanets in other star systems.
So a good place to find any limits on the possible properties of moons that can have magnetic fields is a scientific discussion of the possibilities of life on exomoons.
Like, for example, "Exomoon Habitability Constrained by Illumination and Tidal Heating" by Rene Heller and Roy Barnes, Astrobiology, 2013.
https://arxiv.org/ftp/arxiv/papers/1209/1209.5323.pdf[1]
In section 2. Habitablity of exomoons, they discuss the qualities, including mass, necessary for exomoon habitability.

A minimum mass of an exomoon is required to drive a magnetic shield on a billion-year timescale (Ms ≳ 0.1M⊕,
Tachinami et al. 2011); to sustain a substantial, long-lived atmosphere (Ms ≳ 0.12M⊕, Williams et al. 1997; Kaltenegger
2000); and to drive tectonic activity (Ms ≳ 0.23M⊕, Williams et al. 1997), which is necessary to maintain plate tectonics and
to support the carbon-silicate cycle. Weak internal dynamos have been detected in Mercury and Ganymede (Kivelson et al.
1996; Gurnett et al. 1996), suggesting that satellite masses > 0.25M⊕ will be adequate for considerations of exomoon
habitability. This lower limit, however, is not a fixed number. Further sources of energy – such as radiogenic and tidal heating, and the effect of a moon’s composition and structure – can alter our limit in either direction. An upper mass limit is
given by the fact that increasing mass leads to high pressures in the moon’s interior, which will increase the mantle viscosity
and depress heat transfer throughout the mantle as well as in the core. Above a critical mass, the dynamo is strongly
suppressed and becomes too weak to generate a magnetic field or sustain plate tectonics. This maximum mass can be placed
around 2M⊕ (Gaidos et al. 2010; Noack & Breuer 2011; Stamenković et al. 2011). Summing up these conditions, we expect
approximately Earth-mass moons to be habitable, and these objects could be detectable with the newly started Hunt for
Exomoons with Kepler (HEK) project (Kipping et al. 2012).

Their sources for those mass limits are:
Tachinami, C., Senshu, H., Ida, S. 2011, ApJ, 726, 70
Williams, D. M., Kasting, J. F., Wade, R. A. 1997, Nature, 385, 234
Kaltenegger, L. 2000, ESA Special Publication, 462, 199
Gaidos, E., Conrad, C. P., Manga, M., Hernlund, J. 2010, ApJ, 718, 596
Noack, L., Breuer, D. 2011, EPSC-DPS Joint Meeting 2011, http://meetings.copernicus.org/epsc-dps2011, 890
Stamenković, V., Breuer, D., Spohn, T. 2011, Icarus, 216, 572
So Heller and Barnes beleive that the approximate lower mass limit for a habitable planet or moon or other type of world would be about 0.25 times the mass of Earth.  Note that they suggest that a world with of mass of only 0.12 Earth mass could retain a substantial atmosphere for a long enough time.
This is a somewhat lower mass than the 0.195 Earth mass suggested by Stephen H. Dole in Habitable Planets for Man, 1964), as the lowest possible mass for a planet to retain a dense atmosphere for geologic eras.  I don't know the reasons for that discrepancy.
https://www.rand.org/content/dam/rand/pubs/commercial_books/2007/RAND_CB179-1.pdf[2]
Returning to the subject of the lowest possible mass world to generate a magnetic field, it may be noted that the smallest worlds in our solar system to have detectable magnetic fields are Mercury and Ganymede.
The planet Mercury has a mass of only 0.06 Earth mass, but has an extremely large iron core relative to its size.  It rotates with a period of 58.65 Earth days, much slower than any other planet except for Venus.
The planet Mars has a mass of 0.11 Earth, almost twice that of Mercury, and rotates much faster than Mercury.

Mars lost its magnetosphere 4 billion years ago,[175] possibly because of numerous asteroid strikes,[176] so the solar wind interacts directly with the Martian ionosphere, lowering the atmospheric density by stripping away atoms from the outer layer.

https://en.wikipedia.org/wiki/Mars#Atmosphere[3]
The planet Venus has a mass of 0.72 Earth, about 12 times the mass of Mercury, but a much longer rotation period of 243.02 Earth DAys.

