We don't know. The Multiverse Theory is really just a hypothethis; sure, it makes sense, but it cannot be scientifically proven. We really have no reason to doubt that this is our only universe. Wormholes are just as hypothetical, since they would require infinite energy or negative energy to be created, neither of which we can get. Also, keep in mind that String Theory is, for the most part, unprovable. It cannot be confirmed through scientific experiments and is mostly guesswork.

Also, it will take trillions of years before the Universe gets too cold for life, and we will be long gone by then. However way you put it, it's unlikely that humans will ever reach other habitable planets before we go extinct, should our planet be destroyed. The closest potentially-habitable planet is nearly 480 lightyears from us – as in, it would take us 480 years to get there, if we travelled at the fastest speed in the Universe. We surely won't live to see the end of the Universe.


Expanding our observable universe suggestion

Unfortunately, the limit of the observable universe isn't just "photons from the big bang arriving to us now" - so we won't see anything past it because nothing existed before; anything outside that sphere didn't arrive yet. This limit is called the Hubble Sphere, and is only the "first stage" of unobservability - and indeed, as time passes what we can see will expand and we'll be able to see further.
But there's another limit - called the cosmic event horizon - which, to current knowledge, is quite unconquerable. It's connected with space expansion. Space expansion is a phenomenon of far objects "moving" further apart, with the nasty property of bypassing special relativity with its property of light speed being the same in all frames of reference (which would normally imply no matter how far a light source, the light would reach us sooner or later). Instead of classic motion - change of position in space over time - the "motion" due to space expansion is "production of more space" between the objects - they aren't actually in motion relative to each other (or that motion is irrelevant), it's the very definition of distance between them that is changed.
And as result of that, there is a region distant enough, that light from there - traveling at speed of light towards us - will never reach us as there will be always more space inserted between it and us, per unit of time, than it can cover. And so, the cosmic event horizon describes the actual ultimate sphere of the observable universe, not just "for now" but "forever, unless we discover some way to bypass it, but currently not even a hint of such bypass exists."


If the RA and Dec in question are geocentric, then they are relative to the ECI origin.
The difference between heliocentric and geocentric positions is significant for any solar system object; on a baseline of 1.0 au, a trans-Neptunian object at 40 au has a parallax of 1.4°.
However, for near-Earth objects, the observer's location on Earth is also significant.
An asteroid passing at 0.01 au has a parallax of 0.24° on a baseline of 6370 km.
The Minor Planet Center lists observations and generates ephemerides in topocentric coordinates.


Are the sub-spectral types (1,...9) based on temperature or spectral lines?

The spectral classes (O, B, A, F, G, K, M) and their 10 subtypes (0 to 9) were initially meant only as differentiators of spectral type. Annie Jump Cannon was the creator of this system. Through her work for/with Edward Pickering, she ended up classifying nearly a third of a million stars over a few decades. She (and many others) did not realize that this was actually a temperature scale - the classifications were based on the spectral lines that Cannon observed and sorted into categories. It wasn't until Cecilia Payne's work, and, most notably, her doctoral thesis, that anyone realized that the OBAFGKM system (and the sub-classes it came with) was actually a temperature scale.
Payne's work showed that the difference in spectral type was not due to differing amounts of elements, but mainly due to the temperatures of stars. This also led to the realization that stars were made of far more helium and hydrogen than other elements. However, the professor who reviewed her thesis, Henry Russell, dismissed her findings as impossible, since at the time, the implications went against much of contemporary knowledge. After a few years, Russell realized that Payne was right, and praised her work, crediting her with the discovery.
(Side note: there's an amazing episode of the TV series Cosmos - the new one with Neil DeGrasse Tyson - called "Sisters of the Sun" that dives into this story quite well. If you have the time, I'd definitely recommend it - it's a great watch.)


Why do radio telescopes, but not shorter wavelengths, have this Big Data problem?

Shouldn't it be the other way around, that shorter wavelength telescopes like those in IR and the visible and UV generate more data because they have higher resolution per unit of aperture?

This is a very interesting question!

My answer is no, not currently, but someday it may be the case.

Because we have big fast computer and fiber optic technology, we can generate images from arrays of antennas computationally. We can form images and even implement adaptive optics for wavefront correction at radio frequencies in software (rather than mechanical actuators correcting wavefronts by brute-force bending mirrors).

An order-of-magnitude description of ALMA is 100 dishes each one delivering 1 GHz of bandwidth, and SKA arrays will be far larger. While the Event Horizon Telescope literally records all raw data on boxes of hard drives that's not currently possible for ALMA and larger arrays, so only much smaller, heavily processed intermediate results files can be saved for later analysis and fine-tuning of calibrations and parameters.

For "shorter wavelength telescopes" in this context it's better to think of them as "higher frequency telescopes". Even a 100 nm wide band at say 800 nm (near IR where wavefront correction is working well represents 47,000 GHz! Even with conversion to a baseband frequency we don't have the ability to digitize signals at that rate, much less perform correlations of multiple signals.

So instead we let the light do the interferometry for us!

An optical telescope is just a big, simple, one-trick-pony optical computer. We design the surfaces to perform an integral of the electric field at the aperture. The center pixel receives the integral of all electric fields incident on the telescopes clear aperture without any phase bias. A pixel at one edge of the sensor receives the integral of the same electric fields times a phase function that increases linearly from one side to the other.

One day if and/or when detection and computing speeds and parallelisms are big enough, I am sure an electronic optical computer will probably be built.

For more on optical telescopes versus computational interferometry see answers to

Would “layered” radio interferometry work?
Could mirrors be replaced with CCDs?
How would a Fast Fourier Transform Telescope work without a mirror, dish or lenses?


Could an Earth-like magnetic field protect life at exoplanet Proxima Centauri b?

According to Garcia-Sage et al. (2017) "On the Magnetic Protection of the Atmosphere of Proxima Centauri b", who assume that Proxima b has a magnetic field strength identical to the Earth, the situation does not look promising:

  With an O+ and H+ escape at 1600 km altitude of 4.2×109 cm−2 s−1 as found for thermospheric densities of 3×T,
  the total mass-loss rate is 4.3×105 g s−1 over the dayside
  hemisphere. Using a total atmospheric mass of 5×1018 kg
  (Trenberth & Smith 2005) results in a loss of the entire mass of
  the atmosphere over 365 million years.

(T is the thermosphere temperature of the Earth)

They go on to note that Proxima would have had higher EUV and X-ray emission in the past, which would have made things worse. Ongoing outgassing of volatiles might be possible to create habitable conditions but the resultant situation would be very different from the Earth.


Why did the Comet Shoemaker-Levy-9 fragments cause such large explosions on Jupiter?

Shoemaker Levy 9 was estimated to have released kinetic energy equivalent to 300 gigatons of TNT. That is $1.255 \times 10^{21} {\rm J}$. This release through friction and compression was sufficient to heat the atmosphere to 4000K at first. That in turn should probably be enough to break down the molecular hydrogen and methane, releasing further energy. But that's only my guess. The heating from the impact alone should make a huge fireball.
Sources:
Fragment Size and Mass Estimates
Galileo observations


Why don't Australia/Russia have large optical telescopes?

Optical telescope sites ideally need to be somewhat remote (to avoid light pollution) yet still accessible for construction, engineering, and observing; high in altitude; dry; with exceptional "seeing" (stable atmospheric conditions); and clear weather. Neither Australia nor Russia really have any sites meeting all those criteria: their mountains tend to have some combination of too low altitude, poor seeing, and/or bad weather. (This is also why there aren't any really big, modern telescopes in Europe.)

This site has some discussion of the different criteria that go into choosing the best sites for new, large telescopes.


If an object 1 billion light years away emits light, does it take more than 1 billion years to reach us because of the expansion of the universe?

Yes, in the time it takes light — or, in this case, gravitational waves (GWs) from the black hole merger event GW190521 — to travel from a source to an observer, the Universe expands, thus increasing the distance further.
Various distance terms
In the following, "$\mathrm{Glyr}$" means a distance of a billion lightyears, while "$\mathrm{Gyr}$" means a time of a billion years.
There is a slight confusion, I think, in the quoted distance of $17\,\mathrm{Glyr}$ (Abbott et al. 2020): This is the so-called luminosity distance, which is the distance that satisfies the usual inverse-square law. This is not the same distance that you would measure if you were to freeze time and lay out meter sticks. This physical distance is smaller, only $9.5\,\mathrm{Glyr}$.
These values correspond to a redshift of $z=0.82$. That is, if GW190521 was luminous, its light would be redshifted by a factor $(1+z)=1.82$. In fact, in this case an electromagnetic counterpart was reported, albeit not with a measured redshift (Graham et al. 2020)
The time it took the GWs to travel to us is called the lookback time; it is the quoted $7\,\mathrm{Gyr}$. When GW190521 emitted the GWs we detect today, it was closer to us by a factor $(1+z)$. That is, its physical distance was only $5\,\mathrm{Glyr}$.
For a flat universe (which our Universe is to high precision) this is equal to the so-called angular diameter distance, called so because it is the distance that satisfies the usual relation between distance $d$, size $D$, and angle $\theta$, namely $\theta = D/d$.
Relation between lookback time and distance
In every-day life, all these distance measures are the same, and in the Universe, for small distances they also coincide. But because of the expansion of the Universe, and because the Universe's components (matter, radiation, and dark energy) influence its geometry, as the distance of an object increases they become increasingly different.
You can find the equations here, or use a cosmological calculator such as Ned Wright's (as commented by Alchemista). Alternatively, you can calculate them in Python using the module astropy like this:
>>> from astropy.cosmology import Planck15
>>> from astropy import units as u
>>> from astropy.cosmology import z_at_value

