How fast was the Earth spinning directly after the Moon formed?

We aren't positive that the Giant Impact hypothesis of Moon formation is correct. If we assume it is, we still don’t know how fast the Earth was rotating just after the collision.
Conservation of energy
If we could assume that energy was conserved from then to now, we could calculate the current energy of the Earth/Moon system and then estimate the rotation speed of the Earth when the Moon was in a near orbit.  A previous attempt at this can be seen in the edit revisions for this question. Unfortunately, we can’t assume that energy was conserved since some was certainly lost to tidal heating.
Conservation of AM (Angular Momentum)
If we assume AM is conserved, according to Asphaug 2014:

If somehow brought together into one place, $\vec{L}_{EM}$ would equal the
angular momentum of an Earth-mass planet spinning with a period of∼5
h—an anomalously large value that Cameron & Ward (1976) took to
indicate a giant impact.

Here, $\vec{L}_{EM}$  is the combined angular momentum of the Earth/Moon system.
If the Moon formed just outside Earth’s Roche limit of 2.9 $R \oplus$, presumably Earth’s rotation rate would be slightly slower than 5 hours.
Unfortunately, we can’t assume that AM of the Earth/Moon system is conserved.  There are several ways the Earth/Moon system has shed AM:

The Earth’s rotation is slowing due to its own tidal interaction with the Sun. https://en.wikipedia.org/wiki/Tidal_acceleration#Effect_of_the_Sun
As the size of the Moon’s orbit around the Earth increased, it went through several resonances with other bodies in the Solar System, including Venus at about 49 $R \oplus$.  The Earth/Moon system shed angular momentum as passed through each of the resonances.  Between 6 and 8 $R \oplus$, the lunar perigee was precessing synchronous to the Earth’s orbit around the Sun, which is known as the evection resonance.  Estimates for the amount of AM shed during the evection vary from 10% to 50% (again referenced by Asphaug).
Solar mass ejections and other impacts aren’t thought to significantly alter the Earth/Moon AM since formation.

Is there a range of Earth rotation speeds possible?
Lock et. Al  propose the possibility that the impact that created the Moon imparted so much energy that the resultant system exceeded the corotation limit (CoRoL), which is the hottest, highest AM system possible.  They theorize the result would be a rapidly spinning vaporized torus of gas called a synestia instead of a planet. Here is a figure from their paper:

They talk about the synestia contracting and shedding angular momentum similar to how the accretion disk of our solar system is thought to have shed momentum as the Sun and planets were formed.
It may not make sense to think about the length of a day for a synestia.  However, if Earth coalesced out of a synestia, StephenG calculated the fastest possible rotation rate for the Earth would yield a corresponding fastest day at a little under 1.5 hours: Why doesn't the Earth's rotation throw us off the surface?.
In conclusion, if the Moon was formed from a giant impact, an Earth day was probably somewhere between 1.5 hours and a bit more than 5 hours after the Earth had re-coalesced.


How actively is Deep Learning being used in exoplanet identification?

Also called Astronet-K2, the deep learning A.I is efficient at spotting data that could find exo-planets, but that's were the issue is. That's all it is capable of doing, so it misses or ignores other potentially interesting data that human astronomers would classify as curious or odd, and could lead to other potentially new discoveries. It's missing the human factor, more or less. In both the source linked in the question and mine, it has only found 2 new exo-planets, so it doesn't look like is has free reign In the field. Even if the A.I is currently used automatically, it's work would be monitored and checked by humans.  My source material was from 2019, so if I missed a more recent accounting of the system that has different or upgraded methods, please let me know.
https://www.google.com/amp/s/www.technologyreview.com/2019/04/01/136239/deep-learning-has-found-two-exoplanets-that-human-astronomers-missed/amp/


Subterranean Oceans On Other Planets/Planetoids: How Do Astronomers Deduce This

How they have concluded that there could b.e a subterranean ocean on Ceres is by Spectralscopy.
Spectral Signature can be summarised to like this:

Different elements emit different emission spectra when they are
excited because each type of element has a unique energy shell or
energy level system. Each element has a different set of emission
colors because they have different energy level spacings. We will see
the emission spectra or pattern of wavelengths (atomic spectra)
emitted by six different elements in this lab. We will then identify
an unknown element by comparing the color of the unknown with the
flame color of our knowns.

And another:

When something is hot enough to glow (like a star), it gives you information about what it is made of, because different substances give off a different spectrum of light when they vaporize. Each substance produces a unique spectrum, almost like a fingerprint.

So how the scientists would have concluded that there was water because of the water vapours in the atmosphere. It gives off a certain wavelength that could be compared the closest element that gives off the same wavelength which is water.
Then the scientists would have used Galileo's magnetometer(an instrument which measures the strength and direction of magnetic fields) to conclude that there could be an ocean on the dwarf planet. The strength and response of the induced field would tell the scientists a rough estimate of the dwarf planet's sub surface which in return they could deduce the fact the there is a large amount of water present which equals an ocean.


Sundial (solar clock) in a binary star system

The sun forms a useful basis for measuring time because to a reasonable approximation it stays put, and the rotation of the Earth causes it to traverse the sky once every 24 hours.

Provided the planet is spinning, a normal sundial can be used, but may not be quite as regular as on Earth.

A binary stellar system can have one of three possible configurations: Two stars in a close binary, with a planet orbiting at a much greater distance. This is the "Tattooine" configuration. Everything has a double shadow. A regular sundial can be used but it is not as regular, as the positions of the two stars changes as they orbit each other.

The second configuration is a planet orbiting a star with a second star at a much greater distance. If Jupiter had been a star this would be the second configuartion.  The secondary star would be much weaker, and a regular sundial could be used. The shadow from the secondary would just be ignored.

The last configuration is like the second, except the planet is in orbit around the smaller star. The more distant star is stronger, and compariable in brightness. In this situation it is quite likely that the planet is tidally locked to the star, so a sundial isn't useful for telling the time. The slow motion of the distant star might give a long day (which could last many years) to the side of the planet that faces away from the nearby star, followed by a long cold night. This would be a challenging planet on which to live.


How bright is the Crab Pulsar's 30 Hz modulation in visible light? What color is it?

The optical pulsations of the Crab pulsar have been studied closely since 1969. The observations are actually not that difficult (I did some myself with a photoelectric photometer as a student) and have been achieved with a variety of technologies.

A paper by Fordham et al. (2002) slices and dices the Crab pulsar's pulse shape into fine time and spectral bins in the optical regime. A phase-folded light curve for the full blue-to-red range is shown below. The pulse shape is actually extremely stable and it looks like this in many other papers. The pulse has been "background subtracted" using the signal in the "off" phase of the pulse. The unpulsed component amounts to less than 1% of the integrated brightness in the optical, so essentially "off" does mean off.

The same paper discusses the integrated energy spectrum and the wavelength-dependent pulse shape of the pulsar. The integrated spectrum is almost flat (after correction for interstellar extinction) - as in, if the spectrum $F(\nu) \propto \nu^{\alpha}$, then $\alpha \sim 0$; this result holds for the whole pulse, or if just either of the two peaks are considered (there is a tiny $<1$% variation in the ratio of the two pulses across the optical range).

Postscript - I notice that the Wikipedia page on the Crab pulsar has a slowed-down animation of the pulse as seen at 800nm (very close to optical). The two pulses and their difference in brightness are quite obvious, as is the "off" phase. It would not look much different at shorter, visible wavelengths.


Why has Saturn stopped contracting gravitationally?

The rate at which a planet contracts is determined by how much heat it can radiate from its surface and how close it is to the minimum radius allowed by electron degeneracy pressure (which is independent of temperature).
The contraction process is that heat radiates from the surface and is supplied by gravitational energy released by contraction. The process stops when the central pressure becomes determined by electron degeneracy at high densities. At this point the interior pressure decouples from the temperature (unlike a perfect gas) and the planet can cool at almost constant radius. The same process occurs in white dwarfs and brown dwarfs.
To first order, the radii of Jupiter and Saturn are similar and they have similar atmospheric compositions, so can radiate heat at a similar rate. But Saturn has far less gravitational potential energy to radiate and its centre reaches degenerate conditions earlier. In your equations you can see that $dR/dt \propto M^{-2}$ so Saturn contracts faster.
The equations you write down are essentially correct, but your statement that the terms are all non-zero is not. Effectively, Saturn is not luminous because of contraction, so $L \sim 0$ in these equations, and so is $dR/dt$ . The measured luminosity of Saturn instead comes from the residual heat in its interior and the chemical differentiation of helium and not through gravitational contraction.


Compared with similar stars, does the sun itself have properties that make life on Earth possible?

One thing that is, if not unique, unusual about the Sun is the size of it.
It belongs to the top 5% brightest stars, meaning that 19 out of 20 stars are smaller than the Sun.

Having a Jupiter size planet is certainly not unique, and plate tectonics we know nothing about for exoplanets.


How do we know which elements are in a galaxy spectra?

Because the wavelength ratio of the lines remains constant despite any cosmological red shift.

For example, if the redshift is $z$, all the lines are shifted redward in wavelength by a factor $(1+z)$. This means that a pattern of lines can still be recognisable.

We also have a pretty good idea of what the spectra should look like, which chemical elements will produce visible absorption features with what relative strengths and so on (see below). This usually makes identification of line features straightforward.

Of course if there were just a single line visible in the spectrum (it does happen, usually in high redshift quasars) it can be difficult to pinpoint the redshift.

In terms of analysing what's in a Galaxy, well usually the light is dominated by the mixed spectrum of billions of stars. This spectrum is interpreted and modelled using galaxy evolution models and population synthesis models that predict a spectrum from a given ensemble.

If a Galaxy is near enough, the chemical abundances of its interstellar medium can be estimated from resolved spectra of emission nebulae.

In comparison, interpreting spectra from an individual star is trivial. Hundreds if not thousands of absorption lines can easily be identified and matched with the predictions of very detailed stellar atmosphere models to estimate chemical abundances. These models contain up to millions of possible radiative transitions as well as the various line strengths and broadening processes that affect the spectrum.


What's the influence of a tilted orbital plane, when observing an exoplanet transit?

The depth of the transit tells you the size of the planet (relative to the star) in both cases.

The difference in total eclipse width tells you the impact parameter (how far from the centre of the disk, the transit crosses). This in turn depends on the orbital inclination and the ratio of the stellar radius to the radius of the orbit.