In 1967, Venera 4 found Venus' magnetic field to be much weaker than that of Earth. This magnetic field is induced by an interaction between the ionosphere and the solar wind,[104][105] rather than by an internal dynamo as in the Earth's core. Venus' small induced magnetosphere provides negligible protection to the atmosphere against cosmic radiation.

https://en.wikipedia.org/wiki/Venus#Magnetic_field_and_core[4]
Ganymede, the largest moon in the solar system, has a mass of 0.025 Earth, and has a weak magnetic field.
Callisto has a mass of 0.018 Earth, and Titan, largest moon of Saturn, has a mass of 0.0225 Earth.  They don't have detectable magnetic fields.
Their lack of magnetic fields may be due to their lesser mass compared to Ganymede.  Since all three, and all moons of Jupiter and Saturn which orbit closer than them, are tidally locked to their planets, their rotation periods are equal to their orbital periods.
So the rotation period of Ganymede is 7.154 Earth days, that of Titan is 15.945 Earth days, and that of Callisto is 16.689 Earth days.
So perhaps the slower rotation periods of Titan and Callisto prevent them from having magnetic fields.  All of moons of Jupiter closer than Ganymede, and all the moons of saturn closer than Titan, have rotation periods shorter than 7.154 Earth days, but even the most massive, Io, has only 0.6 the mass of Ganymede.
And of course the internal composition of a world, and other factors, will also influence whether it has a magnetic field and how strong it is.
If it is desired that an exomoon have a magnetic field to shield it from cosmic rays and its star's stellar wind, and so to keep it habitable, it is possible that an exomoon might not need its own magnetic field.  It could orbit within the magnetic field of its planet and be protected by the magnetic field of the planet.
Such a possibiity is discussed in "Magnetic Shielding of Exomoons Beyond the Circumplanetary Habitable Edge". Rene Heller & Joge Zuluaga.
https://www.researchgate.net/publication/256422936_Magnetic_shielding_of_exomoons_beyond_the_circumplanetary_habitable_edge[5]
And anyone interested in the potential habitability of exomoons should check my answer at:
https://worldbuilding.stackexchange.com/questions/198117/what-if-a-moon-orbits-far-away-and-extremely-fast/198135#198135[6]


How to convert luminosity at rest frame wavelength of 1450 A to absolute magnitude at 1450 A?

The absolute magnitude of an object is defined as the brightness of the object observed at a distance of $d = 10\,\mathrm{pc}$. With this distance, you can convert the luminosity density $L_\nu$ in $\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{Hz}^{-1}$ to a flux density $f_\nu$ in $\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}$:
$$
f_\nu = \frac{L_\nu}{4\pi \times (10\,\mathrm{pc})^2}.
$$

From there, you use the definition of the AB magnitude from Oke & Gunn (1983):
$$
M_\mathrm{AB} = -2.5\log f_\nu \,\,–\,\, 48.60.
$$
Note that there's an error in the original equation, as the authors write plus 48.60 instead of minus.

If you want Vega magnitudes or Johnson magnitudes or whatever, you use a slightly different zeropoint. But you don't want that, just stick with the AB magnitudes.

Or you just use astropy's tools for converting between fluxes and magnitudes.


Searching for stars with a given surface temperature and an absolute magnitude?

Though not much convenient, I found one.

https://github.com/astronexus/HYG-Database

This is a database that contains information of a lot of stars.

I may use excel to use this database and derive any needed results.


What exactly is meant by "expanding universe"?

When we say the Universe is expanding, we really mean that space is expanding. The Universe could be infinite in size, but there continues to be more and more space between objects. Essentially, objects are moving away from each other because more and more space is being created between them.

In the early Universe, matter was much closer together than it is now; this density caused extreme temperatures and no hadronic matter could form. The first expansion is thought to have occurred because of how hot and energetic everything was, and spacetime itself was expanded in an event we call the Big Bang.

As objects spread out from each other during the inflationary epoch, the Universe began to cool down. Now, dark energy is the main culprit for its accelerating expansion.

To answer your question, yes, dark energy is making all particles move away from each other, although farther objects move away from each other at a faster rate. Why is that?

Well, the only thing holding all atoms, stars, planetary systems and galaxies together are the four fundamental interactions — the strong force, weak force, electromagnetic force, and gravity. In space, the main attracting interaction is gravity. As such, dark energy will cause farther objects (which are less affected by our gravity) to move away at a faster and faster rate.

Think of it like you are holding a dog with a leash, and the ground begins to spread out from below you. You and your dog are being moved away in opposite directions, but you both stay close together because of the strong leash.

