Does the expansion of the universe increase with distance?

I'm not sure why you say that "expansion is the same" when you also say that "expansion increases with distance", but I wouldn't say the "expansion increases", but rather that "the recession velocity increases".

The recession velocity $v$ of galaxies increases linearly with distance $d$. The constant of proportionality is called the Hubble constant $H_0$. Thus, Hubble's law states that
$$
v = H_0 d,
$$
and currently, $H_0 \simeq 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, but in the past it was higher. That means that, for every Mpc (mega-parsec, $\simeq 3.26\times10^6$ light-years) a galaxy is from us, its speed away from us is roughly $70\,\mathrm{km}\,\mathrm{s}^{-1}$. This is the speed it has right now, but not the speed we see it have, since we see it in the past.

In fact, we don't really see a speed at all. We measure a galaxy's redshift $z$, and $z$ also increases with distance, but not linearly. Thats's because it depends on how much the Universe has expanded since the light was emitted, and the expansion rate changes with time. But for small enough distances, i.e. for galaxies that are not seen that far into the past, the relation is close to linear, so that $v = cz$, where $c$ is the speed of light.

This is indeed approximately true out to a few hundred Mpc. For larger distances, the linear relation becomes inaccurate, and one must instead apply a cosmological model of the expansion history of the Universe, which depends non-trivially on the densities of its various constituents.

Acceleration of the Universe

The term "accelerated expansion rate" refers not to the value of $H$, but to the rate at which the scale factor $a$ — the "size" of the Universe — increases (the Universe may or may not be infinitely large; $a$ is normalized such that, today, $a=1$).

For a given value of $H$, we know the recession velocity of any two points separated by some distance at a given time. At a later time, these two points will have increased their distance, and hence — by Hubble's law — their recession velocity will have increased.

What was higher in the past is the Hubble parameter $H(t)$. That number will, in the future, asymptotically decrease toward a slightly smaller value of
$$
H(t\!\!\rightarrow\!\!\infty) = H_0\sqrt{\Omega_\Lambda} \simeq 56\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}.
$$
The Hubble parameter is defined as $H\equiv\dot{a}/a$, where $\dot{a}\equiv da/dt$ is the Universe's change in size per unit time. If $H$ is constant, and $a$ increases, then $\dot{a}$ also increases. That is, the increase in velocity of any two points accelerates. In fact, we can solve the differential equation and get that, in the future,
$$
a(t\!\!\rightarrow\!\!\infty) \propto e^{Ht},
$$
i.e. the scale factor increases exponentially.

The figure below shows the evolution of $a(t)$ until $t\sim130\,\mathrm{Gyr}$. On the left is a zoom-in of the first $20\,\mathrm{Gyr}$ where you can see how the Universe decelerated the first $\sim10\,\mathrm{Gyr}$, until dark energy started taking over the dynamics.

At the moment the acceleration is actually quite modest, but it will increase without bounds in the future; for instance, in 50 Gyr from now, $a\sim10$, i.e. galaxies will be $10\times$ farther from each other, but in 100 Gyr we have $a\sim150$.


Why is it that we can"t feel how fast the Earth is moving?

Because the rotation of the earth is very smooth and doesn't change, the centripetal acceleration we feel is very nearly constant. This means that the (small) centrifugal force from the rotation gets added to gravity to make up the "background force" we don't notice.

Taken from Physics SE.


Why does the half moon appear to have a "nose"?

Here's an image I captured from a Moon globe app on my iPad.

From your description, I think you may be referring to the southern rim of Mare Tranquillitatis, the Sea of Tranquility, perhaps extending a little into the Rimae Hypatia. This is an area which is lower than the surrounding mountains and craters, into which the shadow is encroaching.

Incidentally, Mare Tranquillitatis is the region where Neil Armstrong and Buzz Aldrin landed on the Apollo 11 mission!


Correct relation between metallicity (z) and iron content ([Fe/H])

$$[\mathrm{Fe}/\mathrm{H}] = \log_{10} \left[\frac{f_\mathrm{Fe}(\alpha)}{f_\mathrm{Fe}(0)} \frac{m_z(0)}{m_z(\alpha)} \left(\frac{Z}{X}\right)_\ast \left(\frac{X}{Z}\right)_\odot \right]$$

You can use this relation to estimate [Fe/H] from [M/H] of a star. From [M/H] you can get the (Z/X) ratio.


Internal heat and planetary differentiation

Your argument (large temperature leads to greater mixing) is correct so long as there are no other large scale forces acting on the system.  This isn't true in planet formation, because gravity plays a very important role.

I'm not an expert on planet formation, but I think the argument goes something like this:  As a planet forms from material from the protoplanetary disc it will begin very homogenous, something like an asteroid, just rock and metal all the way through.  If the planet is heated enough in the core, some material will melt.  Buoyancy (due to gravity) will drive lighter material "higher" in the planet, away from the core.  The more heat you generate the more melting can happen and the more material will start to separate by density.

Of course, there's a lot more to the story.  Fluid material obeys the laws of hydrodynamics, so large scale convective flows can form in some regions, mixing those areas quite well.  Planetary rotation adds centrifugal and Coriolis forces to the mix, which pull more material to the equator.  The surface of a planet is exposed to space and can radiate excess heat, cooling to a solid (like the Earth's crust).  If the core is magnetic and rotating, metals in the planet may be pushed by the magnetic field.

All this is just to say that planetary formation is a balancing act of a lot of competing forces.  These forces can differentiate (or not!) different types of material, but first (for rocky planets at least) they need to be freed by internal heating.


Which kind of properties can we get for cosmic ray particles hitting on an optical ccd?

Cosmic ray "hits" are artifacts caused when high energy particles (cosmic rays, often muons) slam into atoms in the CCD itself and liberate large numbers of electrons, which then show up as bright spots and streaks when the CCD data is read out. They can also occur as a result of radioactive decay processes in the material in the detector and instrument itself and also the immediate surroundings.

They are a tremendous nuisance that can be removed to a certain extent by median stacking images - though this also fails if you get into a situation where cosmic ray hits are found in the same pixels of $>1$ images in your stack. This limits the lengths of exposures that can safely be made with astronomical CCDs. Depending on exact usage, this limit is usually between 30 minutes and an hour.

The flux of muons and the distribution of angles at which they reach the detector is actually very well known already. It is largely unaffected by local shielding (unless your telescope is buried far underground!)

It is possible to distinguish different types of events (muons, locally produced gamma rays etc.) from looking at the pattern made on the detector (e.g. Groom 2004), but I don't think you can tell much about their energies, unless it could be by looking at the flux as a function of how much shielding (lead?) or how far underground the detector was, since often only a small fraction of the cosmic ray energy ends up deposited in the CCD.


How to compare a Catadioptrics light asorbtion with a reflector?

Comparing optical instruments is very difficult. There are numerous variables involved, the performance is not a single number, and the theory is complex. Perfect scenario to baffle amateurs. Also, it sounds like you may have been listening to the wrong rumors.

The architecture of a telescope (refractor, reflector, catadioptric) is one of the least important factors in the performance analysis, regardless of how you define "performance".

Don't forget that how you maintain and where you use the scope are very important factors. Collimation will strongly impact resolving power - so make sure it's always perfect. Light pollution will impact you much more than light absorption within the instrument. Etc.

In terms of amount of light gathered, aperture is the dominant factor. 2x the aperture is 4x the area, so it's obviously very important. The central obstruction may factor in a little, but the effect is small. Going by diameter, a typical newtonian will have a 20% central obstruction; a typical catadioptric scope will have a 40% c.o. But if you translate that into areas, the newtonian is at 4%, whereas the catadioptric is at 16% - both are very small.

In terms of resolving power, again aperture is king. Bigger aperture = more resolving power, no matter what the architecture. But poor collimation will quickly destroy your gains, so keep that scope collimated.

In terms of internal losses, the newtonian will do a bit better - it's two reflections and that's it. The catadioptric will have two or more reflections (depending on design), plus absorption / reflection on the diopters. Again, with modern instruments, the difference is tiny.

Overall, at the same aperture, in theory, the newtonian pure catoptric scope delivers a bit more light to the sensor, compared to a catadioptric scope. In practice, the difference is far outweighed by other factors and you should completely ignore it.

You should make the purchase based on more important things. What's the size of the usable field? What kind of adapters can you use? Does it provide enough back-focus? What's the aberration in different points within the field? Is the telescope too heavy for the driven mount? Is it long focal ratio, or short? (each kind has different applications) Finally but not least important, price?

It's more common to see catadioptrics used as astrographs. That's not to say that a newtonian reflector cannot do the job - but if you're still new to the field, just do what most people do and get a catadioptric, with a good mount.

Final note:

When doing prime focus photography, the brightness of the image is given by both aperture and focal ratio - but not all objects behave the same.

For stars and other point-like objects (smaller than the resolving power), brightness is driven by aperture. Bigger aperture = brighter stars.

For extended objects (bigger than the resolving power), such as nebulae, many galaxies, the Moon, the planets, etc., the brightness of the object is given by the focal ratio. An f/4 scope will make brighter nebulae in prime focus, compared to an f/8 scope. Aperture doesn't matter. It's just that, at the same focal ratio, the bigger scope will also make a bigger image in prime focus.

This is true only for prime focus photography. If you're doing eyepiece projection, or something else, different rules apply.


Astronomy detective question: what part of the sky are these photos of? What are a few of the stars?

Astrometry.net successfully plate-solved both images.

The first image is of the Cygnus region with north to the left.
It is 73° wide and 55° high, with 65 arcseconds per pixel or 55 pixels per degree.
Bright stars include Deneb (α Cyg) left of center and Altair (α Aql) at upper right.

The second image is of the Sagittarius region with north up.
It is 37° wide and 66° high, with 80 arcseconds per pixel or 45 pixels per degree.
The bright object left of center is Saturn, which was at that position around 2019-07-29.


Yes, they can.  The ships that visit the ISS often orbit near the station for several hours before docking.  Further, there are several satellites that travel in pairs or trios with other satellites.  The nearer the satellites are to earth, or the farther apart they are, the more difficult the task of maintaining spatial relationships between the spacecraft. This is because the spacecraft must be in the same orbital plane to travel together.  The most common configuration is where they are in the exact same orbit, with one satellites a few seconds ahead of the other along the same path.


Can all areas on Earth experience a total solar eclipse?

Yes, total Solar eclipses can occur anywhere. On any given place, they occur roughly every 400 years, on average. If you take a look on the picture below (from Wikipedia), you'll see that within the last one thousand year, almost every place on Earth has experienced a total Solar eclipse. The places that haven't experienced one have no special latitude or longitude, so it's just a matter of coincidence.