A second constraint on inclination and the ratio of stellar radius to orbital radius comes from the ratio of total transit duration to the time when the planet is fully eclipsing the star. A more inclined transit has a shorter period of total eclipse but longer ingress and egress.

To get a full solution then requires some estimate of the stellar mass, which leads via Kepler's third law to the orbital radius.

There are also some important additional complications due to limb darkening and eccentricity that mean, in practice, this problem is not solved analytically, but by fitting a detailed model to the light curve. For instance, the transit depth becomes smaller for transits that don't cross the centre of the stellar disk, because of limb darkening.

The best picture I can find is copyrighted material, but you can find it here from Carole Haswell's book on Transiting Exoplanets. The inclinations range from 90 degrees to 80 degrees.


The short answer is: gravitationally bound system will not be ripped by the accelerated expansion.

A longer answer: The current standard model ($\Lambda$CDM model) says that everything started with the big bang. It released a lot of energy that pushed the Universe and is since then expanding. While it's expanding, gravitational attraction has been working to create the structures that we see today (planets, stars, galaxies, cluster of galaxies).

As the Universe expands, the density of matter and radiation within the Universe decreases. At about 5 billions years after the big bang, these densities became smaller than the density of Dark Energy (Dark-energy-dominated era). The strange things about Dark Energy are that, 1) its  density remains a constant regardless of the expansion of the universe, and 2) it has anti-gravitational properties - it pushes rather than pulls. Once Dark Energy came to dominate the density of the universe it began to dominate its overall dynamics, and has caused the expansion of the universe to accelerate.

But in structures which have already formed, gravity is much stronger than dark energy and so they don't feel it.


What is the duration of the era during which it will be possible to perceive a total solar eclipse?

There are a lot of competing models for how fast the Moon is retreating from the Earth, but they all agree the rate of distance increase is non-linear. That means any answer to this question is going to involve a lot of error.  I think there are two different questions here and I will address them separately:
1. How long ago could the whole circumference of the Corona be seen from a hypothetical observer on Earth?
No matter how close the orbiting moon was to the Earth, Bailey's beads and some of the corona would be visible just as the Sun became fully obscured by the Moon.
If you are wondering when the entire circumference of the corona might be visible during a solar eclipse, NASA says the corona extends to about 12 solar radii.

So the whole circumference of the corona presumably wouldn't be visible if the moon was 12 times closer to the earth.  There a few different guesses for the Earth-Moon distance in the past, and Franco and Hinnev show Webb's TD (Tidal Dissipation) model here contrasted against constant extrapolated recession models:

Here, some of the points on the plot denote geological tidal evidence for Earth/Moon proximity. If we believe their TD model, the answer is about 3.9 billion years.
1. How long ago could the angular diameter of the Sun and Moon be the same in the sky as viewed from Earth?
Remember that angular diameter of a sphere is $\delta = \arcsin(d/2D)$, where $d$ is the diameter of the sphere and $D$ is the distance from the observer to the sphere.  The angular diameter of the Sun as seen from the Earth is then $\delta_s = \arcsin(d_s/2D_s)$ and the angular diameter of the Moon is $\delta_m = \arcsin(d_m/2D_m)$, where the subscripts $s$ and $m$ denote the Sun and Moon.  Setting them equal and simplifying yields the common rule of thumb for determining if an exact full eclipse is possible: $D_m/d_m=D_s/d_s$.
Wiki gives Earth's perihelion at $D_s = 1.47095e8 km$, the Sun diameter $d_s = 695,700 km$, the Moon diameter $d_m = 1737.4 km$ so solving for $D_M$ gives us 367,346 km.  This means that the Moon would block the Sun entirely if the Earth was at perihelion and the Moon was 367,346 km from the Earth and all three bodies were colinear.
The current moon's apogee is 405,400 km, so it still would barely eclipse at 38,054 km closer.  This is about 6 Earth radii, so if the above chart is to be believed using their 'TD' model, the earliest time the sun and the moon could have had the same angular size is 1.2 billion years ago.  Of course, this guesstimate doesn't account for the fact that the sun's size isn't constant.  It also doesn't account for the fact that the Moon and Sun aren't perfectly spherical.


There is apparently an option on the STSI (Space Telescope Science Institure) website to try their Spike implementation.

This doesn't clearly answer your question, as it's not open, and I don't know the language that has been used.

However, there is a lot of information in the documents you provided and on the linked website, so maybe you can start your own implementation of the algorithm? I'd be happy to help!


Are there any bright non-stellar objects which make up asterisms?

Most of Messier's objects are too dim to be seen by the naked eye. The few that are visible were mostly listed as stars in older lists.

Edmund Halley lists 6 "luminous patches" which "discover themselves only by the telescope, and appear to the naked eye like small fixt stars" (Halley).

These include M42, in Orion, part of the sword (or what ever that is between the hunter's legs), and Omega Centuari, listed as a minor star in the centaur (roughly on the horse's back) in Bayer's list.

(The other objects on Halley's list were either known to be "cloud-like" [M31] or too faint for naked eye astronomy.]


Mathamatica vs Sympy and Astropy to solve basic astronomy problems: solving linear equations and unit conversion

Unit conversion and handling is good in Astropy, if you need to convert astronomical units I recommend that you do use that package.
Regarding symbolic equations, SymPy is good, but you'd probably be better off doing theoretical work to figure out the units of the outcome and just plugging float numbers into Python. From my experience, having Python solve with units involved is time consuming (especially if you want a scalable solution).


What did I photograph around 6 AM on April 27th, 2021? Supermoon plus comet?

Having seen the extra photos, this is certainly a cloud.
It is a high altitude cloud: a fragment of cirrus composed of ice crystals.  These are aligned roughly horizontally, and they are catching the light of the moon. This is why the region of the cloud is more highly lit than the parts to the side: a sun-pillar (or moon-pillar?) effect.
Its movement over the course of a few minutes also indicates a nearby object. It is not a body in space, as astronomical bodies are so distant that they appear to move only very slowly.  One very simple rule is "if it moves or changes, then it isn't astronomy, it's meteorology or technology" This isn't quite true, but very few astronomical bodies can be seen to change position over the course of a few minutes.


Question about extreme space distortion and creation of a new dimension

Just because there's an extra dimension on a diagram doesn't mean that it's real. No such thing has been created. It's just an artifact of embedding a manifold with a non-Euclidean geometry into a Euclidean space.

One of the biggest stumbling blocks of intuition for people learning general relativity is that the physics only cares about the intrinsic geometry. For example, imagine an ordinary ball in three (Euclidean) dimensions, and take its surface. In jargon, the surface is the two-dimensional sphere $\mathrm{S}^2$, and it has geometric properties that can be described by reference to it alone, e.g., lengths of curves drawn on it, angles between intersecting curves, that if one starts at some point and goes in a single direction, eventually one would be back to the starting point, etc.

One should consider the two-sphere as a valid geometry by itself, and the fact that it can be pictured as the surface of a three-dimensional Euclidean object as purely incidental--it $\mathrm{S}^2$ can be embedded in the Euclidean space $\mathrm{E}^3$ in that simple manner, yes, but that's a fairly arbitrary choice. It can be embedded in other spaces as well, or not embedded in anything at all.

For Riemannian manifolds (the purely spatial geometries), there are some nice mathematical results regarding embedding into Euclidean spaces, but nevertheless, they're usually impractical and not relevant to general relativity. Even worse, general Lorentzian manifolds (spacetimes) have no comparably 'nice' results regarding embedding spacetime geometries into flat pseudo-Euclidean spaces $\mathrm{E}^{n,m}$, so it's doubly useless to worry about it, except perhaps in the cases where the spacetime is extraordinarily simple.

In the end, such extra dimension(s) are just artifacts of making a picture. They're not physically real, so they're not 'created' in any physical sense.


As the Moon and the Earth are predicted to get into tidal lock, how slow would the Earth rotate?