>>> dL = 5.3 * u.Gpc                                 # Lum. dist. in giga-parsec quoted in Abbott+ 20
>>> z  = z_at_value(Planck15.luminosity_distance,dL) # Corresponding redshift
>>> print(z)
0.8174368585313242
>>> print(Planck15.lookback_time(z))
<Quantity 7.11401487 Gyr>
>>> print(dL.to(u.Glyr))                             # Convert parsec to lightyears
<Quantity 17.28628801 Glyr>
>>> print(Planck15.comoving_distance(z).to(u.Glyr))  # Comoving dist. is equal to phys. dist. today
<Quantity 9.53452323 Glyr>

I used this to plot the current distance to GW190521 and other objects as a function of lookback time:

The answer to your title question
To answer the question in your title requires us to define exactly what we mean:

Light from an object which has a physical distance of $1\,\mathrm{Glyr}$ now, is redshifted by $z = 0.070$, its light has been traveling for $0.97\,\mathrm{Gyr}$, and it was $0.93\,\mathrm{Glyr}$ away from us when it emitted the light we see today.
Light from an object that was $1\,\mathrm{Glyr}$ away when it was emitted, traveled $1.03\,\mathrm{Gyr}$ before reaching us with a redshift of $z = 0.076$, and the object is now $1.076\,\mathrm{Glyr}$ away.

As you can see, the difference is not very large, but as you go to higher redshifts, it increases. The hitherto most distant observed galaxy, GN-z11, has a redshift of $z=11.09$. It was only $2.7\,\mathrm{Glyr}$ from us when it emitted the light we see today, but in the $13.4\,\mathrm{Gyr}$ it took the light to reach us (most of the age of the Universe), GN-z11 moved out to a current distance of $32.2\,\mathrm{Glyr}$!


It is a common point of confusion, but an infinite universe can still have been at a unique singularity in space-time at a finite time in the past.

This singularity is not a "point in space", since every point in the universe participated in the big-bang. It is merely a time at which the scale factor, which multiplies the separation between all spatial points in the universe,  was zero.

So in cosmology governed by the Friedmann equation, the age of the universe thus calculated is the same whether you are talking about the "visible" universe or an infinite universe. The light travel time really doesn't come into this, because different parts of even a very large universe can be causally separated yet still have originated at the same singularity.


If an asteroid on an inner orbit catches up with a planet on an outer orbit, and passes close enough to the planet to be strongly influenced by the planet's gravity, it can whip around the planet ahead of the planet and then move in and orbit backwards (retrograde) farther from the sun than the planet.
So the asteroid might be captured by the planet if its retrograde speed is slow enough, and then move in a backwards or retrograde orbit around the planet.  A lot of small asteroid sized moons of the giant planets have retrograde orbits indicating they were captured that way - also Triton, the large moon of Neptune.
I just counted abut 119 natural satelelites with retrograde orbits in our solar system.
https://en.wikipedia.org/wiki/List_of_natural_satellites[1]
And the the asteroid's retrograde speed after passing around the planet is fast enough, it won't be captured by the planet but will continue to orbit the Sun, but in a retrograde orbit.
In fact, a very tiny percentage of all known asteroids have retrograde orbits, only 82 out of more than 726,000.
https://phys.org/news/2017-06-stable-retrograde-orbit-bee-zed-asteroid.html[2]


Why are asteroids with circular orbits rare?

Orbital eccentricities cannot be normally distributed around zero, since the minimum eccentricity is zero.
Thus, in the same way that the Maxwell-Boltzmann distribution for particle speeds is zero at zero speed and peaks at finite speed, then no asteroids have exactly zero eccentricity and we would expect a peak at higher eccentricities.
The analogy can be extended further by asking what distribution might be expected for a "thermal distribution" of eccentricities - a distribution where the energies of the orbits follow the Boltzmann distribution and have been allowed to randomise by some unspecified dynamical or formation processes (e.g. Geller et al.
2019 ). In those case the density of eccentricities should go as
$$n(e) \propto e^2$$
and there should be more high eccentricity objects than low eccentricity objects.
Of course asteroids are not expected to be represented by this random distribution, since their dynamical histories likely have them being perturbed away from very low eccentricity orbits induced by the drag of a gas/dust disc in the early solar system.
Nevertheless, any initial perturbation moves them away from zero eccentricity and subsequent random perturbations might be expected to make them head towards the thermal distribution. I would guess that objects in the top end of the thermal distribution must also then just get selected out - falling into the Sun, hitting planets or getting ejected out of the main asteroid belt.


Diameter of the moon from the diameter seen in the telescope

A telescope wouldn't have a single lens, I'm sure you know. If you were to use the setup in your diagram, focusing the image of the moon on a flat surface, you could find its diameter with the method provided. The distances f in the diagram are the focal length of the lens, where a very distant object will be in sharp focus, and that's where you'd end up placing the flat surface.


How did the moon's orbit become eccentric?

First, your claim that the moon is more eccentric isn't entirely true (lets ignore Pluto):

(Source)

Mercury .205
Mars .094
Saturn .057
Moon .055
Jupiter .049
Uranus .046
Earth .017
Neptune .011
Venus .007

So, it's really not abnormal, though it might be unusually eccentric for a non captured moon, but that's may be due to it being as close to the sun as it is.

Also, our solar system might not be exactly normal.  It's fairly common for solar systems to have a gas giant close to the sun (explained here).  And there are some very odd solar systems - same link.


What is the composition of the Universe in terms of Types Of Stars?

First of all, there is a misconception in your question that red dwarfs and main sequence stars are two completely different kinds of stars.

Red dwarfs are a type of main sequence star. If a star is in the main sequence that just means that it is at a certain stage of its evolution.
On the main sequence, stars can be further categorized based on their color into brown dwarfs, red dwarfs, sun-like stars, blue stars,...

In a comment, you state that the local distribution of stars is enough for your purpose.

To get the local distribution of star types, you can use the Gaia data, which gives the distance, luminosity and color of many stars in our galaxy. From this, you can plot an HR diagram, where the position of a star indicates its type..

This diagram isn't quite correct, because it uses a cut in magnitude to select stars. This means that fainter stars are under-sampled. But if you only took stars up to a certain distance, you could get something more representative.

In this article, you have a breakdown of stars based on their intrinsic luminosity and their spectral class. Most stars (76%) are in the main sequence, 24% are white dwarfs, and the rest (0.6%) are on the giant branch.

The stellar distribution of our Milky Way seems to be pretty typical for this type of galaxy, but there are other types of galaxies where the stellar distribution is rather different. When a galaxy is rather young, there is a lot of gas available for new stars to form, so you get a lot of big bright blue stars. But these big stars are short-lived, so in elliptical galaxies where the stellar population is older, you get more small red stars.

Finding the distribution of star types in these different types of galaxies is the matter of stellar population synthesis. You start out with an initial mass function (IMF) that tells you the number of stars there are with any given mass in one galaxy. This gives you a snapshot of the galaxy at time t=0. Then you use stellar evolution models to see how that initial population evolves over time.

De Marchi et al gives the IMF for a cluster of galaxies (rather than for individual galaxies). This might be as close to a "typical" distribution as you can get.

Further reading


Would you get the same planet-discovering data for our own system at the same distance we're getting from Earth?

There is a lot of detail in my answer to If Alpha Centauri A's solar system exactly mirrored our own, what would we be able to detect? I showed that current Doppler techniques and transit techniques would be unlikely to detect the Earth at the distance of Alpha Cen.

It does not get any easier if you are further away. Stars are point sources at 1pc or 200pc, so these spatially unresolved techniques just care what signal level is being received - further away is worse.

In the case of Doppler techniques we need an improvement in radial velocity precision and stability of a factor of a few.

With transits, Kepler was almost there. So another Kepler with a slightly bigger telescope that observed the Sun for say 5 years would do the trick (if the orbital inclination was about 90 degrees). The reason Kepler 22b was just detected is because it has twice the Earth's radius so eclipses 4 times the area and gives a deeper transit signal for a Sun-like star. It is this contrast that is the key to transit detectability.

Maybe the missing piece of information here is that there is a floor level to photometric precision that can't be bettered even if the star is close and bright. So long as you can reach this level in a time shorter than any transit, then the distance is irrelevant, all that matters is the fractional change in the brightness during transit and the number of transits you observe. The first of these grows as the square of the planet radius (for a fixed stellar radius), the second just depends on the length of your mission divided by the orbital period.

The bottom line here is that the Earth in an Earth-like orbit around a Sun-like star would not quite be detectable by Kepler in a Kepler-length mission, even if it were very close and nor if it were further away.


Do asteroids have a gravitational field?

By definition, gravity is a result of mass. Any body with a non-zero mass (even atoms) will have a gravitational field associated with it. The higher the mass the stronger will be the field. This is basic of classical mechanics. Until we reach quantum scale where the gravitational force is dominated by other 3 forces and the gravitational field becomes irrelevent.

When it comes to graitational field of asteroids, it exists, but is very weak. However over a course of few million years these small asteroids combine together to form large masses of bodies that we now call planets. That is one of the prominent theory of Solar system formation, where the gravity of small dust particles from first our generation disintegrated star over a course of time accumulated to give us what we now know as our Solar system.
Think of it like this, every planet that you see now would once have been an asteroid at some point during its evolution.

