What elements and compounds would be present in a red gas giant?

It's a more complicated question than it appears.   Also, a couple points are in order.

First, you're thinking about gas giant color is wrong.  It's not the ingredients as much as the temperature.   The ingredients don't change all that much.  They change some, but gas giant color has more to do with temperature.

Two, color is a very narrow perspective.  We see, only a narrow band of the spectrum.

NASA brings a variety of cameras on their missions, Infra Red, Radio, UV as well as visible light.   See here.

Also, many photos that we see of planets are doctored to make the variations stand out more visibly, and I think it's great that NASA does this.

And three, the planet's star can affect it's color.   Color is different under a blue star compared to a red dwarf, so there's 2 colors to consider, color under a white light and color under the planet's star.

All that said, color's still interesting and fun and it tells us something.   Using visible light only, many gases are largely transparent, but that doesn't mean they're colorless.  Any gas will reflect and diffract some sunlight and gas giants have atmospheres thousands of miles thick.  There's no such thing as a transparent gas giant planet, with an earth like atmosphere that you can see through.

Hydrogen and Helium are both highly transparent to visible light but bluish when thick enough.  (Neptune gets it's blue color from a combination of it's hydrogen and helium and trace amounts of Methane ice).

Upper-atmosphere temperature is also very important.    Saturn, Uranus and Neptune are cold enough that mostly their color is defined by ice, which tends to be more reflective, giving them lighter and generally considered "prettier" colors.   Uranus and Neptune's methane ice give them their blue/green and more pure blue colors.    Saturn is a little too warm for Methane ice, but it has ammonia ice giving it a lighter color and other gases in it's upper atmosphere giving it some darker yellow and orange colors.

Jupiter's bands can be color-differentiated by where there is abundant ice high in it's atmosphere and where there is more gas.   The darker bands are warmer, generally circulating upwards from deeper in the planet, like Hadley cells on Earth, driven in part by Jupiter's fast rotation and strong Coriolis effect and significant internal heat.   It's lighter bands are colder, where ice reflection is the primary factor in it's color.   Jupiter also has some complex hydrocarbons in it's upper atmosphere due to chemistry, intense lightning and atmospheric mixing.

So there's quite a few difficulties in answering this because a gas giant planet's color depends on many factors, How well it's upper atmosphere circulates, it's temperature and what ices, if any, can form in it's upper atmosphere.   Any trace elements that might carry color, even a 1 part in tens of thousands, can be enough for a dominant color if the other gases are mostly transparent.   Take Earth for example - clouds are a very small part of the atmosphere by mass but they change the color of the sky (viewed from space) significantly.  This makes predicting a gas giant planet's color very tricky and borderline guess-work, because one trace gas can be enough to change it.

Gas giants are mostly hydrogen and helium in their upper atmospheres, but, as noted above, it's often the other gases, ices or trace gases, not the hydrogen and helium that will give them their color.

I always thought Jupiter's "great red spot" was misnamed.  It looks more like an orange-brown to me.    But if that counts as red, then red might be relatively common.

It might be UV rays and a trace gas, ammonium hydro-sulfide, that gives Jupiter's spot it's color.  See article here.   I gather there's still some debate on that question.

Here's a question with a nice answer on the colors of the planets.   Venus is unusually bright of the reflective SO2 high in it's atmosphere.


How (the heck) was this photo of Venus at inferior conjunction (between us and the Sun) taken?

I think you'll have to live with an incomplete answer. It appears this was taken by an amateur astronomer using a simple 4.5 inch telescope. You can find, on the ESO website, a collection of many images taken during the 2004 transit. On this page, is the image you are asking about (along with some other images by the same person). The credit for that image is given as:

  The Phases of Venus

  Statis Kalyvas
  4.5-inch Mizar telescope
  Nikon Coolpix 750
  February 27 - June 8, 2004
  Thessaloniki, Greece

I'm assuming (because I can't find any reference to state otherwise) that the name "Mizar telescope" is just the personal name Statis Kalyvas gave to his telescope. Although you can see that the equipment was nothing an amateur astronomer couldn't buy. A pretty typical telescope with a standard camera. Nothing "professional" or extremely high tech.

We can probably guess at some other pieces of the puzzle though. From the image below, you can see that the 2004 transit began (from Thessaloniki's local time) at 7:13 AM and ended at 1:25 PM.

Given that the observation period of the transit itself happened during the day, Statis definitely would have needed a solar filter for his telescope, which would have blocked upwards of 99.999% of the light from the Sun, allowing him to view the transit safely.

For the other images leading up to the transit, my guess is that he made his observations during in the evening when the Sun was near the horizon and not very bright (due to much of the light being scattered). Venus should've been higher in the sky than the Sun and still visible at that time. An eyepiece which gives a small field of view would've been useful here.

You can possibly guess that the image was taken near the horizon because a 4.5 inch telescope should have a resolution of $1.21\ \mathrm{arcsec}$ at best whereas, near conjunction, Venus is almost an arcminute in size. Venus should be well resolved, but it looks a bit fuzzy. This is likely attributed to the high airmass near the horizon. The drastic difference between the clarity of the image on 3/4 and 13/4 (just ten days later) leads me to believe the telescope was capable of better resolution and was simply impeded by the atmosphere near the horizon.


With a tube length of 46 cm and a focal length of 100 cm, probably the AstroMaster 114EQ is not a classical Newtonian but uses a fast (e.g. f/4) spherical primary mirror and a corrector lens built into the base of the focuser to reduce spherical aberration and increase the effective focal length (e.g. to f/8.8).
This is a variant of the Jones-Bird design of some custom telescopes.
There is some doubt about how well the corrector actually works in budget-priced telescopes.

Most collimation instructions (e.g. Seronik or Carlin) assume a classical Newtonian design.
Jones-Bird collimation is similar in theory but more complicated in practice.
In particular, the corrector hinders the use of a laser device.
Try a collimation cap; anything that fits a 1.25 inch focuser is fine.
You can even get decent results with a homemade one.
If you decide to invest in a Cheshire sight tube, it can be used with any Newtonian reflector.


What Is The Maximum Distance Our Finest Instruments Could See When They're Perfected?

The furthest we can "see" is the cosmic microwave background at a redshift of about 1100.

The proper distance of the CMB-emitting gas that we see today is about 46 billion light years.

If you are talking about galaxies, then the first are thought to have formed at redshifts of about 20 (current distance 36 billion light years) and beyond that are the universe's "dark ages" where no stars or galaxies should exist; the most distant galaxies observed at the moment are at redshifts of about 10 (current distance 31 billion light years).

All current distances depend on the adopted cosmological model. The numbers are for a flat $\Lambda$CDM universe with the Planck cosmological parameters.

There is no chance of seeing our own Galaxy, since it only formed about 12 billion years ago and the universe is a lot bigger than 12 billion light years.


How do tip tilt mirrors correct distortion in adaptive optics?

Not an expert, but offer one solution:

Question 1

The adaptive optics correction is accomplished by a tip-tilt mirror and a deformable mirror. Usually the atmospherical wavefront distortion (or say, the phase screen) looks like this:

Note its ramp-like shape. In other word, the 2nd and 3rd Zernike terms are large.

The deformable mirror has limited correction range: it deals well with the higher-order Zernike terms, but not the 2nd and 3rd Zernike terms. This is why the tip-tilt mirror is here: it introduces a general ramp correction to suppress the 2nd and 3rd Zernike terms, then the afterwards wavefront contains only the higher order Zernike terms, and hence can be corrected well by the deformable mirror.

Question 2

You need a little Fourier optics knowledge to fully understand. Briefly speaking, The image $I$ (this is also called the point-spread-function, the linear system response of a Fourier optical system) captured on the sensor follows:
$$I(\frac{x}{\lambda f}) = |\mathcal{F}\{p(y)*e^{j (2\pi/\lambda) w(y)}\}|^2,$$
where $x$ and $y$ are the coordinates on the sensor and the aperture, $\lambda$ the wavelength, $f$ the lens focal length, $\mathcal{F}$ denotes Fourier transform, $p$ and $w$ are the aperture shape and the distortion wavefront, respectively.

For simplity let $p(y) = 1$ for every $y$ (i.e. infinity aperture size). If $w(y) = 0$, then $I(x)$ is a perfect Dirac function. This simply says your star has no blurring. To the contrary, for example $w(x) = \alpha x^2$ for some $\alpha$, then it results in a defocused blurring point.

The same formula above also helps answering your additional question. Increasing the aperture size (the main mirror will finally be mapped to the aperture) is equivalent to, in the Fourier domain, decreasing the blurring size.


Which would be safer: removing hydrogen or adding hydrogen to our sun?

My kneejerk reaction is that your only option is to remove a chunk of mass from the outer part of the Sun.

The Sun will respond (on a Kelvin-Helmholtz timescale), by contracting and becoming less luminous because the core temperature is lower in a less massive star. This will extend its main sequence lifetime, because only the central parts of the star are involved in fuelling the nuclear burning. The core is separated from the well-mixed outer convective envelope by a stable radiative zone.

Since the default solar evolution involves it getting slowly more luminous with time, it is possible that mass could be extracted at just the right rate to keep the Sun at constant luminosity. This sounds like an interesting calculation to do.

The alternative of adding "fuel" in the form of hydrogen won't work. The Sun's luminosity will increase because of its increased mass and increased central temperature. However, because the new fuel cannot be mixed into the core (only the outer part is mixed by convection), then the main sequence lifetime would be reduced.


How to calculate the angle formed between 2 planets?

The planets are moving along elliptical paths, and we are viewing them from another planet that is also moving along an elliptical path, this makes the motion that the planets make in the sky (relative to distant stars) seem to loop and move with an inconsistent rate. The planets don't all orbit in the same plane, and so may be above or below the ecliptic (the apparent path in the sky of the sun)

So there is no shortcut to finding the angle between two planets: You have to calculate the sky position of the planets and then find the angle between those positions.

The first step is called calculating an ephemeris. The details are beyond the scope of this answer, but NASA have a good ephemeris calculator at https://ssd.jpl.nasa.gov/horizons.cgi. Alternatively, software like Stellarium will be able to calculate the position of a planet on any given date in the near future or past.