The last part of the answer you linked to actually says (right at the end) that tidal locking will never be achieved, with reasoning similar to what I gave in this answer.
That said, even though the Moon and the Earth will never actually achieve tidal synchronization, we can still do the thought experiment and ask, "If there were enough time for the current Earth-Moon system to achieve tidal synchronization, what would the length of the day be at which the rotation of the Earth and the orbit of the Moon would be synchronized?"
To do this, we can assume that the Moon spirals outward due to an exchange of angular momentum between the Earth's rotation and the Moon's orbit.  The Earth's spin slows down as it loses angular momentum, and the Moon moves into a larger (and thus higher angular momentum) orbit as it gains that same angular momentum.  The rotation of the Moon would presumably stay locked to the Moon's orbital period, so it would slow down as well.
So, using $L$ to represent angular momentum, the key equation is
$$ L_{\rm now} = L_{\rm then} $$
where "then" is some time in the future when lock is achieved.  The total angular momentum in the system is constant.
The angular momentum of any object is $L = I\omega$, where $I$ is the moment of inertia, and $\omega = \frac{2\pi}{P}$ is the orbital frequency, related to the orbital period $P$.   For a constant-density sphere of mass $M$ and radius $R$ rotating on its axis, $I = 0.4 M R^2$.  The Earth and the Moon are somewhat more centrally condensed, so their moments of (rotational) inertia are a little smaller than the 0.4 for a uniform sphere.  The leading coefficient is 0.33 for the Earth, and 0.39 for the Moon.
For the Moon orbiting the Earth, it's a good approximation to just treat it as a point mass (since its size is small compared to its distance from Earth), so it has $I = M_{\rm Moon}R_{\rm Earth-Moon}^2$.
Putting all three of these motions (Earth rotation, Moon orbit, Moon rotation) together, we get
$$L_{\rm now} = 2\pi \left( \frac{0.33 M_\oplus R_\oplus^2}{P_\oplus} +  \frac{0.39 M_{\rm Moon} R_{\rm Moon}^2}{P_{\rm Moon}} +  \frac{M_{\rm Moon} R_{\rm Earth-Moon}^2}{P_{\rm Moon}} \right) $$
All of the values there represent current, known values, i.e. $P_\oplus = 1$ day, and $P_{\rm Moon} = 1$ month = 27.3 days.  Similarly, at some point in the (hypothetical) future, we would have
$$
L_{\rm then} = 2\pi \left( \frac{0.33 M_\oplus R_\oplus^2}{P_{\rm then}} +  \frac{0.39 M_{\rm Moon} R_{\rm Moon}^2}{P_{\rm then}} +  \frac{M_{\rm Moon} R_{\rm Earth-Moon,\ then}^2}{P_{\rm then}} \right) $$
or
$$
L_{\rm then}  = \frac{2\pi}{P_{\rm then}} \left(0.33 M_\oplus R_\oplus^2 + 0.39 M_{\rm Moon} R_{\rm Moon}^2 + M_{\rm Moon} R_{\rm Earth-Moon,\ then}^2 \right)
$$
Notice that there's just a single period here, since everything is now assumed to be synchronized.  So we could set this equal to $L_{\rm now}$ and solve for $P_{\rm then}$ – except that we have a second unknown in the equation, $R_{\rm Earth-Moon,\ then}$, the new orbital distance of the Moon from the Earth.   Fortunately, we can use Kepler's third law to relate this distance to the orbital period:
$$ P_{\rm then}^2 (M_\oplus + M_{\rm Moon} ) = \frac{4 \pi^2 }{G} R_{\rm Earth-Moon,\ then}^3  $$
To make life a little easier when we substitute into the equation, we could write this as a proportion with the current values, which makes some of the constants cancel:
$$ \frac{P_{\rm then}^2}{P_{\rm Moon}^2} = \frac{R_{\rm Earth-Moon,\ then}^3}{R_{\rm Earth-Moon}^3} $$
which means that
$$ R_{\rm Earth-Moon,\ then}^2 = R_{\rm Earth-Moon}^2 \left(\frac{P_{\rm then}}{P_{\rm Moon}}\right)^{4/3} $$
Substituting that into our expression for $I_{\rm then}$, we finally end up with
$$L_{\rm then} = \frac{2\pi}{P_{\rm then}} \left(0.33 M_\oplus R_\oplus^2 + 0.39 M_{\rm Moon} R_{\rm Moon}^2 + M_{\rm Moon} R_{\rm Earth-Moon}^2 \left(\frac{P_{\rm then}}{P_{\rm Moon}}\right)^{4/3} \right) $$
So in principle, we're done - we set this equal to $L_{\rm now}$ and solve for $P_{\rm then}$.  It's not a simple equation to solve analytically, but not hard to solve numerically.
Symbolically there's a lot going on, but we know most of these values, so if we plug in numbers for everything we know and simplify, this becomes
$$
P_{\rm then} = 0.16809413\ {\mathrm d} + 0.27626727\ {\mathrm d}^{-1/3} P_{\rm then}^{4/3}
$$
where the "d" represents units of days.  Solving this equation gives a period of 46.9 days, so that is how long the day, the month (i.e. the orbital period of the Moon), and the rotation period of the Moon would be if all were to become tidally locked to each other.
If you want to see the calculation done in Python, I've posted the code in a gist here. It's a nice example of the usefulness of Python's quantities and astropy's constants.


Are there any measurements of other galaxies indicating the existence of exoplanets?

Yes, it is true that our telescopes are not big/strong enough to detect planets in other galaxies. All we know at the moment are some few planets (not even complete systems!) around some of the stars in our closest galactic neighbourhood.
But let me explain why this is so.

Far away objects are very faint and thus send out few light-particles our way. To compensate this, we must build bigger and bigger telescopes to catch more light-particles from those faint objects. And at the moment we are limited until the 30-40metre class telescopes become active in ~2025.

However there is a class of 'cosmic accidents' that allow telescopes to catch more light-particles. We call those accidents micro-lensing events, as they briefly concentrate the light of a background star around a planet's gravitational sphere of influence.
The result is a spike of light in our telescopes. The shape of this spike informs us about the characteristics of the lensing star and planet.

To put this into context with your question about other galaxies, I've plotted for you how far away we have discovered planets so far:

Here we have on the x-axis the distance to the planet-hosting star, and on the y-axis the declination (one of the celestial coordinates or directions).
In red I've painted the planets that we have discovered with microlensing, in blue are all other methods.
The microlensing discoveries lie all on one line around dec = -30°, because this is the direction of the galactic center, where the polish OGLE project is looking.
The famous US-satellite Kepler that uses the transit method has it's field of view around dec = +45°, and this is also the biggest blue bunch of discoveries.
The small bump in between is the fov of the french CoRoT-satellite that also did some transit-discoveries and pioneered the transit searches from space.

So what do we learn from this graph?
We see that we can discover planets on a massive scale up to ~2000pc away.
Further away it's only possible with luck (while OGLE also looks at quite an impressive number of stars at any given time) and lensing. Now to put this into context, our milky way galaxy has a diameter of ~40.000parsec, and the next galaxy, the Andromeda galaxy is 1 million parsec away.

So that's gonna be hard, even with microlensing, and even harder with the other techniques you've mentioned.
Although there is one event from 2009 claiming to have found something that looks like a planet in their lensing-curve, I don't think this technique is mature enough to announce that extra-galactic planet searches.


If we stood on the surface of a Jovian moon, Saturn would appear as a dot. Why doesn't Saturn appear like this in a telescope?

Telescopes magnify, they don't bring you closer.
So if from Earth Jupiter has an apparent radius of 0.01 degrees (measured as an angle because it is the apparent size)
And if Saturn has an apparent angle of 0.005 degrees, then if you magnify 100x then Jupiter will have an apparent size of 1 degree, and Saturn would have a size of 0.5 degrees.  Magnification just increases the angular size in proportion
But if You go to Jupiter you have travelled less than half the distance to Saturn.
So saturn is still small. Travelling closer does not make everything increase in size in proportion.
You can see this simply:  Stand where you can see something 20 metres of so away, and where you can see into the hills (etc) in the background.  Walk towards the thing.  Note that the thing appears to get bigger as you approach, but the distant hills don't change size.  Travelling closer does not magnify everything in proportion.
Telescopes don't "bring things closer" they "magnify".


I may have found a partial answer to my question in the Navigational Astronomy chapter of 'American Practical Navigator' by N. Bowditch . On page 228

  [...] nutation, an irregularity in the motion of the Earth due to the
  disturbing effect of other celestial bodies, principally the Moon

and on page 230

  The precession of the Earth’s axis is the result of gravitational
  forces exerted principally by the Sun and Moon on the Earth’s
  equatorial bulge. The spinning Earth responds to these forces in the
  manner of a gyroscope. Regression of the nodes introduces certain
  irregularities known as nutation in the precessional motion.

So the dominant effects as for predicting the orientation of the Earth's axis seem to be the Earth's ellipticity combined with the attraction of the Sun and Moon.


The average str is about half the size of the Sun. So I get 44 km.

20 km/s is equal to $6.47 \times 10^{-13}$ pc/s, but we've just established that a pc is about 44 km in your sand grain model. So the speed is $2.85\times 10^{-10}$ m/s.

Just a double check - to travel 1 pc at 20 km/s would take $1.5\times 10^{10}$s.

To travel 44 km at $2.85\times 10^{-10}$ m/s takes $1.5 \times 10^{10}$ s.

Check.

I have no explanation for you as to why it doesn't agree wth your answer or with Joshua Winn's. But I suggest you read his problem carefully, because I rather fancy he might have done something like say imagine 1 year is equal to a second or something... perhaps.


"Harvest moon" just means the full moon at around the time of the equinox on September 21.

The full moon is low in the sky during summer (June and July in the Northern Hemisphere) and high during the winter.  So September is a month when the full moon is much higher in the sky than the moon in August. In the past this allowed harvests to continue after dark.

The full moon in September can be seen all over the world, but it may not have the same cultural significance that it had in Europe and North America.


Are other planets' gravity measurable on Earth?

I think the questions you're asking is, "All of the planets exert a gravitation force on everything else in the Solar System, including the Earth. Can we make measurements of that force in a laboratory on the surface of the earth?  (I.e., not by observing events on the other planets or their orbits.)"

If I'm right, the answer is "No, the effects are too small."

First, calculate the gravitational acceleration we feel on Earth from, say, Jupiter.  As it happens, Earth is roughly 10,000 Jupiter radii away from Jupiter.  The acceleration due to gravity follows an inverse square law, so the acceleration due to Jupiter on Earth is about 10-10 that at the surface of Jupiter or about 3x10-10 g.  This is small, but perfectly measurable.

The problem is that we can't measure gravity directly. We're sitting on a freely-falling Earth and (as far as Jupiter is concerned) are freely falling with it. Whatever acceleration Jupiter is imparting, it imparting the same acceleration to the Earth, us and our laboratories and everything in them.  So there's no force to measure.

We can measure tidal forces directly, since they are in essence the differences in acceleration due to objects being a different distances from the gravitating object, but they drop off as the inverse cube of distance and are undetectable at this distance using current technology.


What's the likelihood of the existence of unknown elements in the Solar System?

As far as elements (e.g. on the periodic table) go, I would say the odds are very slim.  We already discovered or produced all the elements of the Periodic Table up to atomic number 112 at least.  As the number increases, the half lives of the elements generally decreases, and is very short for elements above 102. If this trend holds true as the number increases, practically all the "undiscovered" elements should have turned into the lower known atomic number elements.

However, there is hope.  There is a theorized "island of stability" where a narrow range of yet to be discovered high atomic number elements may be stable: http://en.wikipedia.org/wiki/Island_of_stability
I would say there is a slight chance this element could be discovered in the solar system.


Spectroscopy question about a Paschen series absorption

The Paschen series absorption requires a population of H atoms in the $n=3$ level and a radiation field containing infrared photons $\leq 1875$ nm, that can boost them in to the $=4,5,6...$ levels.

If hydrogen gas is at 25,000K, you can use the Saha equation to work out the fractional ionisation. This depends on both the temperature and density of the gas. For the temperature you have been given, the ratio of ionised to netutral hydrogen is $1.7\times 10^{25}/n_e$, where $n_e$ is the electron number density in m$^{-3}$,

The typical particle number densities of an interstellar gas cloud are much, much less (many orders of magnitude) than $10^{25}$ m$^{-3}$, so the gas is totally ionised.

If that is the case then there can be no Paschen absorption and no Balmer absorption due to the cloud since there are no atoms in the $n=2,3$ states. There may be small amounts of Paschen and Balmer emission from the cloud via recombination.

Thus what you basically see is the spectrum of the star and since you aren't told what kind of star it is, then it's impossible to rule out the presence of emission lines or absorption lines of any kind. It is thus a dreadful and unanswerable question.


Gravitational waves and gamma ray burst: how were the error bars determined for this speed of gravity calculation? Was $H_0$ used?

What are the major assumptions and other measurements that went into these error bars?