Another proof to support this is the presence of numerous binary asteroids that orbit each other around a common center of mass, which requires gravitational attraction.


How do you calculate the effects of precession on elliptical orbits?

A good starting point would be <insert name of some scientist from long ago> planetary equations of motion. For example, there are Lagrange's planetary equations (sometimes called the Lagrange-Laplace planetary equations), Gauss' planetary equations, Delaunay's planetary equations, Hill's planetary equations, and several more. The common theme amongst these various planetary equations is that they yield the time derivatives of various orbital elements as a function of the partial derivatives of the perturbing force / perturbing potential with respect to some generalized position.

In general, the only words that can describe the result of this process at first is "hot mess." A hot mess did not deter those brilliant minds of old. Via various simplifying assumptions and long term time averaging, they came up with fairly simple descriptions of, for example, $\left \langle \frac{d\omega}{dt} \right\rangle$ (apsidal precession) and $\left \langle \frac{d\Omega}{dt} \right\rangle$ (planar precession). You can see some of this in the cited 1900 work by Hill below.

While these techniques are old, these planetary equations are still used today. That sometimes you do get a "hot mess" is okay now that we have computers. People are using planetary equations coupled with geometric integration techniques to yield integrators that are fast, accurate, stable, and conserve angular momentum and energy over long spans of time. (Normally, you can't have all of these. You're lucky if you get just two or three.) Another nice feature of these planetary equations is that they let you see features such as resonances that are otherwise obscured by the truly "hot mess" of the cartesian equations of motion.

Selected reference material, sorted by date:

Hill (1900), "On the Extension of Delaunay's Method in the Lunar Theory to the General Problem of Planetary Motion," Transactions of the American Mathematical Society, 1.2:205-242.

Vallado (1997 and later), "Fundamentals of Astrodynamics and Applications", various publishers.
Other than the hole it punches through your wallet, you can't go wrong with this book.

Efroimsky (2002), "Equations for the keplerian elements: hidden symmetry," Institute for Mathematics and its Applications

Efroimsky and Goldreich (2003), "Gauge symmetry of the N-body problem in the Hamilton–Jacobi approach." Journal of Mathematical Physics, 44.12:5958-5977.

Wyatt (2006-2009), Graduate lecture course on planetary systems, Institute of Astronomy, Cambridge.
The results of the Lagrange planetary equations are presented on slide 6.

Ketchum et al. (2013), "Mean Motion Resonances in Exoplanet Systems: An Investigation into Nodding Behavior." The Astrophysical Journal 762.2.


By convention, astronomy uses the Julian Year for the computation of a light year:

  Although there are several different kinds of year, the IAU regards a year as a Julian year of 365.25 days (31.5576 million seconds) unless otherwise specified.

Wikipedia gives the length as

  $31 557 600 s \times 299 792 458 m/s = 9 460 730 472 580 800 m$ (exactly)

The reason that it's exact (and not subject to experimental error) is that the metre itself is defined in terms of the speed of light, so the quantities are fixed by definition.


How would humans with appropriate equipment navigate the surface of Saturn's moon Titan on foot?

I wonder if this would be better for worldbuilding or space travel, but I can touch on the basics.

You'd need oxygen and protection from the cold, but a space suit on Titan would be less restrictive, not more restrictive than the ones used on the Moon.   The pressure isn't too much, so that actually helps and cold temperature is much easier to protect from than a vacuum.    They'd probably need some kind of full body wrap to avoid exposure and frostbite, and goggles but that's not terrible hard.   Electric heating or just insulation would both be options.

But it's safe to assume that even if they carry an electric heating device, their suits would still be less bulky and less restrictive than the ones the astronauts wore on the Moon.

Hop or Walk?

Very likely hop.   I'm going to borrow from this Quora answer, because I agree with it and it makes logical sense.   Walking is kind of a controlled fall.  Walking in 0.14 g, means your body doesn't fall forward in pace with every step, so walking would be weird and hoping would be easier.

Even on Mars, astronauts might hop due to the lower gravity.  If I was to make a wild guess, for comfortable walking I think it would begin somewhere around 50% or 60% Earth gravity.   Less then that, the slower rate of falling with each step and the lighter body weight, I think hopping is the way to go.

The human brain and inner ear begins to stop working somewhere around lunar gravity.   There are videos of astronauts on the moon falling forward and looking clumsy, but they were very fit individuals.  In low enough gravity the human brain loses a sense of which way is down.  (I read that, I'll see if I can find a link).

I would think, if it was for an extended time there, astronauts on Titan would wear lead boots, to help with their sense of which way is down and also for helping maintain musculature, but that's just speculation on my part.  There are other ways to maintain musculature.

Titan's surface might/probably gets soggy as well and humans aren't buoyant in methane lakes the way they are in water, (so maybe the lead boots aren't a good idea).   Falling into a lake or falling through ice into a like might happen on titan, but they'd be wearing full body wraps and have oxygen, so falling into a methane lake wouldn't necessarily mean death, they could probably get fished out and rescued.

Long story short, hopping is likely the way to go, and perhaps, either some kind of snow-shoe feature or perhaps an ice/jagged edges feature if parts of Titan's surface are slippery.   I'm far from an expert in such things, this is just my best guess.


Breakthrough Starshot; could we detect similar probes flying by Earth?

I'm not in a position to toss real numbers around at the moment, but here's a thought experiment:

Iridium satellites reflect sunlight toward Earth's surface, producing bright flares, up to magnitude -8. That's about 1 million times brighter than the faintest light source a fully dark-adapted eye can perceive -- a five-magnitude difference in brightness is a factor of 100, and under really dark skies humans with excellent vision can see down to magnitude 7.

I think these reflections are coming off a panel of about 3 x 1.5 meters, so it's comparable in size to your probe.

These satellites orbit about 800km above Earth's surface. How far away could one be, and still produce a visible reflection? Brightness falls off as the inverse square of distance, so the greatest distance at which you could barely see a reflection would be 1,000 times greater, or 800,000km.

Light travels at 300,000km/s. A probe traveling at 0.2c would cover 60,000km/s. Traveling 800,000km would take about 13 seconds.

That means that something the size of an Iridium satellite, reflecting sunlight directly toward an observer during its entire passage, could be visible for up to almost half a minute -- blue-shifted for the first part of that, and red-shifted for the last part. (Unless it's quite far away at closest approach, which would make it dimmer, the transition from "basically heading straight for you" to "basically heading straight away from you" would take place in the literal blink of an eye, a matter of milliseconds.)

Now, this is probably an upper bound -- probes won't have the same shape and reflectivity as an Iridium satellite, and won't be oriented to reflect directly toward an observer during their entire transit. But, somewhat to my surprise, it looks like you might be able to see one approaching and receding.

On the other hand, it's also possible that an object making 0.2c might have a visible interaction with the solar wind. (That requires more numbers than I'm able to run at the moment, but having thought on it a little bit, it seems unlikely.)


Which came first: Galaxies  Stars  Planets?

The structure we see in the Universe has formed from the gravitational collapse of the matter that was once an almost smooth density field of gas ("baryons") and dark matter$^1$. The word "almost" is important here, for if it had been completely — or even non-completely but much more — smooth, then the collapse would not have had the time to happen before the expansion of space has diluted the matter enough to prevent any collapse, and we would never had come into existence.
That is, the density field was slightly clumpy, and these clumps — or overdensities — existed on all scales.
But calculating which clump sizes collapse first — stellar-sized, galaxy-sized, cluster-sized, etc. — is far from trivial. Analytical attempts has to make several approximations, but can still make quite meaningful predictions which have subsequently been backed up and refined by numerical simulations, and by now, although many loose ends still exist, we have a rather good picture of structure formation:
Gravitational collapse
An overdensity is denoted by
$\delta\equiv\rho/\langle\rho\rangle - 1$, where $\rho$ is the density of the overdensity and $\langle\rho\rangle$ is the average density.
The evolution of overdensities under the force of gravity can be calculated exactly for $\delta \ll 1$ using linear perturbation theory, but when $\delta$ becomes of order unity, the non-linear regime is entered, and severe approximations must be made, so one turns instead to numerical simulations. It turns out that if $\delta\gtrsim1.68$ (i.e. if a region in space has a density which is 2.68 times the ambient density), it will collapse. The answer to your question is then given by what size of clumps first reach $\delta\gtrsim1.68$.
First clumps
Primordial quantum fluctuations$^2$ grew in size during the (admittedly still pretty blurry) inflation, a fraction of a second after Big Bang. In the young Universe dark energy was negligible, and the dynamics of the Universe was dominated by matter. Because dark matter comprises $\sim5/6$ of the total amount, we can initially neglect the presence of gas, but when the density becomes very high, gas pressure builds up and counteracts the collapse.
The overdensities amplified as matter started to collapse. It turns out that the density fluctuations are larger for smaller scales, so the smaller the clump, the sooner it will collapse. This results in the so-called "bottom-up" formation of structure, which is in contrast to what was originally thought; namely that galaxies formed "top-down" in a monolithic collapse (Eggen, Lynden-Bell, & Sandage 1962).
However, this approach neglects both gas and the motion of dark matter particles (treating it as so-called cold dark matter). Taking into account the effects of this puts a lower threshold for the masses of the structures of $\sim10^5$-$10^6M_\odot$ (e.g. Naoz et al. 2006; Yoshida 2009). Hence the first structures that formed are believed to be minihalos roughly the mass of globular clusters.
Stars, galaxies, and clusters
Stars consist of collapsed gas and almost no dark matter, and the formation of a star thus needs the theory of hydrodynamics rather than just gravity. In order for matter to collapse to such dense structures as stars, it must get rid of much of its energy. This is not possible for the collisionless dark matter (at least in the normal sense, but this post), but gas, which can collide and cool by radiating, is able to do so$^3$.
Through radiative cooling, the minihalos thus fragmented further into first gas clouds and then the first stars when the Universe was a few 100 million years old. Subsequently conglomerations of stars merged into galaxies, which eventually formed sheets, filaments and clusters of galaxies.
Dust
The first stars were formed out of pure hydrogen and helium (and small amounts of lithium), and the material for making planets did not exist. But when these stars — which were very massive — exploded as supernovae and polluted the interstellar medium with metals$^4$ and stardust$^5$, the formation of less massive stars became possible$^6$.
Planets
Dust particles stick together to form pebbles, rocks, and planets. Planet formation is probably not possible for too massive stars$^7$, but with the formation of small stars this became possible.
Timeline
We can make the following timeline:

Minihalos
Gas clouds
Stars
Galaxies
Stardust
Planets

However, note that the merging of minihalos to larger "galaxies" (without a significant amount of stars) also may happen at earlier epochs, and in particular that star formation (and dust and planets) is a continuous process which still takes place today (although the bulk of the star formation took place when the Universe approximately 3–6 billion years old (Madau et al. 1998)).