When you have the position of the planets as RA and Dec, and you have converted the units to decimal degrees, you can calculate the angle $A$ between them using

$$\cos(A) = \sin(\mathrm{Dec_1})\sin(\mathrm{Dec_2}) + \cos(\mathrm{Dec_1})\cos(\mathrm{Dec_2})\cos(\mathrm{RA_1} - \mathrm{RA_2})$$

Further details and a calculator is found at http://www.gyes.eu/calculator/calculator_page1.htm


How do astronomers detect the 'metals' in a star? If the atoms are presumably completely ionized?

You are correct that the characteristic emission and absorption lines we see in stars' spectra are from electrons that are bound to atoms making transitions between different energy levels.  That is possible because the elements in a star's photosphere are not fully ionized.  Hydrogen - the easiest element to fully ionize because its nucleus only has a charge of +1 - is pretty highly ionized in the hottest stars (O stars), which is why those stars have weak hydrogen lines in their spectra.
Completely ionized atoms (bare nuclei) do not absorb or emit electromagnetic radiation in the parts of the spectrum where stars emit most of their light.  They can absorb very high energy photons (gamma rays with energies of millions of electron volts) that have enough energy to excite their nuclear energy levels, but the photospheres of stars aren't hot enough to emit this kind of radiation (and as previously stated, the atoms there aren't fully ionized anyway).
Many stars have elements in their photospheres that are partially ionized, but the spectral lines we see from those atoms are still from the remaining bound electrons.  For example, Ca II (singly-ionized calcium, with 19 electrons instead of 20) has strong lines in the violet part of the spectrum of the Sun.


Is having a surname which begins with the letter “A” a big advantage for a researcher in astronomy/astrophysics? Is changing our surname a good idea?

I would say that it matters not-at-all whether your name is Aardvark or Zygote. Astrophysicists are not stupid, we can see when a large author list has been alphabetised. If anything I would say it is a disadvantage, since a casual glance at an author list might lead someone to think you were the lead author of any paper, even one with just two authors, simply because of your surname.

In practice, the tiny minority of author lists are alphabetised in Astrophysics papers. Off the top of my head I can think of the gravitational wave papers by Abbott et al. and the MACHO papers by Alcock et al. Most many-author papers have a nucleus of authors at the start of the work and then a longer list of names, possibly in alphabetical order, of people who contributed to the overall project or mission. Even these papers are a (growing) minority and it is still true that the majority of Astrophysics papers have a relatively small author list ordered by contribution.

Finally, I will add another personal opinion. The presence of someone's name on many long-author list papers does not necessarily impress me unless they have lead some of them. It is more impressive for a young scientist to have forged their own path and perhaps written a few lead-authored papers with small author lists or even a sole author paper. Unfortunately, much "big science" does involve big consortia and though it can be exciting for an early-career scientist to be part of such an enterprise, it can be much harder to make your mark outside of that consortium.

EDIT: As a brief, empirical check on my answer, I looked at arXiv for 22nd November 2016. There were 66 new astrophysics papers listed. Of these, 29 were written by three or fewer authors. Only five were written by $>10$ authors and only one paper, from the Pierre Auger collaboration, appeared to have an alphabetised author list. Try hard not to laugh - the paper is by Aab, A. et al. Good luck in your efforts to beat that.


How are heavier elements such as carbon and silicon distributed within the Sun?

A page from the Institute for Advanced Study links to data from various modifications to the Standard Solar Model. The newest given there is from Bahcall1 et al. (2005), which I'll use as an example. The authors' calculations depend on data from the Opacity project. Data from two models are available through the IAS link: the BS05(AGS, OP) model and the BS05(OP) model; the difference lies in the main thing you're interested in: heavy element abundances. BS05(AGS, OP) has, for instance, a central2 $^{12}\text{C}$ mass fraction of $7.79\times10^{-6}$, while BS05(OP) has a central $^{12}\text{C}$ mass fraction of $1.05\times10^{-5}$. All of the central data from the major thermodynamic variables (e.g. pressure and temperature), as well as the mass fractions, are well within an order of magnitude in each model.

That said, the general heavy element trend is the same in both models. $^{12}\text{C}$ increase by two to three orders of magnitude from the center. $^3\text{He}$ (rather than the more common $^4\text{He}$, which has a mass fraction given by $Y$) increases by one to two orders of magnitude. $^{14}\text{N}$ actually decreases by two orders of magnitude, while $^{16}\text{O}$ remains relatively constant. The model gives no data for silicon.

Other new models agree (though not all), to less than an order of magnitude or so, with the data from Bahcall et al. Graphical representations can be found in e.g. this paper:

I'm unsure of the reason behind the $^3\text{He}$ spike; that may warrant a follow-up questions. My one guess is that this could be due to its role as an intermediate nucleus in the second step of the p-p chain. However, the same trends for each element are followed, although the changes in mass fraction are not constant at all radii $r$.

As for your question about how the central figures were derived, the answer is that helioseismological measurements and neutrino fluxes are some of the best indicators of composition. This was used as the basis for the BS05(AGS, OP) and BS05(OP) models, as well as, in fact, most variants of the Standard Solar Model.

1 This is Bahcall's page, so, naturally, these are some of his papers. However, the results appear to be consistent with the calculations of others.

2 Technically, this is from about 0.0016 solar radii, but this distance step is comparatively small.


Why do things float in space, though the gravity of our star is always present?

It is not true that "objects float around" in the solar system.

Perhaps you have seen video from the space station, and you can see things floating. This is not because there is no gravity, but because everything in the space station going at the same speed in the same direction. This makes it look as if things are floating. In fact the space station and everything in it, is moving at about 7km per second around the Earth.

The Earth and everything on it is is also moving at 30 km/s around the sun. The satellites around the Earth are also affected by the sun's gravity, but because they are affect the same as the Earth, they are also moving at about 30km/s around the sun. If two things are moving at the same speed and in the same direction, it will look as if they are floating. In fact they are "whizzing".

The Sun, Earth and satellites are also whizzing around in the gravity of the galaxy.  The black hole is only a very small part of this; most of the gravity of the galaxy is in the dark matter that we can't see.  But we don't feel this pull because the Sun, Earth, the satellites and you are all being pulled at the same time.


Why do we expect Type Ia Supernovae to be brighter in a decelerating Universe than what has been observed?

Observed brightness (or magnitude) depends on the distance between the source and the observer. The further the distance, the fainter we will observe given the same source.

If the distance r1 is measured between the source and the observer given the accelerating universe, then the distance r2 measured given the decelerating universe would be r2 < r1. Therefore, this answers your second question. For the first question, the answer is directly followed the current model that predicts less matter density given faster expansion of The Universe.


There aren't any.

A 25-30 $M_{\odot}$ main sequence star would have a spectral type of $\sim$ O7 and an absolute $V$ magnitude $M_V \simeq -5$ (see Zombeck 1982). A giant star with this mass would be even more luminous.

At a distance of 30pc, the apparent magnitude of such a star would be $V=-2.6$. Closer examples would of course be brighter. Sirius is the brightest star in the sky with $V=-1.46$.

The nearest O-star may be zeta Ophiuchus which is at about 450 light years (or about 140 parsecs). This might be a bit lower than $25M_{\odot}$.  Betelgeuse, a red supergiant at a Hipparcos-measured distance of $152\pm 20$ pc, probably has  a mass of about $15 M_{\odot}$, but I think this is quite uncertain.

The secondary of $\gamma^{2}$ Velorum has an O7.5V spectral type and reasonably well-determined mass of about $30M_{\odot}$ (Eldridge 2009) and a distance of $336 \pm 8$ pc (North et al. 2007).


Deriving Dark Matter; specifically looking for a table of stellar speed versus distance from center of galaxy to derive dark matter

It is very difficult to measure Galactic orbital speeds of stars in our galaxy for several reasons.  We are moving with them and do not have a reliable measurement yet of even our own orbital motion. We only get one component of motion, the radial component (although Gaia is about to change that).  Dust in the galactic plane prevents us from seeing very far in our own galaxy.  So, when we measure Dark Matter in galaxies, it usually done in other galaxies where we can see both sides of the galaxy and, except for inclination angle, directly measure the orbital motion.


The Earth and moon orbit around a barycentre that is "about 3/4 the radius of Earth" away from the Earth's centre: https://en.wikipedia.org/wiki/Orbit_of_the_Moon. Removing the moon would almost certainly affect the Earth's orbit around the sun due to the change in the centre of mass, and probably a resultant change in momentum of the Earth system (such as it was). That's without even considering the gravitational tug of the planetoid on the Earth, and that of the moon as it gets snatched away. I would think it spectacularly unlikely for the interaction to leave Earth travelling along the orbit of the centre of mass of the old Earth-moon system. But, I haven't actually modelled it, so I might be wrong there.


How to find a place's latitude when you know a star's minimum and maximum altitude?

As barrycarter mentions in the comments, if you take the average of the two altitudes, you'll get the altitude of the celestial pole—north if you're in the northern hemisphere, south if you're in the southern hemisphere.  In this case, the average is just the average in degrees plus the average in minutes:

$$
\frac{38+12}{2} = 25 \text{ degrees}
$$

and

$$
\frac{52+24}{2} = 38 \text{ minutes}
$$

This is equal to the latitude of the observation point.  One can also obtain the declination of the star in question, by halving the difference between the two altitudes, rather than their sum, and then subtracting the result from $90$ degrees:

$$
\frac{38-12}{2} = 13 \text{ degrees}
$$

and

$$
\frac{52-24}{2} = 14 \text{ minutes}
$$

If you subtract that from $90$ degrees, you get a declination of $76$ degrees, $46$ minutes.  I know of no bright star particularly near that declination, but perhaps this is just an academic exercise.

The foregoing neglects atmospheric refraction.  This turns out to be not entirely negligible at the level of precision provided in the question.  At an altitude of $39$ degrees, atmospheric refraction produces an additional elevation of about $1.2$ minutes (barring local and daily effects), meaning that if a star is observed at $38$ degrees $52$ minutes, it's "really" at $38$ degrees $51$ minutes or so.  At an altitude of $12$ degrees, the additional elevation is about $4.5$ minutes, meaning that the star observed at $12$ degrees $24$ minutes is "really" at $12$ degrees $19–20$ minutes.  If the altitude figures given are as observed, then the corrected figures should be used instead.

ETA: Here's a page with information on the extent of atmospheric refraction.