Why do telescopes use hexagonal mirror pieces instead of pie slice shaped ones?

The ideal shape for the mirror is round. It's the easiest to make. It's the best-behaved while in use.

The hex tiles are already harder. The mirror is a revolution surface generated by a conic curve (circle, parabola, hyperbola, ellipse), which needs to be machined with a precision greater than 0.1 microns. That's extremely difficult already with a round mirror, where the surface can be easily distorted near the edge in the process of manufacturing.

With the hex tiles, maintaining that precision near the corners is very hard - corners are like the edge but exponentially harder. Unless you take special precautions and unless you perform extra work, you're guaranteed to end up with surface errors near the corners (the surface is either too raised, or too low, or has a more complex distortion). Based on my experience making round mirrors, I would guess the surface near the corners would tend to be turned down (too low relative to the rest), which is a defect that's extremely hard to correct - one of the hardest, in fact.

With a pie slice, all problems mentioned above increase greatly in magnitude. The blunt angle of the hex tile gives you some support in machining the surface, but the sharp angle of the pie slice gives you almost none. I can't even begin to imagine the difficulties one would encounter while making a high precision surface on a pie slice. The sharp end of the slice would be almost guaranteed to be worse quality than the rest.

What most people don't realize is that maintaining the precise shape of the mirror while in use is difficult too. You can't just lay it on a hard surface - the uneven support would distort the precise shape. Mirrors are supported on the back by a precise mechanism, with a number of support elements precisely sharing the weight in equal amounts (or, if not equal, then deduced from finite element analysis), and located in places carefully calculated.

With a round or hex tile this is, if not easy, then at least a well-understood and manageable procedure. A pie slice would be very difficult to support in the same even manner near the sharp tip. You can't rely on the symmetry of the shape to provide the even load on the support points. The support force would have to be specially adjusted there. It would probably be doable, but harder.

This is only an issue with terrestrial telescopes, which operate under gravity. Space telescopes don't suffer from this issue.

Alternatives to hex tiles could also be square tiles, or triangle tiles. These are worse than hex tiles for all the reasons shown above - the corners have sharper angles which make everything more difficult. The blunt angles of the hex tiles alleviate these issues somewhat.

TLDR: Non-round mirrors are hard to make anyway, but hex tiles are less hard compared to the alternatives.


Why not send a Gaia-like mission to Mars?

The panels need to be about 2.3 times bigger to generate the same power.

The telemetry requirements are limited by distance to some power, assuming that the communications beam cannot be infinitely narrow. Mars is much further away than the L2 point - somewhere between a factor of 50 and 250.

It wouldn't necessarily make it possible to measure parallaxes 50% further. Gaia works by taking approximately 70 (on average) measurements of a stellar position over 5-6 years. This amounts to sampling the 2 au baseline about 10 times, thus reduing the error bar on a single baseline parallax by about $\sqrt{10}$ and making the "equivalent baseline" about 7 au.

To get similar precision with Mars you would need to be able to complete the 3 au baseline about 5 times, which would still take 5 years. So you don't obviously win in precision terms.

Overall I think it is the telemetry problem that is the real killer. Gaia is currently telemetry limited - that is why there is a limit on the number of stars it will measure and on the number of stars it will get spectroscopy for.


What is this pale blue dot in a daytime photo of a Sun halo?

Yes, I believe this is the internal lens reflection. Lens reflections have hexagonal shape, and if you look closely, it somewhat resembles this shape. Also, see other answers in the comments.
Internal lens reflection happens because the lenses don't have 100% transmission. This often happens when taking bright objects, like Sun in your example.


Actually, the core has two parts. The outer core is liquid, while the inner core is solid.

As explained in the Wikipedia article about Earth's magnetic field:

  The Earth's magnetic field is mostly caused by electric currents in the liquid outer core, which is composed of highly conductive molten iron.


We are sure that there are no brown dwarfs closer than ______

I think the answer is roughly 3-5 light years.

The recent WISE survey in the near-infrared should have been capable of detecting even a very cool nearby brown dwarf (and indeed it has detected some very cool brown dwarfs - e.g. a 250K brown dwarf only 6 light years away Luhman et al. 2014). Since WISE was an all-sky survey and there is great interest in finding these ultracool brown dwarfs, then it is reasonable to assume that any similar, object within a few light years would already have been found. The caveat to this is that the point spread function of the WISE survey s not that great an so near the galactic plane it could be that source confusion might have hidden the brown dwarf. On the other hand, nearby objects usually have high proper motion, so it would probably have been uncovered in combination with the earlier 2MASS survey, since as I show below, such an object may have been detected in both.

Some details: As an example let's take a very low mass brown dwarf - say 20 Jupiter masses and 5 billion years old - and place it 4 light years from the Sun. According to the evolutionary models of Saumon & Marley (2008), such an object has an intrinsic luminosity of $10^{-7}$ times (1 ten millionth) that of the Sun and a temperature of 400 Kelvin and would appear to have a spectral type of late T or early Y. This is similar to the coolest brown dwarfs yet found.

From the calibration of absolute magnitudes versus spectral type for cool brown dwarfs in Marsh et al. (2013) we know that at 4 light years, such a brown dwarf would have magnitudes of $H=16$ and $W2=10$. The former is just about bright enough to have been seen in the 2MASS all-sky survey and the latter is easily detected by WISE. The combination of data would also easily have revealed the large parallax of such an object. We can conclude that an object would have to have much lower mass than this to remain undetected and thus would arguaby not be called a brown dwarf, but would really be a "free-floating planet".

Now you might ask, what about a cooler brown dwarf? Well the theoretical predictions of magnitudes and colours become much more uncertain, but I would simply refer to the Luhman discovery. This brown dwarf is about as cool as a brown dwarf could be given the age of the Galaxy, may only be 10 Jupiter masses, yet was still easily detected in the WISE survey. Therefore I think this sets a fairly firm empirical limit.


Corresponding values for the Sun's ultra violet color maps?

It’s hard to tell for sure without seeing the data and playing around, but looking at your image, I would suggest trying to subtract off a constant background level to get the sky as close to black as possible.  You could sample the edges of the image to find the average background level.  Then subtract that value from the whole image array and try plotting it.  Even simpler would be just to subtract the minimum value, but that’s prone to being skewed by just a single pixel.
If you try it, post an image back here and let’s see how it looks.


What effect does the Moon have on the near Earth asteroid population?

I think you answered your own question with your second paragraph.

According to Newton's law of gravity

the Gravitational force is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.
So the biggest body in the neighborhood has the most pulling power provided it agrees with the equation.
As a consequence Jupiter and the outer planets should win out over Earth and keep them from hitting Earth or at least divert the collision course. Especially since the NEA come from their part of the cosmos.
I suppose the same logic would also work with the comets.

Please see the following link for more:
http://earthsky.org/space/is-it-true-that-jupiter-protects-earth


Origin of terms "white dwarfs" and "red giants"?

The terminology is related to Planck's Radiation Law. We have (from Wikipedia):
$$
{\displaystyle B_{\lambda }(\lambda ,T)={\frac {2hc^{2}}{\lambda ^{5}}}{\frac {1}{e^{\frac {hc}{\lambda k_{\mathrm {B} }T}}-1}}}
$$
where $ k_B $ is the Boltzmann constant, h the Planck constant, and c the speed of light, $ \lambda $ is the wavelength, and T is the temperature.

A typical white dwarf is very hot, with $ T > 10^5 K $ even. Here let's assume it is about 50,000 K. A typical red giant is of lower temperature than the sun. Let's take it to be about 4,500 K (it will usually be a bit lower). Then we can plot the Spectral Irradiance of both the sources and compare them (I had an older Matlab code).

(The y-axis is $ W m^{-2} sr^{-1} nm^{-1} $.)

For the white dwarf we see that the peak occurs at a significantly lower wavelength than visible light (at about 0.58 nm). Most white dwarfs are hotter so the peak would shift further to the left. Thus, we expect the white dwarf to appear white with a bluish tinge as blue is still marginally higher than the other wavelengths.

For the red giant, the peak occurs at about 640 nm, which is near the red end of the spectrum. Red giants colder than this would have peaks shifted to the right. So they appear orange-red.

As @AtmosphericPrisonEscape already mentioned, dwarf and giant relate to the physical size as compared to the original star in the main sequence.


Does our universe have to be embedded in a higher dimensional space?

The universe being open, closed, or flat only determines the type of geometry one must use to describe distances (and time). For open and closed geometries, Euclidean Geometry is not what one should use. I would also agree that our universe being open, closed, or flat has nothing to do with the number of dimensions it contains.

At present, there is no direct evidence that there are any additional dimensions over 3+1 dimensions (this just means three spatial dimensions and one time dimension). However, many GUT theories do include additional spatial dimensions in an attempt to unify all of the forces.

Also at present, we have very good evidence that the universe is 'flat'. What this means is that the angles of a triangle have to add up to $180^{\circ}$ and distances are measured in the standard Euclidean way. When we talk about the universe being flat, this is a purely global statement. Locally, however, it is completely possible to live in curved space. We actually do live in curved space. The mass of the Earth is curving space and time in a way that General Relativity predicts, and therefore clocks run very slightly differently depending on where you are on the Earth's surface, and distances are very closely approximated by Euclidean distances, though they're not.

How do we actually determine that the universe is globally flat, you may ask? We use what's known as a standard ruler. Much like a standard candle, if we think we understand the physics, any deviation from what we would predict gives us new information about the universe (in the case of supernovae, it's that they can be used to measure distances, and therefore map out the expansion of the universe). We use the angular size of fluctuations in the Cosmic Microwave Background Radiation (we think we understand the physics behind the fluctuations fairly well) to test how much, if at all, the universe deviates from flatness. The latest results from Planck show very good agreement with the standard picture provided to us by the standard LCDM cosmological model.

Below is the power spectrum of the temperature fluctuations of the CMB. The location of the first peak is what cosmologists use to measure flatness.


If you know how to use python, you can do something like :

import astropy as ap
ap.coordinates.get_sun(time)

It will return  a position in geocentric coordinates for the given time.


Difference of sunset and sunrise of two cities are different

Tehran (35.7$^\circ$ N) has 23 fewer minutes of daylight than Rafsanjan (30.4$^\circ$ N) on Nov 26 because the Sun is 21$^\circ$ south of the celestial equator and Tehran is 5.3$^\circ$ farther north than Rafsanjan.
For the same reason, the north pole has 24-hour night and the south pole has 24-hour daylight at this time of year.
On May 26, when the Sun's declination is 21$^\circ$ north, Tehran has 25 more minutes of daylight than Rafsanjan does, and the north pole has 24 more hours of daylight than the south pole does.