$^1$And radiation, which actually dominated the energy density until the Universe was $\sim50\,000$ years old.
$^2$The term primordial quantum fluctuations might be the coolest term in physics.
$^3$There are hypotheses of so-called dark stars, but that's beyond the scope of this text (primarily because I don't understand them).
$^4$For an astronomer, the term metal means any element heavier than helium. It's easier this way.
$^5$Astronomers just call it "dust", and use only the term stardust when talking to non-astronomers, in order for them not to fall asleep.
$^6$The reason is that with the many possible electronic transitions of metals, the gas has more ways to cool, and hence may collapse at earlier times, before the proto-star clump reaches high masses.
$^7$Because the high radiation pressure of very massive stars blows away and/or destroys the dust (I think, but this is not my expertise).


What is the actual resolution of Event Horizon Telescope?

The resolution of an interferometer at a wavelength $\lambda$ is $\theta\sim\lambda/b$, where $b$ is the longest baseline (distance between two telescopes) in the array. According to the first of this week's six papers, with observations at $\lambda=1.3\text{ mm}$, the longest baseline was $b\approx10700\text{ km}$, giving us $\theta\sim25.1\mu\text{as}$. The team expects this limit to be lowered even further with the addition of more observatories.

The picture, of course, seems much finer. The above limit can be lowered a bit by algorithms like CLEAN or regularized maximum likelihood; this brings the resolution down to about $20\mu\text{as}$. This still doesn't explain why the image seems even finer than that figure; that comes from image synthesis techniques that smooth out the data, rather than leaving it in a coarser form.


An arc spectrum is one produced by a discharge lamp where the discharge is through ionised gas, in the case of He-Ar a mixture of Helium and Argon, which produces a predictable line emission spectrum.

They are often used to provide a calibration spectrum for spectrometers.


The original paper ( and a version on arXiv ) which this relates to does not seem to identify any actual planets.  It's clear from the abstract that it is simply suggesting there is evidence that what they describe as unbound planets seem to exist as an inference of the spectral data they have gathered.

They are not identifying any individual planets at all.  They have not actually discovered any planets and there is nothing to name.

This is the abstract :

  Previously, planets have been detected only in the Milky Way galaxy. Here, we show that quasar microlensing provides a means to probe extragalactic planets in the lens galaxy, by studying the microlensing properties of emission close to the event horizon of the supermassive black hole of the background quasar, using the current generation telescopes. We show that a population of unbound planets between stars with masses ranging from Moon to Jupiter masses is needed to explain the frequent Fe Kα line energy shifts observed in the gravitationally lensed quasar RXJ 1131–1231 at a lens redshift of z = 0.295 or 3.8 billion lt-yr away. We constrain the planet mass-fraction to be larger than 0.0001 of the halo mass, which is equivalent to 2000 objects ranging from Moon to Jupiter mass per main-sequence star.

What they're saying is that, if you accept the hypothesis of their theory and data, there must exist a minimum of about 2000 objects of planet mass per star.  There's no way to identify any individual object and no means to verify their claims that I know of.


How are the world's biggest telescopes cleaned?

One of the ways they clean telescopes is to replace the metal coating on their surface. See, for telescopes, unlike bathroom mirrors, the metal is on the front of the glass. The reason for this difference is that having glass on the front makes the mirror more durable, but it also reduces the mirror's efficiency at creating an image. Because of this every few years the mirror needs to be taken off of the telescope, the old metal removed from the glass by an acid or some other chemical means, and then a new coating of metal is applied. On the tour of the Palomar Observatory they'll tell you that the 5.1 meter Hale telescope uses about the same amount of aluminum to cover it as is in a quarter (not sure if they meant volume or mass - also not completely sure it wasn't a nickle, or 2 quarters).

In the case of telescopes with segmented mirrors, like Keck, they can do this one segment at a time, which makes the process more convenient and spreads it out over time. It's also possible, though I don't know if they do this, to keep segments in backup to put in place of removed segments, because the telescope is made from a relatively small number of segments that have the same curvature and are, therefore, interchangeable.


Relationship between absolute magnitude of a star and its luminosity?

You are specifically talking about absolute visual magnitude; how bright an object would appear to our eyes if the object were 10 parsecs away from us. However, you must remember that stars emit a wide spectrum of light — much of it isn't visible to the naked eye.

Luminosity is a measure of the total amount of energy given off by a star (usually as light) in a certain amount of time. Thus, luminosity includes both visible light and invisible light emitted by a star. So there isn't a precise conversion between luminosity and absolute visual magnitude, although there is an approximation we can do.

Absolute bolometric magnitude is an analogous magnitude system considering all the energy emitted by the star. The formula relating absolute bolometric magnitude with luminosity is as follows:
$$L_\text{star} = L_0 10^{-0.4 M_\text{Bol}}$$

where $L_\text{star}$ is the star's luminosity, $M_{\text{Bol}}$ is the bolometric magnitude of the star, and $L_{0}$ is the zero-point luminosity (the luminosity of a star with $M_{\text{Bol}} = 0$) arbitrarily defined as $L_{0}=3.0128×10^{28}$ watts by the IAU. We can approximately convert between the two magnitudes via a bolometric correction:
$$M_{\text{bol}}\approx M_V + BC$$
The bolometric correction term $BC$ is empirically determined for the spectral class and evolutionary stage (although it can be modeled; see Torres (2010)). A table of values is given on the Wikipedia page.

The idea is that stars of the same spectral class and evolutionary stage have a similar distribution of wavelengths, so the discrepancy between all emitted wavelengths ($M_{\text{bol}}$) and visible wavelengths ($M_V$) should be similar.

Thus, the conversion between luminosity and absolute visual magnitude is approximately
$$L_\text{star} \approx L_0 10^{-0.4 (M_V + BC)}$$
Of course, this is only an approximation, but it can give us a pretty decent idea of what to expect. Trying it on 9 Pegasi, I predicted it would be around $1000 L_\odot$, which is still in the ballpark of the actual value $1950 L_\odot$. Mind you, the approximation would be better suited for less extreme stars than 9 Pegasi.


How do we know the order of the new Trappist-1 planets?

Using Kepler's laws.
We can determine the orbital period, $P$, each planet has by just looking at the time between transits for a given planet. Each planet has a distinct transit so its easy to distinguish which planet is transiting and we can calculate a period pretty easily.
Kepler's law then tells us that if you're orbiting with a specific period, you have to be orbiting at a specific distance (based on parameters of the system). The general equation is given by
$$P^2 = \frac{4\pi}{GM_{star}}a^3$$
Where $G$ is just the gravitational constant, $M_{star}$ is of course the mass of the star and fairly well known, and $a$ is the radius at which the planet orbits. Simply calculate $a$ for each of your planets and put them in the correct order. You could even do this yourself using data from the paper.


How to find Greenwich Mean Sideral Time?

To do so, I'm following what's described in this document and implementing C++ code.

That is a very old document you are using. It's using the 1982 IAU precession model. There have been multiple updates since then. The latest set of changes are working toward eliminating the concept of Greenwich Sidereal Time. (In software terms, GMST and GAST are deprecated concepts. While still currently supported, support will be dropped in some future release of the Earth rotation model.)

