It's simply because the sun and planets were formed out of a big pile of dust.  Originally the dust was spinning.  So once the dust became planets, it kept spinning around.  That's all there is to it - that's why the solar system is spinning around.

It's was once a pile of dust that was spinning a bit; it's still spinning.

You then ask,

"For example, if you turned off gravity, why would the planets carry on moving?"

THat's just the same as asking "if I spin something around on a string, and break the string, why does it keep moving?"

For that matter, it's the same as asking "What is momentum?"  So, if you push something ...... why does it keep moving?

At this stage in history, we have utterly no clue, at all, what the heck time, space, matter, and momentum are.  "Why does momentum do what it does?" is for now just one of those super-deep questions like "What is time" or "So what caused the big bang" or "What is gravity" or "What's the explanation for this quantum stuff?" or indeed ... "What is momentum"?

So, why do the planets keep moving?: answer "momentum".  If you want to know "what causes momentum?", that is one of the basic total mysteries.

For now, nobody has a clue. You may as well ask ... what is space, what is time, etc.

Regarding momentum, you might like to read up about the so-called "Mach's conjecture". The famous Einstein was, like you, fascinated by the question "WTF is momentum anyway?" One sort of general thinking-point originating with this smart guy called Mach is that momentum could have something to do with "all the other mass in the universe". Nobody has a clue about this, and it's just a vague general idea.

Isaac Newton was a pretty smart guy (if a bit whacky), and in the end he found gravity so mysterious, he just put it down to God. He probably found momentum as mysterious, and he was one of the first guys to think about it clearly.

An interesting point: actually everything astronomers look at (all galaxies, all structures of galaxies) in fact does not (!) behave the way small things (our solar system, as in your question) behaves in terms of gravity and momentum.  This is usually explained by invisible unknown matter ("dark matter") or for a few scientists, that gravity works differently than we think presently.  So the fact is with issues like gravity/momentum you ask about ... not only do we have no clue "why momentum works" but when you look through a telescope, issues like "orbits" work totally differently anyway!


It's just called "the Solar System". (Plenty of places and objects have names like that; it's no different from "the Arctic" or "the Moon" or "the Sun".)

("Sol system" is an invention of science fiction writers; it has no general use outside some science fiction contexts. Anything else is going to be something similar, or general crackpottery of one kind or another.)


$$\left( \frac{4 \pi (6.96e10\,\text{cm})^{2}}{1} \right)$$

This is the primary source of your error. Your value of 6.96×1010 cm is the radius of the Sun. The problem specifically said "Assuming at the Earth ...". You need to calculate the flux through the surface of a sphere whose radius is about one astronomical unit rather than one solar radius. The astronomical unit is 149597870700 meters (exactly), or about 1.5×1013 cm. This error alone makes your value low by a factor of about 50000. The remaining factor of two results mostly from using 10-24 grams per amu.

Dimensional analysis can only take you so far. While your result is dimensionally correct, you didn't think enough about the nature of the problem.


Why would the Chang'e-4 lander find lunar far side temp. "colder than scientists expected", when the LRO has already been taking thermal readings?

I would not attribute too much importance to the quote "colder than expected", because that always depends which expectations one had initially. Here the quote you are discussing:

Temperatures on the moon's surface plummeted to minus 190 degrees celsius (-310 degrees Fahrenheit) during the probe's first lunar night, which "was colder than scientists expected," the China National Space Administration (CNSA) said.

As you correctly state, it was scientific knowledge on Moon temperature could get very low. Tim Sharp from Space.com writes in an article What is the Temperature on the Moon? back in 2017:

The Diviner instrument on NASA's Lunar Reconnaissance Orbiter measured temperatures of minus 396 F (minus 238 C [${=\rm 35 K}$]) in craters at the southern pole and minus 413 F (minus 247 C [${\approx\rm 26 K}$]) in a crater at the northern pole.

The conversion to Kelvin was added by myself.
On the other hand, Ramesh B.Malla et al. state in their article Determination of temperature variation on lunar surface and subsurface for habitat analysis and design in 2014:

Aside from the dangerous radiation levels and hypervelocity micrometeoroid impacts, the equatorial temperature on the surface of the Moon can range from ${\rm 102.4\,K}$ to ${\rm 387.1\, K}$.

So if you take the 2nd article as expectation, the value reported by 嫦娥四号 (the Chinese lunar lander) of $-193.15^\circ {\rm C} \approx 80 {\rm K}$ is indeed lower than ${102.4 \rm K}$.
From my time as researcher I remember that the public relation office did indeed sometimes use a more striking formulation than what the researcher themselves would - that's a factor also worth considering.


Examples of astrophysical objects NOT in virial equilibrium?

Not clear what the initial part of your question means. Objects can be in virial equilibrium without being in thermal equilibrium.

A clear exception to the virial theorem would be any system that is gravitationally unbound. So you couldn't apply it to a supernova explosion or a dissolving cluster of stars.

The examples you quote are not self-gravitating, bound systems, so no, the virial theorem would not apply to them.


Observations can be made using different parts of the electromagnetic spectrum. Many of them not within the range of the spectrum that can be made with optical telescopes. Other messengers than photons (e.g gravitational waves, neutrinos,...) can also be observed.

Optical counterparts are observations made in the spectrum observable using optical telescopes that can be matched up with observations made using other observational methods, including radio telemetry, Gamma ray detectors, and gravity wave detectors.

Observers are interested in lining  up observations made up of any combination of observational methods as this provides the opportunities to identify correlations that are a rich source of new discoveries.


Initial research proved to be wrong as intuition does not match up with expectations.

Anyway, the paper referenced by the OP seems to sum up the answer to the question pretty well:

  In  this  explanation,  the  dark  areas  of  our  map represent  thicker  clouds  that  obscure  deeper,  hotter  parts  of  the  atmosphere  and present a  higher-altitude (and  thus  colder) emissive  surface, whereas bright  regions correspond to holes in the upper cloud layers that provide a view of the hotter, deeper interior.

Extract from A global cloud map of the nearest known brown dwarf

Initially I assumed that the metallic clouds would reflect radiation from the nearby star, however as it turns out the hot atmosphere is emitting infrared radiation which affects the brightness much more significantly.

More on Brown Dwarfs at here.


You've answered your own question there, the difference between the two is that asteroids are rock whereas comets are made up of ice and dust.  An asteroid doesn't heat up when it gets close to the Sun and doesn't produce a tail.  Comets can orbit the Sun or any star in a long elliptical (Comet West) or short elliptical (Comet Encke 3.3 Years) period whereas asteroids are found in the Asteroid Belt between Mars and Jupiter and in the Kuiper Belt.  Sedna is a long period asteroid further out any other object in the solar system but it is still an asteroid according to what we know. https://www.universeguide.com/fact/comets  https://www.universeguide.com/fact/asteroids


What can the average Astronomy Enthusiast do to learn more

I recommend that you learn your way around the night sky, so start by buying a book of star charts.  Next try a book that covers interesting objects in each constellation, such as Burnham's Celestial Handbook, or The Constellations by Motz & Nathanson.  This should keep you busy for a while.


Why can't light escape from a black hole?

A black hole has an event horizon which 'marks the point of no return'. So yes, light cannot escape from a black hole.

Why? Well, think of a 'spacetime fabric'. It's the easiest way to understand the physics at work here, in my opinion.

Usually, the fabric would look like this:

(source: whyfiles.org)

However, a black hole has so much gravity that one could say it 'rips' the spacetime fabric:

(source: ddmcdn.com)

When the light hits this area of amazingly intense gravity, it simply cannot get out - the light travels 'along' the fabric, and since there is a rip in the fabric, one could say it simply goes away - it becomes part of the singularity.

This is a simplification, of course, but it's enough to understand at least part of the physics behind this phenonenom.


How do we know so many details about Big Bang, but we do not know if Universe is finite/infinite?

By observing the observable universe we can gather a lot of data about the constituents of the universe now and in previous epochs, right back to the cosmic microwave background. This data places quite narrow constraints on viable models of the Big Bang. This is why cosmologists are confident about the details of the Big Bang, certainly from electroweak symmetry breaking onwards.

However, it is much more difficult to make an observation of the observable universe that could distinguish between a very, very large but finite universe and an infinite universe. That is why this is still an open question.


Formation of "hot Jupiters" before star ignition

Hot Jupiters can and do form before the hot star "ignites", as you put it.

The conventional core-accretion theory, by which gas-giant planets form, requires them to accrete from a disk of gas and dust for about 10 million years. Coincidentally (or perhaps not a coincidence?), the empirically determined lifetimes of circumstellar gas and dust disks around young stars is also about 10 million years.

Meanwhile, the pre-main-sequence lifetimes of newly-born stars (which I define here as the length of time it takes to contract from a large ball of gas down to a much more compact configuration, with a central temperature high enough to initiate hydrogen fusion), range from about 10 million years for stars a little more massive than the Sun to 100 million years or more for stars that are less massive.

I think, though, that the information that you are missing is that there really is no period of time where the "heat has yet to begin" whilst the protostar is surrounded by a gas/dust disk. A protostar is actually more luminous when it is younger and becomes fainter as it contracts towards its main sequence configuration. The energy that powers this luminosity is derived from gravitational potential energy. The inner portions of the gas/dust disk are always too hot to form hot Jupiters in situ and so all current models focus on how hot Jupiters can be "moved" from where they were formed to where they are found now.


If the Higgs field only formed after the Big Bang, how was hydrogen formed?

Hydrogen was not "created at the moment of the big bang".

Particles (leptons and quarks) can attain mass via the Higgs field after the epoch of electroweak symmetry breaking, that occurred about a picosecond $(10^{-12}$s) after the big bang. Only after this was it possible to form the building blocks of hydrogen.

In fact the formation of stable protons comes "much" later at about a millionth of a second. Most of the mass of a proton does not come about from the Higgs field, but from the binding energy of its constituent quarks.

Hydrogen (the atom, rather than the proton nucleus) forms much, much later; about 400,000 years after the big bang.


While such a scenario would be unlikely, all you'd need is an existing ring and an icy moon to enter inside the Roche limit at a measurably different inclination. I wouldn't want to guess how long a set-up like that would be stable, but it's possible to have one.


Its very unlikely for a comet to become a satellite of an inner solar system planet. Much less likely than it is for an asteroid. Most asteroids are on fairly circular orbits, and so the relative velocity between asteroids and planets is quite low. In comparison comets have very elliptical orbits, and their relative velocities to the planets are much larger.