Does Earth changes angular velocity in 24 hours evaluation?

There are a series of resources on the leapsecond.com site which specifically talk about detecting tides with pendulum clocks. It has been done by e.g. the Shortt pendulum clocks which had a master pendulum in a vacuum to remove disturbances synchronized to a slave pendulum. From the first link in the resources at 1, the order of magnitude of the change in pendulum rate caused by the change in gravity is ~1e-7 so your pendulum clock needs to be stable to a few milliseconds per day (few x10^-8) if you hope to see any tidal effects. You will need a more stable source of time such as a GPS Disciplined Oscillator (GPSDO) and normally a time interval counter to measure the stability with.


Are neutrinos affected by gravity fields?

If you buy into General Relativity (you need to throw out an awful lot of modern physics if you don't), gravity is a manifestation of the shape of spacetime around massive objects. It's not that they are being "pulled," but rather that they are just following the straightest path through spacetime that they can. Neutrinos would be the same as any other particles in that respect, and following anything other than this path would require some extraordinary gymnastics and external forces.


What makes an exoplanet to orbit around its star backwards?And why did we consider our orbital movement to be forward?

All the planets in the solar system orbit the same way. This is not a coincidence. Planets all form from the same protoplanetary disc that surrounded the young sun. The particles in the disc orbited the same way as the sun rotates, and the planets continue to orbit in the same direction.

In this context forward means orbiting in the same direction as the star rotates. We would expect most planets to orbit "forwards" for the same reason that all the planets in the solar system orbit forwards.

What probably happened with HAT-P-7b is that it was thrown into an odd orbit following a close encounter with another planet, or perhaps another star. It is also conceivable that it is a captured rogue planet. See a New Scientist article about the planet


How Good Are the Upper Limits on Heavy Elements?

Good question! Normally, the lack of heavy elements is taken as an indicator that something was made from (close-to-primordial) gas.

So an answer could be to ask what are the lowest abundances ever measured with respect to hydrogen.

I am not completely up to date with the current record holders, but stars with iron abundances that are 5 orders of magnitude less than in the Sun have been found (Norris et al 2013). This corresponds to A(Fe)$\sim 2$ on the usual logarithmic scale where hydrogen has A(H)=12.

The constraints on other iron-peak elements are similar. Alpha elements like O, Mg are usually enhanced in very metal poor stars, so the constraints are an order of magnitude higher.


What are the 10 most abundant elements in the universe by number of atomic nuclei?

All right, so I took the first list on wikipedia listing the 10 most common elements by mass in parts per million, and did what Rob recommended and here's what I got.

https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Abundance_of_elements_in_the_Universe

Hydrogen - 739000amu(H)/1amu(H)=739000 H atoms

Helium - 240000amu(He)/4amu(He)=60000 He atoms

Oxygen - 10400amu(O)/16amu(O)=650 O atoms

Carbon - 4600amu(C)/12amu(C)=383 C atoms

Neon - 1340amu(Ne)/20.1amu(Ne)=66 Ne atoms

Iron - 1090amu(Fe)/55.845amu(Fe)=19.5 Fe atoms

Nitrogen - 960amu(N)/14amu(N)=68.5 N atoms

Silicon - 650amu(Si)/28.1amu(Si)=23 Si atoms

Magnesium, - 580amu(Mg)/24.3amu(Mg)=24 Mg atoms

Sulfur - 440amu(S)/32.1amu(S)=13.7 S atoms

So therefore, the 10 most common elements in the universe by atom, with the relative ratios between them, is;

1.Hydrogen (739000)
2.Helium (60000)
3.Oxygen (650)
4.Carbon (383)
5.Nitrogen (68.5)
6.Neon (66)
7.Magnesium (24)
8.Silicon (23)
9.Iron (19.5)
10.Sulfur (13.7)

If anyone sees a mistake that I made with my reasoning or calculations then please point it out.


Problem regarding the absorption lines of the sun

Possibly you are labouring under the misapprehension that the number of photons is somehow a conserved quantity? That isn't true, there are more photons at any given wavelength when you are deeper into the star, because there is a temperature gradient. Cooler material further out is less emissive because fewer atoms are in excited states.
The temperature gradient is responsible for the formation of absorption lines. If the photosphere of the Sun were at a single temperature then we would see a perfect blackbody spectrum, for the reasons you outline.
The filling in of absorption by scattering would only take place if the radiation field that the atoms were in were isotropic. But it isn't isotropic because of the temperature gradient.
A much better way to think about the spectrum of a star is to imagine that you can see to a wavelength-dependent depth into the star. Where there is a strong atomic absorption feature, you cannot see very far into the star at that wavelength.
Since the star gets hotter the deeper you go into it, and the emissivity scales as $T^4$, then the deeper we can see into the star, the brighter it will appear at that wavelength (and vice-versa).
EDIT:
More formally. The radiative transfer equation, if you want to consider the absorption and remission as some sort of scattering process, would be
$$\frac{dI_{\nu}}{ds} = -\sigma_\nu I_{\nu} + \sigma_\nu J_\nu\ ,$$
where $I$ is the specific intensity in the solar photosphere (in this case, directed towards the Earth), $J$ is the mean specific intensity at a point in the solar photosphere averaged over all directions (i.e. $J = \int I d\Omega/4\pi$, where $\Omega$ is solid angle), $\sigma$ is the scattering coefficient (assumed to be isotropic) and $ds$ is a piece of pathlength towards the observer. The $\nu$ subscript just indicates everything is wavelength/frequency dependent.
To avoid creating an absorption or emission line then $dI_\nu/ds$ must equal zero (i.e. nothing is added or subtracted from the beam of light).
This will only happen if $I_\nu = J_\nu$, which would require that the  specific intensity averaged over all directions is equal to the specific intensity emerging from the Sun and heading towards the observer. This will only be true if the radiation field is isotropic and equal to $I_\nu$ in all directions.
Whilst this would be true for a blackbody radiation field at a set temperature,
it isn't true in the solar photosphere. The specific intensity heading towards us (generally outwards) is always larger than the specific intensity heading away (generally inwards and true regardless of which portion of the visible solar disc is considered) because of the temperature gradient in the photosphere, which means it is hotter further into the interior. That means that $I_\nu$ is always greater than $J_\nu$ and hence $dI_\nu/ds < 0$ and we have net absorption.


Can life survive on the equator of cooled and fast rotating white dwarf or neutron star?

I am going to attempt a weak answer, mods feel free to delete it, but I'm fairly certain I'm right.

Shortly - no. It's not possible. Even if you balance gravity and centrifugal force perfectly at ground level at the equator, they will very, very quickly become imbalanced as soon as you move north, south, or up from there. So quickly in fact that the gradients may be too big even for a human being not moving at all. Maybe if you're laying down, with your body oriented along the equator, but even then I think the gradients would be too big.

Maybe bacteria would survive, briefly.

I'm sure the math could be done quite easily to estimate the gradients. This is based entirely on intuition.


What are the differences between matter, dark matter and antimatter?

Matter is the stuff you are made of.

Antimatter is the same as matter in every way, looks the same, behaves the same, except its particles have electrical charges opposite to matter. E.g., our electrons are negatively charged, whereas a positron (an antimatter "electron") is positively charged. The positron is the "anti-particle" of the electron.

When a particle meets its anti-particle, they "annihilate": the two particles disappear, and gamma photons are released carrying off their energy. For this reason, should a lump of matter touch a lump of antimatter, they would annihilate, and a giant explosion would result because of the huge energy released (E=mc^2).

Matter and antimatter are definitely related: same thing, but with opposite signs. Twins, but opposites.

It is not clear why, but it seems like there isn't that much antimatter out there, more like trace amounts. Definitely not as much as regular matter as far as we can tell. This is puzzling to physicists and cosmologists, because you'd expect the Big Bang to make roughly equal amounts of matter and antimatter. Scientists agree that the paradox of "excess matter" will advance physics even further once it's solved.

Dark matter - we don't really know what it is. It's not even sure it's "matter" in a conventional sense, or related to it in any way. We just know that galaxies are rotating in such a way that indicates there's a lot more mass out there, but it is mass that we cannot see and cannot be accounted for in the usual ways. Hence the name "dark" (as in invisible) matter.

Dark matter doesn't seem to interact much with regular matter, except gravitationally. Right now dark matter could be passing through you and you wouldn't notice. Dark matter also does not interact with light, so you can't see it. It doesn't seem to interact much with itself either, so for this reason dark matter cannot form "clumps" such as planets or stars. Instead, it probably exists in a diffuse form. Bottom line, dark matter interacts pretty much only via gravity.

The shape of galaxies is a proof of the existence of dark matter, and is a result of the interaction between matter and dark matter. Without dark matter, galaxies would be much less massive, and the outer parts would rotate much more slowly compared to the center. Due to dark matter, galaxies are quite massive, and they rotate almost as solid objects - the outer parts rotate approximately as fast as the central parts.

Estimates vary, but it seems like there's something like 5x to 6x more dark matter out there compared to regular matter.


Is it possible to extend the life of a star?

If something crashed into it rearranging its content a bit, could its life be extended?

Yes! This is precisely what we think causes the creation of blue stragglers.

When we look at a cluster of stars, we expect them to all be of roughly the same age.  This is usually borne out by observations.  Because larger stars evolve faster, the age of a cluster can then be roughly determined by working out the mass of the most massive star still on the main sequence (i.e. still burning hydrogen in the core).  We call this the main-sequence turnoff.

But in some clusters, we see and handful of stars that are more massive, but still on the main sequence, even though all the evidence says they should have already evolved into giants and other more-evolved objects.  The consensus view is that these stars are at the very least interacting with binary companions, and many are thought to be merger products. i.e. something left over from two stars merging into one.  Either two companions in a binary have orbited so close that they've become one star, or two stars have simply collided (which is quite possible in the dense cores of clusters).  Depending on what types of star collide, you can basically get a new star where everything is mixed up again, allowing the star to be "born again".


Does a black hole become a normal star again?

There's several ways this question could be answered, but they all come down to an emphatic "no" - a black hole will not return to being a main sequence star. The simplest way to see this is probably that a black hole has a much higher entropy than a star or even another type of stellar remnant of even vaguely similar mass and so there simply could not exist a spontaneous process by which a black hole develops back into a star.

A black hole once formed will stay a black hole, however it is believed that the Hawking process will lead to the black hole eventually evaporating. The time scale for evaporation though of a stellar remnant black hole is mindbogglingly long and they will not evaporate until long after stellar formation has ceased in the Universe.