Now, your mention of atoms being torn apart is actually the basis of the Big Rip hypothesis. It states that dark energy will become more and more abundant until even closeby objects separate from one another. First, galaxies will be disbanded as the stars move away from each other in all directions.

Much later, the gravity of planetary systems won't be enough to hold them together. Later still, planets and stars will be disbanded into their molecules. This will eventually continue as the amount of dark energy increases, until not even atoms can be held together by the electromagnetic force, and the subatomic particles will be moved apart.

By the time the expansion of the Universe exceeds lightspeed at the subatomic scale, no particle will ever be able to interact with one another. The Universe would dissolve into countless lonely particles that won't be able to do anything. Now, the Big Rip hypothesis is only one of the three most known fates of the Universe. Currently, the most accepted theory is the heat death of the Universe.


What explains this puzzling radio signature from a fireball?

I checked with the people who run the BRAMS project and they confirmed my reasoning, and added another possible explanation.

Even if the fireball moves away from the receiver AND transmitter, the ionization trail it leaves behind can expand and be blown by mesospheric winds towards the receiver and/or transmitter. It's this trail that mostly reflects the radio waves.

And theoretically it's still possible but unlikely that the observed Doppler signal was coming from another meteor trail unrelated to this particular fireball, but coinciding with the exact time frame the fireball appeared in.


I'm guessing that this misunderstanding is a result of the oft-used rubber sheet analogy. The rubber sheet analogy says that, according to general relativity, mass curves space-time like a heavy bowling ball on a near-taut blanket (or rubber sheet) curves the blanket/sheet. This resulting curve makes other bits of matter/energy move in different ways. I'm guessing that this is your confusion.

The rubber sheet analogy fails massively in one area, any demonstration of it involves gravity on Earth. If I use a bowling ball to deform a sheet, and then role a golf ball along the sheet nearby, the golf ball will move a bit towards the bowling ball because of the force of gravity around me - not "gravity" in the simulation. It thus makes it seem like gravity pulls the golf ball "down" because the bowling ball pulls the rubber sheet "down". This is the result of using a two-dimensional analogy of a three-dimensional universe.

The point is, there is no "pushing" going on in the general relativistic model of gravity. Gravity is attractive (so long as the strong energy conditions holds for the object in question), just like Newton postulated.


Planned telescope to detect alien waste-heat

This would be the SETI's Colossus telescope project, that aims to build a high-resolution, multiple-mirror instrument with ability to directly image the heat generated by other civilizations on planets orbiting stars near us:

    Artist's impression of the proposed SETI's Colossus Telescope (Credit: Innovative Optics/Colossus Corporation)

  As Earth-like civilizations evolve, they use more power. For example,
  in Roman times, we estimate Ω [the ratio of a civilization’s power production to the amount of solar power it receives] was about 1/1000 what it is today.
  Humans’ global power consumption is growing by about 2.5 percent per
  year, even though the world’s population is growing at less than half
  this rate. In contrast, our knowledge base (the combined total of all
  recorded information) doubles in just two years. As cultures advance,
  their information content also must grow, and the power required to
  manipulate this knowledge eventually dominates a civilization’s total
  power use.

  Using a sensitive coronagraph to remove scattered light that would
  obscure an exoplanet, Colossus would be able to find hundreds of
  Earth-sized or larger planets in the habitable zone including any
  civilizations on their surfaces.

  Source: Paul Gilster for Centauri Dreams quoting Jeff Kuhn (University of Hawaii), Svetlana V. Berdyugina (University of Freiburg), David Halliday (Dynamic Structures, Ltd., in British Columbia), and Caisey Harlingten (Searchlight Observatory Network, Norwich, England)

Here is a Jeff Kuhn's, Institute for Astronomy, University of Hawaii, talk on The Colossus Telescope and the search for ET, and a quite descriptive blog post by Brian Wang on Technology for 77 meter primary mirror ground based telescope exists today and it could survey earth sized exoplanets out to 60 light years advocating the project from a technology perspective.

A bit more can also be read on the Innovative Optics Ltd. web pages, that are sadly under construction, but they do give a hint that an article about it was written in June 2013 edition of the Astronomy Magazine, titled How to find ET with infrared light (PDF), by Jeff R Kuhn, Svetlana V Berdyugina, David Halliday, Caisey Harlingten. A nice digest is also available in this Centauri Dreams blog post.


Do telescopes exist that reflect the incoming light more than three times along their length?