For an asteroid to be captured it must lose momentum. This is possible, though rare. For example, a binary asteroid can be captured if it is separated by tidal forces. For an comet with much more momentum, the chance of being captured is much much lower. Asteroids are captured by the Earth moon system, but not into stable orbits, they don't stay long.

If it did occur, the comet would still be active, with a coma of gas, which would be visible just like a very nearby comet. It wouldn't be particularly bright, since the surface brightness of a comet doesn't depend on distance from the Earth.

Over time the comet would run out of volatiles and become more or less indistinguishable from a captured asteroid. If it were in the Earth's orbit it probably wouldn't last that long, as there are not many orbits that are stable in the long term around the Earth, due to perturbations from the moon.

The dust and gas, including water vapour, will initally remain in orbit, forming a faint ring. It will, over time, be disrupted, and either end up in the atmosphere, on the moon, or ejected from the system. A comet doesn't contain enough water to make a difference to the Earth's ecosystem.


When were the horizontal and equatorial celestial coordinate systems invented?

I remember answering a question on the origin of declination last year (Why is declination positive in the northern hemisphere?) which led me to discover this paper on the origin of celestial co-ordinate systems. In it they seem to indicate that the ancient Greeks were the first to use Right Ascension to describe locations in the sky around 300 B.C.
Whereas Alt-Az co-ordinates (or variants of) have been around much longer because the horizon is an easy point of reference to start from. The first documented use comes from Babylonian times but the ancient Egyptians are also thought to have used a similar system as far back as 2000 B.C.

Reference paper: http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1942PASP...54...77W&db_key=AST


Most planets don't have rings.   The ring region is inside the Roche limit which is quite close to the planet.  A ring system outside the Roche limit needs to be either very faint or, it would over time coalesce into a moon or possibly pair of moons at Trojan points to each other.

There's different Roche limits for rocky bodies vs icy, but an icy body couldn't exist in Earth's orbit for long, so it's not really relevant for the Earth.  Saturn's rings are mostly ice, which extends about twice as far as the rocky body Roche limit.

How long a ring system lasts depends on a few factors.   How big it is, whether it's given new material regularly or not, what it's made of and how vulnerable it is to the sun's rays.   Saturn's rings, for example, are losing material, raining down on Saturn as well as slowly adding material from Enceladus, one of it's small and close moon's that gives up ring material by tidal-volcanic eruptions.

Earth's large moon could (not saying it would but it might) play a role in disrupting and Earth-Ring system more quickly than it would with no moon.   A 2 body system with the 2nd body as large as the moon isn't gravitationally friendly to other orbiting objects.

Many man made satellites orbit well within the Roche limit but they are small enough to be held together and not be affected by tidal forces.   Larger objects, just a few miles across, perhaps less, are more vulnerable to tidal forces and tend to break up inside the Roche Limit.


Is the Jupiter-Sun system considered a binary system of some type?

I'm not sure I understand your question entirely, but i'll do my best to offer a decent answer. It's true that the composition of Jupiter is very similar to that of the Sun (very similar approx. $H$ and $He$ abundance and pretty similar in density). The problem is that Jupiter is not nearly massive enough to have the internal pressure and temperature to undergo nuclear fusion. Jupiter doesn't have any particular special status aside from being the King of Planets in our solar system.

As for the last part of your question, all objects orbit around a center of mass. Though because the Sun is much more massive, the center of mass lies very close to the center of the star (except in the case of Jupiter, where the CoM lies outside of the sun and is approx the length of its radius). This is why all planets in our solar system orbit around the sun. This will indeed cause a slight perturbation of orbital alignment, but I don't believe it's significant.


How to measure the altitude and azimuth of a star?

For the second question: If you know the coordinates of the Zenith, your latitude is exactly Zenith's declination. For your longitude you can not rely on the Zenith: the same star will be at the Zenith at the same sidereous time for every place in the same latitude, so you need a clock besides the telescope. cf. http://en.wikipedia.org/wiki/History_of_longitude#Problem_of_longitude

For the first question: The altitude of a star that is crossing the local meridian is also quite easy.

For Northern Hemisphere, a star that is crossing the North local meridian (that is, between the Zenith and the North Horizon) the latitude is
$90-alt=dec-lat$

For Northern Hemisphere, a star that is crossing the South local meridian (that is, between the Zenith and the South Horizon) the latitude is
$90-alt=lat-dec$

Same goes for Southern Hemisphere if you change also North Horizon and South Horizon.

You did not asked specifically for the Azimuth, but again you need a clock for that.


How do I calculate Allan Variance (maybe for pulsars)?

So, after some discussion and paper reading I learned of the sigma_z calculation in this paper by Matsakis, Taylor, and Eubanks. Several astronomers recommended this to me. I was also told to try out Stable32, although I'm not sure if that will work for pulsars. The code for a sigma_z calculation can be found here: sigma_z calculation. I believe this is the best method for doing what I asked.


Catalog of Galaxies in the Visible Universe

There are lots of galaxy catalogs.

If you're only interested in nearby galaxies, there are things like the Updated Nearby Galaxy Catalog; downloadable versions can be found here.

A classic all-sky galaxy catalog is the Third Reference Catalogue of Bright Galaxies (RC3, published in the early 1990s), which you can query here; if you follow
the "FTP" links, you can find links to download the actual catalog.

(The HyperLEDA catalog that Dr. Chuck mentioned started as the RC3, but has various
bits of data and re-classifications added more or less continuously since.)

A more recent catalog, based on the near-infrared observations of the 2 Micron All-Sky
Survey (2MASS) is the 2MASS Galaxy Redshift Catalog, with about one million galaxies.

Even larger catalogs (though not covering the whole sky) can be found among the data
products of the Sloan Digital Sky Survey, e.g. here

Many more catalogs exist, though some do not have accurate redshift information (meaning their distance from us is unknown or uncertain), and many are only for parts of the sky, sometimes very small regions (e.g., catalogs from Hubble Space Telescope observations).

(As for that video, I suspect very little is based on real data once you get beyond the Andromeda Galaxy...)


M31 is actually something like 141,000 light-years in diameter (there has been some variation in our understanding of its size, but 141K is good compromise).  M31 is 2.54 million light-years from us.  The rotational speed of stars near the edge of the galaxy's disk is in the vicinity of 200 km/sec.

Given all that, in one year a star in the vicinity of the outer edge of the galaxy's disk is going to traverse 6.3 x 109 km.  At 2.54 x 106 light-years away, the apparent movement of such a star in the course of a year is going to be about 1.5 x 10-8 degrees, or 0.000054 arcseconds.

The Keck Telescopes in Hawaii, one of the most powerful telescopes in the world, has an angular resolution of 0.04 arcseconds.

As you can see, the best telescope in the world cannot distinguish any change of position of any star at that distance over the course of a year.  It would take about 740 years before such a star would move far enough for a telescope with the angular resolution of the Keck to distinguish any movement at all.

So, it isn't visible through an amateur telescope either.


If water vapor is always blown away into space, how is it able to create chemical compounds on Venus?

Think about the water constantly being both added to and removed from the atmosphere.
Adding water:

Water migrating from deeper layers of the planet crust and mantle into the atmosphere.
Water being formed from sun wind hydrogen and the available oxygen in the atmosphere (the high D/H ratio hints that the primordial water dominates on Venus)
Water being formed from hydrogen and oxygen produced by decomposition of other substances in the atmosphere.

Removing water:

Photolysis - water decomposed to hydrogen and oxygen, both of them engaged in other chemical processes
Escaping hydrogen and possibly whole water molecules into the space.

There is always some equilibrium between the available water and hydrogen and also other hydrogen containing substances like sulphuric acid (that's what Venus clouds are made of) and the widely-discussed today phosphine.

On Earth, we have stratospheric and mesospheric clouds made of water, sulphuric acid and nitric acid in atmospheric layers that are otherwise notoriously dry.  On Venus, the atmosphere has little free oxygen so it is tolerant to much more of a chemical diversity. (E.g. phosphine would self-ignite in Earth's oxygen-rich atmosphere)


Are rising sea levels caused by the moon?

You are correct in saying that the moon is receding from the Earth. However, as you can see, this rate is very very small.

1.48 inches = 0.037592 meters

The Moon is 384400 kilometers away (on average), or 384400000 meters. That means every year, the Moon's distance increases by 0.00000000009779%. That's such a small difference, it is negligible on human timeframes.

Also, although the tides would shrink, this would not mean the sea level would rise. When we refer to rising sea levels, we refer to an average rise in sea levels that have been seen across the world, mostly due to climate change. During high tide, a beach may experience 4 meter higher sea level, but at low tide, it will experience 4 meter lower sea level. If the moon doubled its distance, then gravity would be a quarter its strength (gravity decreases with the square of the distance), so you would have 1 meter above during high tide and 1 meter below during low tide. As you can see, there is still no net change in sea level, it's just that the changes between high and low tide would be smaller. Again, this is negligible on human timeframes.


Inability of Hubble to clearly resolve nearby celestial objects

This has to do with the angular resolution of the Hubble telescope and the ratio between the distance of an object in space and its size in space. The galaxies that the Hubble telescope can see are bigger in size than they are far in light years away compared to pluto from earth.

Take the galaxy NGC 5584 for example:
It spans 50,000 light-years and it's 72 million light-years away which gives us a ratio of 0.00069

Then for pluto:
Pluto spans 2400 km in size and 4675 million km away which gives the ratio of 0.00000051

Take for example a clear view at a mountain with trees on it. You would see the mountain fairly clearly: you can see it has trees on it, perhaps snow, and other aspects. But if you look at any given tree on that mountain you can't make anything other than it's observable color and that it's most likely a tree. The relationship of distance in size is what gives you a good view of the mountain but not a tree on the mountain; the trees size to distance ratio is much smaller than the mountain's size per distance ratio.


Portion of universe visible if gathering image from inflationary epoch

tl;dr: Your field of view would cover roughly one square centimeter of the sky at that time, and you would observe roughly 50 billionths of the observable Universe.

You can't really…

With photons, you will never be able to see further back than to recombination, when the Universe was 380,000 yr old, because until then, the free electrons made is opaque to radiation.

With neutrinos, you can (in principle) see back to 1 sec after Big Bang, but to "see" all the way back to inflation, you will probably need gravitational waves.