Does the cosmic microwave background recede at the speed of light?

Well obviously the CMBR arrives (towards us, not receding) at us at the speed of light as it is a form of electromagnetic radiation and local velocities are not affected by expansion.

What I guess you may mean though is the surface of last scattering, which is the surface from which the CMBR that we receive at a given instant was emitted in the early Universe. This surface of course is no longer emitting CMBR in the present cosmological time and it is not a fixed (comoving) surface either as which surface we receive the CMBR from changes with time, however we can give a recession velocity to the comoving set of points which make up the surface of last scattering at any given instant.

In the standard model of cosmology, the Lambda-CDM model, the instantaneous surface of last scattering is receding from us at approximately 3.2c (i.e. a recession velocity 3.2 times the speed of light).


Possible Reason for the "NASA UFO sighting: Alien ship 25 TIMES size of Earth spotted near the Sun" news (does it fake or ...?)

The Express is notorious for low-quality "science" stories.
In this case, someone who wanted to see a UFO looked at images from the Solar and Heliospheric Observatory until he found a feature he could say something sensational about.
One might as well say, "I like turtles, that cloud looks like a turtle, therefore turtles come from the sky."
The mention of hollow sun "theory" is another red flag.
SOHO and its successor the Solar Dynamics Observatory have recorded many legitimate phenomena including coronal mass ejections and Sun-grazing comets.


working with stellar spectra in fits format in python

The information you need to recreate the wavelength array is in the World Coordinate System (WCS) of the header, specifically:
CRPIX1  =                 1.00
CRVAL1  =            3500.0000 / central wavelength of first pixel
CDELT1  =             0.900000 / linear dispersion (Angstrom/pixel)

which lists the starting/reference pixel of the wavelength array (1.0), the wavelength value at the start point (3500 angstroms (assumed)) and the step per pixel (0.9 Angstrom/pixel). To read this information, it is best to use a WCS library rather than trying to interpret them directly as they can be more complicated and there are many subformats of FITS WCS.
Fortunately astropy has a module to make this easy (starting from your code above and extending it):
import numpy as np
import matplotlib.pyplot as plt
from astropy.io import fits
from astropy.wcs import WCS

hdul = fits.open('s0013.fits')
data = hdul[0].data
h1 = hdul[0].header
obj_name = h1.get('OBJECT', 'Unknown')

flux = data[0]
w = WCS(h1, naxis=1, relax=False, fix=False)
lam = w.wcs_pix2world(np.arange(len(flux)), 0)[0]

plt.plot(lam, flux)
plt.ylim(0, )
plt.xlabel('Wavelength (Angstrom)')
plt.ylabel('Normalized flux')
plt.savefig(obj_name + '.png')

This will produce the following plot:
If you want to do more extended manipulation of spectra, particularly with the bewildering variety of wavelength and flux units, it might be worth looking at synphot and specutils which build on Astropy and add more direct support for spectra beyond simple numpy arrays. For example, you could make a synphot SourceSpectrum from the above by doing:
from astropy import units as u
from synphot import units, SourceSpectrum
from synphot.spectrum import Empirical1D

source_spec = SourceSpectrum(Empirical1D, points=lam*u.AA, lookup_table=flux,
                   keep_neg=True, meta={'header': h1})


Is this a potential planetary setup for the Alpha Centauri System?

The only way to answer the question about stability is to do the integration, because this problem does not have an analytic solution. There are approximate solutions for the stability of two-planet systems (although these are based on a somewhat weaker constraint that allows the outermost object to escape to infinity) but they do not necessarily generalise to more planets. Furthermore, planetary systems tend to exhibit chaotic behaviour so you aren't going to get definitive answers even with an integration, because errors get introduced by the limited precision at which the calculations can be done.

You also need to bear in mind that the parameters you've listed in the table do not give sufficient information to set up an integration:

You need the argument of pericentre ($\omega$) and the longitude of the ascending node ($\Omega$), which together with the inclination form a set of Euler angles describing the orbit's orientation in 3D space.
You need the mean anomaly ($M$) or equivalent (e.g. mean longitude, $\lambda$) to describe where the objects are along their orbits.
You need to specify at what epoch these parameters are specified for.

The integration is going to be further complicated by the fact that it isn't sufficient to just simulate Alpha Centauri AB, you also have to take into account Proxima, which may perturb the orbits of the AB pair on gigayear timescales, and is also on such a wide orbit that you probably need to take the galactic tide into account and the unknown history of stellar encounters along the system's path through the galaxy.


Will the night sky ever be more stars than darkness?

Nope! The universe's expansion is accelerating, which means there's a "horizon" beyond which light can never reach us, no matter how long it travels. Although the light always travels at the speed of light, the "amount of space" between photon and us can increase faster than the speed of light. This isn't a real "speed" - it's not any one object moving faster than the speed of light, it's all of space between us and the photon locally expanding a little bit - but it means the photon will never reach us. The acceleration means that this horizon will eventually shrink, as the expansion speeds up.
This means there's a finite amount of stuff that we can ever reach, and that can ever reach us. So you won't build up more and more distant stars & galaxies in the night sky forever.


It sounds like a tumbling satellite and by your description of the different locations in the sky during the second pass, it sounds like you saw the same one over the course of two orbits.  Maybe not, though, because it'd (most likely) be 90 mins from one orbit to the next and conditions would have to be just right for the geometry of the Earth/Sun/Satellite angle to make it visible for two orbits so brightly. But not impossible!  The Starlink satellites don't vary in brightness that way, so I can safely say it wasn't a Starlink. Tumbling sats can be wild to see, some are spinning very fast and flicker quickly (a couple of times a second) while others are doing a much slower tumble.  Sadly, there is a lot of space junk up there launched without a plan to decommission it (bring it down over the Pacific to burn up harmlessly).


Contradict the expansion of the universe?

No, it doesn't contradict the paradigm of the universal inflation. It would merely mean that the example galactic cluster that you mention has its proper (own) motion at a smaller velocity relative to the proper motion of the Milky Way than the speed of the universal inflation at the distance between them. I.e. it would move towards us, if the Universe wasn't expanding. But since it does, galactic cluster's proper motion only managed to decrease relative velocity at which it moves away from us.

To put it in analogy, say you're on a speeding train that moves at 120 km/h relative to the station it just passed by. You then lean out of its window and throw an apple towards the station that the train is now moving away from, but you can only manage to throw the apple at a velocity of 40 km/h. That apple, since it came from the train moving at 120 km/h and shared its momentum, is now travelling away from the station's platform at 80 km/h, even though you threw it towards the station, because you only managed to reduce its velocity relative to the station for 40 km/h. So it would still move away from it, but for 40 km/h slower than the train does.


What would happen if we stepped on the Sun?

Well first thing's first: You would disintegrate. At the temperature of the Sun, most of the molecules that make up our bodies could not even survive, that is why we would not only fry and die, we would really disintegrate (all the molecules breaking apart, leaving only loose atoms).

Let's now pretend heat doesn't exist. This is what would happen. First, we must remember that there is no solid surface to the Sun just like there is no surface to Uranus.

As soon as you reached the Sun itself, you would sink. The sun's density is less than 1 — and 1 is the density of water. So you would be sucked inside kind of. Except that there are convection currents. If we happen to be just above one, we might be kept close to the surface.

But, eventually, the eddies would entrain us sideways until we came to a "downdraft" that would take us downwards.

Anyway, let's talk about gamma rays now. Gamma rays are a form of light.

So even though heat is out of the way, all of these light rays are now trying to tear up whatever remaining molecules we have. This could trigger nuclear reactions.

Eventually, you (not me, I'm out of there) would cycle between two layers, in a convection cell moving you up and down, above and below the depth where the density of the gas is the same as the density of your body.

So I guess there's one sure thing about all of this.

You're going to need some sunscreen.


Why does the Moon stablize Earth's axis?

One way to look at this problem is to consider angular momentum. The Earth spins around its axis, and has therefore some angular momentum by itself. The angular momentum is proportional to the mass of the Earth, to its squared radius, and to its angular velocity. But the Earth is not alone; it has the Moon rotating around that adds angular momentum to the Earth-Moon system. And, even though the Moon is not as massive as the Earth (it is about 100 times less massive), nor rotates very fast around the Earth (and therefore a lesser angular velocity), it has a large orbit (about 300 000 km) and overall, it adds to the system an amount of angular momentum comparable to that of the Earth itself.

Now, think of a spinning-top: the faster it rotates (and thus the larger the angular momentum), the more stable it is. It is the same for the Earth-Moon system: without the Moon, the angular momentum of the Earth itself would be such that gravitational perturbations could be sufficient, in the long run, to perturbates significantly its axis (exactly like the spinning-top: if it does not rotate very fast, a small perturbation will rapidely increases and the spinning-top axis will start to oscillate more and more). But with the Moon, the global angular momentum of the system is larger, and it is therefore harder to pertubate sufficiently the system to get it oscillate strongly.

Sources:

For those who wants the dirty details, you can have a look at Laskar et al. 1993.


Could a moon be passed from one planet's orbit to another?

Given the nature of 3 body orbital mechanics, it's possible, just unlikely and I hope you'll forgive me if I skip the math.

In our solar-system, it's pretty much not possible because our planets are too far from each other.   Speaking in solar-system distances, Moons have to be close to planets, within the stable regions of their planet's hill spheres in order to have long-lasting orbits.

I'm going to assume that I can ignore temporary moons, which in my opinion, aren't real moons anyway.

The difficulty with one planet giving another planet it's moon is that they'd have to pass quite close to each other, and one of the definitions of planets is that they have their own orbital region around their star, but orbits can move over time, so it's possible for two planets to pass close to each other but there would be other effects as well because, as they got close, the planets would interfere with each other as well.

In a just right scenario a moon could theoretically be passed from one planet to another during a close pass.   This is similar to the theory about how an asteroid in a binary-asteroid pair could get captured if it passes close to a planet.

A planet could also lose a moon and maybe in a few million years that moon could find it's way to being captured by another planet, but I don't think that's what you're asking.

You might think that around a red-dwarf where planets are quite close to each other and much closer to their star, that a transfer could happen due to proximity, but the difficulty there is that planets that close to their star have a hard time having moons at all because they have very short Hill radii.   It's probably more likely, but still unlikely, to happen between two distant planets.