How to Calculate Planetary Equilibrium Temperature in Binary systems

The power radiated from the planet, assuming it is at the same temperature across its entire surface:
$$P_{\mathrm{rad}}=\epsilon\sigma T_{\mathrm{eq}}^4 \cdot 4\pi R_{\mathrm{pl}}^2$$
Where $\epsilon$ is the emissivity (to match the formula in the question, set this to 1), $\sigma$ is the Stefan-Boltzmann constant, $T_{\mathrm{eq}}$ is the equilibrium temperature, and $R_{\mathrm{pl}}$ is the radius of the planet.
The power absorbed by the planet from the stars, assuming it is sufficiently far away from each star that there is negligible illumination from that star beyond 90° from the substellar point:
$$P_{\mathrm{abs}}=\sum_i{f_i\left(1 - A_i\right) \cdot \pi R_{\mathrm{pl}}^2}$$
Where $f_i$ is the flux from the $i$th star, and $A_i$ is the planetary albedo with respect to the $i$th star (this may be different because the reflectivity will be wavelength-dependent, and the stellar temperature affects which wavelengths are being emitted).
So equate radiated and absorbed power, and rearrange:
$$T_{\mathrm{eq}} = \left[ \frac{\sum_i{f_i\left(1-A_i\right)}}{4\epsilon\sigma} \right]^{1/4}$$
The limitation here is that the fluxes are going to be time-variable due to the planet's distance changes from the stars (same applies to a planet in an eccentric orbit around a single star), and it is non-trivial to figure out how to average this out. The real planet will take time to heat up and cool down, and in that time the distances and fluxes will change.
So take the numbers you get from this "equilibrium temperature" calculation with a pinch of salt.

As requested, here are some example calculations. First off, the sanity check: Earth. Taking 1361 W/m2 as the value of the solar constant and using an albedo of 0.3 and an emissivity of 1, the temperature works out as
$$T_{\mathrm{eq}} = \left[\frac{1361\ \mathrm{W\!\cdot\!m^{-2}\,\times \left(1-0.3\right)}}{4 \times 1 \times \left(5.6704 \times 10^{-8}\ \mathrm{W\!\cdot m^{-2}\!\cdot\!K^{-4}} \right)}\right]^{1/4} \approx 255\ \mathrm{K} $$
A bit chilly, but not too bad: the greenhouse effect keeps the actual Earth warmer, which could be taken into account by using an emissivity $\epsilon < 1$.
Now for the example in your question, with fluxes of 1344 W/m2 and 92 W/m2. Let's also suppose that while the first star is Sunlike (so I will keep the 0.3 value for the albedo), the second star is cooler than the Sun. This star emits more of its light in the red where the planet is less reflective. I will use an albedo of 0.25 for this star.
The calculation works out as follows (dropping the units in the calculation step for the sake of space):
$$T_\mathrm{eq}=\left[\frac{1344\times\left(1-0.3\right) + 92\times\left(1-0.25\right)}{4\times1\times\left(5.6704\times10^{-8}\right)}\right]^{1/4} \approx 258\ \mathrm{K}$$
Hope that helps!


Exoplanets, Protoplanets, and planetesimals

Whether or not gas giants form is largely to do with whether rocky/icy cores are able to accrete large quantities of hydrogen and helium, which dominate the mass of the protostellar disc.

Terrestrial planets form from the tiny fraction of the disc that is able to form solids.

There is no indication that systems without gas giants have more "solids" and most of the raw material in the disc is either accreted onto the star or expelled from the system in either case.


Orbital resonances and position of planets

Watched the video, and I thought it was really cool seeing the modeling of different young solar-systems.  The orbital resonances they were talking about were for smaller objects that form debris fields in resonance with a giant planet.   Since they only looked at models for the inner planets, the closest example to the bit in the video with our current solar-system would be Jupiter's Hildas.  For that specific part of the video, they only showed one inner most giant planet.  For more than one, Resonances like that are unlikely to form between giant planets, like in our solar system where most of the resonance objects fall inside Jupiter or outside Neptune.

The mathematics gets too complicated for me to know for sure, but as I understand it, planets in resonance is neither stable nor unstable, unless the aphelion of one planet is in resonance with the perihelion of another, in which case the planets could drawn together I would think.  That's my amateur answer to your question.  I wouldn't think orbital resonance affects stability.

When one planet is dominant and the other quite small, then resonances appear largely stable (Neptune/Pluto for example), provided they never get close to each other, which Neptune-Pluto don't.

Jupiter and Saturn were thought to have entered orbital resonance at one point in the early solar-system and that, from what I've read, affected other objects in the solar system but didn't affect their relation to each other much.  I think the only way to get any kind of answer is to run long term simulations on various model.   In short, planets can be stable both in resonance with each other, and not in resonance and planets can be unstable, again, either in resonance or not in resonance, though if they have unstable orbits, they won't stay in resonance for long.


Mass, Radius, Colour, Size, Type of a Star from the Hipparcos Catalog

The apparent magnitude (in V ~ visual) is in column 42-46. Other magnitudes are BT magnitude (columns 218-223) and VT magnitude (columns 231-236). These are magnitudes recorded with different (colour) filters.
The colour ($B-V$) can be found in columns 246-251, expressed in magnitudes. This is the difference between V (visual) magnitude and B (blue) magnitude also known as the colour index. Blue stars have a low colour index ($B-V<0$) as the star will be brighter in the blue (B) passband (filter), and will therefore have a low B magnitude, while red stars will have a higher $B-V$ colour index.

Mass, Radius, and Temperature are not in the catalogue. Mass can only be directly determined for multiple star systems and requires a lot of observations per multiple star. Radius is determined from stellar models. And Temperature can be determined from the spectrum of the star. A crude indication can be gotten from the colour index of the star. (see for instance this webpage).

Finally the type can be found in the star's spectral type. The spectral type is in columns 436-447. These spectral types are not always easy to read but they consist of a spectral class (e.g. O, or F type stars) and the luminosity class (and some other information). The luminosity class gives the type of class in roman numerals. Below is the list from the wikipedia page on stellar classification.

0 or Ia+ (hypergiants or extremely luminous supergiants). Example:
Cygnus OB2#12 (B3-4Ia+)[12]
Ia (luminous supergiants). Example: Eta
Canis Majoris (B5Ia)[13]
Iab (intermediate luminous supergiants).
Example: Gamma Cygni (F8Iab)[14]
Ib (less luminous supergiants).
Example: Zeta Persei (B1Ib)[15]
II bright giants. Example: Beta
Leporis (G0II)[16]
III normal giants. Example: Arcturus (K0III)[17]
IV subgiants. Example: Gamma Cassiopeiae (B0.5IVpe)[18]
V main-sequence stars (dwarfs), Example: Achernar (B6Vep)[15]
sd (prefix) subdwarfs. Example: HD 149382 (sdB5)[19]
D (prefix) white
dwarfs.[nb 2] Example: van Maanen 2 (DZ8)[20]

The Sun's spectral type is G2V, which means that the spectral class is G2 and the luminosity class is V, which means that the Sun is a main-sequence star.

An example:

The following record is for the star Arcturus (α Boötis):

H|       69673|H|14 15 40.35|+19 11 14.2|-0.05|1|G|213.91811403|+19.18726997|*|  88.85|-1093.45|-1999.40|  0.92|  0.42|  0.74|  0.60|  0.52|-0.52| 0.16|-0.13|-0.07| 0.05|-0.01| 0.03|-0.12|-0.03|-0.31|  7| 2.01| 69673| 1.629|0.010| 0.286|0.009|*| 1.239|0.006|G|1.22|0.02|A|*| 0.1114|0.0020|0.015| 70|*| 0.09| 0.14|       |U|2| |14157+1911|H| 1| 2|C| |A|AB|198|  0.255|0.039| 3.33|0.31|S| | |124897|B+19 2777 |          |          |1.22|K2IIIp      |X

The apparent visual is(V) magnitude $-0.05$. The colour index is $B-V = 1.239$, which means that it is probably a red K type star with temperature $T\approx 4000 \textrm{K}$ (from figure 2 on this webpage). The spectral type is K2IIIp, i.e. the spectral class is K2 and the luminosity class is III, a normal giant star.

An indication for the Radius can be found from the Hertzsprung-Russel Diagram in which radii are added (see Figure here). The luminosity of Arcturus can be calculated as follows:
$$ d = \frac{1}{\varpi}$$
where $d$ is in parsec and $\varpi$ is the parallax in arsceconds. The parallax is in columns 80-86 and is $\varpi = 88.85mas = 0.08885^{\prime\prime}$ for Arcturus. The distance is, therefore, $d=11.25\textrm{pc}$. You can then calculate the absolute magnitude ($M_V$):
$$ M_V = m_V -5(\log{d}-1)$$
where $m_V$ is the apparent visual (V) magnitude. The absolute magnitude of Arcturus is: $M_V=-0.05 - 5(\log{11.25}-1)=-0.31$.
The luminosity is then given by:
$$M-M_{\textrm{sun}} = -2.5\log{\left( \frac{L}{L_{\textrm{sun}}}\right)}$$
where the absolute magnitude of the Sun is $M_{\textrm{sun}} = 4.83$ and $L$ is the luminosity of Arcturus. This gives:
$$\log{\left( \frac{L}{L_{\textrm{sun}}}\right)} = 2.056 $$
$$L = 114  \textrm{L}_{\textrm{sun}}$$
From the Hertzsprung-Russel diagram we get a radius for Arcturus of about $R \approx 10  \textrm{R}_{\textrm{sun}}$.


Hypothetically, Yes. Realistically, No. As James mentioned a singularity isn't a physical object but rather a point in space time. In theory a wormhole is just two singularities which combined.


What do 5-sigma, r, and lowercase-delta mean in the physical descriptions of major telescopes?

The quote appears to be from the Ivezić et al. 2019 paper about LSST.
In this context, σ quantifies the noise level in the images, and r is a photometric passband of ~620±70 nm wavelength.
In single r-band images, they expect stars of AB magnitude 24.5 to stand out 5σ above the background level.
δ refers to declination; they plan to cover the sky from the south celestial pole to 34.5° north of the equator.


When (yyyy-mm-dd) does the north pole of uranus most nearly point towards the sun?

I have made an interactive simulation of Uranus and its seasons at https://ecliptiqc.ca/UranusEN.php. Playing with it, I get the north pole of Uranus pointing sunwards in early 1946 (my steps are 6 months, so I can’t be more precise for now).
The actual date may be slightly different, as I use a “fixed” orbit instead of the osculating elements, but it would be a few days or weeks off at most.


How would the barycenter of a double-planet system affect the two planets' surface gravity?

There's basically no gravitational pull towards the barycenter.  The two objects are in free-fall around each other and objects in free-fall experience (almost) no gravity from the object they are in free-fall around.  Anyone standing on either planet would basically only feel the gravitation of that object alone, as if it were sitting still in space (though nothing actually "sits still" in space), everything in space is flying through it's local gravitational field.