The error bars in the paper are based on the shortest reasonable distance (to the authors) between the source and the Earth and a zero to ten second lag between gravity wave emission and gamma ray emission.
One key assumption is how long it took for the two signals, gravitational wave and gamma rays, to travel from the source to the receiver (the Earth). The 1.74 second difference in timing of the arrival of those signals would mean the speed of gravity and speed of light would be nearly identical if the signals traveled a long time (long distance), but maybe not so little if the signals traveled a shorter amount of time. The authors of the paper intentionally picked what they deemed to be the shortest reasonable light travel time (shortest reasonable distance) so as to magnify the uncertainty bars.
Another key assumption is that the two signals were emitted very closely in time, with the photon burst following the gravity wave by no more than ten seconds. The authors alluded to fringe theories that had the photon burst preceding the gravity wave emission by a non-trivial amount of time, and other fringe theories that had the photon burst lagging behind the gravity wave emission by much more than ten seconds. The paper mentions these only in passing.

Did they use a range of values for the Hubble constant?

The Hubble constant does not come into play here. The authors used what they deemed to be the shortest reasonable distance (the product of travel time and the speed of light) between the source and the Earth, based on luminosity.
Note that at 26 Mpc, the Hubble constant is not particularly relevant.

Was the dispersion of the interstellar medium at optical frequencies included?

Apparently not; this is a simple calculation. Moreover, dispersion at optical frequencies is rather irrelevant as the observations were of gravity waves and gamma rays.


Calculating longitude from star culmination

This is my first post, and I've accidentally deleted almost all I've written!

a later edit with an answer

The RA of a culminating star gives you the local sidereal time and comparison with Greenwich Sidereal Time gives your local longitude by noting the difference between the two.

If local sidereal time is less than Greenwich, your position is west of the prime meridian at Greenwich; if it is more, then your position is east of Greenwich.

Four minutes of Right Ascension = One degree of longtitude.
One minute of RA = 15 arc minutes of longitude.

A worked example:

Greenwich Sidereal time 17:42
          Local Sidereal time 16:36 hours

      Greenwich is ahead (more) than local, therefore local is West.

The difference = 42 minutes plus 24 minutes

= 1 hour 6 minutes of RA to the west of Greenwich

Thus, 60 minutes = 15° longitude

Add  4 minutes =   1°   longitude

Add   2 minutes =   0°30' longitude

= longitude 16°30'W

Which is accurate to 0°05' of the usually given figure for my location.

http://www.abecedarical.com/javascript/script_clock.html will show GST

I use a Sidereal Clock app on my Android tablet, set to show both GST and local sidereal time at my location.

information for background

This is the historical background for the search for a reliable method of calculating longitude when at sea

The Quest for Longitude and the Rise of Greenwich – a Brief History

I've edited this several times because I'm finding my way around how this works and made several editing mistakes! I've also tried to add another link, and got an error message, but no explanation. Is there a limit to the number of hyperlinks permitted?


how accurate are the predictions for coronal mass ejections-(location, timing and magnitude)?

Old question; I hope you’re still active here…
Basically, we CAN’T predict coronal mass ejections in advance. See http://solar.physics.montana.edu/press/ssu_index.html
Other solar events, such as the number of sunspots, are somewhat predictable—not precisely, but at least have an idea about them. For example, sunspots follow a cycle of approximately 11 years. However, we don’t know before a cycle if it will be strong or not—our models are still too imprecise, though, like you wrote, there has been some advancement, but not much.


Regarding the age / size of the Universe

The expansion rate of space is not itself the reason that the radius $R_\mathrm{Uni}$ of the observable Universe is larger than 14 billion lightyears (Gly). Just the fact that space expands is the reason. If space did not expand, then $R_\mathrm{Uni}$ would be the expected 14 Gly, as this is the distance that light can travel in the 14 billion years (Gyr) since the Big Bang. But since space has expanded in the meantime, then the distance to some particle that emitted a photon that we observe today, has been continuously increasing during the 14 Gyr, and hence $R_\mathrm{Uni}>14\,\mathrm{Gly}$.

An analogy to this is a worm on a rubber band that is stretched while it crawls along. The distance between its starting point and it's end point is not just a matter of how fast it crawls, but also of how much you stretch the rubber band.

The term "space expands faster than light" is a bit deceptive. Space expands *homogolously*$^1$, meaning that a given point in space recedes from you at a speed depending on its distance from you. If a galaxy is 1 Mpc (= 3.26 Mly) from you, it recedes at $70\,\mathrm{km}\,\mathrm{s}^{-1}$. If it's 2 Mpc from you, it recedes at $140\,\mathrm{km}\,\mathrm{s}^{-1}$. And so on. At some distance the recession velocity becomes larger than $c$, and in fact galaxies at a distance of $R_\mathrm{Uni}$ recede at more than $3c$. This doesn't violate special relativity which says that nothing can travel through space faster than $c$, because the galaxies do not travel through space. They lie approximately still in space, but space itself simply expands, i.e. the distance between everything increases.

Like three worms on a rubber band, with 1 cm between them. If you stretch the band to double length, the distance from worm #1 to #2 is 2 cm, while from #1 to #3 it's 4 cm.

$^1$Homogolous is not the same as homogeneous. Whereas the latter means "the same everywhere", referring to some physical quantity like density, homogolous means "proportional to distance".


At what density does helium burning start in a star?

The short answer: $\sim10^4$ grams per cubic centimeter.

From this webpage, I have a few statistics regarding required mean density and during each phase of fusion (the fusion of lighter elements may happen in the "shells" outside the core).
$$\begin{array}{|c|c|}
\hline \text{Fusion phase}&\text{Mean density (g/cm}^3)\\
\hline \text{Hydrogen} & 5\\
\hline \text{Helium} & 700\\
\hline \text{Carbon} & 200,000\\
\hline \text{Neon} & 4,000,000\\
\hline \text{Oxygen} & 10,000,000\\
\hline \text{Silicon} & 30,000,000\\
\hline
\end{array}$$
This is data for a star of $25M_{\odot}$.

Note, however, that this is mean density in the star, not the density in the core, where fusion is taking place. That is on the order of $\sim10^4$ grams per cubic centimeter (see e.g. here). There may be a sharp density dropoff starting from the center of the star, showing that only the hot, dense inner regions can fuse the main element in each stage.


Can the asteroid belt be seen from Jupiter?

No. The asteroid Belt is on average about 2.6 AU from the Sun.

Earth is 1 AU from Sun, 1.6 AU from the asteroid belt.

Jupiter is 5.2 from Sun, 2.6 AU from the asteroid belt. I.e. much further away.

Also, it should be easier to see asteroids from inside their orbit since they reflect more sunlight towards us. From Jupiter you would see their shadowed sides. Ceres is to Jupiter what Venus is to Earth. Venus is inside of us just like Ceres is inside of Jupiter. When it is at its nearest, we can only see a thin rim of sunlight on one half of its edge. But when an outer object like Mars is at its closest to us, we see its entire hemisphere lit up by the Sun. Add to that that when Venus is at its closest to us, as when Ceres is at its closets to Jupiter, it is very near to the Sun on the sky and therefor lost in the bright Solar light.

Ceres wasn't discovered until year 1801. That's after the discovery of Uranus 20 AU away. You'd need an even better telescope to find Ceres from Jupiter than from Earth.


Here is a short answer. @barrycarter's helpful comment links to a much larger list of resources. Also, you may find answers below [this question] helpful.

There are Spice kernels for many (but not all) exploratory spacecraft as well, you can view them through the Horizons interface or find them via ftp.

For most objects in Earth orbit, Two Line Elements (TLEs) from Celestrak or Space-Trakck are useful. See this answer as well.


What is the percent flattening in Haumea?

That would be because Haumea is a Triaxial ellipsoid. It's dimensions are different in every cardinal direction. The term 'flattening' only makes sense for the biaxial ellipsoids, because two of the axes are the same and one is different (in the case of Pluto, the equatorial axis is 1% larger, hence 1% flattened). The concept of 'flattened' doesn't make terribly much sense when all three axes are different.

Haumea's axis ratios are approximately 1 : 0.75 : 0.5, if that's the information you're after.


First Confirmed Visual Observation of Gravitational Lensing

The first identified gravitationally lensed object is known as the Twin Quasar.

The Twin Quasar (also known as Twin QSO, Double Quasar, SBS 0957+561, TXS 0957+561, Q0957+561 or QSO 0957+561 A/B), was discovered in 1979 and was the first identified gravitationally lensed object. It is a quasar that appears as two images, a result from gravitational lensing caused by the galaxy YGKOW G1 that is located in the line of sight between Earth and the quasar.

You can see it right in the middle of this picture:

The Twin Quasar's two images are separated by 6 arcseconds. Both images have an apparent magnitude of 17, with the A component having 16.7 and the B component having 16.5. There is a 417 ± 3-day time lag between the two images.

The quasars QSO 0957+561A/B were discovered in early 1979 by an Anglo-American team around Dennis Walsh, Robert Carswell and Ray Weyman, with the aid of the 2.1 m Telescope at Kitt Peak National Observatory in Arizona, United States.
[...]
Critics however identified a difference in appearance between the two quasars in radio frequency images. In mid 1979 a team led by David Roberts at the VLA (Very Large Array) near Socorro, New Mexico/USA discovered a relativistic jet emerging from quasar A with no corresponding equivalent in quasar B. Furthermore, the distance between the two images, 6 arcseconds, was too great to have been produced by the gravitational effect of the galaxy G1, a galaxy identified near quasar B.
Young et al. discovered that galaxy G1 is part of a galaxy cluster which increases the gravitational deflection and can explain the observed distance between the images. Finally, a team led by Marc V. Gorenstein observed essentially identical relativistic jets on very small scales from both A and B in 1983 using VLBI (Very Long Baseline Interferometry). The difference between the large-scale radio images is attributed to the special geometry needed for gravitational lensing, which is satisfied by the quasar but not by all of the extended jet emission seen by the VLA near image A.


Satellites traveling in single file across the sky

In January SpaceX launched into space a string of 60 satellites into space. These are satellites for internet communications. Astronomers say the proliferation of the bright metallic satellites could seriously degrade the night view, interfering with both optical and radio astronom$y^1$

1.https://www.sciencealert.com/spacex-just-launched-a-third-batch-of-its-starlink-satellites

Could this be what you saw?  I don't know, I'm just saying this is maybe the case.


How strong is the gravitational stretch we experience from the edge of the universe?