If your willing to accept more than 2 discontinuous mirrors, the Three Mirror Anastigmat has 4 passes along some/most of the overall tube length. An early working prototype example (which I've actually seen in person many years ago) was built at the University of Cambridge by Dr. Roderick Willstrop. The Institute of Astronomy has a page on the Three Mirror Telescope which includes the following optical diagram:

The 0.5 meter 3 Mirror Telescope (3MT) has a focal length of 0.8 meters, in an overall tube length of 1.2 meters producing a very compact telescope with a large field of view (5 degrees in diameter) with good image quality (<0.33" across the whole field of view). The quality of the site at Cambridge was not ideal for building on the progress of the prototype and a proposal for building a larger version at a better site was declined by the UK science funding agency of the time (SERC, which became PPARC and then STFC) in favor of buying into the Gemini collaboration of two 8 meter telescopes.
Some constructed or in construction examples of three mirror anastigmat include:

James Webb Space Telescope (see Figure 2 of this page on the JWST Telescope)
The Simonyi Survey Telescope of the Vera C. Rubin Observatory which will carry out the Legacy Survey of Space and Time (LSST) has M1 and M3 ground into the same piece of glass (see Figure 7 of the LSST overview paper) with a separate secondary M2 mirror and prime focus camera
The planned ESO Extremely Large Telescope will use 5 mirrors in a folded three mirror anastigmat design along with 2 flat mirrors which will have 5 bounces along all or part of the overall "tube" structure before exiting to a Nasmyth focus; see the not very clear diagram below:    from the ESO ELT optical diagram page, somewhat better diagram in  Figure 4 of Hippler 2018.


Asteroid locations at the time of the crucifixion of Jesus

The computer program you are asking for is literally impossible:  Even ignoring the affect of the greater universe and transient visitors such as comets on hyperbolic orbits, the Solar system is a chaotic complex system.  We have a pretty good handle on the motion of the bodies we are, and have been aware of, during the time modern astronomy has been studying the skies, i.e., over the past few hundred years.  (Remember that in the earlier years of telescopic astronomy, we were unaware of most of the currently known members of the Solar system, including all of the asteroids, and every massive object beyond Saturn.)  As you deviate further from the present (either past or future), the chaotic nature of the system comes more into play, especially on smaller bodies such as asteroids.  As a result, stating orbital positions either far in the past or far in the future requires a larger margin of error as the time differential grows.  Eventually you get to a point where the error is so great that we cannot say reliably where an object was or will be.  Perhaps if someone had come up with the laws of planetary motion sooner than Kepler, we would have more reliable data further into the past, and therefore more accurate orbital elements over a longer range than we do, but that didn't happen.

The asteroid belt (white) and Jupiter's trojan asteroids (green)

From Wikipedia, linked from the Asteroid page

In addition to the chaotic nature of the Solar system, there's also the fact that the vast majority of the asteroids revolve around the Sun outside of the Earth's orbit, so the chances of any asteroids we know of having "a gigantic collision ... that could have produced a shadow on the entire Earth" (i.e., between the Earth and the Sun) are between slim and none.  (Note also that the amount of material present in said "dust cloud" would have to be immense in order to have a noticeable affect on the amount of sunlight falling on the Earth.)  The only way such a collision could have occurred would be if two objects traveling through the Solar system happened to collide in such a manner that the "dust cloud" the collision produced happened to fall between the Earth and Sun - and also left the Solar system without a trace afterwards.  This would have required the objects being in similar trajectories crossing the plane of the ecliptic at a high angle and speed - a chaotic event if there ever was one!

So, basically you're looking for a way to prove that a chaotic event happened sufficiently far in the past that chaotic errors make our knowledge of most of the Solar system (asteroids and such) and outside influnences questionable - no, that's not going to happen.  Math doesn't work that way.


Is the angular size of the black hole in the movie "interstellar" completely overblown?