Take the planet Neptune, as an example.  Neptune has a Hill radius of a bit over 100 million km, about the distance between Venus and the Sun, and it's true region of stability, between 33% and 50% of that, so Neptune could in theory have a moon in a stable orbit as distant as perhaps up to 50 million km.  At that distance, it's moon would orbit it very slowly, about .37 km/s with an orbital period of about 25 years.

(A fun mathematical quirk, which has nothing to do with your question, but if you calculate the maximum orbital period a moon can have inside the stable orbital region around it's planet, it works out to about 1/6th the period the planet has around it's star, it doesn't matter the mass of the planet or the distance to the star that 1/6th rule roughly applies to every scenario).

So in theory, if Uranus passed close to Neptune, say within 100 million km, the theoretical distant moon, especially if it had a somewhat elliptical orbit, could get transferred from Neptune to Uranus and as is passed.  In the process, it would become a wrong-way moon around it's new planet.  (insert wacky theory about Triton having once orbited Uranus here).

Uranus and Neptune never get that close, though perhaps they did when our solar-system was young and Jupiter was migrating.

It's more likely, if two planets got that close that they'd alter each other's orbits rather than trade a moon.  The moon could also be ejected from one planet and not captured by the other.  That's probably a more likely scenario.   If one of the planets was considerably smaller than the other, the entire planet could be tossed around significantly, perhaps tossed out of the solar-system.

What probably does happen, is binary systems like Pluto-Charon, when they pass close enough to a planet, one of them can be captured.  This article suggests that Triton, Neptune's wrong way moon, was captured by that method.  A capture involving 3 gravitational bodies isn't that strange, but the criteria that two of those bodies are planets makes it more rare.

3 body orbits allow for a variety of scenarios, and trading a moon is probably one of the least likely options but theoretically possible.  This just might happen when solar systems are young and chaotic and full of planetesimals.

On a related note, I've read an article a while back that stars in the center of our galaxy can trade planets, but I looked, I couldn't find that article.


Get date and time by position of the sun and the observer position

I think you can very easily use the transformations available in the wiki link to obtain the Julian day, then exploit the position of observer to get local sidereal time.
Lets start by computing the Julian day
you have a given longitude $L$ which gives you $n$ number of days since Greenwich noon as
$$n = \dfrac{L-280.4860}{0.9856474}$$
giving the corresponding Julian day as
$$JD = n+2451545$$
once you have Julian day you can use the following link to obtain UT
http://ssd.jpl.nasa.gov/tc.cgi


Are the Jupiter-Saturn conjunctions and winter solstice related?

Not in any way, no.
The December solstice is the moment when the Sun reaches its southernmost point in its daily path in the sky (the June solstice, when the Sun reaches its northernmost point). It only depends on the tilt of the Earth on its orbit and the Sun.
On the other hand, Jupiter and Saturn being in conjunction is a phenomenon that doesn’t depend at all on the Earth’s axial tilt. It does involve the Earth, though, as it’s our vantage point. From Earth, Jupiter and Saturn appear close in the sky then. Seen from Mars, for example, they may or may not appear close together in the sky at the same moment.
The last time a Jupiter–Saturn conjunction happened near a solstice was in June 2000. You can find a list of Jupiter–Saturn conjunctions happening within ±24 hr of a solstice at https://astronomie.quebec/conjsols.html.
Solstices, on the other hand, happen twice a year (one in June, one in December).


Calculate right ascension in Zenith at this moment in my location

Right ascension at Zenith should always be your local sidereal time.


Is an elliptical orbit around a black hole, or an orbit consisting of three bodies (including the black hole) possible?

Elliptical Orbit

Yes, an elliptical orbit around a black hole is possible.  Provided the orbit doesn't cross too near the event horizon, the orbital mechanics are the same as orbiting any massive object.  Gravitational time dilation is a function of distance only, so the time dilation experienced at any point in the orbit should be easily calculated and presumably predicted ahead of time or calculated for past times.

Three Body Orbit

When you mention an orbit consisting of three bodies, do you mean the black hole, another massive object, and a much smaller object?  Or do you mean three massive bodies?

If the orbit consists of the black hole, another massive object, and a smaller body, the system should be predictable as a restricted three body system.  This is made even easier if the smaller body finds its way to a Lagrange point, or resides in a stable repeating orbit.  There are limits to the predictability of the restricted three body cases, which differ based on requisite constraints (e.g. circular vs. elliptical orbit for the smaller body).  If the constraints are not met, or near their limits (not hard limits), the predictability of the orbit will suffer and thus the predictability of the time dilation will suffer as well.  Constraints for the elliptical restricted three body problem include the mass of the small body being "negligible" relative to the two larger bodies, and the ability to approximate the co-orbit of the larger bodies as Keplerian between point masses.  Thus if the mass of the smaller body is some appreciable fraction of either larger body, or if the larger bodies orbit very closely to one another, this may not hold.  The circular restricted three body problem adds the constraint that the co-orbit between the larger bodies must be circular, so if it is near-circular that condition may or may not be said to be met.  Whether constraints are met would be most easily judged by setting a time over which predictability of the system may be maintained.  So if the goal is to be able to predict the system for 1000 years, and the smaller mass is small enough that the system is predictable under the restricted three body problem, the system can be said to meet the constraints for those purposes.
If the orbit consists of the black hole and two other massive bodies, the system will not be perfectly predictable (this may not be totally true, apparently solutions exist in the form of convergent power series though they may not be practical due to their size).  If this is the case, time dilation will not be predictable, but will be calculable for the present and past (provided there is position data from the past).


Auroras: is there any bump in frequency at certain times of night, or times of year?

Magnetic midnight occurs when the sun, the observer and the magnetic poles are in the same plane. This occurs around local midnight: In the USA it is about 1 hour before local midnight, in the UK it is about 2 hours before local midnight (local time is the time as shown on a sundial, with no adjustments for daylight saving etc)

The aurora season is the winter, mostly because at the high latitudes where aurora are visible, the sky is not fully dark between April and September.

Aurora are most likely during solar maximum, which occurs every 11 years (with a good deal of variability). Solar activity remains high for about 4 years. There have been maxima in 2000 and 2014 (the last cycle has been long and weak) The next is due around 2023-2025.


Yes ... and no.
Experimentally, there's no indication that the laws of physics are changing (c.f. conservation of energy: if the laws of physics change with time, energy would not be conserved via Noether's theorem). So if you're an empiricist, you'd conclude that the universe is stable (or at least its mean lifetime must be more than ~14 billion years, the age of the universe).
However, if we believe string theory (which has no compelling experimental evidence), then we expect roughly $10^{500}$ "vacuum states", which correspond to $10^{500}$ different possible universes, each with their own laws of physics. It's possible one of these other universes have a lower "total energy" than us (this word's meaning is vacuous since if one varies the laws of physics, the word 'energy' itself might lose meaning). This means it's in principle possible that our universe will quantum tunnel to a universe at lower energy, in which case our universe is metastable.
Of course, this leads to other problems. How would one prove the multiverse exists? If a generally accepted way exists, I've not heard of it. But if you can't prove the multiverse exists, how would you prove our universe is metastable? For that matter, how do we know the electron is stable? Why can't it have a mean lifetime of, say, $10^{100}$ years (so long that we don't observe it decaying)?
At this point, we're purely in the realm of speculation. We're talking about things so far above the experimental threshold that some people will just dismiss it as pointless. Suffice to say that if the universe isn't stable and quantum tunnels to a lower energy, we won't be able to see the tunneling coming, and we're all instantly dead anyway. Whether the question is worth thinking about further depends on the individual.
tl; dr:

Experimentally, universe is stable.
Theoretically, maybe it isn't, depending on what your theory is because we don't have a widely-accepted, experimentally-backed theory.


The ecliptic is in periodic wobble wrt the celestial sphere, but that's because of nutation, not precession. Precession (nodal precession) changes where the Sun is during the equinoxes and solstices. The Spring Equinox, for example, has moved from Aries (it's still called "the first point of Aries") to near the Pisces-Aquarius border


Typical sunspots have a dark region (umbra) surrounded by a lighter region, the penumbra. While sunspots have a temperature of about 6300 °F (3482.2 °C), the surface of the sun which surrounds it has a temperature of 10,000 °F (5537.8 °C).
From this NASA resource:

Sunspots are actually regions of the solar surface where the magnetic field of the Sun becomes concentrated over 1000-fold. Scientists do not yet know how this happens. Magnetic fields produce pressure, and this pressure can cause gas inside the sunspot to be in balance with the gas outside the sunspot...but at a lower temperature. Sunspots are actually several thousand degrees cooler than the 5,770 K (5496.8 °C) surface of the Sun, and contain gases at temperature of 3000 to 4000 K (2726.9 - 3726.8 °C). They are dark only by contrast with the much hotter solar surface. If you were to put a sunspot in the night sky, it would glow brighter than the Full Moon with a crimson-orange color!

Sunspots are areas of intense magnetic activity, as is apparent in this image:

You can see the material kind of getting stretched into strands.
As for the reason it is cooler than the rest of the surface:

Although the details of sunspot generation are still a matter of research, it appears that sunspots are the visible counterparts of magnetic flux tubes in the Sun's convective zone that get "wound up" by differential rotation. If the stress on the tubes reaches a certain limit, they curl up like a rubber band and puncture the Sun's surface. Convection is inhibited at the puncture points; the energy flux from the Sun's interior decreases; and with it surface temperature.

All in all, the sunspots appear dark because the are darker than the surrounding surface. They're darker because they are cooler, and they're cooler because of the intense magnetic fields in them.


What theories are there in explaining the formation of the Pluto-Charon system?

The favored theory is a giant impact, at the end of the final stage of planetary formation (Canup, 2005). A "giant impact" is a collision with an object of a comparable size of the planet. That is also the most probable scenario for lunar formation (Canup, 2004).


Frequency of gravitational wave detection

Harry (2009) cited several different sources stated that, as far we know, the rate of detectable events will be

40 neutron star mergers per year
30 10 M$_\odot$ black hole mergers per year
10 neutron star/black hole mergers per year

This is within a radius of about 200 Mpc. This cannot, however, be used to extrapolate the total rate of such events, because of detection bias - the more massive the objects, the more easy they are to detect. The same thing happens with exoplanets, but for different reasons (e.g. planets that are more massive or closer to their stars are easier to detect via transit or by radial velocity methods).