Other than using an outdated version of a deprecated concept, what did you do wrong? The expression you are using to compute GMST is at the top of page A-26:
$$\Lambda = H_0 + \Delta H + \omega^{\ast}\,(t-\Delta t)$$
The $\Delta H$ term takes one from GMST to GAST. Dropping that term, the expression for GMST is
$$H = H_0 + \omega^{\ast}\,(t-\Delta t)\tag1$$
where
$$\begin{align}
H_0 = &\,24110.54841 + 8640184.812866 T_u + 0.093104 {T_u}^2 - 6.2\times10^{-6} {T_u}^3\tag 2 \\
&\,\text{is the Greenwich sidereal time at midnight, in seconds of time} \\
\omega^{\ast} = &\,7.2921158553\times10^{-5} + 4.3\times10^{-15} T_u\tag3 \\
&\,\text{is the Earth's sidereal rotation rate, in radians}/\text{second} \\
t\phantom{\Delta}\quad&\text{is UTC time of day, in seconds, and} \\
\Delta t\quad&\text{is the difference between UT1 and UTC.}
\end{align}$$

Note that there's a units mismatch in equation (1). Equation (2) yields a result in units of sidereal seconds while equation (3), when multiplied by time of day, yields a result in units of radians. The latter needs to be converted to sidereal seconds. To do this, simply multiple both terms in equation (3) by $\frac{86400}{2\pi}$:

$$\begin{align}
\omega^{\ast} = &\,1.00273790935 + 5.9\times10^{-11} T_u\tag4 \\
&\,\text{is the Earth's sidereal rotation rate, in sidereal seconds}/\text{UT second}
\end{align}$$

Applying this to 2019 Jan 01 08:00:00 UTC yields the following:

The Julian date at midnight UTC 2019 Jan 01 is $2458484.5$, or $2451545+6939.5$. See footnote 1 for an explanation of this convention.
Dividing the residual 6939.5 by 36525 yields $T_u$: $T_u = 6939.5 / 36525 = 0.1899931553730322$.
Applying equation (2) yields the Greenwich Mean Sidereal Time at midnight: $H_0 = 1665686.527373333$ sidereal seconds.
Applying equation (4) and multiplying by $t=8*3600$ seconds yields another 28878.85178960284 sidereal seconds to the above. What about $\Delta t$, aka dUT1? I've ignored it (i.e., treated it as if it was zero.) See footnote 2.
1665686.527373333 + 28878.85178960284 = 52965.37916293584 (modulo 86400). This is the mean sidereal time, in seconds. Converting to hours, minutes and seconds results in 14:42:45.4. The 0.4 seconds is bogus because I ignored dUT1.

Footnotes:

There's a problem with using Julian dates to express time when high precision is of concern. The problem is that adding 2.45 million and change to the time since noon on 1 Jan 2000 results in potential truncation problems. The next representable Julian date after midnight 1 Jan 2000 is 40 microseconds after midnight. I myself try to stay a couple of orders of magnitude away from the quantization level, which means problems for dates at the millisecond level.

Almost every scientific package that supports Julian dates either uses a pair of double precision numbers or uses some sort of modified Julian date, such as time since 12 Noon 16 Nov 1858 (omits the leading 2.4 million in Julian date), time since 12 noon 23 May 1968 (omits the leading 2.44 million), or time since J2000.
The term $\Delta t$ in equations (2) and (4) is better known as dUT1. This is the difference between UT1 (time using the mean rotating Earth as a clock) and UTC (time that runs at the same rate as an atomic clock, but with leap seconds occasionally inserted to keep UT1 and UTC in sync). Those leap seconds keep $|\text{UT1}-\text{UTC}| < 0.9\,\text{seconds}$. Simply ignore the $\Delta t$ term if that's the case -- but remember that fractions of a second are meaningless.


Since when are we able to discover interstellar objects similar to 1I/ʻOumuamua?

In general terms we have had the observational capability to discover interstellar objects like this for quite some time - before digital imaging, photographic plates were used, which allowed accurate and verifiable measurements to be made. The issue is not so much in the observational and recording capability, but in collecting sufficient data to be able to compute an orbit.

With a few hours of observation you can compute a path that allows further study on subsequent days. This usually assumes a circular orbit as a first approximation, but with increasing elapsed time after the observations the computed path increasingly diverges from the actual orbit.

With observations over a span of days you can compute a better orbit, but again with increasing elapsed time after the observations the computed path increasingly diverges from the actual orbit.

With observations over a span of weeks you can compute an orbit that allows you to find the object again in subsequent years... and so on. Increasing elapsed time correlates to increasing divergence from computed orbits.

So, with a moderate span of observations, we would find that an interstellar object is diverging from its predicted path (orbit around the Sun) and that it is passing through the Solar System instead of being part of it.

But with the greater number of automated surveys (especially compared to pre-digital imaging) that routinely flag asteroid and comet detections, the likelihood of interstellar objects being detected is greater now than in the past.

But someone has to be looking for it. The Minor Planet Center likely has a large number of objects in its database which are actually interstellar objects. But it's set up to correlate detections based on Solar System orbits and the interstellar objects will exist as unconnected short-path observations that don't fit to an orbit. Maybe someone could find more by mining the data, but unless they are still near enough to follow up with further observations these will be unverifiable. But it would be an interesting project!


Was there ever a meteorite with a measurable gravity?

Was there ever a meteorite with measurable gravity?

At risk of stating the obvious:   Yes.  All mass, as far as every test I know about, has measured gravity.  No test has ever measured no gravity within the range of accuracy, so even low mass objects can have their gravity measured.
Rubble pile asteroids couldn't form if they didn't exert gravity.  Asteroids have been observed in orbiting pairs, so there's little question that they exert measurable gravity, if you can get close enough and have good enough equiptment to one to measure it.

Did we ever find a meteorite (a meteoroid that has fallen down to Earth) whose mass, density and gravity were high enough to demonstrate
their gravity similar to the Cavendish experiment?

Cavendish measured gravity using 158 kg lead balls and some sensitive equiptment.   These larger lead balls were stationary sources of perpendicular gravity on two smaller, 0.73 kg lead balls as the suspended objects on a wire upon which the gravity acted.   By the twisting of the suspension wire, and from a distance, so his mass wouldn't affect the experiment, Cavendish measured gravity.
A 158 kg lead ball is about 15 cm in radius and the smaller balls about 2.5 cm.  The distance between them (and I'm assuming this was center to center as that's standard in physical notations of this type, is listed as about 9 inches or 29.5 cm.    That means, surface to surface the distance was about 12 cm, which doesn't change the equation, but you wouldn't want to calculate gravity with two surfaces inside each other, even if the math lets you get away with that, so, I thought I'd mention it.
Some quick math and 29.5 cm, 158 KG, the gravitational force is easy to calculate. F=Gm/r^2, which comes to about 2.7/10,000,000 meters per second squared about 35 million times weaker than earth gravity.   A meteor could certainly have that force of gravity pretty easily.  While less dense than lead, greater overall could accomplish 1/35 million G forces easily enough.
What does that mean?  1/35 million G forces?   Lets call that amount of gravity one Cavendish - for obvious reasons.
Have you ever walked up a flight of stairs, one step at a time?  By just taking one step up a flight of stairs, you've reduced the gravitational force acting on you by one Cavendish.  A bit more than one, actually.   By taking an elevator up one floor works out to about 33 Cavendishes drop in gravity, for every 3 meters.    That's how gravity works.   Mass increases (roughly speaking) with the cube of the radius.  Gravity falls off with the square of the distance, so larger objects not denser objects, usually win in the gravity contest.   There are exceptions if there's enough variation in density, obviously.  Earth, for example has higher surface gravity than Uranus and Mercury has very nearly equal surface gravity to Mars, but as a general rule, when the overall mass is much different, density matters less.   The Sun and the Moon appear about the same size to our eyes and as a result have similar tidal forces, but the Sun exerts hundreds of times more gravity on the Earth than the Moon does.
The weakness of this force indicates how sensitive Cavendish' experiment was for it's time.
How did Cavendish know that there wasn't a mountain range in the distance that would exert equal pull on his test, or maybe some part of the Earth that was heavier on one side of where he lived than the other side?
Lets imagine that there was a mountain 12 km from Cavendish's laboratory, lets pretend this mountain was a 1.5 km high and weighs about 4 trillion kg.  That mountain would exert nearly 7 times the gravitational pull on Cavendish's test balls that the his 158 kg stationary lead balls did, so, did Cavendish need to find a flat region of equal gravitational pull in all directions?
Actually, the answer is no, because a mountain in the distance would have generated a gravitational field that would have been part of the initial conditions and he was measuring the change from the initial conditions.
And even if somebody had decided to mess up Cavendish's experiment by putting a mountain 12 miles from his house during his experiment, the effect would have been nearly equal on the two balls so it wouldn't have made much difference. It would just have been a tidal force, and a very small specific force on the torque of the wire that Cavendish was interested in.   Cavendish's experiment only works under very specific conditions, one 158 kg weight pulling both suspended balls on the wire in the same clockwise direction so the twist can be measured.  That doesn't work with a single large mass. It's the wrong kind of setup.
But to answer your question, forces greater than 1 Cavendish are easy for any relatively large body and a meteor could do that.  Bennu, which you mentioned in a comment exerts 6 micro-G, or about 22 Cavendishes.  It's radius is about 250 meters, so within about 1100 or 1200 meters, an object that size would exert 1 Cavendish of gravitational force.
Ofcourse, if Bennu was to actually strike the Earth, it would likely decrease, not increase gravity at the spot it hits because it would eject material over a wide area, leaving a crater.  Likewise, the effect would be very small, but on the edge of the crater, probably several Cavendish's worth of decreased gravity.
Perpendicular forces like this happen on Earth.  As land rises and falls.  As glaciers shift, the perpendicular gravitational force near those massive objects changes and this is most noticeable on the ocean.  The gravity of the Greenland ice sheet or the Antarctic ice sheet is measurable on sea level and calculating sea level rise if those ice sheets melt (which will take some time), but accurate calculation has to take into account the gravitational pull of the mass of glacial ice on the ocean's sea level.
The mass of a meteorite is generally quite small in the grand scheme of things and it's gravitational force on a nearby blade of grass, for example, while perhaps several Cavendishes, is negligible for all practical purposes, but measurable if you have sensitive enough equiptment.  You can now tell your friend when you drive past a mountain range that your car is experiencing a force greater than the force Cavendish first use to measure the gravitational constant and they probably won't know what you're talking about.
And if you're talking a much smaller, mostly Iron meteorite fragment, then you need to be quite close.  The largest meteorite fragment or solid piece is the Hoba, about 60,000 KG.  You'd need to be about 4 meters from it's center to experience 1 Cavendish of gravitational force.
It's fun to think that walking past a very dense, very heavy object, that you might feel it's gravity, but most things aren't nearly massive enough.  You don't feel the gravitational pull of a mountain range when you walk past it because the pull is negligible and mountain ranges are much more massive than most asteroids that are likely to hit the Earth.   Mountain ranges also are generally made up of lighter rock than the surrounding area, so there's also that, but even if they weren't, you still couldn't feel the gravitational pull of even the largest string of mountains.
Hope that wasn't too long.  I got answering and it came out longer than probably necessary.