…but let's assume you can anyway

Anyway, lets's do a thought experiment and assume that you have a RLT (Ridiculously Large Telescope) with a field of view (FoV) of the HST. What fraction of the Universe can you see? The answer might surprise you:

The farther an object is from you, the smaller an angle it spans on the sky, i.e. the smaller it looks. This is true for birds and planets and even for the nearest galaxies. However, for distant galaxies a strange effect counteracts this: Because of 1) the finite speed of light and 2) the expansion of the Universe, distant galaxies were closer to you when they emitted the light you see today, and therefore spanned a larger angle. Hence, if you compare galaxies of the same physical size, they look smaller and smaller until around 15 Glyr (giga-lightyears), after which they'll look larger and larger.

The Hubble Ultra Deep Field has FoV of 2.8 by 2.5 arcminutes. The region of the Universe that this FoV spans is thus largest at a distance of 15 Glyr, where it span $4.8\times4.3\,\mathrm{Mlyr}^2$ (i.e. square-mega-lightyears). But then your FoV begins to span an increasingly smaller region. At recombination, it spans only $34\times30\,\mathrm{klyr}^2$ — in other words, if a Milky Way-sized galaxy were present there (it weren't), it would be larger than the FoV.

It you could look all the way back to 4 hours after Big Bang, your FoV would span roughly a square-lightyear. Three minutes after BB, when all the Universe's hydrogen and helium had just been created, it would span $\sim0.1\times0.1\,\mathrm{lyr}^2$. A few microseconds after BB, it would span roughly a square-AU (the distance from Earth to Sun). Approximately $10^{-22}\,\mathrm{s}$ after BB, it would span a square kilometer.

And at $10^{-32}\,\mathrm{s}$ after BB, your FoV would span $0.8\times0.7\,\mathrm{cm}^2$, i.e. roughly one square-centimeter!

Fraction of the Universe observed

The part of the observable Universe that you would observe is the volume "behind" your FoV. That can be obtained by integrating the area along the distance, but an easier way is to simply note the fact that, if your FoV were the whole sky, you'd observe the whole Universe. Since your FoV covers a fraction of the sky of
$$
\begin{array}\\
f & = & \frac{2.8'\times2.5'}{\mathrm{full\,sky}}\\
  & = & \frac{8\times10^{-4}\,\mathrm{rad}\,\,\times\,\,7\times10^{-4}\,\mathrm{rad}}{4\pi\,\mathrm{rad}^2}\\
  & = & 5\times10^{-8},
\end{array}
$$
this is the fraction of the Universe that you would observe.

At $t=10^{-32}\,\mathrm{s}$, the part of the Universe later became our observable Universe today, was about ten meters in radius. Another way to calculate the fraction is to realize that the above mentioned observed square-centimeter roughly comprises a fraction $f$ of the surface area of a sphere with a radius of ten meters.


That is RapidScat, described in this answer in the Space Exploration SE.


It is a bright star. You can see the "diffraction spikes" that are the signature of a point-like object.

The star will be in our Galaxy, so is certainly closer than the galaxy in the image. However, this is not why it appears to be big.

The image appears big because the starlight will be spread by the telescope objects over a finite area - the point spread function. In addition, when a star is very bright, the central CCD pixels saturate and charge can spill into adjacent pixels. Even if that were not the case, there is enough light in the "wings" of the point spread function to produce a large image.


Do hypervelocity stars have Oort clouds?

There is no problem with orbiting a star that moves very fast. Remember that space is mostly empty, so there is nothing like air resistance or friction that would tend to deform the orbit.

Stars that become hypervelocity stars likely tend to lose their Oort clouds. They gain their velocity by passing very close to a massive body and getting a velocity increment, for example by having a secondary star or planet ripped away by tidal forces. That already doesn't bode well for the stability of any Oort cloud.

Simulations show that planets tend to be ejected from hypervelocity systems unless they are very tightly bound, since they have lower mass and respond more strongly to the velocity changes. This suggests that the lightly bound and small bodies in the Oort cloud will be scattered - likely at even higher velocities.


7-21mm is a pretty common zoom eyepiece option, available from the companies you identified as well as others like SVBONY and Orion. (8-24mm is another popular option).
Doing a quick search for the Meoptex, it appears to have the same apparent FOV's as the Astromania, as well as the same eye relief.  It's not impossible that they are the same design, repackaged for another company.
Another feature to look for in telescope eyepieces is "fully-multi coated" which will minimize reflections and stray light (Astromania, Orion, and SVBONY list this feature, but Meoptex does not).
IN my limited experience with telescopes, it generally holds true that the more expensive options have better performance, but your total performance will be limited by your poorest component.  (A \$200 eyepiece won't make a $25 telescope any better)
You can also look for used eyepieces, advice, and reviews on cloudynights.com.


How significant are the recent results that Scholz’s Star has perturbed several observed hyperbolic objects?

Not an answer, but just on radiants:

Radiant means angle of approach.   It's applied to meteor showers because the Earth intersects the dust trail that causes the meteor shower at a specific place in orbit and dust trail has a specific angle of intersection relative to Earth and Earth's orbit.

Source of picture.

  What is the "radiant" of a minor body with a hyperbolic orbit in a
  simulation

Radiant can't be applied to elliptical orbits though extremely elliptical orbits kind of have a radiant, but it can be applied to objects approaching the solar system.  A hyperbolic orbit generally implies an object that came from outside the solar-system though it could also be a gravity assisted object from inside the solar system that was accelerated from an elliptical to a hyperbolic orbit.

Here's an article that mentions radiant in association with an exo-solar object

  astronomers looked for the radiant it seemed to originate from.
  Calculations determined that A/2017 U1 came from a radiant located
  close to (R.A. = 18h 41m 25s ; Dec=+34° 15′) coordinates, which is
  near the actual position of Vega

exo-solar objects approach the solar system in a mostly straight line until they get close enough for the sun to bend their path.   Where they came from before their direction was curved by the sun's gravity is their radiant.


Are there any stars that orbit perpendicular to the Milky Way's galactic plane?

The Sun and most of the other stars are in the bulging disk of the Milky Way galaxy, but about 1% of the galaxy's stellar mass is in the galactic halo.  The halo also includes 50 globular clusters and about 20 satellite galaxies according to Helmi 2008: The stellar halo of the Galaxy.  Here is a nice graphic:

Note that the Sagittarius Stream of stars is extremely close to passing the galactic poles.  If you were on a planet next to one of these stars as it passed a pole, you would probably have a glorious view of the Milky Way stellar disk.
Bodies in the galactic halo don't necessarily follow the elliptical paths predicted by Kepler, so their orbits may not be consistently perpendicular to the galactic plane.  Some stellar streams have extremely odd orbital paths like the Phlegethon stellar stream, referenced in this answer:

There is some speculation that the Halo stellar streams are the remnants of dwarf galaxies the Milky Way has absorbed.  Indeed, Liang et al. say that:

the Galactic halo has complicated assembly history and it not only
interacts but also [is] strongly mixed with other components of the Galaxy
and satellite dwarf galaxies


Two body orbit dynamics for orbital periods approaching light cone limit

As far as I know the only option here is numerical general relativity. There are no known exact solutions for the general two-body problem in GR, since the field equations are very non-linear. Even there there are limits, since when the fields get strong enough the bodies will start to deform in complex ways that also depend on the equation of state of the matter they are made from (and magnetic fields!). This is

However, current models of merging neutron stars do not usually produce toroids. The stars become elongated, splashing out matter that either forms tails or a disk. Some examples:

https://www.youtube.com/watch?v=Dyn9KbB_zeo
https://www.youtube.com/watch?v=AW_W-Fbrdu8&list=PLSYkic-CsfSGFb3KY2cjLA1EZyNDadK_w
https://www.youtube.com/watch?v=ow9JCXy1QdY

The reason for this is that maintaining a toroid against self-gravity is fairly hard, making it tend to elongate into an ellipse and then a spiral.


What are the problems that need to be taken in consideration when trying to find planetary position by using jpl ephemerides?

You don't mention what you are trying to access this with but I would strongly recommend AstroPy for handling the ephemeris and handling the co-ordinates and time systems. For your questions:

The Moon and the Earth are both given relative to the Earth-Moon barycenter (center of mass). Separately you can get the vectors from the Solar System Barycenter to the Earth-Moon Barycenter. So if you want SSB->Moon you will need to do two calls to the ephemeris and do vector addition.
It depends how accurately you want things and how fast the thing you want is moving. You can probably use Terrestrial Time (TT) with the ephemeris and you want be horribly off for anything other than Moon. The offset TT-UTC (currently 67.184 seconds) consists of two parts; a variable part from UTC to TAI, currently 37 seconds, which consists (mostly) of the number of leapseconds and a fixed part (32.184 seconds) from TAI to TT. TDB varies around TT by about 2 milliseconds and it's somewhat periodic (dominated by Earth's orbit ellipticity and Jupiter and Saturn) but doing it properly requires a 791 term expansion (see e.g. http://star-www.rl.ac.uk/docs/sun67.htx/sun67ss154.html#Q1-158-946)
Yes, if you want to do anything related from the Earth such as predicting positions of something in the sky, you will need to take the frame bias, precession and nutation into account along with Earth rotation/sidereal time to align the celestial reference frame (ICRS/J2000) with that of the Earth (ITRS).

I recommend USNO Circular 179 (http://aa.usno.navy.mil/publications/docs/Circular_179.php) which has a lot of details of how to do this in the new IAU framework. Meeus's book is very good (I have it also) but is based on the old IAU1984 equator and equinox of date methods.


How is the maximum density in a circumstellar disk determined?

If your $r_{peak}$ is not known a priori then you have two unknowns and only one equation: this cannot be solved. At least not without a second equation or a measured $(\rho,r)$-pair (say the central density $\rho_0 = \rho(r=0)$).

In any case, this definition of radial density looks like an ad-hoc assumption to me, and not something that came out of a proper hydrodynamic model. So without knowing why you chose this form or the context of your disk model I can't comment on what equation is best suited to determine $r_{peak}$. A well-informed guess seems like the way to go.


The American Astronomical Society (AAS) answers this question on their How to Tell If Your Eclipse Glasses or Handheld Solar Viewers Are Safe page.

Basically they say that since anyone can claim ISO 12312-2 compliance even if they haven't been certified, you should make sure you obtained your eclipse glasses from a vendor whose ISO 12312-2 compliance has been verified by the AAS Solar Eclipse Task Force:

  We used to say that you should look for evidence that they comply with the ISO 12312-2 international safety standard for filters for direct viewing of the Sun. But now the marketplace is being flooded by counterfeit eclipse glasses that are labeled as if they're ISO-compliant when in fact they are not. So now we suggest that you make sure you get (or got) your eclipse viewers from one of the suppliers listed on our Reputable Vendors of Solar Filters & Viewers page.