Neutrino Modelling in Friedmann Equation

The energy density of a Fermi gas is
$$\rho_{\nu}= \int \rho(p)\ dp = \int E(p)F(p)g(p)\ dp $$
$$\rho_{\nu}  = \int \left(\sqrt{p^2 c^2 + m^2 c^4}\right)\left(\exp (E/k_BT) + 1\right)^{-1} \left(g_s 4\pi p^2/h^3\right)\ dp$$
in units of energy per unit volume.

Before neutrino decoupling at $k_B T \sim 1$ MeV, the neutrinos are ultrarelativistic with $pc \gg m_{\nu}c^2$. After decoupling, the shape of the occupation index function $F(p)$ does not change - so $F(p) = \left(pc/k_BT_{\nu} +1\right)^{-1}$ in subsequent evolution.

Thus
$$\rho_{\nu} = \frac{g_s c}{h^3} \int \frac{ \sqrt{p^2 + m_{\nu}^2c^2}}
{\exp(pc/k_BT_{\nu}) +1}\ 4\pi p^2\ dp$$

I don't understand where your $(2\pi)^3$ comes from, other than to suggest that the unit system is actually $\hbar = 1$.


What is the closest an asteroid or comet has passed the Sun and survived?

Wikipedia offers some useful info on sungrazers:

  As of October 21, 2017, there are 1495 known comets that come within ~12 solar radii (~0.055 AU).[2] This accounts for nearly one third of all comets.[3] Most of these objects vaporize during their close approach, but a comet with a nucleus radius larger than 2–3 km is likely to survive the perihelion passage with a final radius of ~1 km.

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

There's a bit about pyroxenes sublimating off at 7 solar radii, but no mention of whether any such object made it around the sun. Pyroxenes generally melt at around 1000°C.
An Iron/Nickel body melting at 1538°C/1455°C might well make it through a 7 solar radii passage.


Is there a clear-cut difference between rocky and gaseous planets?

The currently-detected planets do not show a clear distinction between rocky and gaseous planets. While there seem to be somewhat two ensembles, the rocky planets of Earth-size and "super-earths" and the gaseous planets (Jupiter-size), there is a broad transition between them. The transition is roughly where we find Neptune and Uranus. Play around with plotting various parameters at exoplanet.eu

It's still subject to debate whether this apparent depletion in the 'ice giants' (cold, gaseous, but with heavier gases like Methane etc instead of Hydrogen and Helium as main constituents) is real or due to observational bias.

In brevity:

The argument for "is real" is that it is more likely that in the inner nebula, inside the snow zone you have enough material to form terrestrial planets, but being too hot and too few to grow to a size that a sizable atmosphere can be retained. Outside the snow zone you have plenty of more solids (all the ices), so that you form cores which become big enough to also accrete an atmosphere.

However, the counter-argument is that you do find planets of all sizes in all kind of stellar distances, you have various initial conditions with different densities of solids in protoplanetary discs, and thus any final result is possible with possibly various probability. The actual observations seem to support this.

Generally, in current literature and on conferences, planets are often distinguished further than just rocky and gaseous: you have terrestrial, super-earths, neptune-likes and the gas giants, and often then somewhat separated even the so-called hot jupiters and hot neptunes, the gas giants which migrated inwards after formation.

EDIT: one possible distinction which could be made between rocky and gaseous is looking at the atmosphere : if the atmosphere gets dense enough at the lower levels that the distinction between gas and liquid vanishes (thus the critical point reached) , it's a gas planet, otherwise a Rocky planet with atmosphere.


How much oxygen did the Warwick/Cambridge study find in the rocky debris around white dwarf GD 61?

In Farihi et al. (2013) (it's a Science paper, unfortunately I'm not sure its content is freely accessible), they actually measured metal excess in the white dwarf GD 61 (for an astronomer, everything that is nor hydrogen neither helium is a metal). Due to high surface gravity in white dwarfs, any heavy element should sink rapidely in its atmosphere; therefore, they infer that, if you find traces of metals, it means that it comes from accreted material, "polluting" its atmosphere.

So if you find oxygen in a white dwarf spectrum, you know that it has to come from somewhere else, and if the white dwarf has a circumstellar disk, it has to come from planetesimals that are accreted from the disk.

Now, what they did was to measure abundances (or upper limit for their abundance) for a bunch of chemical elements (O, Mg, Al, Si, Ca, Fe) and carbon. Then, they could infer different things to determine the total oxygen budget:

There is a carbon deficiency, that means that carbon abundance has no impact on the total oxygen budget;
Elements Mg, Al, Si, Ca, are assumed to be carried as MgO, Al2O3, SiO2, and CaO at their maximum abundance, so once they had these abundances, they could substract it to the total oxygen budget;
The remaining oxygen excess, found in FeO and debris, is interpreted as water-rich (if you want to form FeO, you need water somehow).

That's how they were able to determine this water abundance in debris around the white dwarf GD 61.

I'm not sure the oxygen abundance would be very meaningful here, but since you asked for it, it is of -5.95+/-0.13 dex.


This expression is valid for low frequencies, including the case of the 1420 MHz hydrogen line. It arises from treating the source as a black body with temperature $T(\theta,\phi)$$^{\dagger}$, and assuming $h\nu\ll k_BT$:
$$I_{\nu}=B_{\nu}=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/k_BT}-1}\approx\frac{2h\nu^3}{c^2}\frac{1}{h\nu/k_BT}=\frac{2k_BT\nu^2}{c^2}$$
where we have made the approximation
$$e^{h\nu/k_BT}\approx1+\frac{h\nu}{k_BT}$$
and I drop the explicit $(\theta,\phi)$ dependence. The flux density is then, by definition (and assuming the source is small, so $\cos\theta\approx1$),
$$S_{\nu}=\int I_{\nu}\cos\theta\;d\Omega\approx\int I_{\nu}\;d\Omega=\frac{2k_B\nu^2}{c^2}\int T(\theta,\phi)\;d\Omega\tag{1}$$
If we calculate the mean of the function $T(\theta,\phi)$ over the source, we get
$$T_{\text{mean}}=\frac{\int T(\theta,\phi)\;d\Omega}{\int d\Omega}$$
and some algebra and noting that $\int d\Omega=\Omega$ should convince you that we can make the desired substitution. We can essentially always take the mean of a well-behaved function when integrating over a patch of sky.

$^{\dagger}$ The brightness temperature $T_b$ is typically defined as the temperature of a black body which would yield the same specific intensity.


What if the atmosphere of Titan were like the one on Venus?

If the pressure were the same as the current pressure on Titan, the planet would be slightly warmer for a while, but not for long. The CO2 would soon be frozen out. It is probable that 4 billion years ago the atmospheres of Venus,Earth,Masrs and Titan were much more similar. A hotter Saturn may have warmed Titan in those days. It probably,like the others, had a lot of CO2 in its atmosphere,but that has long ago been frozen out. There is plenty of frozen water on Titan, and if gelogists ever dig down far enough,they are likely to find a layer of frozen CO2 there as well.


Energy deposition in planet Earth due to gravitational waves

Don't know about the energy transfer but from a rough assumption we can calculate the energy amount that actually passed through Earth depending on the mass converted and the distance of the event. The energy absorbed by Earth would be far less than that obviously.

The measured event was two black holes merging with a total mass loss of $3 M_{\odot}$. While the distance for that event was calculated to $1.3*10^9 ly$.

With comparing the total energy released (assuming full conversion of mass to gravitational wave energy) over the released area:

$$
E_{Earth} = \pi r_{Earth}^2*\frac{m_{conv}c^2}{4\pi r_{dist}^2}
$$
$$
E_{Earth} \approx (6378km)^2*\frac{3M_{\odot}*c^2}{4*(1.3*10^9years*c)^2}
$$
$$
E_{Earth} \approx 4.0*10^{13}m^2*\frac{5.7*10^{30}kg}{6.72*10^{33}s^2}
$$
$$
E_{Earth} \approx 3.39 * 10^{10}J
$$

So the maximum amount of energy passing through earth would have been roughly $34 GJ$. That is about the kinetic energy of three Boeing 747 in flight - or neglectable on a planetary scale.

So even IF the total energy would have been absorbed by Earth, the amount would still have been - not much.


The statement "They're only emitting infrared light" is wrong, or at least poorly phrased.

The Ly$\alpha$ forest

The Ly$\alpha$ forest (LAF) is caused by the spectrum of the quasar being redshifted along its way; wherever a cloud of neutral hydrogen is located, the part of the spectrum that at that at that position has been redshifted to $\lambda_0 = 1216\,\mathrm{Å}$ has a high probability of being scattered. As the probability of being scattered back into the line of sight is negligibly small, the spectrum effectively suffers absorption, but in fact the light is not absorbed.

Because the lines are caused by the Ly$\alpha$ line, it only affects the part of the spectrum blueward of the quasar's "intrinsic" Ly$\alpha$ emission (well, similar but weaker spectral featured are seen e.g. at Ly$\beta$, and for very strong absorbers such as damped Ly$\alpha$ absorbers you also see metal absorption).

The Gunn-Peterson trough

The further back in time you go — i.e. the higher the redshift you observe — the higher the fraction of the gas of the Universe was neutral, and the denser the Universe was. As you approach the Epoch of Reionization, the LAF "absorption" lines begin to overlap, until eventually, around $z\sim6$, everything blueward of the quasar's Ly$\alpha$ line is erased. This is the so-called Gunn-Peterson trough.

A quasar seen at $z \geq 6$ has its Ly$\alpha$ line redshifted to
$$
\lambda_\mathrm{obs} = (1+z) \lambda_0 \geq 8\,500\,\mathrm{Å},
$$
which is in the infrared. But it is incorrect to say that the quasar only emits in the infrared.

Reionization

The reionization started with the first light-emitting sources, most likely hard UV from the first galaxies; a non-negligible fraction may also be from quasars, but this is probably only significant at later epochs. That is, it started already at $z\sim12$ or higher, and by $z\sim6\text{–}10$ the Universe was almost fully ionized.

So if most gas is ionized at $z=6$, why is all light blueward of Ly$\alpha$ erased? This is because the Ly$\alpha$ cross section is so large: it doesn't take more than a neutral fraction of $10^{-4}\text{–}10^{-3}$ to create the GP trough.