Future redshifting and effect on the 'pitch' of CMB radiation

The wavelength of the CMB increases linearly with the scale factor $a$, which is defined as 1 today, so the "sound" of the CMB has dropped an octave when the Universe has doubled in size (in all three directions), i.e. when $a = 2$, which will happen when it is roughly 25 billion years old (see e.g. Fig 1 of Davis & Lineweaver 2004).

The magnitude of the dipole of the CMB is $3.4\,\mathrm{mK}$ (Kogut et al. 1993), i.e. $\sim10^{-3}$ of the peak value. The primordial fluctuations are even lower, roughly a factor $10^{-5}$ of the peak value. The dipole is partly due to our motion round the Milky Way, and so will fluctuate somewhat with the period of a Galactic year, which is 225 million years.

The figure shows how the peak frequency evolved in the past. Its current value is 160.4 Ghz. To calculate how a human would perceive the sound at a given time, I suppose you'll have to multiply the spectrum with the frequency-dependent transmission function of the human ear, but that's physiology which I know nothing about.

The figure was produced by calculating the frequency at a given time as
$$
\nu(t) = \frac{160.4\,\mathrm{GHz}}{a},
$$
where $a$ runs from $1/1100$ to $4$,
and calculating the age by integrating the Friedmann equation, assuming a Planck 2015 cosmology.


Are there clouds/clusters of gas in space that have red and blue color adjescent to one another?

Images that you see of nebulae show more intense colour than you could ever see with your eyes. Most nebulae are too dim to for our colour vision to be very active when looking at them. Cameras are designed to try and mimic human sight, to produce an image with colours that is close to what we would see. Telescopes are doing science and the colours are there to enhance features that we couldn't otherwise see.

A telescope like Hubble will image a region at several wavelengths. Some of these wavelengths might be in the infrared or ultraviolet, beyond human sight. Then a colour is assigned to each wavelength, and the images are mixed. (See this description of how to achieve this with photoshop) This brings out details that would be lost in natural colour.

Now there are two main reasons that dust and gas in a nebula might shine. One is that it is reflecting the light of nearby stars, the other is that it has been energised and is emitting its own light.

Hydrogen gas will emit light with a red colour (in the right conditions), and some stars shine in blue-white light. If an emission nebula is next to a reflection nebula then you will get blue-on-red. A good example of this is the Trifid Nebula in Sagittarius. As in this photo by Scott Rosen

  The lower, reddish part of the nebula contains the emission nebula — hydrogen gas which is emitting red hydrogen light since it is being heated by the hot young stars within the nebula. The upper blue portion of the Trifid is a reflection nebula — it consists of dust particles that are reflecting the light from the hot stars contained within the emission nebula. Because the hot stars are mostly blue, we see the reflection nebula as having a blue hue.


Why there is no axis measurement when telling distance between objects in space ?

You might benefit from watching this Minute Physics video on Why the Solar System is Flat. A summary is due to the way objects in our solar system and particular galaxy are drawn around a central massive item by gravity. Through the conservation of this angular momentum, and the loss of vertical motion though collisions, the solar system and galaxy "look" flat.

Talking of "upper" and "lower" doesn't really make sense in space. Gravity feels like it's pulling us down on Earth, and that is why we have this concept, but in space we lack this reference. For our own purposes, humans have created an artificial means of determining points of reference for spacial coordinate systems. Due to the rotation of the Galaxy we have established a base plane called the Galactic plane, from which a coordinate system and sense of "up" and "down" might be inferred (I think our solar system tilts at a 63 deg to the galactic plane of the Milky Way). In the context of Earth and our solar system, we use the plane that the Earth "makes" as it circles the Sun as the point of context for the Earth's axial tilt; we call it the ecliptic plane.

As for the rest of the Universe you may wish to check out this blog post from Cornell called Why is the Universe flat and not spherical?. Basically the Universe has a slight curvature to it but since we are on the surface, and have a limited view, it is difficult for us to see anything but flatness. This is related to why graphical views appear flat. Graphical views are generally limited due to our point of view and the sheer astronomical size of distances in space. Even in "dense" super clusters there are vast distances between objects. Overall it's easier to linearly show things rather than as they truly might be.


Chris Peat maintains a very sophisticated database of satellites and their visibilites at http://heavens-above.com, and there is an app available.


Guess inclination angle from radial velocity measurements

Radial velocity is a measurement in the change of speed of the star towards us, due to the planet's rotation around it. Speed variation change is detected thanks to the Doppler spectroscopy.

If the planets rotation plan is orthogonal with our direction of view, then we cannot see the radial velocity change, because the star speed towards us is constant. This would mean the planets you are talking about in your assignment were discovered using another method.

Here is a picture to illustrate.

By User:Zhatt - Own work, Public Domain, Link

In the above, the star speed changes in the plan that is 90° with our own, so there is no radial velocity variation we can detect with Doppler spectroscopy.

edit:
In your case, you are not telling if inclination is 90° to the line-of-sight or to the line perpendicular to the line-of-sight. If it is 90° to the line perpendicular to the line-of-sight (which is the usual in astronomy), then speed changes are maximum.


Is expanding universe adding potential energy?

Yes, in an expanding universe there is an increase in potential energy (with same caveats).

The (Friedmann) equations of a homogeneous and isotropic universe with no spatial curvature or cosmological constant can, in fact, be derived very easily from nonrelativistic Newtonian gravity. (This derivation is shown, e.g., in Mukhanov's Physical Foundations of Cosmology.) So it is actually valid to think of such a cosmos in Newtonian terms, using randomly but approximately homogeneously distributed point masses. At early times, the point masses are flying apart at high speed but the average distance between them is small. At late times, they will have slowed down, but the average distance between them will have increased. All that kinetic energy is converted into gravitational potential energy. This also remains true if we take into account Newtonian perturbation theory (the growth of overdense and underdense regions from initial perturbations).

As for the caveats: First, as I mentioned above, when spatial curvature is present, the Newtonian solution is no longer valid. Second, there is a century of literature with a variety of attempts to define a meaningful stress-energy tensor for the gravitational field in general relativity, none completely satisfying. Last but not least, a cosmological constant/dark energy term adds its own interesting twist to the story, as its energy density remains constant in an expanding universe, causing the expansion to accelerate: in this case, both the kinetic and the potential energy increase, thanks to the role played by dark energy's negative pressure.


There is no obvious astronomy error, but your use of math.pow is wrong

You write constgrav = math.pow(6.67,-11).  That means $G = 6.67^{-11}$.  You mean constgrav = 6.67e-11 or $G=6.67\times 10^{-11}$.

With completely different initial conditions, it is not surprising that the Earth flies off into space.

As a suggestion, try not using SI units for this, as it makes for some very small and very large intermediate values, the computers have difficulty in rounding.  Instead measure distance in Astronomical Units, Mass in "solar masses" and time in years. You can convert back to SI units when printing or plotting.


Why are Uranus' satellites named after characters in Shakespeare?

From W. Lassell (1852) in Astronomische Nachrichten vol. 34: "Beobachtungen der Uranus-Satelliten" (Observations of the satellites of Uranus):

In these tables I have for facility of reference employed a nomenclature of the satellites of Uranus proposed at my request by Sir John Herschel and selected by him from fairy mythology. The most distant of the two bright satellites discovered in 1787 by Sir W. Herschel is denominated Oberon, the other Titania, and pursuing still the order of distance, the interior ones now discovered are named Umbriel and Ariel.

This shows that it was John Herschel (son of William Herschel, the discoverer of Uranus, Titania and Oberon) who originally decided the nomenclature scheme at the request of William Lassell, the discoverer of Ariel and Umbriel. Interestingly it appears that the scheme was originally intended to be fairy mythology, which is not how the names are currently assigned. Nevertheless there is some debate around who exactly was responsible because the actual correspondence has been lost.
This pattern was broken by Gerard Kuiper, the discoverer of Miranda, see Kuiper (1949) in Publications of the Astronomical Society of the Pacific, vol 61, "The Fifth Satellite of Uranus". Unlike the previous satellites, Miranda is named after a human character (the daughter of Prospero) in Shakespeare's The Tempest:

Miranda was chosen as the name for the fifth satellite.
Uranus' own children, the Titans, are not suitable for mythological reasons; they have been assigned to the son of Uranus,
Saturn (Kronos), who gained supreme power after wounding
his father. Sir John Herschel named the four bright satellites
Ariel, Umbriel, Titania, and Oberon.
Oberon and Titania are
the king and queen of the fairies in Shakespeare's Midsummer
Night's Dream; Ariel and Umbriel occur in Pope's Rape of the
Lock, while Ariel is also found in Shakespeare's Tempest. In
the Tempest Ariel is "an airy, tricksy spirit, changing shape at
will to serve Prospero, his master," while Miranda is "a little
cherub that did preserve me" (Prospero).