They say that checking eclipse glasses to see if they meet the ISO 12312-2 standard would require a spectrophotometer, which most people do not have access to:

  Unfortunately, you can't check whether a filter meets the ISO standard yourself — doing so requires a specialized and expensive piece of laboratory equipment called a spectrophotometer that shines intense UV, visible, and IR light through the filter and measures how much gets through at each wavelength.

So they have assembled a task force to put together a list of manufacturers and vendors of eclipse glasses that they have verified to be ISO 12312-2 certified (their list is longer than NASA's list of 5 manufacturers):

  The AAS Solar Eclipse Task Force has been working diligently to compile a list of such vendors, now posted on our Reputable Vendors of Solar Filters & Viewers page. We've checked manufacturers' ISO paperwork to make sure it's complete and that it comes from a recognized, accredited testing facility, and we've personally examined manufacturers' products. We've asked manufacturers to identify their authorized resellers, and we've asked dealers to identify the source of the products they're selling. Only when everything checks out do we add a vendor to our listing. If we don't list a supplier, that doesn't mean their products are unsafe — only that we have no knowledge of them or that we haven't convinced ourselves they are safe.

They also give advice on how to tell if your eclipse glasses are not safe to use:

  How can you tell if your solar viewer is not safe? The only thing you can see through a safe solar filter from a reputable vendor is the Sun itself. If you can see ordinary household lights through your eclipse glasses or handheld viewer, it’s no good. Safe solar filters produce a view of the Sun that is comfortably bright (like the full Moon), in focus, and surrounded by black sky. If you glance at the Sun through your solar filter and find it uncomfortably bright, out of focus, and surrounded by a murky haze, it’s no good.


A moonless night is, as you suspect, a night in which the Moon does not appear visible in the sky. This happens once per month, when the Moon is near the Sun. Due to the proximity of the Moon and the Sun in the sky, at that time the Moon is the smallest sliver possible, and therefore not a full moon.

This is because it is actually the Sun that illuminates the Moon, and when the Sun and the Moon are in the same direction in the sky we are seeing the non-illuminated side of the Moon. Note the direction of the sunlight in this image:

Obviously, the direction of the sunlight is the direction of "up" during the day. If you look at the horizon slightly after sunset or slightly before sunrise, you might actually catch a glimpse of the sliver of Moon before it set or rises slightly after or before the Sun.


Why was Dawn's (the spacecraft) flight path circular?

Dawn was not exploring L3. But instead it explored both Vesta and Ceres. In order to do so in a cost-saving manner, Dawn was also a mission demonstrator for an interplanetary mission equipped with ion propulsion.

Because of the low specific impulse of ion propulsion, Dawn cannot fly in a straight line, as this would imply a high initial velocity and the spacecraft couldn't deaccelerate in time. Therefore, Dawn started with low $v_0$ at Earth, and raised its orbit slowly and continuously, creating the spiral you see.


Why do objects appear blue while on a neutron star?

Your suspicion is right, the term is gravitational blueshift.
Going down a gravity well, photons gain energy, pushing their wavelengths in the blue direction. The opposite would be gravitational redshift observed when photons climb out of a gravity well.
In terms of general relativity, when you observe the world from near the neutron star, time appears to go faster outside the gravity well, and therefore arriving wavelengths are blueshifted.


Which binocular should I buy to be able to see Saturn and its rings?

In your choice, you should also consider the binocular stand, your experience and how good your sight is. Also, with that money, you can also buy this telescope. With it, you can also see all the Messier objects. I have SkyWatcher 10 ", and I am quite satisfied with it. For cheap, but good telescopes, look at this site.
I haven't quite answered your question, but I think, that this will open more options.
Also, as @Mick has mentioned, this question is quite similar to yours, but you've said how to see Saturn and other objects in the night sky, so I searched for a link for these other objects. If I'd have 400 bucks, then I'll definitely go for a telescope.
Hope this helps!
P.S.
Try to buy as fast as you can because Saturn can be seen earlier and earlier in the evening. Now (middle of Juny 2020) it is observable at approximately 00:00—morning, but soon, it will be observable at 21:00!


Is there an online planetarium where the observer is on another celestial body?

Yes, the free software Stellarium (also Wikipedia) can do that. It has a list of celestial bodies and you select which one you want to be on. When you open Stellarium go to Location Menu (or the button F6 for Windows OS) and that is where you set your viewing point.

There might be additional options in Where can I find/visualize planets/stars/moons/etc positions?


Why weren't the Hubble light-echo images of V838 Monocerotis supplemented by ground telescopes?

Adaptive optics does not work (or at least did not 10+ years ago) on ground based astronomical telescopes at visible wavelengths. No images remotely comparable to HST could be obtained for faint$^*$ objects from the ground in the BVI wavebands.

I checked for papers discussing adaptive optics at other wavelengths and couldn't find any. The field of view is a factor. Back in 2003 there wouldn't have been instruments that could have covered V838 Mon at IR wavelengths with NIR AO observations. It can just about be done now at Gemini South, but field of view is still an issue for AO visible wavelength observations.

$^*$ NB "Lucky imaging" or speckle imaging can work from the ground for resolving binaries etc, but is not so good for imaging faint, extended objects. Having said that, I have seen some images of things like planetary nebulae taken with Lucky Imaging, but that is now and that was then. Nobody was doing it in 2003.


Moving object seen through telescope, what is it?

Comments formed into community wiki

Satellites can and have been seen through telescopes. Possibly it is just a very small or faint satellite which requires a telescope to see and is too difficult to detect by the naked eye. The description of a "moderate speed in a straight line and eventually dimming out" sounds like every satellite. A piece of space junk is also possible. Lots of second stage boosters and other rocket parts floating around up there that are small enough that you would only notice them through a telescope

You might be able to identify it by going to the Heavens Above site and providing your location, the date & time.


What factors makes a star big in size(Physically)?Is the size of the nebula a relevant factor?

The size of the nebula doesn't matter at all, what matters to the size of a star is the mass and age of that star.  Even the mass of the nebula doesn't really matter, because giant molecular clouds fragment dramatically as they make many many stars.  It's a very complicated and detailed process that controls how they fragment, so usually one merely assumes a statistical distribution of stellar masses, called the "initial mass function," which applies more or less universally to any sized nebula.

Once the mass of the star is determined (statistically), the star contracts with age, so very young stars have all kinds of different sizes while they contract.  Eventually fusion initiates in the core (which is sometimes considered the "birth" of the star, though it was pretty much already a star before that), and then the size of the star is determined by its mass because a higher mass star doesn't need to contract as small to reach fusion temperatures.  Later, when core fuel runs out, stellar envelopes expand dramatically into giants or supergiants, which to a point reverses the process for the envelope as what happened in the original contraction of the star.  The size varies a lot after that.


Is it Possible for Planetary Bodies To Exist In Close Proximity Without Adverse Effects?

Quick calculation of Roche limit for two identical earthlike (fluid) bodies.
Roche distance is about equal to 2.44 X radius X(density1/density2)^1/3.
For identical bodies the density conveniently cancels, so we get
Rd = 2.44 X radius.
Earth radius is 6378 km.
So Rd = 2.44 X 6378 = 15500 km (center to center).
The distance surface to surface (15500- 6378) is 9184km.

At that distance, by trigonometry, another earth would subtend about 44° of sky, pretty big!, and just at the bleeding edge of stability.

The nearest planet in the provided picture looks like it might be at about that distance. Evacuation might be in order as the other three planets or moons will degrade the close binary's orbital stability.


Is there a way to know how the magnetic field was at the very beginning of our universe?

I don't know of one. But then again I think there was no magnetic field at all at the very beginning of the universe. I can't point to any material that proves this beyond a doubt. But I can say that a quiescent black hole "has no hair", so it has no magnetic field, and the evolution of the universe is said to be something like pulling away from a black hole. See Hawking saying the universe is like a time reversed black hole here.

  I recently read that "Gamma-ray bursts are catastrophic events, related to the explosion of massive stars 50 times the size of our sun. If you ranked all the explosions in the universe based on their power, gamma-ray bursts would be right behind the Big Bang". I'm asking this question in the tentative to compare the Big Bang to a gamma-ray burst. I keep in mind that a key difference between both is that Big Bang was Not an explosion, but want to know similarities to point out.

There aren't many similarities. A gamma-ray burst is where some of the infalling matter is directed by magnetic fields generated by infalling matter and converted into gamma rays. The Big Bang is when the space of the universe started expanding.

  Near the formation of a black-hole they tell in this same article that matter and magnetic field play a key role the gamma-ray bursts.

True enough. But as far as I know matter and magnetic fields played no role in the Big Bang.

  I know magnetic fields are induced by charge displacement. Is it possible there was a kind magnetic field before protons and electrons formed, or during the first expansion phase?

It's possible I suppose. A magnetic field is associated with curl aka "rot" which is short for rotor. The early universe could have had an overall magnetic field, or gravitomagnetic field, if it was somehow rotating. See Mark Hadley's Galaxy sized twist in time pulls violating particles back into line. Also see the universe may have been born spinning. But I think it's rather speculative stuff myself.

  Do we have a way to know how magnetic fields were at that time?

I don't think so. The early universe was said to be maelstrom. Hence I think information about the initial state got scrambled.


Solar Eclipses for dummies: Step 1 - moon equatorial orbit around a planet without axial tilt

May, or may not.
No.
Q1: A formula: yes, a simple formula: no.
Q2: Not for each point on Earth. This is e.g. due to different distance quotients.
Q3: Depends on distances, sizes, angular velocities, position on Earth.
Q4: Yes, but it's different for each point, not just by latitude. Even at the equator, there may exist points with, and points without totality. That's since the shadow of the Moon isn't cylindrical, but a cone. And because the Earth isn't a plane perpendicular to the Sun - Moon axis. And the distance Moon - Sun is varying due to its orbit around Earth.


How much does the equivalent width of a line change by the introduction of 5% scattered light?

Imagine your line as a rectangle of width $w$ and depth $d$ relative to a normalised continuum.
Without scattered light, the area blocked off by the line is $wd$ and if the continuum level is normalised to 1, then the equivalent width is $wd$.
Now add 5% scattered light. The height of the continuum is 1.05 (but we're going to renormalise it) and the depth and width of the line are still $d$ and $w$.
Now renormalise by dividing by 1.05. The width of the line is the same, but the depth is smaller by a factor 1.05 and thus so is the equivalent width.