There can be small variations to the local objects gravity, generated by any rotation.  Tidal locking might be common with a dual-planet system and Tidal locking is usually a relatively slow rotation unless the objects are very close, but while it's a slow rotation, it's still greater than zero, which creates a tiny tidal bulge and a very slight centrifugal force which would effectively lower the gravity.   This has nothing to do with the barycener, it is a product of the objects rotation.

As the objects get closer there is a tidal bulging which has some effect on gravity.   Parts of the planet that bulge outwards experience lower gravity and parts that are effectively stretched in closer to the center experience an increase in gravity.   This effect is pretty small and it's an indirect effect, caused by tidal forces from the other object, not caused by the barycenter.

At the Roche Limit, gravity is effectively cancelled out by the tidal force and objects can drift off the surface both on the near and far side of the smaller of the two tightly orbiting objects.

The amount of gravitational variation would be small, unless the orbit was very close.

I'd like to post a map of the change in gravity on Pluto's Moon Charon, or Jupiter's innermost Moon Io due to the tidal forces and tidal bulging, but I couldn't find any written results on that and I'm reluctant to try to do the math myself.  Even under very strong tidal forces like those two, the effect is likely still quite small, perhaps in the range of 1% or less.

Here's a question where the tidal force and it's effect on Phobos' gravity is discussed.  You need very strong tidal forces before the gravity gets significantly effected.


Short answer: 1) Yes and no; 2) Yes, there is a supercritical fluid, of hydrogen.

Long answer:

It's fairly hot deep down in Jupiter; estimates range from 10,000 K to 24,000 K. You would think that anything in the core would be liquefied, and you could be right. Many models predict that Jupiter's core is rocky, but others predict that it is liquid. Still others say that it is "liquefying" - i.e. parts are liquid and parts are solid. We don't know for sure which are right.

In the early solar system, Jupiter would indeed have to have had a large core of rock (and potentially ice). It would have accreted gases, and eventually the planet we know today would have formed, with thick cloud layers complete with hydrogen, ammonia, and lots of other elements. But after Jupiter accumulated its mantle, there would have been a lot of pressure on the original core. It seems like much of the matter inside it would have been liquefied and carried away ("redistributed") into other parts of the planet. Still, some might have remained.

Surrounding the core, there is a layer of liquid hydrogen. Yep, liquid hydrogen, just as you said in your question. Temperatures and pressures are so high that hydrogen has passed its critical temperature (33 K) and become a supercritical fluid.

So, in summary, we don't know for sure the state of Jupiter's core is, but parts or all of it may be liquefied, and we also think that there is indeed a layer of hydrogen as a supercritical fluid surrounding the core. The same may also be said for other gas giants, although, still, we don't know for sure.

I hope this helps.

Sources:

http://www.dailygalaxy.com/my_weblog/2013/02/jupiters-dissolving-core-sheds-light-on-alien-planets.html
http://www.universetoday.com/47966/jupiters-core/
http://www.nasa.gov/audience/forstudents/5-8/features/what-is-jupiter-58_prt.htm
http://phys.org/news/2011-12-jupiter-core-liquefying.html
https://en.wikipedia.org/wiki/Jupiter#Internal_structure
http://arxiv.org/pdf/1111.6309v1.pdf (You might find this particularly interesting, as it considers gas-giant exoplanets)


How far from Betelgeuse is its habitable zone?

Nobody knows the limits of the Sun's habitable zone, or how broad or narrow it is.
Here is a link to a list of various estimates of the inner, or outer, or both, edges of the circumstellar habitable zone of the Sun:
https://en.wikipedia.org/wiki/Circumstellar_habitable_zone#Solar_System_estimates
Note that one well known estimate, that of Hart et al. in 1979, makes the habitable zone very narrow, while another well known estimate, that of Kasting et al in 1993, produces a conservative habitable zone that is several times as wide as Hart's and an optimistic habitable zone that is wider still.
And there are other estimates with the inner edge of the habitable zone closer to or farther from the Sun than Kasting's and with the outer edge of the habitable zone closer to or farther from the Sun than Kasting's.
So if you study all of the original papers where those limits of the habitable zone were proposed, you can decide which ones are most convincing to you, and then use the habitable zone for the Sun to compare with Betelgeuse to determine Betelgeuse's habitable zone.
Taking the safe route, I will assume that a planet would have to receive exactly as much radiation from Betelgeuse as Earth receives from the Sun in order to be habitable.
According to Wikipedia, Betelgeuse has a luminosity about 126,000 times that of the Sun.
https://en.wikipedia.org/wiki/Betelgeuse
Since the square root of 126,000 is 354.96, a planet orbiting about 354.96 AU from Betelgeuse should receive the same amount of radiation from Betelgeuse as Earth receives from the Sun and thus should be within the circumstellar habitable zone of Betelgeuse, no matter how broad or narrow that habitable zone is.
But Betelgeuse is a variable star. Its luminosity varies between about 90,000 to 150,000 times the luminosity of the Sun, so the distance from Betelgeuse where a planet would receive exactly as much radiation from Betelgeuse was Earth receives from the Sun would vary between 300 AU and 387.298 AU.
Possibly a planet could orbit Betelgeuse with a somewhat elliptical orbit so it was closest to Betelgeuse when Betelgeuse was least luminous and farthest from Betelgeuse when Betelgeuse was most luminous and so constantly receive the same amount of radiation as Earth receives from the Sun.

Betelgeuse is classified as a semiregular variable star, indicating that some periodicity is noticeable in the brightness changes, but amplitudes may vary, cycles may have different lengths, and there may be standstills or periods of irregularity. It is placed in subgroup SRc; these are pulsating red supergiants with amplitudes around one magnitude and periods from tens to hundreds of days.[8]

https://en.wikipedia.org/wiki/Betelgeuse#Variability
The semiregular nature of Betelgeuse's variability means that nobody could even begin to design an orbit around Betelgeuse that would enable a planet to constantly receive the same amount of radiation as Earth receives from the Sun.
So planets in the habitable zone of Betelgeuse would receive significantly varying amounts of radiation from Betelgeuse as the star varied.  Whether that would make life impossible on those hypothetical planets is unknown.
I may add that there are still more problems with having a planet with conditions suitable for life orbiting Betelgeuse; the answer by antispinwards mentions some of them.
So the probability that there will be lifeforms on any planets orbiting Betelgeuse when Betelgeuse becomes a supernova and destroys all its planets seems to be very, very, very low.
I note that humans can not survive in the majority of the biosphere of Earth.  Humans can't survive in most places where other Earth lifeforms can survive.
A discussion of the conditions necessary for humans or similar beings to survive can be found in Habitable Planets for Man Stephen H. Dole, 1964, 2007.
https://www.rand.org/content/dam/rand/pubs/commercial_books/2007/RAND_CB179-1.pdf
Here is a link to a list of the nearest stars and brown dwarfs to the Sun, stars within a distances of 5 parsecs or 16.3 light years:
https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs
There could be worlds with some kind of life in some of those star systems.
And what would happens to those hypothetical worlds with life orbiting stars within 5 parsecs or 16.3 light years of the Sun if the Sun became a supernova?  The Sun will never become a supernova, but if it did the planets of those nearby stars would probable receive so much radiation during the supernova event that  their oceans and atmospheres would boil away and their surfaces turn to red hot lava, and all life on them would die.
And a supernova might be deadly to life on worlds at a much greater distance.  I am not very familiar with the distances at which a supernova would wipe out all life on a planet.
Betelgeuse is expected to have a supernova explosion in less than 100,000 years.  During that time many stars will get closer and closer to Betelgeuse and then start to get farther and farther from Betelgeuse, as those stars and Betelgeuse orbit around the center of the galaxy.
Here is a link to a list of stars which, according to the calculations of astronomers, have passed within 5 light years of Earth in the past three million years or will pass within 5 light years of Earth in the next three million years.
https://en.wikipedia.org/wiki/List_of_nearest_stars_and_brown_dwarfs
And it seems to me that any alien astronomers on a planet orbiting a star near Betelgeuse would be very displeased to learn that their star is getting closer and closer to Betelgeuse and is likely to be only a few light years from Betelgeuse when Betelgeuse supernovas.


What did I see streaking across the sky last night?

It could have been an Iridium Satellite.  Their solar panels are at an angle such that the sun reflects off them causing them to look like a bright flash that crosses the sky for a few seconds.  I use a site called Heavens Above that allows you to enter your exact location so you can see what objects appear above, at what time and in what direction.  Check that site out and see if any Iridium pass over there.


Ok, here's my take on calculating the color of a blackbody, or any spectrum in fact:
Disclaimer: I'm not a color theorist, and there may be more accurate methods. But the result, shown in the bottom, looks about right.

Spectrum
First note that since color is a function of the relative intensity in various wavelength bands, it doesn't matter whether we express our spectrum in flux, flux density, intensity, luminance, or something else that only differs by constant values.
Let's consider a spectrum $S(\lambda)$ as a function of wavelength $\lambda$. If you're interested in a Planck spectrum, first calculate this from the temperature $T$:
$$
S(\lambda) = B(\lambda,T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda k_\mathrm{B}T} - 1},
$$
where $h$, $c$, and $k_\mathrm{B}$ are Planck's constant, the speed of light, and Boltzmann's constant, respectively.
CIE 1931 color space
The CIE 1931 color space was the first attempt to quantify the relation between the distribution of wavelengths and the color perceived by humans. It makes use of the so-called color matching functions $\bar{x}$, $\bar{y}$, and $\bar{z}$, which are functions of $\lambda$, seen here:

Credit: Wikipedia
You can find tabulated values of these functions on the internet.
The color (which is 3D because humans have three different cones) of any spectrum can be mapped to a 2D position $\{x,y\}$ in the CIE 1931 color space chromaticity diagram seen here:

Credit: Wikipedia
Tristimulus values
The pair $\{x,y\}$ are normalized versions of the tristimulus values $\{X,Y,Z\}$ defined by
$$
X = \int S \, \bar{x} \, d\lambda\\
Y = \int S \, \bar{y} \, d\lambda\\
Z = \int S \, \bar{z} \, d\lambda.
$$
(Actually they're defined using the spectral radiance), but this is equal to the intensity to within a constant factor.)
The lower-case $\{x,y\}$ are then given by:
$$
\begin{array}{rcl}
x & = & \frac{X}{X+Y+Z}\\
y & = & \frac{Y}{X+Y+Z}\\
z & = & \frac{Z}{X+Y+Z} = 1-x-y,
\end{array}
$$
and the last of these relations is the reason only two values are needed.
XYZ to RGB
To go from $\{x,y\}$ or $\{X,Y,Z\}$ to RGB is a linear transformation which can be conducted by matrix multiplication. There are many variations here; my understanding is that is has to do with whatever device you're using to display you colors, but there might be more to it.
The sRGB color space assumes that the $\{X,Y,Z\}$ set has been scaled to $\{0.9505, 1, 1.0890\}$ for the CIE illuminant D65 ("white"); that is, $Y\equiv 1$.
With the above values, this new set can be calculated as
$$
\begin{array}{rcl}
Y & \rightarrow & 1\\
X & \rightarrow & x/y\\
Z & \rightarrow & z/y,
\end{array}
$$
and the RGB values are then, using the conversion matrix from here which cites this,
$$
\left[
\begin{array}{l}
R\\
G\\
B
\end{array}
\right]
=
\left[
\begin{array}{rrr}
 1.656492 & -0.354851 & -0.255038 \\
-0.707196 &  1.655397 &  0.036152 \\
 0.051713 & -0.121364 &  1.011530
\end{array}
\right]
\left[
\begin{array}{l}
X\\
Y\\
Z
\end{array}
\right]
$$
Gamma correction
I'm not 100% sure, but I think this set of RGB values needs to be gamma-corrected, which is a non-linear correction that account for the fact the the device displaying the colors shows different relative luminances than the human eye perceives. At least my result, shown in the bottom, looks about right, whereas it doesn't if I don't perform this correction.
The gamma correction is given by the function
$$
\gamma(u) = \left\{\begin{array}{ll}
12.92 u                 & \mathrm{if\,} u \leq 0.0031308 \\
1.055 u^{1/2.4} - 0.055 & \mathrm{otherwise}
\end{array}\right.
$$
and the perceived RGB values should then be
$$
\begin{array}{rcl}
R & \rightarrow & R\times\gamma(R)\\
G & \rightarrow & G\times\gamma(G)\\
B & \rightarrow & B\times\gamma(B)
\end{array}
$$
These values are mostly in the range $[0,1]$, but if they fall outside, they're usually clipped. If you want your values to be in the range $[0,255]$ you just multiply them by 255.
Result: the color of the Universe through time
I implemented the above algorithm in Python, as a part of a larger code called timeline that calculates various properties of the Universe as a function of time after the Big Bang. You can find it here on GitHub.
With that code I calculated the color of the Universe as a function of time after the Big Bang, and hence as a function of temperature. Until star formation kicks in around 200 million years after the Big Bang, the color is given by the temperature of the cosmic microwave background.
When stars arrive, the average cosmic spectrum becomes more complicated and can't be calculated (yet) from first principles, but must be observed. I haven't found such observations at early times, but Baldry et al (2002) presents the spectrum at various times after an age of 11 billion years, at which time the Universe is a dull "cosmic latte". So in this figure, I grayed out the color between $t=200\,\mathrm{Myr}$ and $t=11\,\mathrm{Gyr}$:

But please note the disclaimer in the top of this answer :)


Origins and most frequently used; perinigricon vs peribothron?

I don't work with black holes, but I've been in the field of astrophysics for several years now, and until I read this question, I had never heard of either the two terms.

But a query on NASA/ADS, the primary database for "professional" astronomical papers (both refereed and non-refereed) yields 54 hits for "peribothron" vs. only 10 hits for "perinigricon". Not a lot.

As your links reveal, the etymological origin of peribothron/-nigricon is "near the black hole" in Greek and Latin, respectively. The reason has been to conform with terms such as perihelion, meaning "near the Sun". The first to use the Greek version in the literature was Frank & Rees (1976) who write in a footnote:

  $^\star$We are grateful to W. R. Stoeger for suggesting this word, derived from the Greek bothros, a pit.

William Stoeger was the second author Martin Rees' student (and Stephen Hawking's classmate).

The first to use the Latin version seems to have been Schödel et al. (2002), but they don't offer any explanation for the term.


What will happen if a black hole appears near our solar system?

The way not to fall into a black hole is essentially the same as the way not to fall into the Sun.  It's no more dangerous than any other body with the same mass, which is typically a few times the Sun's mass.

The danger of a passing black hole (or other star) is not that it could swallow us but that it could perturb our orbit, with unpredictable effects on climate; as well as stir up the comets of the (hypothetical) Oort Cloud, sending some of them toward the inner planets, with the possibility of an impact that could ruin your whole day.


What about BSC5p for stars brighter than magnitude 6.5? The "alt_name" "is usually the Flamsteed and/or Bayer constellation-based name for the star".

Here a list of more advanced catalogs.

Here a list of some common star names for Hipparcos numbers. Probably not much different from your query result.

On Wikipedia a probably more complete list of more or less wide-spread common star names.

Those common names often are not unique, and not all common names are widely accepted. That's probably why they aren't provided in the star catalogs.

The only way I could imagine to get a solution is to convert the Wikipedia table to a csv file, and to import it to a database table used as a relation from the various catalog names to the proper names. That will probably need quite a bit of manual work.

None of the naming systems is complete, not even necessarily unique. Stars frequently turn out to be not one, but several stars, or no star at all. At some point it's easier to use 'scientifics'.


Intuitive explanation for why the universe is flat

The CMB lets us measure how close to flat the universe is right now.

On the other hand, inflation tries to explain how we got from whatever the early universe was to right now.

The motivation for the latter being that even extremely small deviations from perfect flatness in the early universe should have resulted in very obvious deviations from flatness today.  That we don't see obvious deviations then suggests the universe was amazingly close to perfectly flat early on, and such perfect tuning seems bizarre. Inflation basically lets a (in some sense "very") non-flat early universe get flattened to this precision, thereby resolving the "problem" of having an almost perfectly flat early universe.  It takes the situation from an extremely narrow range of initial conditions for the flatness of the universe and gives us a broad range of equally valid possibilities, instead.

Inflation is also commonly used to explain the remarkable homogeneity of the CMB, which suggests that parts of the universe that cannot ever have received light speed signals from each nevertheless achieved thermal equilibrium (an at-most light speed process).  Inflation throws in a super-luminal expansion to allow these regions to be within light speed communications of each other in the very early universe, with just enough time to achieve thermal equilibrium, and then inflates them beyond light speed communication.


Why can you see the space station on some days but not on others?

Two basic conditions need to be met for anyone on the globe to observe the International Space Station (ISS) from any location:

ISS needs to pass roughly overhead of your location, and
it needs to do that during night so it's visible to the naked eye

Now, obviously there are other requirements, like e.g. weather (if it's overcast we won't be able to see it through the clouds), but those screens you attach don't account for that, since it would be nigh impossible to predict weather so far in advance.

Let's discuss the two basic conditions mentioned a bit more. ISS has an orbital inclination of 51.65°. Why is that is explained in this answer on Space.SE, but to quickly recap, this orbital inclination was chosen because it's near ideal for launches from the Baikonur Cosmodrome, where most launches towards the ISS are made from, it is still achievable from the US launch facilities and most US built launchers, and it allows the ISS to go over majority of the populated areas on Earth. So it was decided both for economic reasons, as well as to allow ISS to do as much science as possible.

But what this inclination means is that the ISS will make a ground track (path along the surface of the Earth directly below it) that moves towards the West with each new orbit, even though the ISS itself moves West to East (or saying it otherwise, it's in a prograde Low-Earth orbit). Its ground track could look something like this:

                  Ground track of the International Space Station with an area where it is observable from the Earth's surface                  shown in the white circle (Image source: ISS Tracker)

ISS orbital period, the time it takes it to circumnavigate the globe and complete one orbit, is roughly 93 minutes. As you can see in the screen grab from the ISS Tracker above, its ground track moves westward on each new orbit, which is due to Earth's rotation on its axis, so this shift towards the West will be equal to how much the Earth rotates in the meantime, which in those 93 minutes comes out at roughly 22,5° or a good 2,505 km (1,556 mi) of the Earth's total equatorial circumference.

But ISS's orbit isn't Sun-synchronous, meaning it won't be over any given Earth latitude at the same local mean solar time, so when it will be observable depends also on its orbital position relative to the position of the Sun. We require that it is dark, remember? ISS isn't bright enough (doesn't reflect enough light with its truss and solar panels) during the day to be brighter than the day skies are. So we have another parameter, which is local time. As you probably anticipated by now, the length of the day isn't really equal at any day of the year, and moving further away from the equator only increases this difference of the length of the day in winter, or the length of it during summer. This is another range or values that your software you're attaching print screens from has to account for.

So, when your software collects data from you regarding your geolocation and current date and time (might do that automatically if you're using it on a device that has GPS enabled), it will establish the times (timetable) of when the ISS ground track is over or around your area, and if that happens during night. The ISS doesn't have to be exactly overhead, at zenith, so it will also calculate angle to the zenith for any area that the ISS goes over, at which you would still be able to see it.

Understanding this, even if we don't look at your timetable, we could establish that if you're at latitude to the equator that enables you to see the ISS, you'd be able to do that roughly a few times in a row separated by around 93 minutes, then wait for the ISS ground track to come back towards your longitude (16 orbits or about one day and an hour), and do that during night when you'd be able to observe it. Nights are long now during December at Northern latitudes, so you see a long row of couples of such chances during each night. As long as the weather holds, of course.


Does bending of light around the Sun depend on the wavelength?

The amount of "gravitational light bending" is independent of the photon energy (light wavelength).
The reason is that the light follows a path through spacetime that is appropriate for a massless particle and this is unique for a given set of initial conditions.
That this is so is amply demonstrated by the consistent angular displacement of "stars" near the limb of the sun whether observed at optical or radio wavelengths.
As pointed out in comments - there are small effects that must be taken into account, associated with the well-understood phenomenon of refraction in the corona of the Sun. However, these do not affect observations of lensing taken well away from the solar limb - which is easily possible at radio wavelengths and now becoming possible for the same sources using Gaia data.
Further evidence comes from the wavelength-independent nature of gravitational lensing and microlensing seen outside the solar system.


Is there a standardized "Astronomical Alert" system?

In general, no there is not a standardized alert system (unless you count email...). Most subareas of time-domain astronomy that deal with alerts tend to have their own central clearing houses and alert distribution mechanisms. Most of the high energy/explosive (supernovae, novae, blackhole transients (stellar or massive)) transient community tends to announce things via Astronomers Telegram which have basically replaced the older IAU Circulars and the Central Bureau for Astronomical Telegrams for announcing things. More recently, those working on comet outbursts have also sent notifications via ATel e.g. the recent Comet 29P outburst.