In the past, the universe was very hot and dense, and we are seeing the light from this time period as the Cosmic Microwave Background.

This statement is (partially) correct.
However....

  Since gravity apparently travels at the speed of light, we should be experiencing a gravitational effect from this shell around us as well. As I understand, the universe was extremely dense back then, and it is around us in every direction, which would intensify the gravity greatly.

This statement perhaps indicates that there are some misconceptions about the present state of the universe and also about what Cosmic Microwave Background Radiation really is (emphases added to identify the problem areas). So, in order to give a reasonable answer to the question asked, it would require addressing these misconceptions too.

The Big Bang model suggests that everything we see in our Observable Universe was once all concentrated in an infinitesimally small, hot and dense region of space. However, the Big Bang happened everywhere, which was followed by a super-luminal expansion of the space by a process called Inflation. The universe remained too hot still for any photons to roam around freely. It wasn't until the next 380,000 years that it continued to expand and eventually got cooled enough for the first hydrogen atoms to form. This is when the light first became visible. The remnant of this light is what we now now call Cosmic Microwave Background Radiation.

Again, it's important to note that the space was expanding in all possible directions continuously when the first light was emitted everywhere. In other words, the "dense" region from which the spacetime had begun, had now become comparatively much less dense overall. All through these years, this expansion hasn't stopped and every region where the light was first emitted, has now receded to great distances (46 billion light years in any direction). All the regions have cooled down further still to evolve into stars and planets, galaxies, clusters and any other celestial objects imaginable.

So, to answer the questions asked (for simplicity, we'll stick to CMB as the "edge")....

There is no space outside our Observable Universe which is surrounding us as a concrete shell. To be precise, it is merely a limit to how far back into the past and how deep we can peek into the space with our current technologies. Also, this has got nothing to do with the present state of the universe there. The universe is homogeneous and isotropic — meaning, you could go to any other place in the universe and it would still appear the same as it does from here — stars and galaxies everywhere.  Your vantage point would determine which galaxy appears as a distant blob to you and which region of space would merely appear as a CMB radiation.
The regions which we now see as CMB, 46 billion light years away, are indeed one of the farthest reaches of space from where anything, including gravity, would have had any effects on us. By now, everything that we see as CMB would have coalesced to form celestial objects. But, there's a catch.... Remember, if the Sun was to disappear from the Solar System now, it would still take approximately 8 minutes for the Earth to be set free from its orbit. Meaning, the state of the Sun 8 minutes ago determines what gravity we feel because of it now.
So, to compute the gravitational tug from CMB (46 billion light years away), we need to look at the state of the matter from 13.8 billion years ago minus 380,000 years (when the CMB was emitted). We can roughly assume that, back then, it was just Hydrogen atoms all over. The overall mass of all those atoms combined would have been equivalent to the mass of the Observable Universe, i.e. $10^{53}kg$.
But again, the CMB was emitted everywhere and the isotropic nature of CMB tells us that departures from uniformity are only up to seven parts in a million! Meaning, when the CMB was emitted 13.8 billion years minus 380,000 years ago, the mass of the Observable Universe was distributed almost perfectly uniformly in every direction possible. From the distance of 46 billion light years then, this translates into almost equal gravitational pull from everywhere that light (and gravity) has reached us since then. Hence, the gravitational effects of the matter that is (i.e. was) spread uniformly in a radius of 46 billion light years from us, would be equivalent to what one would feel if they were to somehow reach the center of the earth — i.e. experiencing a nearly perfect equilibrium, with a net force of zero.

PS: Should there be any discrepancies above, I'd wholeheartedly welcome the suggestions on correcting or improving the answer.


Which direction do the constellation revolve in?

The sky appears to rotate around the Poles. Stars that are in the northern Hemisphere will move in circles around the Pole. Some Southern hemisphere stars are also visible from the North, the move in circle around the southern pole (Although the star is sometimes visible the southern pole is always below the horizon from the Northern hemisphere)

A star that rises due East, will reach its highest position in the sky when it it is due South. It won't enter the Northern half of the sky.

Astronomers don't tend to talk much of Capricorn in December. (That leads to astrological nonsense) But it means that the sun in in the same region of the sky as Capricorn at that time (which means that you certainly can't see it) In fact the best time to see Capricorn is in Early Summer, when it is as far from the sun as possible.


Is it possible to tell if a certain extrasolar planet produces its own magnetic field?

There are three approaches with which people have looked for this, and not yet been too successful:

A transiting planet planet with a strong magnetic dipole and/or very strong host star winds might produce a visible signal when the magnetospheric bow-shock passes in front of the star as well. The idea is that at the bow-shock the streaming hydrogen would be heated up, ionize and produce a detectable ultraviolet photometry signal upon recombination at a characteristic distance (the stand-off distance) from the planet.
HD189733b was for a time the first planet to have measured its magnetic field in that way, however the discovery was retracted because the ultraviolet opacities that the group used to determine the bow-shock position were wrong.
It turns out there is no signal from the bow-shock.
The second idea is to look at a hot Jupiter that has an evaporating atmosphere. The shape of the ionized hydrogen tail in the magnetosphere will be influenced by the strength of the planetary dipole. But then, the amount of hydrogen that is in the planet's shadow is also a function of the magnetic field strength.
Hydrogen that comes into the planet's shadow will recombine with magnetospheric electrons due to the lack of direct irradiation from the star and produce a signal in the Lyman-$\alpha$ recombination line.
Going further, only the particles that stream away from the planet, and towards us will be part of this process. Thus the blue-shifted side of the Lyman-$\alpha$ line can inform us about the strength of the magnetic field.
So far the theory. The practice is that our last remaining UV-spectrometer in space that could perform such a thing (mounted on the Hubble) is even only barely capable of observing this process. But it has been done for the planet HD209458b
The data quality is lousy, and the interpretation that has been done in Kislyakova et al. 2015 has more parameter than datapoints and is therefore questionable.
But if you choose to believe the authors, then this planet has a magnetic moment of $0.1$ Jupiter magnetic moments.
The third possibility that could yield real results are radio observations with the LOFAR array. LOFAR observes at the edge of observable frequencies from Earth (at very low frequencies the ionosphere interferes with radio observations) at the for the moment highest achievable spatial resolution. The idea here is to look for analogues of Jupiters decametric radiation, which is simply synchrotron radiation produced when ionized particles start gyrating around a magnetic field line.
This already has been tried at higher frequencies / different observatories with the planets around $\epsilon$ Andromedae  and $\tau$ Bootes, as yet to no avail.


The $b$ in your equation, brightness, is what we call a "Flux".  That's the amount of energy per unit area at a certain distance.  Flux can be calculated using the Stefan-Boltzmann law.
$$
J=\sigma T^4
$$
where $\sigma=5.670373\times10^{-8}\space W\space m^{-2}\space K^{-4}$is the Stefan-Boltzmann constant, This is measured in Watts per square meter.  That means that we will need our distance to be in meters, and we will get a luminosity in Watts.  We can go a little further and divide our answer by $3.828\times10^{26} \space W$ to get our final value in solar luminosities (L$_\odot$).

That will make our final equation:
$$
L=\frac{4\pi R^2 \sigma T^4}{L_\odot}
$$
Because we're using the Stefan-Boltzmann equation, instead of the distance to the star, we have to use its radius.  Vega's radius is $2.362 \space R_\odot$, which is $1.64\times10^9$ meters. Its surface temperature is 9,600 K.  Plugging in those numbers yields a luminosity of:
$$
L=42.7L_\odot
$$
This value is slightly different from the value quoted by Wikipedia, due to rounding, but is well within the acceptable range.


Solar maximum
Solar maximum or solar max is a normal period of greatest solar activity in the 11 year solar cycle of the Sun. During solar maximum, large numbers of sunspots appear and the sun's irradiance output grows by about 0.07%.1 The increased energy output of solar maxima can impact Earth's global climate and recent studies have shown some correlation with regional weather patterns. source

Solar minimum
Solar minimum is the period of least solar activity in the 11 year solar cycle of the sun. During this time, sunspot and solar flare activity diminishes, and often does not occur for days at a time. The date of the minimum is described by a smoothed average over 12 months of sunspot activity, so identifying the date of the solar minimum usually can only happen 6 months after the minimum takes place. Solar minima are not generally correlated with changes in climate but recent studies have shown a correlation with regional weather patterns. source


Replenishing hydrogen in the core of the sun

There is very little mixing in the core of the Sun, where the stratification is fixed by radiative (rather than convective) heat transfer. The heavier helium does "fill the core", but takes about 12 billion years to do so, during which time, the concentration of helium gradually increases.

During its main sequence lifetime, most of the energy generation takes place in the central 20% of the Sun, where approximately 600 million tonnes of H is turned into He every second. (You get this by dividing the solar luminosity by $c^2$ and note that the pp H-burning chain is 0.7% efficient.) If we take this as a good average value over the main sequence lifetime (the Sun was fainter in the past, but will be brighter in the future), we calculate that $2\times 10^{29}$ kg of H is turned into He, which is 10% of the solar mass, and about 13% of its initial H.

Once the Sun evolves away from the main sequence, it burns more hydrogen in a shell around the core, and the convective envelope can mix fresh material into this shell. Given that the Sun is expected to leave behind a $\sim 0.5 M_{\odot}$ C/O white dwarf at the end of its life, then at least half of its mass will have been processed by nuclear reactions (and probably more, since some processed material is lost in the solar wind during the red giant phase).


Are there actual asteroids in the Oort cloud?

Tl;Dr: It is a matter of consideration if you want to call Kuiper belt bodies/Oort cloud bodies "asteroids" or not.
Long answer:
The Oort cloud defines the cosmographic boundary of the Solar System and the extent of the Sun's Hill sphere and hence it is loosely bound to the Solar System, and thus is easily affected by the gravitational pull both of passing stars and of the Milky Way itself. These forces occasionally dislodge some bodies from their orbits within the cloud and send them toward the inner Solar System. This is how comets are originated. Based on their orbits, most of the short-period comets may come from the Kuiper Belt/scattered disc and the long period comets are thought to originate from the Oort cloud.
So, the general assumptions is that maximum comets(if not all) are Kuiper Belt/Oort cloud bodies but all Kuiper belt/Oort clouds bodies are not comets.
So, are they asteroids?
You "may" consider most of the KBOs and OCBs to be more like asteroids because most of the the them have fairly circular orbits around the Sun, and most of them don't come close to the Sun at all (only those which comes close to sun are considered comets). So if an object needs to have an icy composition and a highly elliptical orbit to be considered a comet, then KBOs and OCBs are more like an "icy asteroid belt and bodies" than a group of comets.
However, such classifications were done based on what we knew 60 years ago, and it's now clear that the Solar System is much more complicated. So, we need to have a better classification to avoid confusion. Now, we name celestial bodies in a very specific manner considering various factors like position, behavior, size, age, composition etc., for instance: "Main Belt Asteroids", "Kuiper Belt Objects", "Near-Earth Asteroids", "Long-Period Comets" etc., as opposed to just "asteroids" or "comets".