Turbulence sources:

There are numerous sources of turbulence in the interstellar medium, at all scales:

at large scales, there is the shear from galactic rotation. One way to sustain turbulence and to couple large and small scales would be the magnetorotational instability (MRI).
at large scales, gravitational instabilities can also play a significant role, through spiral structures.
outflows and jets from forming stars play an important role, releasing a lot of energy in the ISM.
in star forming regions, massive stars are also important. Radiation and stellar winds from massive stars are an important input of energy in the ISM. And eventually, the most massive ones will blow up in supernovæ, releasing even more energy.

Therefore, one could then regards separatly three processes related to massive stars:

stellar winds
ionizing radiation
supernova explosion

Importance for star formation:

They are all relevant to star formation, one way or another. One key property of turbulence is to cascade from large to small scales; therefore, even if you inject turbulence at large scales (galactic scale) you'll get turbulent motions down to the scale of a molecular cloud.

A nice illustration of the tubulent cascade is the Larson's relation (Larson 1981):

The Larson's relation shows the evolution of the velocity dispersion with the size of the structure you're looking at. Velocity dispersion is an indicator of turbulence. Indeed, these dispersions are non-thermal: knowing the typical temperature of the MIS (about 10 K), one can estimate the thermal velocity of, for example, the CO molecule ($v_th = \sqrt{2kT/\mu m_H}$, with $k$ the Boltzman constant, $T$ the temperature, $\mu$ the mean molecular wright and $m_H$ the hydrodgen atom mass) that is about 0.07 km s$^{-1}$. Measured velocity dispersions are of the order of 1 to 10 km s$^{-1}$, and these is interpreted as a turbulence signature (and estimate).

Details:

Energy rates: values are given (roughly) for the Milky Way

MRI: $ \small \dot e = 3 \times 10^{-29} {\rm erg\ cm}^{-3}\ {\rm s}^{-1} $;
Gravitational instabilities: $\small \dot e = 4 \times 10^{-29} {\rm erg\ cm}^{-3}\ {\rm s}^{-1}$;
Outflows: $\small \dot e = 2 \times 10^{-28} {\rm erg\ cm}^{-3}\ {\rm s}^{-1}$;
Ionizing radiation: $\small \dot e = 5 \times 10^{-29} {\rm erg\ cm}^{-3}\ {\rm s}^{-1}$
Supernovæ explosions: $\small \dot e = 3 \times 10^{-26} {\rm erg\ cm}^{-3}\ {\rm s}^{-1}$
Stellar winds: it strongly depends on the type of star: it varies as the power of -6 of the star's luminosity. It therefore ranges from an energy comparable to a supernova explosion (or even more for Wolf-Rayet stars) to almost nothing.

Sources:

Mac Low & Klessen 2004


Why do different instrumental filters use different magnitude systems (Vega vs AB)?

I'd guess it's largely a matter of historical accident and inertia, mixed with a certain amount of taste.

Vega magnitudes are the traditional system, based on how bright the star Vega is within the filter (+ detector response, etc.). The Johnson and Cousins filter systems were defined back in the early 1950s when Vega magnitudes were the default, and were to some extent meant to reproduce earlier photographic magnitudes. The AB system was proposed a bit later (1960s) and was originally used for narrow-band photometry and spectrophotometry, not broad-band photometry like the Johnson-Cousins and similar filters. Some more modern filter systems -- such as the SDSS filter system -- have been defined using AB instead, because the creators preferred it.

Some filter systems have both definitions. For example, the near/mid-IR filters of the IRAC camera on the Spitzer Space Telescope have both Vega and AB definitions; some papers use one, some the other. (Sometimes papers that are part of the same overall project.)

If I do aperture photometry on an I-band image and an i'-band image, will the first one return Vega magnitude of the object and the second return AB magnitude of the object, by default?

By default, no. Doing aperture photometry implies choosing a calibration: that is, a conversion between observed counts and magnitudes. (Such a calibration might already be provided in an image header, e.g., from SDSS.) It's certainly possible to do aperture photometry on an I-band image calibrated to i-band, and vice-versa. I have on occasion done things like aperture photometry on R-band images calibrated to K-band, though I certainly wasn't expecting things to be very accurate.

if I wanted to do SED fitting ...

I think the answer depends on what form your model SED is in. If for some reason your SED is in AB magnitudes, then, yes, you'd want to convert all your observed data to AB magnitudes. If, on the other hand, your model is in flux units, then you'd want to convert your data to flux units (either Vega mag to flux units or AB mag to flux units).


Is the difference in time between the sideral and synodyc month - constant?

The time difference (synodic month) - (sidereal month) will not be constant. The reason for the variation is due to the eccentricity of the Moon's orbit: the Moon moves faster at perigee (closest to the Earth) than when at apogee (farthest from the Earth). If the Moon is near perigee around the time of New Moon, it takes less time for the Moon to "catch up" with the Sun, so the synodic month is shorter.

Also, the Moon's perigee and apogee distance changes from month to month due to the influence of the Sun. This also affects the speed of the Moon in its orbit. (The apparent speed of the Sun also changes because of the Earth's elliptical orbit around the Sun, but the variation is smaller.)

It may be helpful to take a "geocentric" view; that is, look at the motion of the Sun and Moon on the sky. In the following figure, a rare annular eclipse occurs with the Sun and Moon occulting the star Regulus ($\alpha$ Leonis). The days listed are the average sidereal and synodic periods. After 27.32 days (one sidereal month), the Moon is again aligned with Regulus. The Sun is farther east in the sky, so it takes a few more days for the Moon to catch up with the Sun and complete a synodic month (39.53 days).

Note: The distances travelled are approximate in this figure. The apparent size of the Sun and Moon are greatly exaggerated!

The two months are related by the speed of each object:

average speed of Moon = 360 degrees/27.32 days
average speed of Sun = 360 degrees/365.26 days
distance travelled by the Moon in one synodic month = T*360/27.32
distance travelled by the Sun in one synodic month = T*360/365.26

The two distances are not equal: the Moon moves an extra 360 degrees. So the length of the average synodic month, T, can be calculated by solving this equation:
$$ T*\frac{360}{27.32}-360=T*\frac{360}{365.26}$$

It should be apparent from this equation that if the speed of the Moon varies, then the duration of the synodic month T will change as well.


Given a date obtain latitude and longitude where is the sun zenith

What you're looking for is the latitude and longitude of the subsolar point.

Calculating it with accuracy is somewhat complicated.  Here is a python library that has routines to do it for you:
http://rhodesmill.org/pyephem/

You could also possibly query this information from other sites.

If you really want to work it out on your own, I suggest "Spherical Astronomy" by Robin M. Green published by Cambridge University Press as a starting point.


Relation between black hole mass and radius, and our universe's

According to the standard ΛCDM cosmological model, the observable universe has a density of about $\rho = 2.5\!\times\!10^{-27}\;\mathrm{kg/m^3}$, with a cosmological consant of about $\Lambda = 1.3\!\times\!10^{-52}\;\mathrm{m^{-2}}$, is very close to spatially flat, and has a current proper radius of about $r =  14.3\,\mathrm{Gpc}$.

From this, we can conclude that the total mass of the observable universe is about
$$M = \frac{4}{3}\pi r^3\rho \sim 9.1\!\times\!10^{53}\,\mathrm{kg}\text{.}$$
Sine the universe at large is nonrotating and uncharged, it's natural to compare this to a Schwarzschild black hole. The Schwarzschild radius of such a black hole is
$$R_s = \frac{2GM}{c^2}\sim 44\,\mathrm{Gpc}.$$
Well! Larger that the observable universe.

But the Schwarzschild spacetime has zero cosmological constant, whereas ours is positive, so we should instead compare this to a Schwarzschild-de Sitter black hole. The SdS metric is related to the Schwarzchild one by
$$1-\frac{R_s}{r}\quad\mapsto\quad1 - \frac{R_s}{r} - \frac{1}{3}\Lambda r^2,$$
and for our values we have $9\Lambda(GM/c^2)^2 \sim 520$. This quantity is important because the black hole event horizon and the cosmological horizon become close in $r$-coordinate when it is close to $1$, a condition that creates a maximum possible mass for an SdS black hole for a given positive cosmological constant. For our $\Lambda$, that extremal limit gives $M_\text{Nariai} \sim 4\!\times\!10^{52}\,\mathrm{kg}$, smaller than the mass of the observable universe.

In conclusion, the mass of the observable universe cannot make a black hole.

  Well, we don't fully comprehend black matter, do we? And it was just "yesterday" that we discovered the "black energy", wasn't it?

If GTR with cosmological constant is right, we don't need to "fully comprehend" it to know its gravitational effect, which is what the calculation is based on. If GTR is wrong, which is of course quite possible, then we could be living in some analogue of a black hole. But then it's rather unclear what theory of gravity you wish for us to use to try to answer the question. There's no remotely competitive theory that's even approaching general acceptance.

  From the perspective of our huge ignorance, I think that 14.3Gpc and 44Gpc are not even one order of magnitude apart, which I consider a good approximation.

Actually, the point of that calculation was to show that it's at least prima facie plausible. The Schwarzschild radius calculation doesn't rule out the black hole--quite the opposite. However, it's also not appropriate for reasons I explained above. The more relevant one actually does have mass more than one order of magnitude apart, and shows inconsistency. So if GTR with Λ is correct, it's unlikely because the ΛCDM error bars aren't that bad.

However, even if we still treat it as "close enough", that does not by itself imply what you want. The question of what kind of black hole all the mass of the observable universe would make, if any, is quite different from whether or not we're living in one. The black hypothetical needs to be larger still.

The biggest point of uncertainly, though, is the cosmological constant, even if GTR is otherwise correct. If we're allowed to have very different conditions outside our hypothetical black hole, then we could still have one, but then we get into very speculative physics at best, and just complete guesswork at worst.

So treat the above answer as conditional on the mainstream physics; if that's not what you want, then there can be no general answer besides "we don't know". And that's always a possibility, although not a very interesting one.


Why can't we detect water on Mars in the same way as in exoplanets?

We can look up for water in the atmosphere of the planet, not at the planet itself. We can do that when a planet passes in front of a star. That tehnique is called transmission spectroscopy.


Could someone see anything while being inside black hole?

The answer is most definitely yes, or at least yes, as far as our current understanding of how gravity works goes. It is observationally untestable (let's be more specific - nobody could report the results of an observational test!) since no signal can emerge from inside the event horizon.

The scenario is treated in some detail by Taylor & Wheeler ("Exploring Black Holes", Addison, Wesley, Longman - highly recommended) in terms of what an observer would see on a direct radial trajectory into a non-rotating "Schwarzschild" black hole. I won't bore you with the maths - it is fairly complex.