The convention subsequently seems to have been formalised with the discovery of the inner satellites by the scientists working on Voyager 2. The IAU set up a nomenclature committee in the wake of the naming of the Mare Moscoviense (which had the somewhat dubious justification of declaring that Moscow is a state of mind). The IAU nomenclature committee decided that the common factor behind the existing nomenclature was the source works (Shakespeare's plays and Alexander Pope's The Rape of the Lock) and named the satellites according to this pattern. Some details of this can be found in this podcast from the Folger Shakespeare Library that includes Tobias Owen, who was involved in the naming process.


What are practical considerations for backyard radio-astronomy detection of black holes?

The resolution/error box. Radioastronomy has always been hindered by the resolution, because it is inversely proportional to the size of the telescope and making larger telescopes (even with interferometry) is not always easy. No amount of modern technology can substitute for a large effective diameter.  (When I say effective I'm including interferometry here; either way one needs a large area to build on).

Let's take a look at Reber's initial paper1 where he first mapped the sky:

In the diagram on the left, from top down, the three peaks in the contour map are Cas A, Cyg A, and finally Sgr A. The latter two have black hole origins, the former is a supernova remnant.

The resolving power of Reber's telescope here seems to have been 6 degrees, and it was 31.4 feet in diameter (and he focused on a 1.9m wavelength).

Now, by the Rayleigh criterion, the angular resolution is proportional to the wavelength divided by the diameter. As mentioned before, this is the major limiting factor for radioastronomers, and will be what holds amateur radioastronomers from making great telescopes — amateurs usually don't have acres or land in which to build a good interferometer (let alone the precision), and single-dish telescopes can't be made too large by an amateur. One may notice that I'm citing rather old observations here, on old telescopes; however given that radioastronomy technology has not changed nearly as much as size has, it should be OK to compare amateur telescopes with the smaller telescopes of the past.

Now, Cyg A was the first to be identified as a black hole, even though the radio-brightness of Sgr A was discovered at the same time. I'm focusing the rest of my analysis on Cyg A for this reason, as it stands to reason that the first confirmed BH out of the brighter radio sources would have more prominent indicators that it is a black hole.

Let's have a look at Cyg A with a better resolution:

(From this paper2, using the 5km array)

Note that the black blob in the center is the actual galaxy (probably an optical photograph superimposed on the contour map).

We can see that the lobes are less than a minute wide. (The actual galaxy is around 50 arcseconds wide)

To me, the most interesting thing that one would like to see here are the gas jets coming from the central galaxy. As mentioned in my answer here, these radio-emitting gas jets are in a steady line over thousands of light years, indicating that they come from some sort of cosmic gyroscope which has been steady for very long. However, even with the Ryle telescope the 1969 people could not get a picture of them; just a slight hint of their existence from the shape of the lobes.

Alright, so no gas jets. What else can indicate a black hole? They could try to look at the lobes themselves. They do not directly indicate the existence of a black hole, but their shape hints towards them being formed from jets (this is pretty much in retrospect).

However, with lobe dimensions of less than an arc minute, there's not much an amateur can get here, either. It's possible that a really good amateur telescope would just manage to notice that there are two lobes, but not much else as far as I can tell.

The other interesting bits would be the central galaxy itself, but it's too small. In the optical region one may have a chance of seeing Baade's "colliding galaxies" (it only looks like a pair of colliding galaxies). The gravitational effects (lensing, etc) are really only visible in the optical and beyond, for it to be visible in the radio we would need to be very lucky and have a huge radio source pass behind Cyg A — not happening anytime soon.

I'm pretty sure that a similar analysis would work for Sgr A or any other black hole candidate; the gas jets would be too small for an amateur radiofrequency resolution, and the gravitational effects of the black hole would only work well in the optical and X ray frequencies.

1. Reber, G. (1944). Cosmic Static. The Astrophysical Journal, 100, 279.

2. Mitton, S., & Ryle, M. (1969). High resolution observations of Cygnus A at 2. 7 GHz and 5 GHz. Monthly Notices of the Royal Astronomical Society, 146, 221.


How to understand this multi-day exposure photograph of Jupiter and the Moon

Looking toward the western horizon from 33°N latitude, hour angle increases in the 11-to-5 o'clock direction, and declination increases in the 8-to-2 o'clock direction.
The Moon was at -2.8° declination on Jan 8 and +2.2° on Jan 9, while Jupiter was near -2.0° on both nights.
Where they overlap in hour angle, the Jan 9 Jupiter images appear closer to the Jan 8 Moon images.

For the Moon to appear at similar hour angles on consecutive nights, the photographer must have started taking images ~40 minutes later each night, increasing Jupiter's hour angle ~10° per night.
On Jan 7 and 8, Jupiter was probably still above the ~20° high frame at the ends of those short series of exposures.
When the Jan 10 series began, Jupiter was likely already behind the building.

The scenario is reproducible in Stellarium if the location is in the southwestern US and times are evenings in January 2011.


An exceptionally bright star in the morning or evening is usually the planet Venus.

In May 2017, Venus is rising about two hours before the sun, and is by far the brightest star-like object in the morning sky. As it moves around the sun, it will get less bright. In a telescope, you would see it change from a crescent at the beginning of the month, to a semi-circle by the end. #

Planetarium software, such as Stellarium, can show the position of stars and planets at a given time and so help identify stars and planets.

Look out for Venus on the Morning of May 22, when it will be close to the crescent moon.


Term for the period between planet alignments?

The time between oppositions of Mars (or another planet) is called its synodic period. That is the length of time it takes for Mars to be in the same position in Earth's sky relative to the sun, and is about 780 days.

It contrasts with the sidereal period, which is the length of time the planet takes to orbit the sun.

There isn't a fixed number of days between conjunctions, so I don't think this has a name.


landing on a comet is more important than landing on an asteroid?

This is a very brief answer but I hope it helps.

Distance

Hayabusa collected samples from a near-earth asteroid while Philae landed on a comet after traveling a distance of over 6.4 billion kilometers.

The Type Of Mission

Hayabusa's did not literally land on the asteroid. It just touched down and collected samples. While, Philae had to land, steady itself, drill and study the composition. Plus, the asteroid on which Hayabusa landed was a near-earth asteroid and so it's nature could have been known easily. On the other hand, the Rosetta team came to know about the exact nature of 67P just weeks before the landing.

The Complexity

The Rosetta mission used about 6 gravity assists and performed 2 flybys. It was not an easy task to plan a journey of 10 years. Plus, when the probe reaches the comet, radio signals would take a lot of time to reach it and thus it would be very difficult to control the probe if something went wrong. Plus, the surface of an asteroid is vastly different from the surface of a comet. The Hayabusa Mission was not the first to touch down on an asteroid. It had been done once by NEAR Shoemaker and so we knew what to expect. See this for a detailed report.

I think this sums up how the Rosetta mission was a major breakthrough.


Could pure iron from a star make it to Earth?

Sure. Iron is not made in significant quantities in the Big Bang, so 100% of the iron on Earth today was synthesized in a star. There's no other source.

It escapes in many ways:

Supernovae
Mergers between compact stars like white dwarfs and neutron stars
Stellar winds from more ordinary stars

Supernovae are the main source of modern-day iron, with r-process nucleosynthesis during the explosion producing a lot of iron and elements near iron.


The real reason of conservation of information in quantum mechanics is that the von Neumann entropy $S(\rho) = -Tr(\rho \log_2 \rho)$ of an isolated quantum system with density operator $\rho$ does not change in time. This happens because $S(\rho)$ only depends on the spectrum of $\rho$, and the unitarity of time evolution preserves the spectrum.

If you start out with a perfectly certain state it will evolve unitarily, and remain perfectly certain - but realistically, there are going to be subspaces you did not measure and their uncertainty are going to "leak out" and get mixed with the state. This is basically the no hiding theorem which states that when your system interacts with the outside world and decoheres the information about the original state ends up on the outside and cannot be recovered by operating on just the internal state (but it is indeed conserved for the whole system).

Note that none of this depends on uncertainty per se nor the issue of what measurement does, but rather that the deterministic effects of the Schrödinger equation/unitary evolution makes things dependent on degrees of freedom you have little control over.


How do astronomers at SETI know if they received intelligent radio signals from other planets?

I'm not sure what you mean by "wave modulations".

Loosely speaking, there's 2 things to look for regarding alien intelligence/technology.   One is a signal of some kind and the other is a very large or very energetic structure.

The most obvious place to look for a signal is in the water-hole.  Good article on that here.  Because of the relatively low interference in that bandwidth, it's a good region to send a signal from one solar-system and hope it gets picked up by another.   Stellar distances are so large that this still requires a very large output and very sensitive receiving devices.   People have been listening for water-hole signals for decades, with no luck, outside of the extremely inconclusive "Wow" signal.

The other thing they can look for is mega-structures but that's much harder.   We have a hard enough time seeing entire planets in other solar-systems and the idea of an alien technology building something as large as a planet, but distinguishable . . . a bit of a stretch.

Planets are mostly those are observed by seeing wobbles or changes in the star's orbit or brightness as the planet passes in-front of them.   To see an alien structure and recognize it as something manufactured, not natural, it would need to be very large.  See silly article.  Outside of a structure as large as a dyson-sphere, we don't have the means of seeing an alien structure.