Three times the Schwarzschild radius corresponds to the closest stable circular orbit around a black hole. The general idea is that as matters moves in towards the black hole it gets stuck in an accretion disc where angular momentum has to be moved outwards in order to allow the matter to move inwards. The generic mechanism is some sort of viscosity, which heats the gas and hence you get radiation.

However, once the matter gets inside $3r_s$, that problem disappears. There are no stable orbits, no angular momentum loss or viscosity is needed and the material is able to flow (rapidly) straight into the black hole.

Thus when we observe black hole accretion discs we expect them to be truncated at $3r_s$.

So I think the argument then is along the lines of - the gravitational potential energy of unit mass falling to $3r_s$ is converted into an orbital kinetic energy of $0.5v^2 = GM/6r_s$ per unit mass and the rest is converted to radiation. Thus
$$L = \left[\frac{GM}{3r_s} - \frac{GM}{6r_s}\right] \frac{dM}{dt}$$
$$ L = \frac{GM}{6r_s} \frac{dM}{dt} = \frac{1}{12}c^2 \frac{dM}{dt} .$$


Cannot calibrate “upgraded starpointer finderscope” for Celestron AstroMaster 114EQ

I do not have one of these scopes, so my answer reflects general knowledge of astronomical telescopes.

Yes, the lens should be at the far end, away from your eye.

As for the alignment question, you can probably get a bit more movement in "left/right" direction by loosening the screws that hold the base of the finder to the scope, and twisting the mount and then re-tighten it.  For the up/down adjustment, you may need to use a shim (a thin bit of metal) to adjust the angle, but the adjustment built into the finder is normally adequate.

I doubt if there is a problem with the scope - if you can get a good focus, that's fine.


How much solar energy concentrates at Jupiter's focal point via gravitational lensing?

Solar energy - from our Sun at Jupiter's focal distance?    A negligible amount.    Gravitational lensing doesn't have a focal point so much as a focal region that begins roughly at a point, and that point (which really shouldn't be called a focal point) can be calculated, ProfRob gives the formula here.
Jupiter's gravitational lensing distance is so far (about 10 times the Sun's or about 5500 astronomical units), that Jupiter essentially appears 1/10th the size of the Sun from that distance, so it doesn't really work at focusing the Sun's rays.  You still get a lot more light direct from the Sun, which at that distance, would just be the brightest dot in the sky, with Jupiter blocking more light than it lenses in the observer's direction.   You get more light from the Sun when Jupiter isn't blocking the sun at that distance.
Gravitational lensing works best when the lensing object is either very massive, so it pulls in light from a sizable area, or if it's larger than the object that's being studied.  That is, You could use Jupiter's gravitational lensing to study an exoplanet, which is incredibly tiny to our observation light years away, but the gravitational bending of light by Jupiter gives a very distorted image of the incredibly tiny/distant object, but that distorted image would have a great deal more information.  In that scenario, Jupiter would focus much more light from the very distant, very dim object, but only because the object is effectively so small by our observation.
Maybe somebody here knows how to run the math for gravitational lensing brightness, but it varies on how pin-point the object you want to get a better look at is, so it's not a simple one line formula.
With the case of the Sun and Jupiter, the Sun being larger, it doesn't really work.   A black hole could image something larger than itself, but a black hole bends a lot more light than Jupiter does and it's so called "focal point" begins much closer.
A magnifying glass, which focuses sunlight to a point (and this only works because the sun is so distant that the rays of light are essentially parallel), that focused sunlight can never be hotter than the surface of the Sun itself.   That's a law (xkcd did a nice bit about that)
The focused (but very blurred) image that a planet makes at it's gravitational lensing point wouldn't come close to hot enough to cook ants, if that's your goal.   A better way to think about it would be looking through a telescope, but the image is spread out into a ring but not that bright.   Maybe with a perfect scenario, like a black hole or neutron star and a star in the distance, you might get some brightness, but still much less concentration than a magnifying glass can give you because it doesn't focus at a point, it's more of a range.
If you make your question more specific, somebody might be able to give you a more exact mathematical answer.


Changing Appearance of the Moon During a Single Night

If you take a line from the middle of the terminator, but at 90 degrees to the termination, it more or less points towards the sun.  Now consider the moon at first quarter: as the moon rises, the sun is due south, so imagine a line from the moon to the sun, and the terminator will be at 90 degrees to this, so pointing from bottom right to top left.  At mid-day, the sun is in the west, and the terminator will be sloping bottom left to top right.


The Moon orbits the Earth in such a way that it sometimes can be on the opposite side of the Earth as seen from the Sun. In other words, it’s possible to have Sun-Earth-Moon along an imaginary straight line, in that order.
However, by Ptolemy’s time, it had already been noticed that Venus is always seen close to the Sun (maximum angular distance about 47°). So it can never get opposite to the Sun as seen from Earth.
Ptolemy didn’t know why (the reason is that Venus orbits the Sun, not the Earth, which was inconceivable to Ptolemy), but his system made it so that Venus (and Mercury, for that matter) were orbiting the Earth closer to it than the Sun, and in such a way as to always be near it.
So imagine moving your two hands at arm’s length, but one closer to your face than the other. Imagine the distant one is the Sun. Then your closest hand would never get completely lit. Unless you put it behind your back, where Venus (and Mercury) never go.
So that’s why in the Ptolemaic system, Venus (or Mercury) would never get fully lit.


How many wandering black holes are there in the milky way?

Yes as StephanG said there are thought to be 100 million black holes in our galaxy composed of approximately 100 billion stars. Most of these are thought to be stellar mass black holes. Not the super massive black hole assumed to be at the center of our galaxy. Some references are:http://hubblesite.org/explore_astronomy/black_holes/encyc_mod3_q7.html and https://www.sciencenews.org/article/we-share-milky-way-100-million-black-holes.


I think she should have clarified the question by putting velocity and radius in the same choice.

By looking at her answer, it seems reasonable to think that velocity is more important. However  without knowing the radius we cannot determine the mass of the stellar system. She argued this idea by saying that the size is approximately known for stellar systems. But I don't think that is a solid argument.

Theoritically we should know both R and V to calculate M.

Note: I did not understand why you wrote kinetic energy= gravitational force ? Thats not possible.

In stellar bounded system you should use virial theorem. By stating that $-2<E>=<U>$  and from there you get

$$M=v^2R/G$$


Do meteorites streak across the sky in the same direction?

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

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

Here a more complete list of meteor showers.


Why is frequency different for gravitational waves coming from SMBHs vs small compact binaries?

The reason for the upper limit is that gravitational force scales as the product of masses over separation squared and that the closest approach of black holes is limited by their event horizons. The lower limit is because the amplitude of the waves gets smaller and becomes undetectable at lower frequencies.

In more detail:

The frequency of gravitational waves is twice the orbital frequency.

Kepler's third law tells us that the orbital period $P$ is related to the total mass $M$ and orbital separation $a$ as
$$ P^2 \propto M^{-1} a^3 $$

The orbital frequency $\omega =2\pi/P$ so
$$\omega \propto M^{1/2} a^{-3/2} \tag{(1)}$$

The frequency of gravitational waves therefore depends on both the mass and separation of the binary system.

However, the maximum frequency occurs at closest approach, which for black holes is when their event horizons come into contact. The Schwarzschild radius is $2GfM/c^2$, where $f$ is the fraction of the system mass of each black hole. Thus the closest separation $a_{\rm min}$ is the sum of the Schwarzschild radii and is simply $2GM/c^2$, since $f_1+f_2=1$.

Using $a_{\rm min}$ for $a$ in equation (1) we see that the maximum frequency
$\omega^{\rm max} \propto M^{1/2}a_{\rm min}^{-3/2} \propto M^{-1}$
Thus the maximum frequency gets bigger as the mass gets smaller, and as supermassive black holes are a million to a billion times more massive than stellar black holes, the maximum frequencies are correspondingly smaller.

The lower limits to frequency are not really limits at all. The frequencies can be anything smaller than the maximum. However the amplitude of the gravitational waves depends on the acceleration of the masses, which in turn decreases as they become more widely separated. This effectively gives a window between where the waves become detectable and the maximum frequency.


How to determine zenith coincidence from date time data sets

Technically, the location of the zenith is constant: 90 degrees altitude, undefined azimuth. :-) I assume that you want it in terms of the coordinates declination and right ascension instead of altitude and azimuth. Declination and Right Ascension are the coordinate system used to locate objects in the sky. They are similar in concept to latitude and longitude. The difference is that one is "locked" to the sky, and the other is "locked" to the Earth.

The declination of the zenith is the same as the observer's latitude. If your location is the same, then the declination is the same. (If the latitude is different, the coordinate of the zenith is different.)
The right ascension of the zenith can be calculated by using the formulas for the local sidereal time (LST or LMST for the mean local sidereal time.) See the post Local Sidereal Time.


A model is fitted to the signal that depends on various input parameters. The input of the model that fits the signal best is then considered the "most favored" model, while the input of models that fit better than a given threshold are used to infer confidence intervals on the various parameters.

The model has to simultaneously reproduce as well as possible the frequency and the amplitude of the signal. Among the input parameters are the black hole masses ($36_{-4}^{+5}\,M_\odot$ and $29_{-4}^{+4}\,M_\odot$ ) and (luminosity) distance ($410_{-180}^{+160}\,\mathrm{Mpc}$), corresponding to a light travel time of roughly 1.3 Gyr, depending on the assumed cosmological density parameters (they used those of Planck 2015).


For a small region not far from the equator,
an equirectangular projection will do.
If you want celestial north to be up,
declination should increase from bottom to top,
and right ascension should increase from right to left.
To correct the aspect ratio, scale the RA axis by cos(dec).
Then the sun and moon are circles of the respective angular radii, without the RA scale factor.

In the equation for $z$, $\Delta$ is the declination of the sun, $\delta$ is the declination of the moon, and $a$ is the difference in their right ascensions.


How are the gaps filled in images captured by CCD arrays?

In general, the common practice is to dither exposures and fill in the gaps. The dither serves multiple purposes:

provides some exposure in the gaps between the CCDs
smooths out pixel-to-pixel variations in the response.
prevents one bad pixel from ruining an entire observation

The dither pattern varies from telescope to telescope. For example, the following image shows the dither pattern of Chandra telescope.