A star situated at exactly 180 degrees from the observer's radial path will always appear in that position as the observer looks back - right down to the singularity. The light will be gravitationally blueshifted by an ever-increasing amount - essentially tending towards an infinite blueshift at the singularity.

For stars at an angle to the radial path, their positions will be distorted such that they appear to move away from the point at which the observer has come from (and are also blue-shifted). In the final moments (it takes less than $1.5\times 10^{-4}$ seconds of proper time to fall from the event horizon to the singularity of a 10 solar mass black hole, but a huge $\sim 60$ seconds for the black hole at the centre of our Galaxy) the light from the external universe will flatten into an intense ring at 90 degrees to the radial direction of motion. So you would end up seeing blackness in front of you, blackness behind and the sky split in two by a dazzling ring of light (almost seems worth it!).


Does a replacement exist for Dexter? (A tool that reconstructs data from a pdf of a graph)

You question reminded me of a tool I first encountered in the 80’s (I know right!) called “data thief”. Back when Mac applications had a four letter “creator code” this one had the code “DIEF” (which is the Dutch word for “thief”).
I appears to be around still - and as it doesn’t rely on a browser it may well still work (I haven’t tried in years though).
Take a look here


What causes Titan a moon to have a denser atmosphere than that of a Mars?

Mars and Titan differ markedly in distance from the Sun, composition, and possibly geological activity.

Titan is about 6.3 times as remote from the Sun as is Mars, which means Titan receives about 1/40th solar radiation that Mars does, with about the same reduction in solar wind. Mars' small size coupled with its much closer proximity to the Sun allowed the solar wind to rip much of Mars' atmosphere away. Despite the marked differences between the atmospheres of Venus, Earth, and Mars, the atmospheric loss rates (in terms of mass loss per unit time) from these three planets are estimated to be about the same. This rate represents a pittance for Venus' very large atmosphere of Venus, a small amount for Earth's somewhat large atmosphere, but a huge proportional loss rate for Mars. Titan is spared to some extent.

Mar's average density is 3.93 g/cm3. Mars is a terrestrial planet. Titan's average density is less than half that, 1.88 g/cm3. Titan is an icy moon: some rock, but a lot of volatiles. There's a lot of ammonia mixed in with the water ice (and possibly liquid water deep down, if Titan's core is warm). Some hypothesize that Titan regenerates its atmosphere via releases of ammonia. Sunlight splits ammonia near the top of Titan's atmosphere into hydrogen and nitrogen. The hydrogen escapes, but not the nitrogen.

Finally, Mars is most likely dead geologically. It's covered with craters that may date back to the Late Heavy Bombardment. Titan exhibits some signs of being somewhat active, geologically. It too has craters, but it's not as plastered with them as is Mars. It's mountains appear to be younger, at least according to some. The geological activity may help to regenerate Titan's atmosphere.


Are there any unmanned large telescopes located on Earth?

The largest "robotic" (i.e., unmanned) telescope I'm aware of is the 2.4-meter Automated Planet Finder. Other large robotic telescopes include the 2.0-meter Liverpool Telescope and its copies (Faulkes Telescope North and Faulkes Telescope South). You can read about the automated control system of the Liverpool telescope here.

These are all located at existing observatories (e.g., APF at Lick Observatory, Liverpool Telescope at Observatorio del Roque de los Muchachos, La Palma, Canary Islands, Spain), which means there are people on site for potential maintenance work, etc., though said staff are really devoted to working on other telescopes.

The closest thing to an isolated, automatic telescope in an uninhabitable setting might be the small (0.5-m) Antarctic Survey Telescope at Dome A in Antarctica, which I believe is serviced annually by Chinese expeditions. It can apparently be remotely controlled via Iridium satellite communications, though actually retrieving the full data sets is part of what the annual expeditions are for.


There are several ways to look at this.
One rather simple way is to compute the Rayleigh number of Mars' mantle. The Rayleigh number is a dimensionless number defined as:
$Ra = \dfrac{\rho g \alpha \Delta T d^3}{\eta \kappa}$
where $\rho$ is the mantle density, $g$ is gravitational acceleration, $\alpha$ is the thermal expansion coefficient, $\Delta T$ is the difference in temperature over the convective layer of thickness $d$, $\eta$ is the viscosity, and $\kappa$ the thermal diffusivity. It is generally considered that convection happens when $Ra > 1000$, and higher $Ra$ mean more vigorous convection. $Ra$ for Mars is estimated at $10^6$ by Mulyukova & Bercovici 2020 (compared to $10^7$ for the Earth), meaning that there is convection in Mars' mantle, although less vigorous than on Earth.
Another approach is to look at some geological features of Mars that can provide constraints on the internal dynamics of the planet. For example, recently active point-source volcanoes, such as Olympus Mons, suggest the presence of a mantle plume—thus convection—rising from beneath (Li & Kiefer 2007). Likewise, the presence of mantle plumes/convection can explain variations in lithospheric thickness (Kiefer & Li 2009).
A study by Stern et al. (2017) used volcanisms, faults and impacts to define a "Tectonic Activity Index" (TAI) for 26 planetary bodies, ranging from 0 to 3. They found that Mars has a TAI of 2, less than the Earth and Venus (3), but still inferred to be "tectonically and convectively active" (bodies with TAI < 2 are inferred to be tectonically dead). They think that Mars actually represents the transition between active and dead bodies:

The present tectonic style of Mars represents an almost completely
stabilized stagnant lid [SL]. Here, a single upwelling mantle plume,
pouring lava out at the same sites for hundreds of millions of years,
produced the immense Tharsis rise and volcanoes (Zuber, 2001). When
these last few mantle plumes are extinguished, tectonic style
transitions to the final SL stage – planetoid death. This stage is
typified by Mercury and Earth's moon, bodies with ultra-thick,
ultra-stable lithospheres

Possible evolution of magmatotectonic styles for a large silicate body, like the Earth. Plate tectonics requires certain conditions of lithospheric density and strength to evolve and is likely to be presaged and followed by stagnant lid tectonics. After Stern et al. (2017) (CC BY-NC-ND 4.0)


What radio frequency ranges are most beneficial for astronomy?

Radio astronomy involves a wide range of frequencies, covering the range from $\sim$10 MHZ to $\sim$100 GHz.$^{\dagger}$ With four orders of magnitude to work with, the most valuable band depends strongly on what sort of object you're observing, as well as what receivers are actually available on a specific telescope. Given the diversity of sources out there, it's difficult to narrow down the list to any small number of important bands, so this answer is far from comprehensive - my apologies to any extremely high-frequency spectral line observers out there. A truly comprehensive answer would be several pages long, so I'll forego that in favor of a briefer overview.
Spectral lines
The most well-known radio line is the famous hydrogen line arising from the spin-flip transition. Its rest frequency is approximately 1420 MHz, but as is the case with all spectral lines, extragalactic HI sources will show the line at significantly lower frequencies due to cosmological redshift. The same holds true for clouds within the Milky Way due to their motion relative to us, albeit to a much weaker extent. Neutral hydrogen is an important tool both locally (e.g. mapping the structure of the Milky Way) and for high-redshift sources (e.g. providing a handy tool for probing the early universe around the time of reionization).
There are four major hydroxyl radical lines between 1612 MHz and 1720 MHz, including the 1665 MHz line that led to the first detection of masers. They remain one of the key molecules detected in masing sources, and hydroxyl megamasers remain the dominant type of megamaser currently known. Notable, the hydroxyl lines lie on the opposing side of the water hole from the hydrogen line.
Carbon monoxide (12CO) has rotational transitions at multiples of 115 GHz, with the most notable lines being those at 115 GHz and 230 GHz (13CO also has a line near 110 GHz). They are quite useful for probing molecular clouds, meaning that CO surveys can reveal structures that might be absent on HI maps.
These are just a few of the key spectral lines in the radio spectrum; a 1991 National Academy of Sciences report came up with a far more comprehensive list of spectral lines of varying priorities. Formaldehyde, water (two more key maser molecules), methanol, ammonia, methine, and deuterium are all included, along with many others. Almost all lie between 1 GHz and 300 GHz (only two orders of magnitude!). An even more detailed list can be found here.
Continuum observing
Some broadband sources may be observed across several bands, and observations may take advantage of existing receivers often used for other purposes. For instance, the Arecibo L-Band Feed Array, or ALFA, included the 1.4 GHz hydrogen line and has been important for neutral hydrogen surveys (e.g. ALFALFA), but it turned out that its capabilities were also quite useful for pulsar searching (e.g. PALFA). Therefore, continuum observers might use some of the bands mentioned above, depending on available receivers and backends.
That said, different types of continuum emission peak at different frequencies, so we can certainly divide the spectrum up into chunks. The Handbook of Frequency Allocations and Spectrum Protection for Scientific Uses has some nice figures illustrating this on pages 13 and 14 of chapter 4. I won't reproduce them because I'm not sure about copyright status, but the gist is the following:

Pulsars are the lowest-frequency continuum sources, observable below a few GHz. The 406-410 MHz band is often used, but again, 1.4 GHz receivers are also commonly used due to their availability. As pulse profiles sometimes show different features at different frequencies, it can be quite useful to observe the same pulsar in different bands.
Supernova remnants (SNR) produce radio emission through synchrotron emission from relativistic electrons, and are actually visible across quite a wide range of frequencies (with synchrotron emission also in other parts of the electromagnetic spectrum). I'm not as familiar with radio SNR observations.
The same holds true for radio galaxies, which are primarily detectable through synchrotron emission across a similarly large frequency range.
The cosmic microwave background is a blackbody; at its current temperature of 2.7 K, it peaks close to 300 GHz, which again places it near the boundary of radio astronomy.

Protected bands and unusable bands
Many spectral lines (e.g. the HI line) lie in protected frequency bands in many jurisdictions, with widths which ideally take into account redshifts or satellite lines. Some key continuum bands (e.g. 406-410 MHz) have also received protection. Unfortunately, human activities are not the only thing that can make some portions of the spectrum unusable. For example, pressure-broadened oxygen lines in the 50-70 GHz range can be problematic. Water vapor, in addition to the strong line at 22 GHz, also produces strong lines at 557 GHz, 752 GHz and 970 GHz.${\ddagger}$

$^{\dagger}$Stricly speaking, there are indeed observations in the THz range, but at this point, the boundary between radio astronomy and infrared astronomy becomes a bit shaky.
$^{\ddagger}$See Essential Radio Astronomy by Condon & Ransom.