What's to be gained from the New Horizons mission once it's beyond the Kuiper Belt?

Possible rendezvous with a Kuiper belt object and Voyager type data on the heliosphere:

  "We should have power until the 2030s, so we can get into the outer part of the heliosphere," says Spencer. "As long as we can continue to get good data—and persuade NASA to pay for it—we will keep getting the data, because we will be in a unique environment that we've never been in before."

Added:
NASA just picked the object for flyby.

  a 30-mile-wide chunk of ice known as 2014 MU69 that orbits nearly a billion miles farther out.

Not much is currently known about the object. We'll no doubt learn more from the January 2019 encounter.

More info from Emily Lakdawalla and Casey Dreier at the Planetary Society:

New Horizons extended mission target selected


Do parsecs scale as an inverse proportion with parallax angle?

For angles of magnitude 1/50 arcsecond (of the order $10^{-7}$, the difference between $\tan\alpha$ and $\alpha$ is truly negligible (if alpha is in radians) and only a matter of a choice of units if you don't. It is quite correct to say: distance (in AU) = 1/angle (in radians)  to get the distance from the angle in radians.

Using parsecs a means that we don't need to use a different unit for the angle, and the point of parsecs is that we can use "arcseconds".
So it is correct to use distance (in parsec) = 1/parallax angle (in seconds)


How is it determined that the X-ray and radio intensity come from a magnetic field bridge between two clusters of galaxies?

I think the answer is that it's primarily a matter of association on the sky, though there are some (weak) additional constraints.

The first thing to note is that there are three observations that are associated on the sky: the X-ray emission (which has been known about since the mid 1990s), the newly detected 140 MHz LOFAR emission, and evidence for a Sunyaev-Zel'dovich effect in the same region, from Planck data (e.g., Bonjean et al. 2018). From the caption to Figure 3 in the new paper: "The Planck data show a bridge of gas between the pair of galaxy clusters Abell 0399 – Abell 0401, in the same location as the LOFAR ridge." and from the main text: "The radio ridge is located along the filament of gas connecting Abell 0399 and Abell 0401 detected with Planck (9, 10)."

(The "Planck $y$-parameter", also known as the "Compton parameter", refers to the measurement of the Sunyaev-Zel'dovich effect due to inverse Compton scattering of cosmic background radiation photons, derived from Planck satellite data. I believe it's proportional to the line-of-sight integral of the gas pressure.)

The Suzaku X-ray spectrum of this region (e.g., Figure 2 of Fujita et al. 2008; see below) does have some spectral features, including the Fe K emission line. Due to the low spectral resolution, these features don't constrain the redshift of the X-ray-emitting gas very strongly (and I think Fujita et al. fixed the redshift at the mean of the two clusters in their analysis), but it would probably be noticeable if it differed by more than $\sim 0.1$ in redshift.

From Figure 2 of Fujita et al. (2008): Suzaku X-ray spectrum of part of the "bridge" region between the two clusters. Black is observed data, red is the combined model, including Galactic and extragalactic background (dashed blue), and emission from the inter-cluster gas (green).


Venus-Saturn conjunction in Sagittarius on Dec 11, 2019 - How often does this happen?

To answer How to calculate conjunctions of 2 planets earlier, I computed all planetary conjunctions from -13200 to 17190 and you can search the results at:

http://search.astro.barrycarter.info/index.pl

Prefilled with the values you want (assuming two fingerwidths is 2.5 degrees):

http://search.astro.barrycarter.info/index.pl?p1=Venus&p2=Saturn&lowyear=-13200&highyear=17190&sep=2.5&sunsep=0&submit=SEARCH

and a direct link to the results (with the "LIMIT 200" removed, so you can see all 26,671 results):

http://ae2bcbd33ae927c8e17c62ff155325f3.astro.db.mysql.94y.info/

Note that there are 26,671 2.5 degree conjunctions in 30,390 years, which works out to about 0.88 conjunctions per year.

The http://search.astro.barrycarter.info/index.pl page also provides information on bulk downloading the data, and you may also want to visit:

https://github.com/barrycarter/bcapps/tree/master/ASTRO/CONJUNCTIONS/

Read the README first, but the file you need is the 26.out.bz2 files, since Venus is the second planet and Saturn is the 6th planet.

As always, I'm not perfect, so please don't rely on my results, but do tell me if you find errors, thank you.


Surface height distribution for Jupiter's moon Io?

This blog by Butcher and Conwey on Geomorphology (or should we say io-morphology?) was a good starter. It seems usual to deal with the distribution of mountains and volcanos on Io. There are height maps (DEM) which give the distribution colour-coded. The closes to a hysometric curve found for Io is in this paper by White et al (2014) which gives a hypsometric curve where they exclude the volcanos in figure 9.
Noteworthy might also be this poster by Wiliams et al with geological map but also a height distribution given single measurement "points" (where the points are areas of various size).


In astronomy Is there a name for the exact time between solar noon and sunset?

No, there is no such word better than "afternoon". I can't think of a situation that would require technical language for this period of time, the exact point halfway. When astronomers want to be technical about time they use UTC or UT1 or AT1 or any of the other "times". When they want to talk about more loosely defined periods they use "morning", "afternoon", "evening" just like everybody else.

"Mid-afternoon" is an ill-defined period, from about 2pm to about 4pm


Why do some pulsar planets have designations with a capital letter at the end?

I will just cite this from Wikipedia:

On their discovery, the planets were designated PSR 1257+12 A, B, and C, ordered by increasing distance. They were discovered before the convention that extrasolar planets receive designations consisting of the star's name followed by lower-case Roman letters starting from "b", in order of discovery, was established. However, they are listed under the latter convention on astronomical databases such as SIMBAD and the Extrasolar Planets Encyclopaedia, with A becoming b, B becoming c, and C becoming d.

In a nutshell, this is the older nomenclature still kept because of historical reasons.


What does this information about "upper limit column densities" and "upper limit relative abundances" mean?

A column density is just the integral of a (number) density along the line of sight.

Often, you would get the same observational absorption or emission signature from a number density of $n$ molecules cm$^{-3}$, spread along a line of sight through a cloud of diameter $x$, as you would with a density of $0.1n$ spread over $10x$.

In other words, all we can infer is the product or, more accurately, the integral of $n\ dx$, and this is known as the column density.

In the paper you refer to, I doubt they know the thickness of the shock they are investigating, so a column density is all that can be determined.

The relative abundance is more straighforward; it is just the ratio of the number densities of the two labelled species.


Any body with an irregular mass distribution will experience those wobblings. And more if it is not just irregular but fluid.


Telescope in Sun's gravity lens focus - pointing, gain, distortions

The pointing is not a fundamental problem with the suggested design: The suggested trajectory is designed to include a Sun flyby as the last flyby. This ensures an asymptotically radial trajectory away from the sun after the flyby, hence maintaining the pointing relative to the sun. The proper motion of the observed object may be some challenge, but the trajectory could be adjusted appropriately by an additional burn.

114 dB are an amplification of a factor of about $2.51\cdot 10^{11}$.
It doesn't refer just to the magnification, but to the intensity of the signal.
Therefore interferometry with a long baseline isn't the same; the latter provides a high resolution.
Whether those numbers are achievable in practice is a different question; a factor of 1000 for naturally occuring gravitational lenses would mostly be regarded as excellent.
Theory allows arbitrary amplification for perfect alignment of observer, lense and observed object.

Most of the distortions can be deconvoluted, revealing the field of gravity of the sun, and the shape of the observed object.

Gravitational lensing is used today, here some lecture notes, and here quasars as an example.


What does Eddington Limit (Eddington Luminosity) Show?

The Eddington limit represents the maximum luminosity that can be achieved by a body (such as the star) when there is hydrostatic equilibrium (http://astronomy.swin.edu.au/cosmos/H/Hydrostatic+Equilibrium). For luminosity greater than Eddington limit, the radiative force of the luminosity on matter exceeds the gravitational force on the matter. If the luminosity radiated by an accretion disk exceeds the Eddington limit, the matter falling towards the supermassive black hole can be blown away.


What color distribution does a protoplanetary disk have?

The answer seems to be a clear "it depends":

Homogenity of the disks depends on the age of the protoplanetary disk.
Temperature is a function of the radius and the color-scale used in (old) artists' impressions is indeed correct.

On Wikipedia on the topic there is a gallery of different observations, also in the visual spectrum. An impressive picture of the density fluctuations within the protoplanetary disk is provideded by the Disk Substructures at High Angular Resolution Project (DSHARP) as explained e.g. by the ESO.
Edit Remarks following a suggestion by Rob Jeffries:

Please note that the DSHARP images are not from the visual part of the spectrum.


Is the Astronomy community still concerned about the lumpyness of matter distribution in the universe?

The "lumpy" distribution of galaxies in the universe is now well understood.  In the current cosmological paradigm, the very early Universe was seeded with fluctuations in the distribution of matter and energy.  Some regions of the universe happened to be slightly more dense than average, and some happened to be slightly under-dense.

Over time, these perturbations evolve.  Gravity, for instance, will cause the initially over-dense regions to become more over-dense.  When galaxy formation occurs, it usually happens in the dense regions where there is more material with which to form galaxies.  Therefore, galaxies will tend to trace (to some degree) the underly fluctuations in the matter density.

In short, the lumpy distribution of galaxies results from (a) the initial density fluctuations set down in the early universe, and (b) the evolution of these fluctuations according to well known laws of physics.  Cosmologists frequently model the lumpy distribution of galaxies using, for instance, numerical simulations or theoretical models; we find that these models agree extremely well with the observed distribution of galaxies in our Universe.  In fact, measurement of the galaxy distribution is used as a powerful tool to test and constrain our cosmological models.

Of course, the above answer doesn't explain the process responsible for laying down the initial density perturbations.  One theory that attempts to address this question is called Inflation.