Image from havard.edu

The effect of dither is removed during high-level ground processing of the data.


Is there any positive N on which it's agreed among astronomers that 10^-N is accepted as 0 (zero)?

No, there isn't.

Astronomy overlaps with cosmology and there can be some pretty big and small numbers.

However most physical parameters also have units and the units chosen affect the numerical values.

A great example is the Jansky:

  The jansky (symbol Jy, plural janskys) is a non-SI unit of spectral flux density, or spectral irradiance, used especially in radio astronomy. It is equivalent to 10^−26 watts per square metre per hertz.

In some cases one might even imagine going the other way and using L☉/ℓP² myr (solar luminosities per square planck-length per million years) which would require numerical values about 1x10^-110 times smaller.

Instead, astronomers and other scientists usually decide is something is small enough that it doesn't matter within the context of the current discussion. THey use words like negligible and small to indicate they're going to ignore it.


If two black hole event horizons overlap (touch) can they ever separate again?

You have already got some good answers, but I'll just try to provide one more intuitive solution on why the event horizons will never separate again if overlapping each other:
First, imagine a speck of dust that comes inside the EH of a black hole. I believe we'll agree this speck can never escape the black hole, because nothing can come back from behind the event horizon.
Now, imagine the same speck of dust, but inside the overlapping parts of the EH of two black holes passing each other. This speck of dust will never escape any of those two black holes, because it is inside the EH of them both. If these black holes would be able to separate again, the speck caught between them would obviously escape at least one of the black holes, after being behind it's event horizon.
Since this can not happen, the two black holes will be united from the point their event horizons are overlapping, no matter their speed.


What if you 'placed' the Chicxulub asteroid onto the Earth?

An asteroid resting on Earth would be a mountain. Or, for smaller asteroids, a pile of gravel.
Mountains are limited in altitude by the strength of stone to resist compression: a too tall mountain would sink down as the base crumbled and spread out. The limit on Earth is about 10 km. Besides the strength issue mountains are also floating ("isostasy") on the fluid but denser interior of Earth, so even if it doesn't fall apart it may sink down partially.
So the impactor would not hold together that well. A big ball of rock 10 km in diameter would first definitely sink down several kilometres into the crust, and almost definitely break. That means that something like $10^{15}$ kg will descend several kilometres, giving an energy $mgh \approx 10^{19}$ J (and there are estimates 100 times larger for the mass). That is "just" like a big hurricane, but will mostly be released as localized seismic energy and heating. The end result would be a massive pile that likely destabilizes the crust in the vicinity a lot: far less than an actual high velocity impact, but still a major disaster.
If you do the same with the moon you will break up the crust, boil off the oceans and wipe out all life on the planet.


Milky Way stellar number density : is the stated equation in this paper incorrect?

It may make more sense to convert the exponential into the product of the individual exponential terms:
$$
\exp(-\frac{R}{L}) \exp(-\frac{Z}{H}) \exp(\frac{Z_{\odot}}{H})
$$
This makes it clearer that the $Z_{\odot}$ term is a constant, which is just part of the overall normalization (similar to the $\exp(R_{\odot}/L)$ term earlier in the original equation).

And, yes, $Z$ is assumed to be always positive, so $|Z|$ would be slightly more correct.


Can I see Jupiter, Saturn, Mars and Venus with the Spectra Optics Telescope 600x50?

Generally speaking, you should never buy any telescope which advertises itself in terms of magnification.
Telescopes at this level are, unfortunately, not much more than toys which will serve more to frustrate than to delight.
These telescopes are made almost exclusively from plastic parts - the optical tube, the focuser, even the eyepieces are a polymer and not a glass. The tripods are flimsy aluminium.

If you only have €50 to spend then you could buy a pair of 8x42 or 10x50 binoculars. At this price they won't be a great pair of binoculars, but they'll show you a lot of objects in the night sky which you can't see with the naked eye alone.
More positive points in favour of binoculars:

you can use them during the day
they're very portable
they'll continue to be useful as you continue in the hobby --- I still use my €40 10x50s to sweep the sky even though I have a reflector and a refractor; each instrument gives a different view

Alternatively, you could put that €50 into a savings account and then try to add to it at every opportunity you get. Your goal would be to save up around €200
A telescope which is generally accepted as being one of the best starter telescopes is a 130mm F5 reflector sold in the USA under the name "Astronomers Without Borders OneSky" and in Europe under Sky Watcher brand with the name "Heritage 130p".
It's a Newtonian reflector on a Dobsonian mount. It has an aperture which is large enough to show you details on Jupiter, Saturn, and Mars (around opposition, at least).

In between the €50 binoculars and the €200 reflector there are many achromatic refractors, but I hesitate to recommend any of these because you mainly want to observe planets:
When light passes through a refracting telescope's objective lens, different wavelengths are diffracted different amounts, and this results in some colour separation which is most noticeable with bright objects.
Planets are some of the brightest objects we observe.

Also, many of the aforementioned refractors also suffer from less-than-brilliant build quality, which is another reason to go for the the 130mm reflector. Here's how I think about it:
At this price point you want to make sure that as much of your money as possible is going into what's important: the optics.
Anything else: bundled eyepieces, diagonals, tripods, equatorial mounts, ...all of this costs money, and money going into these components is money that isn't going into the optics.
This is why a Newtonian reflector on a Dobsonian makes the most sense: the optics are mirrors and the mount is a couple of bits of wood. Telescopes don't get any more simple than that :-)

You mentioned in a comment that it was hard to find the Heritage 130p in Europe.
Right now it's hard to find any telescopes anywhere - the pandemic has created a lot of new amateur astronomers and has also affected supply from China.
Here's a short list of European shops I can think of which you might want to check out:

Astroshop
Bresser
First Light Optics
Teleskop Service

Of these, if you're looking for good deals then keep and eye on Bresser's and FLO's special offers pages. (If you're in the EU then be careful ordering from FLO because they're based in the UK and you'll have to pay additional taxes on your orders). Bresser is probably the place where I've managed to get the best deals.
Finally, consider buying second-hand. Your money will go a lot further.


How to estimate age of asteroid family Erigon

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

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

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

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

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

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

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

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

You can check out more details in my source below.

Source:

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


Why could Quasi-stars ("black hole stars") have only existed when everything was hydrogen and helium? (no metal "contamination")

The presence of heavier elements makes the medium absorb more radiation. This means a nascent star that would have been able to collapse in the absence of metals, will lose its outer layers to radiation pressure as the core starts to heat up.


My interpretation would be that these are the planets Jupiter, Saturn and Mars.

The bright star is Jupiter. It is rather lower than 20 degrees above the horizon, it is at about 10 degrees, but rising.  The two other stars, one brighter than the other are Saturn (brighter) and Mars.  Mars is really red, Saturn may be reddened by being low in the sky.  Saturn is about 5 degrees East of Jupiter and Mars is a couple of degrees more.

To get your bearings, locate the "summer triangle" of Vega, Deneb and Altair. This triangle (currently) points towards the area containing the three planets. (Draw a line between Deneb and Vega, and through Altair and continue on a great circle in the sky)

I used Stellarium to view a simulated sky from Eindhoven at this time.


I must say that I agree with you, this way of derivate the Kepler law is not the most intuitive, it's maybe for this reason they specify: "revisited".
R1: The reason is stated at the very begin of the chapter, where you derive the relation between $r$ and the angle to the perihelion $\theta$ for the general case of an ellipse (Eq 3 in your book in Cap 2.1 Elliptical orbit, but I can see I have an older version than yours), where:
$$ r = \frac{a(1-e^2)}{1+e\,cos(\theta)}$$
here, the eccentricity of a ellipse is then define as:
$$ e = \sqrt{1- \frac{b^2}{a^2}}$$
in the case of Kepler's law you can define the eccentricity as $e = D/\mu G M$ and  $a$ as:
$$ a = \frac{L^2 G\, M}{\mu^2\,G^2\,M^2 - D^2}$$
and you can work out in equation (29), obtaining the equation in the same form as for an elliptical orbit.
The reason to derive $r$ is because it indicates the distance of the object from the focal point (centre of mass of the system), which in the case of the solar system, is the distance of the planet to the Sun (actually the focal point of the solar system, or better to say of the Sun-Jupiter system, is a few kilometre above the solar surface).
R2: the eccentricity $e$ quantify the shape of a cone cross-section Eccentricity (math) Wikipedia.
For example, in the case of the an ellipse you have $0 < e < 1$, and it indicates how much the ellipse is "squashed". Another example, if you have $e=0$ than you have a circle.
In the case of $e=1$ you have a parabola, from the equation of $e$ above it means that $a\gg b$. In a more physics sense, it means that the system is not in a gravitationally bounded state, and the less massive object $m_1 \ll m_2$ is on a parabolic trajectory, deviated by the gravitational pull of $m_2$, and will not stay in orbit.


How do pressure, temperature, and gravity affect a smaller body than Earth with the same bulk composition?

If you know for sure and can prove it: write a paper, submit it to Nature or Science. They will accept it and will give you the cover image.
We can only look at the planetary interiors superficially. We know the bulk mass, we can do a bit tomography with seismics to determine boundaries in the speed-of-sound, with magnetotelurics in the magnetic and electric permeability and maybe a bit with gravity and rotational information on the distribution of masses inside and their surface compositions.
We can make reasonable assumptions on the chemistry and our knowledge how these materials behave. We can even make experiments where we try to re-create the conditions within the interior - but that's where the difficulty starts: temperatures of a few 1000°C and especially pressures over long duration are something which concurrently are hard to re-create in a laboratory (e.g. Dorogokupets et al) , thus many information are from shock-experiments where the pressures only persist very shortly.
In essence the equations-of-state for the materials inside the planets cores are only approximately known, deduced as well as possible, interpreting available information as consistently as possible. This level of detail, in getting all the internal structure right is of course tremendously easier on Earth with all our seismic, magentotelluric and gravity stations and plethora of satellites measuring all things also from the exterior. Data on other planets is MUCH more sparse, so information on the internal structure must be taken with many more grains of salt than the ones of Earth. Many is assumed by analogy, taking into account the few information available while matching the physical models we can make of the chemistry and their assumed equations of state.


What is the probability that there is life on other planets?