Has NASA confirmed earth now has a second moon orbiting the planet?

It's not true; this object is not a moon of Earth's.

Here's a NASA/JPL announcement of it. 2016 HO3 is the "name" being used for it at the moment.

They're calling it a "quasi-moon". It's in an orbit that is in resonance with Earth's but is not permanently bound. (It's not the first object to be found with such properties too.)


Bismuth is produced via the s-process.
It is the slow absorption of neutrons that can produce heavier elements beyond the peak of the binding energy per nucleon curve at iron. The alpha process isn't capable of producing much of anything beyond nickel.


Temperature of the intracluster medium (ICM)

The answer to your first question is "Yes, the temperature referred to is the 'normal' temperature, reflecting the average kinetic energy of the gas particles".

The answer to your second question is a bit more complex:

Cooling function

Gas cools by various processes, with an efficiency depending on the temperature, the density, and the composition of the gas. At "low" temperatures, most cooling happens because particles collide and their kinetic energy is used to excite or even ionize atoms. The atoms subsequently recombine or de-excite, emitting radiation which can carry energy away from the system.

Most of the gas in the Universe is hydrogen, which cools most efficiently around $10^4\,\mathrm{K}$. At higher temperatures, helium becomes the major contributor. Additionally, various metals contribute at various temperatures depending on their density and ionization state. At very high temperatures ($\gtrsim10^7\,\mathrm{K}$) where the gas is highly ionized, cooling mainly happens through bremsstrahlung, i.e. deceleration of charged particles by other charged particles.

The figure below (from Mo, Bosch & White 2010) shows the cooling function for various metallicities (with my own annotations):

Collapse of gasous halos

So what makes the intracluster medium (ICM) different from the interstellar medium (ISM)? In the expanding Universe, overdensities try to contract. These overdensities can reach hydrostatic equilibrium only if radiative cooling is small. If the cooling timescale $t_\mathrm{cool}$ is much smaller than the free-fall timescale $t_\mathrm{ff}$, it can collapse and form stars. From the virial theorem, which gives the relation between potential and kinetic energy of the system, you can calculate the relation between the two timescales. It's a bit too much math for this post, but not difficult. I recommend reading chapter 8.4 in Mo, Bosch & White (2010). Their Fig. 8.6 shows a cooling diagram with the locus of $t_\mathrm{cool} = t_\mathrm{ff}$ in the density-temperature plane:

Above the locus, cooling is effective, and the cloud can collapse. The tilted dashed lines are lines of constant gas mass. Now you see that even for solar metallicity (the $Z = Z_\odot$ line), clouds of gas masses larger that $\sim10^{13}\,M_\odot$ are unable to cool. This is the fundamental difference between a galaxy and a cluster. They are both overdensities that resisted the expansion, but the gas in the cluster is not able to cool and form stars.

Voids

When people talk about "voids", they usually refer not to the ICM — which is the hot and dilute gas between the galaxies of a cluster — but to the immense regions of even more dilute gas which are… well, void of galaxies. Take a look at this image (from here). Here, I marked clusters with green and voids with purple. I also marked some filaments with cyan.

The cosmic microwave background

As for your last question, the ICM doesn't interact much with the CMB, so they're not in thermodynamic equilibrium, and there's no conflict. A small fraction (<10%) of the CMB photons does interact. This doesn't change the state of the gas, but it tends to polarize a fraction of the CMB, and this we can use to study the ionization history of the Universe.

EDIT: I forgot about the Sunyaev-Zel'dovich effect, which is the interaction of hot electrons with CMB photons. Rob Jeffries discusses this in his answer.


What is Earth's average temperature as a whole?

Sealing the planet or using $E = mc^2$ isn't a practical means to measure the temperature. Here's how the average is usually determined:

Source: Tufts University - Global Warming - Heating by the greenhouse effect

  "... the average surface temperature of the planet is 288 degrees kelvin (15 degrees Celsius or 59 degrees Fahrenheit).".

Source: Space.com - What Is Earth's Average Temperature?

  "According to GISS, the global mean surface air temperature for that period was estimated to be 57 F (14 C). That would put the planet's average surface temperature in 2017 at 58.62 F (14.9 C).".


Is there anything special about our position in the galaxy thats necessary for life?

Certain regions of the galaxy are more favourable for life, others less. Life takes some time to develop, and it needs chemical elements heavier than helium, e.g. carbon or silicon.

To get sufficient time to develop, the solar system must not be disturbed too much by neigbouring stars. Therefore inner densely populated parts of the galaxy are less suitable than the periphery we are located. (The central star of the planetary system must not be too young, as well, to allow evolution.)

Heavier chemical chemical elements are sufficiently abundant mainly in parts of the galaxy with relatively young stars (population I). This kind of stars is found mainly in the thin disk close to the galactic plane.

Very old stars tend to expand, and may eventually destroy potential life on their planets, if they are surrounded by rocky planets at all.

More discussion on a potential galactic habitable zone on Wikipedia.


What are criteria to compare and evaluate different 12 inch truss Dobsonian telescopes?

The cooldown fan is purely to speed getting the primary mirror to the local ambient temperature. It is not used - and should not be left on during observation, as it'll undoubtedly be a vibration source.

That shouldn't be a significant factor in your final decision.  I would go with total weight, and perhaps adaptability to an equatorial mount for star-tracking.


JPL HORIZONS can do that.
With settings like these:

  Ephemeris Type : OBSERVER
  Target Body : Moon [Luna] [301]
  Observer Location : Geocentric [500]
  Time Span : Start=2019-09-12, Stop=2019-09-19, Step=1 d
  Table Settings : QUANTITIES=10,23,24

it returns results like these:

 Date__(UT)__HR:MN       Illu%    S-O-T /r    S-T-O
***************************************************
 2019-Sep-12 00:00      95.688 155.9731 /T  23.9640
 2019-Sep-13 00:00      98.620 166.4750 /T  13.4888
 2019-Sep-14 00:00      99.814 175.0444 /T   4.9422
 2019-Sep-15 00:00      99.231 169.9061 /L  10.0670
 2019-Sep-16 00:00      96.874 159.5743 /L  20.3723
 2019-Sep-17 00:00      92.791 148.7668 /L  31.1541
 2019-Sep-18 00:00      87.074 137.7510 /L  42.1469
 2019-Sep-19 00:00      79.858 126.5419 /L  53.3370

where Illu% is the illuminated fraction $k$, and S-T-O is the sun-target-observer phase angle $i$.
These quantities are related by
(Meeus 41.1)
$$ k = \frac{1 + \cos i}{2} $$

An observer location on the surface of the Earth instead of the geocenter gives slightly different results due to parallax.

Tanner's rendered images appear to be indexed by the difference between lunar and solar ecliptic longitude.
The sun-observer-target elongation angle S-O-T is a close approximation in most cases.
Where the /r column is /T for trailing, use that value directly.
Where /r is /L for leading, subtract S-O-T from 360°.

A single cycle of Moon images cannot capture libration effects as the NASA SVS images do (see SpaceBread's answer), so Tanner's images look a little different from the actual Moon.


Can someone calculate the age of the KOI-4878 star?

According to Wikipedia on KOI-4878.01 nobody did yet calculate the age of that object, but says exactly what you quoted:

its low metallicity and fairly high space velocity suggest that KOI-4878 is older than the Sun.

This supposidely relies on the following sources:

https://exoplanetarchive.ipac.caltech.edu/cgi-bin/DisplayOverview/nph-DisplayOverview?objname=KOI-4878&type=KEPLER_TARGET
https://www.drewexmachina.com/2019/09/05/koi-4878-01-keplers-most-earth-like-planet-candidate/

A search with the keyword KOI-4878.01 on different publication databases like arXiv or scholar.google did not help either.
Related

How do scientists determine the age of stars? focuses on the general question how to determine the age of a star.


Based off of the study that found the universe has a positive curvature, how big would the whole universe be?

Surprisingly small!

(To me, at least.)

The paper referred can be found on the arXiv as Di Valentino et al. (2019).

As is customary with Planck results, the exact values of the cosmological parameters depend on how much confidence you put in auxiliary data, such as baryonic acoustic oscillation data (from surveys such as the 6dF Galaxy Survey, SDSS, and BOSS), supernova data, and Big Bang nucleosynthesis models. In this answer, I'm going to assume that the inferred value of the curvature parameter is given by their "$\Lambda\mathrm{CDM}$$+$$\Omega_K\!$" model, with their 99% confidence limits of $-0.007>\Omega_K>-0.095$, i.e. I'll use
$$
\Omega_K = -0.0438^{+0.0368}_{-0.0512}.
$$
The dynamics of the Universe is given by the Friedmann equations, which can be rearranged like this:
$$
\Omega_K = -\frac{kc^2}{R_0^2\,a(t)^2 H(t)^2},
$$
where $k=+1$ for a closed universe, $a\equiv1$ today, $H=H_0$ is the Hubble constant today, and $R_0$ is the radius of curvature, and $c$ is the speed of light. The value of $H_0$ depends somewhat on the assumed value of $\Omega_K$; here I'll use $H_0 \sim 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$.

I this case, we get that the radius of the Universe is roughly
$$
R_0 \sim 67_{-21}^{+100}\,\mathrm{billion\,light\text{-}years}.\qquad (99\%\,\mathrm{C.L.})
$$

Accepting the "mainstream" cosmological parameter the observable Universe has a radius of $46.3\,\mathrm{Glyr}$. This result will change somewhat in the case of a closed Universe, but I can't seem to identify the author's preferred set of cosmological parameters. If we just use this value nonetheless, it means that we are currently able to observe a volume fraction of
$$
f = \frac{V_\mathrm{obs}}{V_\mathrm{tot}} \simeq \left(\frac{46.3}{67_{-21}^{+100}}\right)^3 \sim 33_{-31}^{+67}\%
$$
of the total Universe, i.e. we see somewhere between a few percent and everything, but most likely "a third".


How bright would Alpha Centauri appear (in apparent Magnitude) if it were 1 light year away?

The change in magnitude is given by:
$$m_2-m_1=-2.5\ \log_{10}(B)$$

Where B is the change in brightness.

Alpha Centauri is about 4.4ly so inverse square says it would be 19.4x brighter at 1ly, so a magnitude differential of -3.2.

Its apparent magnitude would therefore go from -0.27 to -3.47. This is brighter than Sirius (-1.46) but not close to the Sun (-26.7) nor Venus. Venus doesn't cast a shadow, so neither would Alpha Centauri at 1ly.