The NEO community tends to coordinate through the Minor Planet Center NEO Confirmation Page (NEOCP) for newly discovered NEO candidates found by the surveys. In recent years, this has been supplemented by the JPL/CNEOS Scout system which performs risk analysis on the NEOCP candidates and sends out email alerts for close passing objects or where they may be issues with the measurements - follow-up and discussion on these is done via email.

The Gamma Ray Burst (GRB) community has had the GCN Circulars as previously mentioned which started around the end of Compton Gamma Ray Observatory and the start of BeppoSAX when better location of GRBs become possible. An issue with both ATels and GCN Circulars is that they are written in English text and are sent via email and are mainly designed for human consumption and reaction. This has been recognized as an issue for many years and lead to the development of the VOEvent 2.0 which became an IVOA Recommendation in 2011. Quoting from the Abstract of the document:

  VOEvent [20] defines the content and meaning of a standard information packet for representing, transmitting, publishing and archiving information about a transient celestial event, with the implication that timely follow-up is of interest. The objective is to motivate the observation of targets-of-opportunity, to drive robotic telescopes, to trigger archive searches, and to alert the community. VOEvent is focused on the reporting of photon events, but events mediated by disparate phenomena such as neutrinos, gravitational waves, and solar or atmospheric particle bursts may also be reported. Structured data is used, rather than natural language, so that automated systems can effectively interpret VOEvent packets. Each packet may contain zero or more of the "who, what, where, when & how" of a detected event, but in addition, may contain a hypothesis (a "why") regarding the nature of the underlying physical cause of the event.

It's fair to say that adoption of VOEvent has been slow and limited, driven in part by a lack of easy-to-use libraries or toolkits, but usage is picking up with the LIGO/VIRGO gravitational wave alerts going out in VOEvent XML format (Alerts User Guide) and a standard for reporting Fast Radio Bursts. Sending of VOEvents from the LSST brokers is also planned and so the standard is likely to see greater adoption.


Can you recommend a book about big bang nucleosynthesis and chemical abundances?

I highly recommend Nucleosynthesis and Chemical Evolution of Galaxies by Bernard Pagel. It contains the basics of nuclear reactions andstellar evolution, chapters on big bang nucleosynthesis and light element production, as well as covering the broad swathe of stellar nucleoynthesis and how these link together into predicting the chemical evolution of galaxies.


Astronomical data convertion from Jy/pixel to MJy/sr?

For both conversions, you need a bit more information about the data.

How can we switch from Jy/pixel to MJy/sr?   The Jy -> MJy part is just a factor of $10^{6}$ Jy/MJy.  The pixel -> sr (steradian) part requires that you know the angular size of a pixel on the sky.  For PACS it looks like the pixel size is different for different wavelengths; see Table 3.1.  Once you know the pixel size in arcseconds, then the conversion is 206265 arcseconds / radian (or more precisely, 3600*180/pi).  Square the size in radians to get the area in steradians.   You also may be able to find the pixel size for your data by looking in the FITS header, if you have FITS images.

Is "Jy/beam" equivalent to "Jy/deg^2" or "Jy/sr"?  No, but you can convert between the two.  Both are flux per angular area on the sky.  You need to know the angular area of the telescope beam (e.g. in square degrees, or steradians) and then you can convert between the two.

The README for the ATLAS survey says this about the conversion in their SPIRE maps:
"To convert the maps from Jy/beam to Jy/pixel, in order to carry out aperture photometry,
the values in the maps should be divided by the ratio between the beam area and the pixel
area in arcsec^2 (469/36, 831/64 and 1804/144 at 250, 350 and 500 microns respectively).
The mean of the maps should also be subtracted. Users interested in the most accurate
aperture photometry of extended sources should consider scaling the SPIRE fluxes using the
K4 corrections for extended sources. They should also consider making an aperture
correction."


In other phases, the angular distance of Sun from Moon is equal to the phase of the moon, in the direction of the lit side. One of the methods could be to estimate the position of Sun based on this, and then you can, assuming time is known, estimate North.

If you don't know time, use constellations (which can form a separate question) to estimate North and the reverse of above method for time.

Now, if there are clouds and you cannot see any constellations, well, you need to be able to calculate cosines and sines and their inverse in your heads. This method, which I will not describe depends on the rate of change of moon's position, which you'll have to estimate using your hands.

I will have to search for my notes for actual calculation of time, let's see if someone else answers it. But let me tell you how to find direction just using Moon, though this is less accurate than if you already know the time.

From the center of the Moon, draw a line in the sky and follow it towards the lit side of the Moon, the spot where it cuts on the Horizon is either East or West. It is East if you are observing this in Waning Phase i.e. after a Full Moon, else it is West (if after the days of a New Moon). You can get the rest of the directions this way.

Here's how it works. The line you just drew was approximately the ecliptic, the apparent path of Sun in sky. It is at an angle of 23.5° from the equator, so that is the maximum inaccuracy that you might have for your cardinal directions. This would actually lead you off in a wrong direction initially, but the ecliptic keeps on moving, so keep on doing this for the whole night, the positions of your North will be varying but the averaged out value will be better than your initial value.


Why is the universe full of spinning objects?

The MinutePhysics channel on YouTube made a wonderful short animation that answers your very question. It explains how a cloud of gas and dust with zero net angular momentum collapses down to form a flat spinning object such as a galaxy or our Solar System.

The answer involves a few things:

The universe conserves angular momentum in every interaction between
physical objects.
Mathematics says a big cloud of gas and dust must have exactly one plane in which the overall angular momentum is zero.
When particles of gas and dust collide, any angular momentum they have outside that plane is cancelled, so over millions of years and (countless numbers of collisions) the cloud ends up flattening.
But as they collide, their angular momenta within that one plane are
conserved, so the flattened system ends up spinning.


Why don't Neutron Stars form event horizon?

As Francesco Montesano points out, using the wrong mass leads to the wrong answer.  Also, using the density here seems a complicated way to get to the answer; you could compute the Schwarzschild radius for the NS, and see whether it's smaller than its actual radius.

Since the density scales as ρ ~ M/R^3 and the Schwarzschild radius as Rs~M, the density of BHs scales as ρ~1/R^2; more massive BHs are less dense and simply testing whether a NS is denser than a BH alone is not sufficient - they must be of the same mass, which means that you are in fact comparing radii.


Comparing orbits between a planet and Red Dwarf

earth like planet orbiting around a sun like star (365 days)
red dwarf on an elliptical orbit around the star that passes close to the planet (1896.59 days, eccentricity 0.866)

If we work in AU and years we can use $GM_{Sun} = 4 \pi^2 (\text{AU}^3 / \text{year}^2)$ for the standard gravitational parameter of the Sun. If we also assume the red dwarf's mass is small enough that it can be ignored we can leave the Earth in it's circular 1 AU orbit and assume that the Sun doesn't move. Then we can check the relationship between period and semimajor axis as follows:
$$T = 2 \pi \sqrt{\frac{a^3}{GM}} = a^{3/2}$$
$$a = T^{2/3} \left(\frac{GM}{4 \pi^2}\right)^{1/3} = T^{2/3}$$
which gives $a=1$ AU for 1 year, and $(1896.59/365)^{2/3} \approx 3 $ AU for the red dwarf so this checks out.
However your drawing is off. The periapsis and apoapsis of the red dwarf will be $3(1-e)$ and $3(1+e)$ or 0.4 and 5.6 AU and this plus this comment makes it clear that your drawing is wrong.
If the ratio of the two periods were a rational number, say $T_2/T_1 = n_2/n_1$ where $n_1, n_2$ are integers, then their relative motion would be periodic with the synodic period
$$T_S = \frac{1}{\frac{1}{T_1} - \frac{1}{T_2}} = \frac{T_1 T_2}{T_2 - T_1} = \frac{n_1 n_2}{n_2 - n_1}$$
but the distance of closest approach would depend on phasing; the initial positions at time $t=0$.
However in your case you have $a_2/a_1 = 3$ and so $T_2/T_1 = 3 \sqrt{3}$ which is irrational and not expressed as the ratio of two integers. That means that if these two orbits are in the same plane and the two bodies do not interact gravitationally with each other, sooner or later they may collide.

Python:
import numpy as np
import matplotlib.pyplot as plt

twopi = 2 * np.pi
a1, e1 = 1.0, 0.0
a2, e2 = 3.0, 0.866

theta = np.linspace(0, twopi, 201)

r1 = a1 * (1. - e1**2) / (1. + e1 * np.cos(theta))
r2 = a2 * (1. - e2**2) / (1. + e2 * np.cos(theta))

x1, y1 = [-r1 * f(theta) for f in (np.cos, np.sin)] # minus to match plot in question
x2, y2 = [-r2 * f(theta) for f in (np.cos, np.sin)]

plt.figure()
plt.plot(x1, y1, '-b', linewidth=2)
plt.plot(x2, y2, '-k', linewidth=1)
plt.plot([0], [0], 'or')
plt.gca().set_aspect('equal')
plt.xlim(-3.2, 6.8)
plt.ylim(-4, 4)
plt.xlabel('Distance (AU)', fontsize=14)
plt.ylabel('Distance (AU)', fontsize=14)
plt.show()


Are there any galaxies visible in the night sky around the Orion constellation?

The short answer is there are no galaxies in or around Orion that are visible to the naked eye.

The International Astronomical Union (IAU) has some simplified star charts for each constellation. The charts show galaxies the brightest galaxies that are visible in amateur telescopes as red ovals. The chart for Orion and the surrounding constellations show no galaxies. Of course, larger telescopes are able to show fainter objects, so you are bound to find a galaxy in every constellation with a large enough telescope.

The galaxies that are easily visible to the naked eye are as follows: (There are a few other that are visible to the naked eye, but they are not as obvious.)

Large Magellanic Cloud in the constellations of Dorado and Mensa
Small Magellanic Cloud in the constellations of Tucana and Hydrus
Andromeda Galaxy in the constellation of Andromeda


how to find the approximate [stellar radius]? (distance from center to most distant star) of a circular flat disk galaxy based on its stellar mass?

What you are looking for is called the "galactic size-mass relation" for galaxies dominated by disks.
Theres an interesting research paper by Rebeca Lange and others where this relation (equation 3 in the paper) has the form
$R = \gamma M^{\alpha}(1+M/M_0)^{\beta-\alpha}$
Where $R$ is the radius of the galaxy in kiloparsecs ($1$ kpc $= 3260$ light year), $M$ is the mass of the galaxy measured in solar masses ($1$ solar mass $= 1.988\cdot 10^{30}\; kg$), and $\gamma$, $\alpha$, $\beta$ and $M_0$ are parameters that have been tweaked to adjust the relation to the real data.
In their paper they propose values for $\gamma$, $\alpha$, $\beta$ and $M_0$ for different kinds of measurements and galaxies. For the late-type galaxies (which are the ones that you want to study), these parameters are shown in table 2 with values that have a certain range.