Assuming the spin parameter of the black hole is close to the critical value, it could be quite large.
The scenario of heating a planet orbiting close to a black hole via blue-shifted cosmic background radiation has been explored. The shadow of the black hole would take up a large fraction of the sky on such a planet, see for example figure 1 in Bakala et al. (2020). They find that for an Earth-like planet, the region where this can be achieved is located beyond the Roche limit (where the planet would be broken up) for black hole masses $M_{\rm BH} > 1.63\times10^8 M_\odot$. As shown in figure 6 of the paper, the bulk of this radiation will be in the ultraviolet, but there's probably enough visible light in the mix to make a bluish patch near the edge of the black hole shadow. Bear in mind that this would need to be a very isolated black hole, because stars would also be blue-shifted and therefore be extremely bright sources of high-energy radiation, which would be an extremely unpleasant environment.
The scenario in the movie includes an accretion disc, this scenario is considered in Schnittman (2019), who notes that for accretion rates low enough to produce the Sun-like colour of the radiation depicted in the movie, you'd have to be a long way out:

Regardless of the specific details—whether the planet is tidally locked or not, embedded
in the disk or not—it is clear that for any planet to be habitable while also receiving its
primary heating flux from a Novikov-Thorne type accretion disk, the planet either has to
be extremely far from the black hole, or the mass accretion rate has to be an infinitesimal
fraction of Eddington. This leads to an important inconsistency: The Novikov-Thorne model
is predicated on an optically thick, radiatively efficient disk. At accretion rates down near
$10^{-12} \dot{M}_{\rm Edd}$ the optical depth is unlikely to be large enough to allow the gas to cool, thus
leading to a very hot, low-density accretion flow.


No.

Besides the 13 Jupiter-masses required to ignite deuterium burning, and make Jupiter into a Brown Dwarf, there is a clear difference between the formation pathways of Brown Dwarves and Gas Giants.
Gas Giants are planets, that form via processes in their parent protoplanetary disc. Contrasting this, Brown Dwarves form via direct fragmentaion of the parent giant molecular cloud, possibly as binary fragments of a more massive companion star. The distinction between those two processes is supported by the finding and characterisation of the Brown Dwarf desert in i.e. Grether & Lineweaver (2006).

The 'Brown Dwarf' desert is a sharp break in the number distributions between low-mass stars and high-mass planets. This is commonly interpreted as those two classes of object being distinct in origin. Furthermore we understand distinct theoretical pathways to form each of those object classes, as mentioned above.

Note that if you're talking about the other elements, or as astronomers would call them, metals, then there is data on those as well: The sun, and any star possesses a certain amount of metals, see i.e. Asplund et al. (2009). For example the ratio of Oxygen to Hydrogen in the solar atmosphere is about $n(O)/n(H)\sim 5\times 10^{-4}$, which is the most abundant 'other' element in the sun.
The values of heavier elements found in the gaseous planets are elevated compared to the solar values. This is thought to represent a contribution of solid material that planets need to form according to current theories. If the giant planets would have formed through direct collapse from the same cloud material as the sun did, the abundances of metals in the giant planets would not be elevated compared to the solar values.


Is the universe moving through infinite space time as it expands?

I think what you're asking is a very fundamental question that doesn't really have a definitive answer. I interpreted your question slightly differently that Sir Cumference, so I'll answer it differently.

For people just learning about the expansion of the Universe, they are often taught about the analogy of the expanding balloon. The concept here is that you have a deflated balloon that you draw two dots on. These dots may represent two galaxies in the Universe defined by the surface of the balloon. The important concept here is that the surface of the balloon represents the entire universe in this scenario. The balloon is then blown up, and you notice that because of the "expansion" of the universe, the dots (i.e., galaxies) are moving farther apart. This is shown in the diagram below.

This concept is usually used to teach an important fact about our universe, namely that there is no center. At no point on the balloon can you saw that you are the center of the expansion because the expansion is happening everywhere in equal magnitude.

What you seem to be doing though, is trying to take this analogy one step further. Even though the balloon universe (ballooniverse?) is defined on the surface of the balloon and is thus two dimensional, there still exists a three dimensional space that the 2D ballooniverse is expanding into. So you question, what does that say about our 3D universe? Are we expanding into some 4D space?

That, in my opinion, is a distillation of the fundamental concepts behind your question. Assuming I'm interpreting your question correctly, then I would have to say the answer is I don't know.

There is a somewhat (unsatisfying) tautological answer in that the Universe is, by definition, everything that exists. Therefore, there can't be a 3D Universe and also a 4D space into which our 3D Universe is expanding into, because that 4D space would necessarily be a part of our Universe, by definition.

A more reasoned approach to this answer is to look at it from a mathematical perspective. If you begin to learn higher math (and by higher I mean upper undergraduate/graduate level math) you'll learn about manifolds. Manifolds are basically mathematical objects that can define things like the surface of a balloon. What you'll find is that the definition for the surface of a balloon, using the concept of manifolds, does not require that the surface exist in 3D space, even if that manifold is expanding. It is perfectly possible to define a 2D manifold that has the same topology (fancy word for structure) as an expanding balloon, without "embedding" it in 3D space. This concept makes your question kind of moot. It allows for an expanding 3D universe, without requiring the existence of some 4D space to expand into. So ultimately, I'm claiming that it is possible for a 3D Universe to exist which both has no center and is expanding, just not into a 4th dimension.