On gravity, Jupiter is roughly one thousandth the mass of Alpha Centauri but would be about 63,000x closer than the star at 1ly. It would therefore have a gravitational effect on the Sun of:
$$63\,000^2/1000=\mathrm{4\ million\ times}$$
that of Alpha Centauri at 1ly!

Sounds like this works for your story.

Data brightness sources: https://www.space.com/21640-star-luminosity-and-magnitude.html

Mass and distance sources: https://en.wikipedia.org


Is this render of a ringed planet's shadow accurate?

This might happen if you model the ring as opaque but of very little thickness and in the same plane as the vector between the planet and the sun  Seen edge on from the equator, the ring would block none of the sun's light, but from a little north or south of the equator, the ring would block more of one hemisphere of the sun.

So I would check first if your geometry has things lined up too perfectly before considering a fault in the rendering engine.


What is beyond the observable universe?

There's no way to know because we can't see it.  No information from the unobservable part of the universe can ever get to us.  If we can't see it, we can't know what's there.

Over time, the observable universe grows and we see more of the universe, though this is actually a pretty complicated due to the rate of expansion and I'd rather not open that can of worms, but for further reading, see this question or this one.

But avoiding the complexities of dark energy expansion, we have a pretty good idea that the unobservable universe is very much like what we can see.   This makes logical sense because it likely formed the same way at roughly the same time and out of the same stuff.   So, we expect it's just more galaxies and more space.

In addition to this, if there was a point where the galaxies stopped being there at a theoretical 'edge', not far from the edge of the observable universe, we'd expect to see that lack of gravitation resulting in the galaxies moving away less fast and that's not observed.   The galaxies appear to be moving away from us equally fast in all directions, so we have good reason to believe that the universe is uniform out considerably beyond what we can observe.   It's at least several times larger than what we can see.

If we think of the big bang at happening at a point, which is often depicted in sciency-tv shows, then we shouldn't expect uniformity due to gravity and faster moving parts spreading out more, but what we see when we observe the universe is uniformity.   That means that if it did explode from a singular point, it must be so large that we can't see any variation.

The more popular approach is that the big bang didn't happen at a point but it happened everywhere.   Google Everywhere Stretch for some explanations of this.

It's worth noting that not all shapes (or manifolds) have centers.   The surface of the Earth has no center on the surface.   So even if we could see the entire universe, finding it's center might still not be possible.

Determining the center of the observable universe, however, is easy.   We can see in all directions just about equally so we're the center.  But we can't get a good grasp on the size and shape of the entire universe, or just universe is more correct.   Since we can't see all of it, finding the center is impossible.

  The bang must have happened in the center of the universe.

Keep in mind that the big bang isn't well understood.  It's what physicists call a singularity which means, the math breaks down.     The point exploding outwards is often represented, but shouldn't be considered accurate.

Mathematical approaches can come up with a variety of possibilities like hyperplanes intersecting or 4 or 5 dimensional spheres.   (Stephen Hawking had a 5 dimensional sphere argument where no big bang happened, it's just all a circle - that's in one of his books, I think blackholes and baby universes).   He was able to create a universe without a big bang.   That the universe is infinite and the big bang happened across infinite space is also a popular idea today.    The singular point that blew up into the universe isn't big right now, though that viewpoint still applies to the observable universe, but it's fallen out of favor for the entire universe.

My two cents anyway, as a layman.


What you have to keep in mind is that the Grand Tack was devised as a mechanism to generate a truncated planetesimal disc between 0.4 and ~2 AU during the protoplanetary disc phase, in order to explain the small masses of Mars and Mercury (Walsh et al., 2011).
The smallness of Mars and Mercury might be explained differently by pebble accretion (Lambrechts et al., 2019), which wasn't fully worked out yet during that time.
A direct test of whether Jupiter/Saturn indeed migrated during their growth phases, is their Trojan number asymmetry, which is the basis of the paper by Pirani et al., which you quoted. They find that the Trojan-to-Greek numbers should be equal only in the case of in-situ formation, but they are not.
The observed asymmetry is consistent with migration inwards during Jupiter's growth. However, from their work, no clear statements can be made on

where Jupiter's core formed;
how fast the inward migration proceeded; or
whether there was a later outward migration.

Furthermore, the migration rates required for the Grand Tack to operate might be wrong. Recent work coming from people working on migration rates (McNally et al., 2019, Lega et al., 2020) seem to indicate that migration in low-viscosity discs works very differently than what was envisioned in earlier decades. Outwards migration, or stopping, is then easier for giant planets.
Summarizing, many of the puzzle pieces on why the Grand Tack was envisioned and how it worked are currently under active investigation. This is not an issue I expect to be solved soon, as certainty into any of those models would additionally require Trojan asymmetry numbers for Neptune and Uranus, and those are very challenging.


At that time (20:50), and that location (Houston, Tx), Mars was closer to the horizon.
You can see that in the sky-chart provided by heavens-above.com (attached below).

I cannot tell what's the other object you describe, but if you are not sure whether something is a planet or a star, the planets doesn't twinkle whether the star does.


What would happen if an astronaut orbiting Earth was exposed to radiation?

I guess that depends on how much radiation they're getting. I should also point out that astronauts are not 100% protected from radiation, as discussed in this question. One of the most common affects on actual astronauts is the development of cataracts. They can also receive damage to their nervous system over time, among other effects.

But it appears that you're asking what would happen to a person who was exposed to a huge amount of radiation, such as if they went into space with a suit that did little to protect from radiation. I think to answer that, we just have to look at cases where humans have actually been exposed to huge amounts of radiation.

One interesting example that comes to mind is Anatoli Bugorski who was a Russian particle physicist who accidentally had his head inside a particle accelerator when it was on. His head was subjected to huge amounts of radiation. The results of this are described as

  The left half of Bugorski's face swelled up beyond recognition and,
  over the next several days, started peeling off, revealing the path
  that the proton beam (moving near the speed of light) had burned
  through parts of his face, his bone and the brain tissue underneath.
  As it was believed that he had received far in excess of a fatal dose
  of radiation, Bugorski was taken to a clinic in Moscow where the
  doctors could observe his expected demise. However, Bugorski survived
  and even completed his Ph.D. There was virtually no damage to his
  intellectual capacity, but the fatigue of mental work increased
  markedly. Bugorski completely lost hearing in the left ear and only
  a constant, unpleasant internal noise remained. The left half of his
  face was paralyzed due to the destruction of nerves.1 He was able to
  function well, except for the fact that he had occasional complex
  partial seizures and rare tonic-clonic seizures.

Now imagine what could happen if a person were to be exposed in a similar fashion to highly energetic particles in space, but throughout their body and not just their face. I expect that a person could not survive such an exposure and that their body would be destroyed. Even if they survived, it is highly likely they would live with crippling medical conditions and very likely develop cancer later in life.


How feasible would it be to get to a planet orbiting directly opposite Earth?

Venus has roughly the same gravitational potential as earth and the Venera probes successfully and softly landed there. Similarly, the Apollo space craft landed safely on earth on return. In both cases aerodynamic friction with the atmosphere can be used to get rid of excess velocity in order to enter orbit and eventually land. Thus landing will work cheaper, if there is atmosphere. But at the expense of much more fuel also without (see moon landing again... you'd need stronger descent rockets than Apollo for an exclusively rocket - driven landing)

Getting there is not a big deal either. Look at the Stereo mission of two satellites which trail earth's orbit more and more to get a better understanding of the sun, gaining more and more separation from earth while (nearly) staying at 1 AU: You get to another position in earth's orbit by just lifting your orbit slightly and waiting for earth to go and the other to catch up (or go slightly towards the sun and be faster yourself).


Units in optical and radio spectral data cubes: Flux vs Brightness

The answer is partially historically and partially scientifically. Flux, i.e. energy of radiation integrated across a solid angle (steradian) is basically counting all photons coming from a solid angle, while specific intensity (units of specific intensity are $erg/s/cm^2/Hz/sr$), is the energy per frequency along a line of sight! A surface brightness is flux divided by solid angle, i.e. the mean intensity within this solid angle. This evenly smears out all photons in this solid angle.

As a consequence, flux is the preferred unit if one observes a point source, i.e. a source with a solid angle smaller than the angular resolution of the detector. For point sources, the beam size does not matter. If we observe the same point size with two instruments with different beam sizes, they both detect the same flux because they just count all photons coming from the source. A larger beam does not change this quantity. On the other hand, the brightness (mean intensity) of the object goes down with a larger beam size because we average the flux over a larger area.

If you observe an extended (larger than the beam!) object, e.g. a molecular cloud in radio, then surface brightness is the preferred quantity. Assume it is equally bright across its surface, then the surface brightness becomes independent of the beam size because the flux is proportional to the beam size. A telescope with a larger beam (smaller dish) observes a larger flux (always assuming the object is even larger) compared to a larger telescope with a smaller beam. Comparing the two measurements becomes difficult. The surface brightness of both observations is the same though.

Since optical astronomy was primarily observing stars, i.e. point sources, they used flux. Radio astronomers observed the interstellar medium, i.e. extended clouds of gas and dust, so they used surface brightness.


What does the Sun look like from the heliopause?

0.004° converts to about 14.4 arc-seconds.
That's within the range typical for Mars; 5-25 arc-seconds.
You'll see the sun as a point with the naked eye.
Heliopause is about 14 billion miles out. Absolute magnitude of sun is 4.83.
14 billion miles = 0.0007297pc.
Apparent magnitude will be -15.8543. Magnitude of the full moon, from earth, is about −12.74
So you'll see a point source of light, brighter than the full moon.


What would happen if Jupiter and Earth were at the same distance as the Moon is from Earth?

That's no good idea. Earth wouldn't necessarily fall into Jupiter in the short run, provided it orbits Jupiter fast enough (within about 1.7 days), and on a circular orbit, but we would risk to collide with Io, destroy it by tidal forces, or change its orbit heavily.
The other Galilean moons would get out of sync and change their orbits over time.

Tides would be severe on Earth, not just limited to oceans, but also for "solid" ground, as long as Earth isn't tidally locked. This would result in severe earth quakes and volcanism.

Our days would be dim due to the distance to the sun. After tidal locking of Earth and ejection/destruction of Io the tides as heat source would be lost, oceans would freeze, temperature would fall to about -160°C mean temperature. During the polar night oxygen would probably condense from the atmosphere and form lakes, may be even nitrogen. By this atmospheric pressure would drop.

Since Spock is smart enough to know these consequences in advance, he wouldn't do it.