Why does the moon 'peel' away from the horizon in this photograph?

Looks like an inferior mirage to me.

This:

https://aty.sdsu.edu/explain/simulations/inf-mir/Kaplan_photos.html

Has a work through for the disk of the sun but I think the same general principles apply here as well.

The specific effect will be somewhat different since in the image from space rays are moving from space down into and through the atmosphere, and then back up again, whereas in the image below the rays terminate at the Earth's surface.

Simulation and photograph from Andrew T. Young's page Photographs of an Inferior-Mirage Sunset:

  The inferior mirage is very clearly shown in a series of sunset images taken by George Kaplan, of the U. S. Naval Observatory. Here, I describe what's seen in these images, which are actually on the Naval Observatory's website. They are such a “textbook example” of the inferior-mirage sunset that I begged for some details, and permission to show the images here.


What calculations show that Comet Shoemaker-Levy 9 orbited Jupiter for several decades before its spectacular impact? (Chodas, Sekanina & Yeomans)

Chodas and Yeomans published a paper called The orbital motion and impact circumstances of Comet Shoemaker-Levy 9.  In it they describe the algorithms and techniques they used to reconstruct the orbit backwards from the discovery of the comet.  SL9 had an incredibly chaotic orbit in the mathematical sense in that small perturbations in the position would result in large effects in reverse-time orbital simulations.  In the words of the paper authors:

Unfortunately, SL9'sorbit about Jupiter was among the most chaotic of
any known solar system body, with an effective Lyapunov time on the
order of 10 years (Benner and McKinnon 1995). As a result, a single
backward numerical integration does not provide definitive answers on
the orbital history of this object. A better approach is to account
for the uncertainties in the initial conditions of the backward
integrations, and to investigate the motion in a statistical manner
using a Monte Carlo analysis (Chodas and Yeomans 1995).

For comparison, the Lyapunov time for the Solar System is on the order of millions of years.  An orbit trajectory plot for SL9 is the most complicated I've ever seen (excluding human-built craft)!

I don't know if precovery data was ever used to refine the orbit, though papers are still being published on SL9's orbit.  Unfortunately, Invariant Manifolds and the Capture of Comet Shoemaker-Levy 9, published in 2019, is behind a paywall.


Why is the Boomerang Nebula colder than the CMB?

The Boomerang Nebula (or Bow Tie Nebula) is a cloud of gas being expelled from a dying low-mass star, at $164~\mathrm{km}~\mathrm{s}^{-1}$ (cf. Raghvendra Sahai and Lars-Åke Nyman: The Boomerang Nebular: The coldes region of the universe?). In general, when a gas expands, it cools (see extended explanation below). If the gas were optically thin to the CMB — that is, if it were sufficiently dilute that CMB photons could easily penetrate — it would quickly get reheated to the temperature of the CMB, i.e. 2.7 K. However, the Boomerang Nebula is optically thick (dense), so the CMB hasn't yet had the time to heat it. The temperature in its outer parts, however, is higher, and as the nebula expands further, it will eventually (on timescales of, say, 1000s of years) be heated not only by the CMB, but also by the central white dwarf, i.e. the remnant of the star that produced the nebula.
Why does an expanding gas cool?
The usual approach to explain this is to consider a gas in a piston. When the volume is increased, the gas molecules do work on the piston, and hence lose energy, so temperature decreases. However, in the case of the Boomerang Nebula, there are no walls on which the gas can do work.
Cosmologically speaking, the nebula expands rather quickly (it has "only" been expanding for ~1500 yr). Assuming that it doesn't have the time to exchange energy with its surrounding, the expanding is thus adiabatic. For an ideal gas undergoing a reversible adiabatic expansion (or contraction), we know that
$$P V^\gamma = \mathrm{constant},$$
where $P$ and $V$ are the pressure and volume of the gas, respectively, and $\gamma$ is the adiabatic index. For a monoatomic gas, $\gamma = 5/3$, but here there are probably also molecules, so it's probably somewhat higher. At any rate, it's higher than $1$, which is the important thing for us, as we shall see below.
Now the temperature $T$ of the gas can be obtained from the ideal gas law:
$$P V = N k_\mathrm{B} T.$$
Here, $N$ is the total number of particles, and $k_\mathrm{B} = 1.38\times10^{-16}~\mathrm{erg}~\mathrm{K}^{-1}$ is a constant (Boltzmann's, to be specific). Even for non-ideal gasses, this relation is usually a pretty good approximation. Combining these two equations, we see that
$$T V^{\gamma-1} = \mathrm{constant},$$
and since $\gamma \gt 1$, it is evident that if $V$ increases, $T$ must decrease.


Calculating effective SSN (sunspot number)

The effective sunspot number $R_{\text{eff}}$ is calculated through a mix of observations and model calculations. It focuses primarily on some parameter called the frequency of the ionosphere F2 layer, denoted $\text{foF2}$, the highest frequency of electromagnetic radiation which can be reflected off the F2 layer; it can be observed as well as computed based on a certain model (real-time observational data from stations around the world is available here). More information on $\text{foF2}$ can be found on this page.

The mean-squared error between the observational value $\text{foF2}_{\text{obs}}$ and the calculated value $\text{foF2}_{\text{calc}}$ over observations by $n$ stations is
$$\Delta=\frac{1}{n}\sum_{i=1}^n\left(\text{foF2}_{\text{obs},i}-\text{foF2}_{\text{calc},i}\right)^2$$
where $_i$ denotes the value at station $i$. $R_\text{eff}$ is defined to be that chosen sunspot number such that $\Delta$ is minimized.

This is just an application of a least squares method to fit a model framework to match observations. An example of this using the Simplified Ionospheric Regional Model (SIRM) is Zolesi et al. (2004), although the procedure for calculating $R_\text{eff}$ was first proposed by Houminer et al. (1993). This was introduced to augment or replace something called the 12-month smoothed sunspot number, $R_{12}$, which is entirely based on observations. Zolesi et al. write

  It is therefore expected that during periods of relative ionospheric quietness, when the observed $\text{foF2}$ value is much closer to the model predicted median values, the $R_{\text{eff}}$ does not differ much from the $R_{12}$. On the contrary, during periods of ionospheric activity, the $R_{\text{eff}}$ should have large difference form [sic] $R_{12}$.


Infrared Astronomy for non transiting objects?

Yes it is. As long as light is produced by a source and then passes through a medium, we can perform spectroscopy on it. When considering a planet like Uranus, most of the light that it emits is actually reflected light or re-emitted light from the Sun. Uranus produces very little light by its own merit. For light that is directly reflected (by that I mean the light from the Sun goes into the atmosphere, hits molecules, and gets scattered out of the atmosphere again), this is directly equivalent to the standard transiting planets you mention in your question. For light that is re-emitted, you still see absorption spectra, but the black body background has changed. In this scenario, the light is absorbed lower in the atmosphere and re-emitted at a longer wavelength, but that emitted light must still try to leave the atmosphere and will experience some absorption along the way, creating absorption lines.


Observatories with public access in Europe

There are plenty of possibilities - and often, even when not publicly advertized, many observatories will be very open to find an arrangement when approached, especially by a group of interested audience. So this list IMHO would be the complete list of possible observatories found in central Europe.
The wiki page on astronomical outreach resources in Europe has such list of observatories (well, for the whole world, but filter it) - mostly the professional ones; of course it's not complete, e.g. the Thuringian observatory at Tautenburg is missing (who also offer regularily public tours) as does the GEO600, the gravitational wave observatory near Hannover.
There are various lists for amateur associations, many have their own observatory, e.g. this one lists many in the DACH countries sorted by post code.


What is visible light colour output of different stars?

Compare the effective temperature of the stars with Planck's law, and with a light bulb (2000 to 3300 K): The color of a light bulb corresponds to spectral class L (red brown star) or class M (red star, including dwarfs and giants); violet, and some blue are missing.

Light of Brown Dwarfs (spectral class T) would look red from their black body radiation, but due to their chemical composition the actual color may vary a bit over different tones of red. They would mainly feel hot, and look dull red at the same time, similar to glowing iron. Colors can hardly be distinguished under these lighting conditions.

Some of the very hottest stars (spectral class O) may ressemble "black" light, looking bluish, and causing fluorescence due to to the shift of the spectrum into the ultraviolet, including x-rays.
Without protective ozone/atmosphere layer, this wouldn't be healthy.

This may apply to some degree also to spectral class B (blue white stars); plasma cores of lightnings are of a similar temperature (up to 30,000 K or more) as the effectve temperature of class B or class O stars.

Any spectral class in-between (A, F, G, K) would look almost white, a little bluish for hot stars, yellowish for cooler stars, after some adaption of the eyes.


Can you view the Perseid meteor shower anywhere on Earth?

You can see perseid meteors from all the Northern Hemisphere, and the Northern part of the Southern Hemisphere. Provided the sky is clear, the moon has set, and Perseus is above the horizon you will see meteors. This shower has a wide peak, so you can see them over several nights.


What if a primordial black hole went through the Sun?