Look up the laboratory wavelengths of the Calcium H and K lines
Calculate the redshift implied by the observed wavelengths of the H and K lines.
$$ v/c = (\lambda - \lambda_0)/\lambda_0,$$
where $v$ is the implied recession velocity, $c$ is the speed of light, $\lambda$ is the observed wavelength and $\lambda_0$ is the laboratory wavelength.
Plot recession velocity versus distance and see what you get.
Read about the Hubble redshift-distance relation.


Is HR 5171 a Spectroscopic Binary or a Symbiotic?

I found arXiv:1401.2628 which says

Modeling the light curve with the NIGHTFALL program provides clear evidence that the system is a contact or possibly over-contact eclipsing binary. A total current system mass of $39^{+40}_{-22}$ solar mass and a high mass ratio $ q>10$ is inferred for the system. The low-mass companion of ${\rm HR5171 A}$ is very close to the primary star that is embedded within its dense wind. Tight constraints on the inclination and vsini of the primary are lacking, which prevents us from determining its influence precisely on the mass-loss phenomenon, but the system is probably experiencing a wind Roche-Lobe overflow.

Does the need for simulation indicate that it is a spectroscopic binary?


When did we discover that our sun does not orbit its own axis but a barycenter?

Newton explained the motions of planets in the solar system in Principia in 1687.

This amongst many other things "defines the very slow motion of the Sun relative to the solar-system barycenter;" (https://en.m.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica ).

Since Newton came up with the ideas of gravity and barycentres, surely this is the answer? (Though, I note there is some debate that the inverse square law may have been around and discussed during the previous decade).

Ideas that the Sun moved were of course commonplace before then.


Solar series 142 and 147 started same year - series 147 and 152 will be ending same year. Why? Coincidence?

The reason Saros Cycles begin and end is down to the positions of the orbits of the Sun, the Moon and the Earth. From the main Wikipedia page on Saros:
For an eclipse you need the 3 bodies to line up at a full or new moon,

and repeat occurrences of these lunar phases result from solar and lunar orbits producing the Moon's synodic period of 29.53059 days...however for the majority of these, the shadow of the Earth or Moon fall to the north or south of the other body.

It can only happen when a full or new moon is around the ecliptic.

The period of time for two successive lunar passes through the ecliptic plane (returning to the same node) is termed the draconic month, a 27.21222 day period. The three-dimensional geometry of an eclipse, when the new or full moon is near one of the nodes, occurs every 5 or 6 months when the Sun is in conjunction or opposition to the Moon and coincidentally also near a node of the Moon's orbit at that time, or twice per eclipse year.

Also:

Two Saros eclipses have the same appearance and duration due to the distance between the Earth and Moon being nearly the same for each event because the Saros eclipse cycle is also an integer of the anomalistic month, the period of the eccentricity of lunar orbit, 27.5545 days.
After one saros, the Moon will have completed roughly an integer number of synodic, draconic, and anomalistic periods (223, 242, and 239) and the Earth-Sun-Moon geometry will be nearly identical

The reason we get varying start and ends is because of this. Those roughly integer periods vary.
This diagram from that Wikipedia page illustrates perfectly why they can be different lengths, stop and start at changing intervals etc.


How do I find stars within milky way using galactic latitude and longitude

Stars beyond our own galaxy and its immediate neighbors are indistinct without a large telescope, so I assume the question is about the band across the sky, rich with faint stars.

The Milky Way's boundary is fuzzy, so its extent is expressed in terms of isophotes: lines of constant brightness per unit solid angle (e.g. square degree).
It is wider near the central bulge (galactic longitudes around 0° or 360°) than away from it (around 180°).
Most stars within galactic latitudes ±10° are in the Milky Way; most outside ±15° are not.
At galactic latitudes in between, galactic longitude and the choice of isophote determine whether a star is in or out.


How does one go about submitting a name for a newly-discovered lunar crater?

You use this from https://planetarynames.wr.usgs.gov/FeatureNameRequest. You need to follow the rules as set out in https://planetarynames.wr.usgs.gov/Page/Rules as well as following the themes for naming Moon based features https://planetarynames.wr.usgs.gov/Page/Categories.