$\alpha$ between $0.13$ and $0.17$
$\beta$ between $0.91$ and $1.00$
$\gamma$ between $0.08$ and $0.18$
$M_0$ between $19\cdot 10^{10}$ and $48\cdot 10^{10}$ solar masses

To make an approximate guess let's take the average values they present for each parameter. Then you will have the relation:
$R = 0.12\cdot M^{0.15}\cdot(1+3.8\cdot 10^{-12}M)^{0.81}$
Again with $R$ in $kpc$ and $M$ in solar masses.
With everything into account we are ready to plug-in some numbers. For example, the Milky Way galaxy is known to have a mass of $M = 10^{12}$ solar masses. This means that according to our model the Milky Way disk has a radius of $R = 26.97 \; kpc$. Well it turns out that we think the disk to extend between $170,000$ and $200,000$ light years in diameter, which translates to $26 \; kpc$ to $31\; kpc$, so we proved that the model works at least for our galaxy, even if our prediction is in the lower bound of the range.

Now, the devil is in the details; $R$ is not actually the radius of the disk as you might think. The radius is an ambiguous term here since disk galaxies don't have a sharp edge where stars stop to exist. The $R$ in the formula is actually the so-called "effective radius", wich is in fact just the radius which contains half of the light emmited from the galaxy (this is an unambiguous way to measure the size of the galaxy). So, when talking about the radius $R$ I would take this into consideration, and use maybe a $1.2R$ estimate (20% larger) as my "true" radius for the galaxy.
Another important thing to consider is that the mass of the milky way has been measured including the mass of the dark matter and the mass of the gass. If you only want stellar mass, then the calculation will yield a galaxy that might be smaller.
These two are unimportant points if you are making just a highschool project and they don'tchange drastically your models either way, but I have to warn you about a consideration here. A galaxy that has a mass of $x$ solar masses is not a galaxy that has $x$ stars (planetary systems). That would certainly happen if all the stars where like the Sun, but as it turns out the vast majority of stars are less massive than the Sun, thus an $x$ solar mass galaxy will have much more than $x$ stars. If you are going to play with the Drake equiation that might be important. Finally if you search for the Milky Way data you will see that we think it has a mass of $M = 10^{12}$ solar masses, as I said, and we think it has around $2\cdot 10^{11}$ stars, so what's happening here? Shouldn't a galaxy with mass $x$ solar masses have more than $x$ stars? The problem here is, as I said previously, that we are considering the mass of the entire galaxy including gas, dust and dark matter, but we are only concerned about stellar mass.

I'm very interested in your project. Would you mind to tell me more about it? I'm just curious.


Can a Ritchey–Chrétien telescope be collimated using a standard Newtonian laser collimator?

No, the fundamental designs of the two types of telescope are in this respect quite different.

The Newtonian laser collimator works by shining a laser through the eyepiece holder to the secondary mirror which reflects the beam onto the primary. The primary, usually a parabolic mirror but always concave, is adjusted so that the beam is reflected back to the secondary mirror such that the reflected beam returns to its point of origin.

A RCT, or other Cassegrain-like telescopes have a convex, possibly hyperbolic, mirror as the secondary. While it is possible to shine the laser through the eyepiece holder to the secondary mirror, and the mirror can be adjusted to reflect the laser back to the point of origin, it cannot be reflected onto the primary and then back to the secondary*. Any adjustment of the secondary to reflect the laser back to the origin only ensures that the incident surface of the secondary mirror is perpendicular to the line-of-sight through the eyepiece holder. It does not ensure that the secondary mirror is aligned correctly with respect to line-of-sight through the eyepiece, and neither with respect to the primary mirror.

* Notwithstanding that the two mirrors could be misaligned so badly that this could occur at some particular point.


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


What is a typical polar angle of a pulsar beam?

Typical angles can span the range from several degrees to close to $90^{\circ}$ in some extreme cases. For pulsars with narrow emission cones, the opening angle of the emission cone is approximately
$$\rho\approx1.24^{\circ}\left(\frac{r_{\text{em}}}{10\;\text{km}}\right)^{1/2}\left(\frac{P}{\text{s}}\right)^{-1/2}\propto P^{-1/2}$$
with $r_{\text{em}}$ the emission height and $P$ the period (Lorimer & Kramer, Handbook of Pulsar Astronomy). The emission height is the distance from the center of the pulsar at which the radio emission occurs; it is typically much larger than the radius of the pulsar. The expression can be derived by making some assumptions about the location of the last open magnetic field lines. The relation works quite well for long-period pulsars with $P\gg100\;\text{ms}$, which, accordingly, fall in the range $1^{\circ}\lesssim\rho\lesssim10^{\circ}$, but fails for millisecond pulsars, some of which have opening angles quite a few times larger.
To give a better visual perspective: The angle of the emission cone, $\rho$ depends on several things. In the classical Julian-Goldreich model, the pulsar is surrounded by plasma, which is forced to corotate with the neutron star by its magnetic field. If we take this to its logical extreme, there must be a radius at which rotating plasma would have to travel faster than the speed of light; this radius defines a surface we refer to as the light cylinder.
Any magnetic field lines which cross the light cylinder's surface do not close. Assuming that the emission cone must fall within the larger cone of open field lines, we can determine the largest allowed angular radius of this cone-within-a-cone if we know the emission height - the point above the pulsar at which radio emission occurs. Not surprisingly, the size of the light cylinder is determined by the rotational period; from the argument above, we can estimate $\rho$ from that and the emission height.


Future of CMB observations: How will our knowledge of the early universe change?

This is a great question. I know of a couple of really big things about inflation people want to be able to nail down by using the cosmic microwave background.

The first is measuring what are known as E- and B- modes, which are the curl-free and divergence-free components to the modes of cmb radiation:

Essentially, measuring large scale Gaussian B-modes from primordial gravitational waves will help constrain the energy scale of inflation. It may also be able to rule out most ekpyrotic and pure curvaton/inhomogeneous reheating models (same source).

The other thing people are looking at is this idea of primordial non-Gaussianity, which are second order corrections to the Gaussian fluctuations present in the cmb (review article; early planck results). Measuring a parameter called $f_{nl}$ (deviation from Gaussianity) has been a fairly crucial part of current and future studies and will also help rule out various inflationary models. This $f_{nl}$ parameter is defined as follows:

    In this case the multipole coeﬃcients $a_{lm}$ of the CMB
    temperature map can be written as
    $$a_{lm} = a_{lm}^{(G)} + f_{nl} a_{lm}^{(NG)} $$
    where $a_{lm}^{(G)}$ is the Gaussian contribution and $a_{lm}^{(NG)}$ is the non-Gaussian contribution.


Can we see the stars at the edge of the Milkyway?

Take a look at the HR-diagram. You have forgotten to take into account that stars come in vastly different luminosities.

For example, we can barely see all red dwarves out to a distance of 10pc, while we know that they vastly outnumber the more luminous stars.

The luminous red giants however we'll be able to see on side of the edge of the disc, but even Gaia will not be able to look through to the other edge. It's limit magnitude is 20, while extinction through dust in the direction of the galactic center is often cited to be up to 100th magnitude.
Hubble deep field galaxies for comparison are around 30th mag.

However, there is Baade's window, a region just above the central galactic plane that has low dust content on our line-of-sight there. We'll be able to spy quite some stars there.

If you're interested in the Gaia mission, you can find information here.


Does the size of the atom limit the focal length of telescopes?

Ok, finally I can explain it. Not sure why it took so long.

Bottom line, for visible light atoms don't matter. Let's say the blue side of the visible spectrum of light has a wavelength of 400 nm (round number for simplicity). Let's assume the distance between two atoms in glass is 0.2 nm.

The thing is, light is a wave. Anything smaller than about 1/4 of a wavelength becomes essentially invisible to that wave. It's not a sharp threshold, but as things get smaller and smaller, the wave "sees" them less and less. It just goes around the object as if it's not even there.

1/4 wave of visible light is 100 nm, which is 500 glass atoms.

I make telescope mirrors. I can still see the effects of surface errors down to about 1/20 ... 1/25 of a wave. Beyond that the image is not really distinguishable from a geometrically perfect mirror, either in real-life tests under the stars, or in the lab using the equipment. Maybe an interferometer could push that limit a bit further, but not a whole lot.

Let's say the absolute measurable limit is 1/50 wave, which is 8 nm, which is 40 atoms. That's plenty of atoms. Any defects smaller than that simply do not change the image created by the mirror - when talking about regular telescope mirrors.

The surface level difference from a single layer of atoms does not even register in visible light. It takes far more thickness than that to make a visible difference.

So your ultra-long focal length mirror would be almost flat, slowly adding extra atomic layers towards the edge. As long as it approximates the desired ideal surface (paraboloid, etc), then it will be fine.

I see one exception. If the focal ratio is indeed so big that there isn't enough curvature in the surface to require even a single extra layer of atoms at the edge. Then, indeed, the mirror would not work, because it's flat down to the atomic level. Keep in mind though, that would be a mirror with a humongous focal ratio - and indeed with a cosmic-scale focal length. There are many issues with such a system before you even begin thinking about surface errors. Probably not feasible with current technology.

But then what you would do is make a segmented mirror, increasing the apparent diameter and reducing the focal ratio. You could even have multiple flat segments (flat down to the atomic level) but far from each other to massively increase the total diameter, and slightly tilted according to the shape of the ideal surface. If you can afford a cosmic-scale focal length, then you can surely afford a cosmic scale diameter, and that's how you reduce the focal ratio. Maintaining that precise tilt angle between segments would be a tremendous feat of engineering.


(How) Can one determine if an exoplanet is synchronously tidally locked or not?

No. Not with the limitations of our current technology. 99% of anything you've heard describing the state, makeup, nature of or appearance, of any exoplanet, is wishful thinking and conjecture. Our current technology is still rather limited. Today, we are able to come up with reasonable guesstimates of the sizes of exoplanets and their distances from their host stars, all with large degrees of uncertainty. That's it - size and distance - nothing else - at least for today (2017).

When we look at a star and see that its light has regular periodic dips, we measure the length of, the strength of, and the period of those dips (that is all we can see). Then we do a lot of math and guess at the size (mass) of the exoplanet and its distance from its host star. Even the size (both mass and diameter) of these planet-hosting stars are mere guesses. Only 16 of all the visible stars have ever been directly imaged (resolved to a disc).

However, people do love to dream. Some scientists LOVE to embellish. Some scientists like to walk around like peacocks with their plumage spread in display, hoping folks won't see their behind behind their tail feathers.
(I got to tell you - I really enjoyed thinking up that last sentence.)