The answer is that there is a limited rate at which matter that can be crammed into a black hole of this size. That rate is small enough that the black hole will traverse through the Sun hardly picking up any mass (when expressed as a fraction of its original mass).
Details:
You could use the Bondi-Hoyle accretion rate:
$$\dot{M} = \frac{dM}{dt} = \pi \rho \frac{G^2 M^2}{c_s^3}\ ,$$
where $\rho$ is the density of the solar material and $c_s$ is its sound speed.
The biggest values will occur at the centre because the density is greatest at the centre and increases more rapidly than the cube of the sound speed. Using rough values of say, $c_s \sim 500$ km/s and a density of $10^5$ kg/m$^3$, we get, for $M=6\times 10^{25}$ kg,
$$ \dot{M} = 4\times 10^{19} \left(\frac{\rho}{10^3\ {\rm kg\ m}^{-3}}\right)\left( \frac{c_s}{300\ {\rm km\ s}^{-1}}\right)^{-3}\ {\rm kg\ s}^{-1} $$
The black hole would be travelling at around 600 km/s, so would traverse the central 10% of the Sun in around 100s (the accretion rate gets orders of magnitude lower in the outer parts of the Sun, so can be neglected), so might increase its mass by $4\times 10^{21}$ kg (or 0.007%). This would be insufficient to slow it appreciably and it would shoot out of the other side and disappear into space again. Note that a larger black hole would pick up more mass as a fraction of its original mass because $\dot{M} \propto M^2$.
All this assumes that the accretion can occur at the Bondi-Hoyle rate. This is very unlikely for two reasons - the first is turbulence which reduces the accretion efficiency. The second is radiation pressure.  There will be some sort of super-heated region immediately around the black hole (just imagine trying to compress $4 \times 10^{19}$ kg of material into a 20 cm diameter hole every, the event horizon diameter for a 10 Earth-mass black hole, every second) and the radiation from this region will exert an outward pressure on the surrounding gas. The limiting Eddington accretion rate, where the radiation pressure from the hot gas immediately around the black hole, exerts enough pressure to halt further infalling gas, is given by
$$ \dot{M} \leq \frac{4 \pi G M m_p}{\epsilon c \sigma_T}\ ,$$
where $m_p$ is the proton mass and $\sigma_T = 6.6\times 10^{-29}$ m$^2$ is the Thomson scattering cross-section for free electrons in the solar plasma (i.e. the effective area an electron presents to radiation). The $\epsilon$ parameter is the radiative efficiency of the infalling gas (i.e. what fraction of its kinetic energy is radiated away) and is usually assumed to be of order 0.1. Taken together, this gives a maximum accretion rate (for a ten Earth-mass black hole) of
$$ \dot{M} = 4 \times 10^{10} \left(\frac{\epsilon}{0.1}\right)^{-1}\ {\rm kg\ s}^{-1}$$
Whether the Eddington limit scales down to micro black holes and dense environments like this, I am unsure. But it seems clear that the accretion rate will be way lower than the Bondi-Hoyle rate. I conclude that the Bondi-Hoyle accretion rate is probably a vast over-estimate, and that the black hole would shoot through the Sun and leave on the other side, having hardly accreted any mass (as a fraction of its own mass).


Who or what will the Arecibo message reach?

The Arecibo Message was not broadcast with the aim that it would intercept with a notable astronomical object. Neither do we realistically expect that anyone will ever listen to the Voyager Golden Records.

Such symbolic gestures are simply vectors for public relations and education. We send these "time capsules" into space because we can, as a showcase of our technological achievements, and as a token of hope.

  "As the choice of frequency, duration of message, and distance of the
  target clearly shows, the Arecibo message is very unlikely to produce
  interstellar discourse in the foreseeable future. Rather, it was
  intended as a concrete demonstration that terrestrial radio astronomy
  has now reached a level of advance entirely adequate for interstellar
  radio communication over immense distances."

  -- NAIC staff (1975), http://adsabs.harvard.edu/abs/1975Icar...26..462

  "The spacecraft will be encountered and the record played only if there are advanced spacefaring civilizations in interstellar space. But the launching of this bottle into the cosmic ocean says something very hopeful about life on this planet."

  -- Carl Sagan


It's a rather difficult concept to grab but, in short, the image of expansion you have in your mind is wrong. It's not your fault, it's everywhere out there, the inflatable balloon etc. And it's terrible :)

Another way to look at it is that the universe properties are changing. Whether it is infinite or finite, its size (as measured within itself) changes. Things are pulled appart. You could imagine that free vacuum keeps being generated between all objects but that would be a misrepresentation of the theory.

The most simple way I see it is actually what the equations say. It's lame but sometimes the math are the simplest way to understand a phenomenon. The equation say the metric is expanding. It means two point in the fabric of the universe are measured at increasing distances over time. As a result, unless objects are pulled to each other with sufficient force, they tend to drift appart. It's a property of the universe that has not been explained yet (at least not completely), and we still are on the fence as to what influences the expansion rate.

Ultimately, this concept is very easy to accept when you cease to represent the universe from outside of itself. The universe is everything that is so if any image in your mind represents the universe from outside, it won't give you a good picture. Balloons are especially bad :). If you need an image, draw a grid or take a notebook, draw some points here and there on it and imagine the grid keeps getting bigger and bigger. But don't look at the whole page, forget the page limits, focus on the grid. And keep in mind that the points do not expand, only the grid does...


Will I age slower in space or on other planets?

If a human average life span is 60-80 years on earth, on mars will the average life span of a human be 62-82 years of earth years? Or will it be 60-80 years of Mars years?

Relativity is about the relative state of the thing observing and the thing they observe.

Locally (e.g. on Mars) things do not go slower for you, but someone observing (e.g. from Earth) sees time passing more slowly for you.

Locally your life on Mars is not lengthened.  Measured from Earth, more time will pass on Earth (as measured by someone local to Earth) than passes on Mars (as measured by someone on Earth observing Mars).

This is more complicated than it might seem at first glance because it works both ways.  You (on Mars) see me (on Earth) moving, so from your point of view you also see my time frame slow down relative to yours.  From an everyday human point of view that seems like a paradox, as we would not expect both of us to see the other's person's time pass more slowly than ours, but relativity does work that way.

There's a question (or two) on Physics SE dealing with the "why" that happens and why it's not a mistake or a fault with the theory.  Here is one.

It's even more complex when you have planets in orbits around a star, because that uses General Relativity and there are even more subtleties and additional effects on the rate of time we each measure for the other.  Gravitational time dilation does not work both ways - that depends on the strength of the gravitational field (from everything) so it's different for each observer.  Wikipedia's Time Dilation page is a reasonable place to start reading from, I think.  Note that general relativity is considerably more complex mathematically than special relativity.

We have measured these effects - they are real, but very small in human experience of them.


Could a distant black hole (detected due to its mass) be in reality a Dyson sphere?

The thing here is that a Dyson Sphere doesn't "block all the emitting light" -- that would violate the conservation of energy.  It does absorb the radiation from whatever body it surrounds and then re-radiates it.

As such, it's likely to radiate something like a relatively normal-looking blackbody spectrum, rather than the plasma X-rays from matter falling into a black hole, or the almost-nothing (thus "black") from a black hole with no infall.

As pointed out in @SteveLinton's comment the Dyson sphere would be much larger than the star or a black hole of the same mass in order to not melt, so at that distance the deflection would be far less, since it scales as inverse radius:

$$\approx \frac{m}{r}$$.


How to calculate sunrise and sunset on on a given date and location?

Assuming you mean this in practical sense:

Just use one of the numerous free websites that give this data:

Eg. https://keisan.casio.com/exec/system/1224686065

Assuming you want to know about the math:

Check out the sunrise equation


Collimating with a home-made collimation cap

The Gary Seronik website that you cite is a good one - it's not leading you astray!

There are a couple of important things to look at:

Look straight down the main tube of your telescope at the primary (largest) mirror. Is there a small paper ring or another mark dead in the centre of that mirror? If so, good. If not, read through this webpage http://garyseronik.com/centre-dotting-your-scopes-primary-mirror/
What are the lighting conditions when you're performing this? The photo you included is helpful but seems to imply that you might be pointing the scope at a bright light. This will wash out all contrast and make the procedure difficult. I collimate my scope just using ambient light during the daytime

Don't worry about only having one support rather than four. The key will be to identify your centre dot (you can just see it on Seronik's webpage, first picture ) and to see how it is centred when you look through the collimation cap onto the secondary.

Maybe read this article: http://www.skyandtelescope.com/astronomy-resources/how-to-align-your-newtonian-reflector-telescope/

Or search for Newtonian Collimation on Youtube?


Would drinking melted ice water from Mars taste like sparkling water?

Any ice that forms from water will be free of salt -- the process of crystallization does an excellent job of removing impurities.

But that's only when freezing out of liquid water, which is not a process that happens today at the Martian surface. When ice forms out of the air, it forms as tiny crystals -- "hoar frost" we call it on Earth -- and forms on substrates, such as grains of sand.  Since the soil of Mars is laden with salts of various sorts (including things a lot nastier than NaCl, such as perchlorate salts), it's hard to believe that the Martian hoar frost wouldn't be contaminated with salty sand and dust.

And, of course, the Martian atmosphere has dust suspended in it which is continually precipitating out and dust which is blown up by the not-infrequent dust storms and dust devils and all of this falls on any newly-formed frost.

It's very hard to see how there could be any macroscopic ice deposits on Mars that don't have dust and salt as impurities.


Is it possible to calculate distance between two objects?

If you know the distance of each star from earth, and the angle between the stars, then you can us the cosine rule to find their distance from each other.  See this for detail
https://en.wikipedia.org/wiki/Law_of_cosines


How does the length of the minor semi-axis of the moon's disc's inner terminator vary through the cycle?

The moon is rotating wrt to the sun. As the moon rotates, the longitude of the terminator increases linearly. Treating the moon as being at "great distance, so the moon appears as an orthographic projection, the position of the middle of the terminator varies sinusoidally, it moves fastest when the moon is at first or third quarter.

In the orthographic projection, the longitude and latitude $(\lambda,\varphi)$ are mapped to (x,y) by
$$
\begin{align}
x &= R\,\cos\varphi \sin\left(\lambda - \lambda_0\right) \\
y &= R\big(\cos\varphi_0 \sin\varphi - \sin\varphi_0 \cos\varphi \cos\left(\lambda - \lambda_0\right)\big)
\end{align}$$

For points on the equator, and treating the central longitude $\lambda_0$ as the lunar prime meridian, and a scale of R=1, that simplifies to
$x = \sin(\lambda)$. As the terminator moves, its longitude increases linearly, so x varies sinusoidally.

Correcting for elliptical orbits, librations of the moon and a general perspective view would only marginally alter this.


What resolution is needed to see a quasar through the galactic center?

Whilst awaiting clarification, I'm going to assume that your question is asking about calibrating the reference frame within which the motions of stars at the Galactic centre are measured.

The process is described by Ghez et al. (2008); does not involve observing quasars at infrared wavelengths; and the study does not suffer as a result.

The Galactic centre adaptive optics images are 5-10 arcseconds across, but contain thousands of stars. Only a fraction of these are close to the Galactic centre. The ensemble can be used, iteratively, to define a relative reference frame by minimising the displacements of thousands of stars, excluding those very close to Sgr A* and those with large motions.

This is really all that is required to do the orbital analyses of the black hole.

To put the coordinates in the International Celestial Reference System, a wider field is observed that contains a few infrared-bright giant stars, that are also bright maser sources at radio wavelengths. The positions of these are used to bootstrap the coordinates onto the ICRS, and this is important if, for example, we are interested in the relative motion of the Sun around the Galaxy.

The maser sources have ICRS positions by virtue of their radio source positions versus the radio coordinates of distant (but still bright and point-like) background quasars.The issue of background sources, extinction and confusion does not arise at radio wavelengths.