The probabilites are unknown at the moment (March 2014), since there is only one known planet (Earth) harboring life. This doesn't allow any meaningful probability estimates for the occurence of life, based on empirical data.

The overall formation of life is too complex to allow simulations based on current technology.
Although some intermediate steps can be simulated or performed by experiment, e.g. by the Miller-Urey experiment.

The Drake equation is a simplified attempt to decompose the probability into factors. Some of the factors aren't known yet. But there has been much progress in estimating the frequency of exoplanets, and the probability of planets to be habitable. From this news release one could estimate (order of magnitude), that there is one habitable planet around a Sun-like star per about 1,700 (about 12³) cubic light years in our region of the Milky Way ("... the nearest sun-like star with an Earth-size planet in its habitable zone is probably only 12 light years away...").

We don't know the probability for habitable planets to develop life, nor to be colonized. Probabilities of complex life forms are even more difficult to estimate (yet). More discussion about habitability of exoplanets on Wikipedia.


How were the designations of "North" and "South" applied to the hemispheres of Mars?

The research in other answers helped me come across the actual International Astronomical Union (IAU) standard:

  The north pole is that pole of rotation that lies on the north side of the invariable plane of the solar system.

The north of the invariable plane is the side that Earth's North pole points to. In fact, one of our other users has observed:

  The North Pole of any system being the pole on the same side as earth's north is indeed the correct definition. This applies to the Milky Way galaxy as well, in which case the North Galactic Pole is the pole towards the hemisphere of the galactic sphere which also contains the celestial north pole.

Dwarf planets, minor planets, their satellites, and comets do follow the right-hand rule mentioned in other answers (see Axial tilt on Wikipedia).

Sources:

Report of the IAU/IAG Working Group on cartographic
coordinates and rotational elements: 2006 - P. Kenneth Seidelmann, B. A. Archinal,
M. F. A’hearn, et al.
Poles of astronomical bodies - Wikipedia


Comprehensible full manual for bolometric light curve reconstruction

If you want a bolometric light curve from photometric points, you don't need to reconstruct its SED inside the range you already have data points. Use effective central wavelength of each band with its magnitude. Do trapezoid. This is an approximation, but it is better than reconstructing the SED that makes you rely more on assumptions for most circumstances, especially that you have UBVRI.

If, under some circumstances, you really need some more data points in between, or you really want a SED, blackbody is the first-order approximation, and is commonly used in SNe (at young ages). Less known issue in optical VBRI. There might be some issues with U like UV excess/suppressed. In this case, just fit the data to a blackbody.


This is too vague a term to be useful, because the Sun does not have any kind of sphere of influence (beyond the solar system) and nor do the nearby stars affect us with their light or gravity (Oort cloud perturbations aside).

Instead it would normally refer to the stellar population in the solar vicinity. This is a mixed bag of objects, with few shared characteristics. For example, objects within 10 light years of the Sun now, will not be so in 10 million years time.

A working definition might be those stars that are close enough to the Sun that we think the census is complete to that distance. Unfortunately, incompleteness remains in all such samples at low masses.

The RECONS programme has been working hard for the last decade to complete the census of objects that are within 10pc (33 light years).

The definition of a distant star is even more vague - all stars are distant. The Sun is 5 orders of magnitude closer than the next stars, so if anything, that is the only natural division between local and distant!


Is it just coincidence that the Moon follows the ecliptic?

No, this is not entirely a coincidence, though perhaps not in the way you expect (and other answers have indicated): it may have been coincidal at formation, but is required for long-term stability of the Moon-Earth system itself and, possibly, Earth climate.

The formation mechanism of the Moon (or the Moon-Earth system) is still debated. The favoured model involves a collision (between a pre-Earth and a Mars-sized object), when the resulting orientation of the Moon-Earth orbit depends critically on the impact parameters, which are essentially random, i.e. coincidal. Conversely, if the Moon-Earth system formed by some form of capture (possibly involving some surrounding gas to absorb the excess energy), a orientation in the eccliptic is more likely.

Howerver, the formation mechanism is not all there is to this. It is also required that the Moon-Earth system remains stable over 4.5 Gyr. This excludes any orbit initially circular and inclined by more than $i=\arccos\sqrt{3/5}\approx39.2^\circ$, when the Kozai mechanism would result in a collision of the Moon with Earth within a few years. The critical incliniation for a non-circular orbit is even smaller.

Even an inclination $i\approx30^\circ$ may not be viable, since it may imply that the Earth's spin axis (which is after all tidally coupled to the Moon's orbit) would be/have been tilted considerably stronger than by merely $\approx23^\circ$. A strong tilt of the Earth spin axis would have significant consequences for the Earth's climate and may well have prevented the formation of higher live forms. This is essentially a weak form of the antropic principle.


To understand this, you have to understand what it means to have negative or positive energy.

Following the example given by the text, the gravitational attraction of the Earth and Moon is a system with negative energy. This is because it requires an energy input to undo the system. One must exert energy to separate them, thus requiring energy input to the system to break it apart.

The matter comprising the Earth however is all positive energy. You can get an energy output from this matter and thus it must be positive. An example of energy output might be a gravity assist around the Earth, radioactive decay of nuclear materials in the Earth, or matter-antimatter annihilation of Earth particles. Any process that involves a matter interaction producing energy in the form of heat/light/whatever takes that energy out the Earth's matter and represents an energy output. Note that energy in the matter may take many forms such as heat, motion, or simply the intrinsic energy comprised in matter (via the famous $E=mc^2$ equation)

The black hole, much like the Earth, is simply comprised of matter (albeit in some unknown form at the singularity) and as such energy can be extracted from this matter, making it a system with positive energy.


How to form Copper from Calcium in a supernova explosion?

Copper is not thought to be primarily made in a supernova. It is thought to be mainly produced by the s-process of slow neutron capture onto iron-peak nuclei that already exist inside a star. These reactions are endothermic.

The source of the neutrons is still somewhat debated, it could either be from the decay of $^{13}$C in relatively low-mass asymptotic giant branch stars or is more likely from the decay of $^{22}$Ne in more massive evolved stars (e.g. Pignatari et al. 2010).

Below you can see a typical route for producing copper from $^{56}$Fe. The axes of the plot are neutron number on the x-axis and proton (atomic) number on the y-axis. Three neutron captures are followed by a beta decay, a neutron capture, a beta decay, then 3 more neutron captures followed by another beta decay to form $^{63}$Cu.

The net process is
$$ ^{56}{\rm Fe} + 7n \rightarrow \  ^{63}{\rm Cu} + 3e + 3\bar{\nu}_e$$

The copper is then distributed into the interstellar medium by a later supernova explosion in the same star.

If there are no "seed" iron-peak nuclei (e.g. in the first/second generation of stars), then copper can be inefficiently produced by explosive r-process neutron capture during supernovae explosions. However, this would not contribute very much to the copper we see on the Earth.


Does a lunar occultation of Mars happen twice a year?

No, in the sense that the number of lunar occultations of Mars per year is quite variable and not always two.
I do not have a reference handy, but I am sure that there are some years with no occultations.
And we can see that in other years the number can be both less than and much greater than two:

In 2019, there was one lunar occultation of Mars according to the Astronomical Almanac Online 2019 Lunar Occultations.
In 2020, there were five! See 2020 Lunar Occultations.


How to calculate galaxy bolometric luminosity?

The bolometric correction is the difference between a bolometric magnitude and the magnitude in some band.
$$BC = M_{\rm bol} - M_{5100} = -2.5\log\left(\frac{L_{\rm bol}}{L_{5100}}\right)
$$
$$\log L_{\rm bol} = \log L_{5100} - 0.4BC \ ,$$
$$ \log L_{\rm bol} = \log L_{5100} +0.4\log L_{5100} - 21.2  \ ,$$
$$ \log L_{\rm bol} = 1.4(1.4 \log L_x -16.8) - 21.2\ , $$
$$ \log L_{\rm bol} = 1.96\log L_x  -44.72\ .$$


Are there stars with an average density greater than the central density?

$$\frac{dP}{dr} = - \rho g,$$
is the equation of hydrostatic equilibrium, where $\rho$ and $g$ are the local density and gravity, $P$ is pressure and $r$ is the radial coordinate. This can be rewritten as
$$\frac{d\rho}{dr} \frac{dP}{d\rho} = -\rho g.$$
Since $\rho$ and $g$ are positive numbers, the pressure gradient is negative.
For all types of matter $P = f(\rho)$, such that $dP/d\rho$ is also positive (i.e. pressure increases with density).
Thus $d\rho/dr$ is negative and the density increases towards the middle.
EDIT: This argument only obviously works if you ignore the dependence of pressure on other variables - such as the temperature or number of particles per unit mass in your gas. It takes more work to establish that $dP/d\rho>0$ in a more general case. For example you might consider a situation where a sufficiently negative temperature gradient might counterbalance a positive density gradient.
i.e. if $P = f(\rho, T)$ then
$$ \frac{dP}{d\rho} = \frac{\partial P}{\partial \rho} + \frac{\partial P}{\partial T} \frac{dT}{d\rho}$$
and $dP/d\rho<0$ if
$$ \frac{dT}{d\rho} < -\frac{\partial P/\partial \rho}{\partial P/\partial T}$$
Leaving this as a placeholder for the moment.


Were there any images of Sanduleak -69 202 (progenitor of SN1987A) before it exploded?

Yes, it is clearly visible in photographs of the LMC from the 1980s.  It is shown in this comparison from the Australian Astronomical Observatory
We know it was this star in the most simple possible way: The supernova occurred exactly at the same location as this star, and when we look now, that star has gone (replaced by a little bipolar supernova remnant)


Can dark matter decrease the Jeans length?

This is an area of active research. The current Cold Dark Matter (CDM) paradigm predicts bottom-up formation of halos, whereby smaller halos coalesce in to the larger halos we observe indirectly today through X-ray measurements of clusters. This suggests that small dark matter halos wouldn't survive to the present day.

There has been research in to the possibility of mini haloes surviving, but interactions with dense baryonic structures, such as stars, as well as streaming by an irregular galactic potential tend to wipe out any small overdensities in the distribution.

However, if such mini haloes did exist in the early universe they could have contributed to the collapse of primordial gas to form Population III stars. Simulations have been carried out to investigate such a model, with Population III stellar masses comparable to those expected $(\sim100 M\odot)$.