Inverse of the sunrise equation - finding locations with a given sunrise time on a given day

To avoid the headaches associated with finding roughly two dozen similar solutions in the zig-zagging time zone boundaries , you might want to stick with UTC. If you want to deal with local time - it's going to get a little crazy if you also try to include daylight savings toggling on and off.
The points on the earth where the center of the sun coincides with the horizon (ignoring topography, oblateness, atmospheric refraction, the finite speed of light and other small effects) is just the circle on the earth where a cone drawn from the Sun intersects a spherical Earth.

above: A sphere inside a cone, from http://mathcentral.uregina.ca.
But the next step is hard - getting from that circle to the lat/lon coordinates on the surface, because the axis is tilted, and because the motion around the sun speeds up and slows down as the Earth revolves closer and farther from the Sun. These can be approximated by periodic functions with some coeficients, and that's what you will find in the math behind the first answer. For the second answer, I'll list two Python solutions - PyEphem and Skyfield. Both are easy to use, but you are separated from the actual math (in one case it's an ephemeris/table. The third answer is really a collection of NASA/JPL routines that are highly regarded but may involve more time for you to get up to speed compared to the Python packages.
Answer 1: Astronomical Algorithms
This is something you'll have to dig into a bit, but if you like to program, it may be exactly what you're looking for. The Gaisma website is one of my favorites on the internet - easy to use and presents a bunch of information in easy-to-understand graphics. Click around!
I believe that this site uses algorithms from the collection found at this NOAA site. Click around there as well. They provide Excel spreadsheets which contain the algorithms and other resources. The "main" resource is a collection of algorithms published in the book Astronomical Algorithms - Jean Meeus. From that Amazon page you can see that there are many similarly titled books. I'd recommend going to a library if possible, because it's (in my opinion) always good to go to libraries. However parts of these can be found on-line. For example, a few pages shown from the book Astronomical Formulae for Calculators (1988) include an interesting table of contents.
Answer 2: Python packages PyEphem and Skyfield
I'll copy a bit of the text from this answer:

The Python package PyEphem has been around and well supported, and is the pythonic reincarnation of XEphem. I haven't used it, but I believe it keeps enough information about orbital parameters at certain epochs to generate an ephemeris, including some gravitational perturbations. In other words, it's much more than planets moving on fixed elliptical orbits around a fixed sun. So I believe it runs without internet connection.

I never used it because I was recommended to look at Skyfield and it's exactly what I needed. It downloads a standard JPL ephemeris that you choose, and then just uses it from your hard drive after that. However, in order to deal with leap seconds and other time related effects, it occasionally needs to check the internet for leap second information updates, since these are arbitrary.

I don't know if Skyfield has a mode to avoid that. Actually that's a good question. If you work with a timescale that doesn't have leap seconds, I am not sure if it will run in its current version.
Both Skyfield and PyEphem Python packages have been written and are maintained by @BrandonRhodes.

I've included a simple Python script just to illustrate how Skyfield can be used. If you are comfortable with coding without curly braces, I would strongly recommend you give this a try. It's incredibly powerful and Pythonic.
This is just a starter - you need to add some better housekeeping to detect sunrises vs sunsets, and maybe a more global-type search in some cases. Actually there is some slightly tedious housekeeping necessary to make this work robustly.
note: you can turn on atmospheric refraction using the arguments in the apparent() method. See the Skyfield API documentation for more info, and for a discussion about iterating using Skyfield methods - especially solving for times, see this helpful answer.

def alt_lonlat(lon, lat, t):

    topo = earth.topos(lat, lon)

    alt, az, dist = topo.at(trise).observe(sun).apparent().altaz() ## apparent() args for atmospheric refraction

    return alt.degrees

from skyfield.api import load
import numpy as np
import matplotlib.pyplot as plt
import scipy.optimize as spo

data  = load('de421.bsp')
ts    = load.timescale()

# your example:  '2014-08-27', sunset = '10:00 PM', sunrise = '7:30 AM'

trise  = ts.utc(2014, 8, 27,  7, 30, 0)
tset   = ts.utc(2014, 8, 27, 22,  0, 0)

earth = data['earth']
sun   = data['sun']

zerozero = earth.topos(0.0, 0.0)   # gotta start looking somewhere!

alt, az, dist = zerozero.at(trise).observe(sun).apparent().altaz() ## apparent() args for atmospheric refraction

print "at trise, JD = ", trise.tt
print "at (0N, 0E) Sun's altitude: ", alt.degrees, "azimuth: ", az.degrees
print "at (0N, 0E) Sun's distance (km): ", dist.km

# Find points on equator where sun is on horizon (rise or set) at t=trise

limits   = ((0, 180.), (180, 360.))
lonzeros = []

for a, b in limits:

    answer, info = spo.brentq(alt_lonlat, a, b,
                              args=(0.0, trise),
                              full_output = True )

    if info.converged:
        lonzeros.append(answer)
        print "limits ", a, b, " converged! Found longitude (deg): ", answer
    else:
        print "limits ", a, b, "whaaaa?"
        lonzeros.append(None)

# make some curves

lats = np.linspace(-60, 60, 13)

longis = []
for lon0 in lonzeros:
    lons = []
    for lat in lats:

        answer, info = spo.brentq(alt_lonlat, lon0-90, lon0+90,
                                  args=(lat, trise),
                                  full_output = True )
        if info.converged:
            lons.append(answer)
        else:
            lons.append(None)

        lons = [(lon+180)%360.-180 for lon in lons]  # wraparound at +/- 180

    longis.append(lons)

plt.figure()

for lons in longis:
    plt.plot(lons, lats)

for lons in longis:
    plt.plot(lons, lats, 'ok')

plt.xlim(-180, 180)
plt.ylim(-90, 90)

plt.title("at trise, JD = " + str(trise.tt))

plt.show()

Answer 3: SPICE Kernels
As pointed out by @barrycarter in this comment below this answer, the JPL SPICE Kernels are available. I'm not familliar with them but it's what NASA uses so it must be pretty good :)

Appendix:
Here are some screenshots for London, UK (Gaisma from here):


Why do they think that WDJ0551+4135 is "snowman shaped" i.e. a contact binary?

It depends who the "they" is in your question.

If by "they", you mean the astronomers who discovered it, they don't. The paper (which is available on arXiv) clearly states it formed from a merger, and doesn't mention the term "contact binary" at all. The merger has already happened long ago, what we're seeing is the end result of it.

If by "they" you mean the BBC, it looks like their correspondent has gone and read far too much into the artist's impression attached to the press release. The artwork depicts the final stage of a merging white dwarf, rather than the object as it appears today. It's probably understandable since a contact binary does look quite bizarre compared to the usual spherical star, and the title was "Astronomers find a bizarre, dying star that 'didn't make any sense'".

There's definitely an argument to be had about whether artists' impressions are misleading when attached to this kind of science communication, but that's something for another website...


What's the significance of finding neon on lunar atmospehere? new explanation of moon light?

The moon's "atmosphere" isn't much of an atmosphere, and is sometimes called an exosphere.  The particle density in the moon's atmosphere is low enough to be considered a very good vacuum for Earth experiments.
The discovery is mentioned in this article on phys.org, which specifically says

There's not enough neon to make the moon visibly glow because the moon's atmosphere is extremely tenuous, about 100 trillion times less dense than Earth's atmosphere at sea level.
...
The behavior of a dense atmosphere is driven by collisions between its atoms and molecules. However, the moon's atmosphere is technically referred to as an exosphere because it's so thin, its atoms rarely collide.

As far as the question "what's the importance" in your title, it seems mainly to be a confirmation of something that has been predicted for a while, as well as providing more data concerning exospheres: the most common type of "atmosphere" in the solar system.


Site to see direction to the Voyager by entering time and coordinates of my position on the Earth?

TheSkyLive has charts showing the positions of a few interplanetary spacecraft in the sky for a given time and place on Earth.
Their positions for Voyagers 1 and 2 agree with the authoritative JPL HORIZONS, which can generate a table of equatorial coordinates for a range of dates.

Voyager 1 is in northern Ophiuchus.
This is in the August evening sky for most of the Earth.
As of August 2018, it is 143 AU away from the Sun and receding 3.6 AU/year.
Voyager 2
is in northern Pavo, in the September evening sky for the tropics and southern hemisphere.
Its current distance from the Sun is 118 AU, receding 3.2 AU/year.

Since both probes are so far away (compare Neptune at 30 AU) and moving away from the Sun, their apparent positions among the stars will not change much in the next few years.
Of course they are not visible in any telescope, and a very large dish is needed to receive their radio transmissions.

If you like using desktop planetarium software, you can try importing hyperbolic orbital elements for the Voyagers.