Finally, and to your question.  There is no technological "wavelength" to look for, light is light.  We transmit our cell-phone signals at about 2 gigahertz but an alien species might use 5 or 8.  There are ranges that are best for traveling between atmosphere and satellite and regions of comparatively low interference that are best for communication across a solar-system, but at the distance of several light-year's it's difficult to overcome the background radiation.   We could recognize an intelligence signal if we saw one.  Any information or language carrying signal would have a different pattern than natural signals, but the hard part is finding the signal in the first place, over such vast distances.

Basically, Seti is limited to looking for the extremely large and so far, (at least as far as I know), nothing even remotely close to definitive has ever been observed.   They don't have the means to observe most technology, assuming there is technology to be seen.  Only the most enormous mega-structure could maybe be observed, and only under the right conditions.

As our telescopes get better they might find something.  The atmospheric signature of a technological alien society might become visible to the James Webb telescope, but only for planets that transit their stars, not all planets and that might only offer possibilities, nothing definitive.

Signals like our television, cell phone and satellite broadcasts, those would be extremely hard to pick up from light-years away.  The signals are too weak and over light-years of distance, the background radiation of space would grow too loud by comparison.   Likewise the visible light from the night-side of Earth, while we can see that from the space station orbiting the Earth, observing that kind of light from light years away would require a huge telescope, many times larger than anything we have.   I don't have the knowledge to give you all the specifics on how hard to see or how big a telescope but much bigger than anything we have.   An Alien race with a telescope a few miles across might be able to observe the light from our cities on the night side of Earth.   We don't have the ability to observe anything like that ourselves.


The ninth planet can absolutely be a "virtual", in the way you describe it, meaning observed data indicating the gravitational influence by an object is not actually caused by such an object.

A simple visualisation of this is the case of a two-body system, where we observe two objects orbiting a common barycentre. From the observed data, one can get the impression that there is an object in the barycentre, pulling both of your observed objects against it.1

However, a barycentric way to explain the observations gets less likely as you add more bodies. Be aware that the current indications of "planet 9" is based on a mere correlation of some not-so-accurate data, from a small sample size, heavily biased by observation constraints.

  how could we distinguish a "real" 9th planet from a virtual one,
  except by the obvious answer if we can observe it directly?

That is only by its gravitational influence alone.

As a rule of thumb, incomplete and inaccurate data always imply the risk of causing virtual objects.

It should also be noted too that observations in the Kuiper belt generally can not yield how a orbit changes, due to the extremely long revolution times. observing "Interactions" are actually mostly about tracing an orbit back from its current state vectors, to find probable interaction events in the past. That is literary to ask for data artefacts and virtual objects.

1Before anyone objects about how easy it is to falsify that: Yes it is easy to determine that there can not actually be an object there, based purely on observing the two bodies. But, that requires 1. accurate data about the motion of the two objects, and 2. their mass. Our current data about the Kuiper belt fails to give good enough data on both points.


How does time "add up" for observers outside of and travelers inside of a Krasnikov tube?

I think one of the most important things you missed in this article is the bit where they say the energy requirements are "unphysical".  This is no different from saying it's impossible from our point of view.
A couple of quotes from the article :

Here as well, it was found that the wall of the “warp bubble” surrounding a spaceship must be unphysically thin compared to the bubble radius
Estimates of the total negative energy required to construct Krasnikov tubes of even modest dimensions were shown to be unphysically large.  Similar difficulties have been recently shown to plague warp bubbles and wormholes.  The Krasnikov tube suffers from some of the same drawbacks as these other pro
posed methods of faster-than-light travel, and is hence also a very unlikely
possibility.

The model you're discussing is really best thought of as a mathematical exploration, rather than a practical proposal of any kind.  I would suggest the authors are disproving the viability of Krasnikov tubes rather than anything else.  What was done was to try and fix problems with the previous model, but in doing so the authors ( to their credit ) show that they cannot bypass the fundamental problems making such a system unrealisable in practice.


The speed relative to Earth is greater than the speed of light. This is consistent with Relativity as the space between us and the distant galaxies has stretched.

The speed is given by

$$(\mathrm{Hubble\ constant}) \times (\mathrm{radius\ of\ the\ observable\ universe\ in\ Mpc})$$

Which, as noted by Pela in a comment, is roughly 3.3 times the speed of light.


Mars is dark, surrounding is bright, what am I doing wrong?

Sounds like you are very out of focus. Dial the focuser all the way out towards you. The little knob on the right.


How to figure out what something in the sky is?

It's hard to say 100% for sure without more info.  What is your location and what date/time was this?  How fast did it seem to move?

To me, it looks like Venus.  If it looked like a bright star, but much bigger and brighter than other stars, in the west around sunset, and moving slowly toward the horizon, then it was most likely Venus.


What's the useful magnification for a scope with 70mm objective lens?

I make telescopes and telescope optics (mirrors). The question is actually pretty complex. I get it a lot, because most people believe magnification is the end-all-be-all of telescopes, but reality is a little different. Here's what I mean by that:

You are correct that the actual magnification of the instrument (how much bigger the image is, compared to the object) is given by the formula:

M = F / f

M = magnification
F = focal length of the objective
f = focal length of the ocular (eyepiece)

The formula seems to suggest that you could get near-infinite magnification from any telescope by just reducing f, the focal length of the ocular. And it's technically true.

But the question is - how much can I increase magnification before it's too much to be useful?

When you zoom into an image with Photoshop, at some point you start seeing the pixels; any zooming past that point does not reveal more details. Telescopes do not have pixels, but the image the objective provides does not have an arbitrary amount of details - just like with Photoshop, arbitrarily increasing magnification with a telescope past some point will only give you a messy blur (not pixelated, just blurry).

So how much is too much? It depends.

If you had perfect optics, used in perfect conditions, then the maximum useful magnification would be:

Mu = 2 * D

Mu = maximum useful magnification
D = diameter of the objective in mm (a.k.a. aperture)

So for an instrument with a 75 mm aperture, the maximum useful magnification is around 150x. If you increase magnification past this point, the image is bigger, but it also gets progressively more blurry, so no additional details are revealed. But those are ideal numbers.

In practice, optical elements (lenses, mirrors) are not perfect. Only the most expensive instruments have optics that are anywhere near perfection. Imperfections add to the blur.

Even with perfect optics, if the telescope is not perfectly collimated (the lenses and mirrors are not perfectly aligned), that reduces performance too. Most refractors are collimated from factory and need no adjustment, but some cheap refractors may not be perfectly collimated. Other instruments such as dobsonian reflectors need periodic tuning to redo their collimation. When the instrument is miscollimated, the maximum useful magnification is reduced.

The atmosphere is turbulent. This is called "seeing" in astronomy. When seeing is bad, extra blur is added to the image, and you have to dial down the magnification.

Finally, magnification is not the ultimate parameter of a telescope. When you buy a car you don't ask the dealer "what's the fastest car you have?" You adjust the speed of the car to the conditions: drive at one speed on the freeway, at a different speed in a residential neighborhood, and at yet another speed in the parking lot at the local grocery store. Same with telescopes.

Some objects require a lot of magnification - things like Mars at opposition, or certain double stars, or small craters on the Moon, these all benefit from the highest magnification you could get. But objects such as Jupiter are better seen at more like a medium or medium-high magnification (Jupiter is low contrast so too much magnification washes out detail in most cases - but not always). And then there are nebulae and clusters which are better seen at low magnification. Horses for courses.

So what's the actual answer?

Well, you have the upper limit set in stone: 2 * D, or 150x for a 75 mm scope. But that's more like a theoretical bound. In practice the actual value will rarely be equal to that, usually will be a little below it, and sometimes much below it.

And then you have to choose your magnification based on the actual object you're observing: high, medium, or low.

This is why for experienced astronomers magnification is just a number. They've learned to adjust the instrument to the local conditions, to the object being observed, and even to the characteristics of the observer (younger folks tolerate very high magnification better than older ones - something to do with the narrow size of the beam of light exiting the scope at high magnification, then hitting floaters and other defects that accumulate with age).

This is why most people end up with a handful of eyepieces, and choose the best one depending on each case.


How do stars affect the orbits of moons?

Short answer : yes.

Long answer : it's complicated.

This is a classic three body problem, and a real example would be the Sun-Earth-Moon system.

Unfortunately there is no general solution of the 3-body Newtonian gravity problem, so we have to rely on special methods for special cases and even these turn out to be quite complex.

The best answer I could find to your specific question applied to the Sun-Earth-Moon is this one on Yahoo.  As you can see the resulting effect is quite a complex set of periodic effects.  It's small but not negligible.

You can intuitively visualize the effect in such a system by noting that typically to a good approximation the planet-moon system will orbit the star at a considerably larger distance than the planet-moon distance.  The planet and moon themselves orbit around their center of mass.  The relative difference in gravitational forces between the position where the moon is furthest away and closest will give you a rough guide to the order of magnitude of the effect.  A back of the envelope calculation for Earth makes it about a 1% order of magnitude for the Sun-Earth-Moon.

Other systems will be different, but there will certainly be an effect.


What would we notice if the Sun stopped fusing atoms?

As stated in my answer to Would we have more than 8 minutes of light, if the sun "went out"? , the Sun would cool at roughly constant luminosity for some tens of millions of years. That means "we" would not notice anything change on human timescales (apart from the lack of neutrinos, which has no measurable effect on anything apart from the odd atom in a vast vats of cleaning fluid or water, underground in South Dakota and Japan).

If $dL/dt$ was roughly $-L_{\odot}/30$ in solar luminosities per million years, then it would be 1000 years before the luminosity of the Sun fell by 0.1%, which is roughly the kind of variations we see over the course of the solar cycle.

However, the most likely evolution of the Sun is for it to contract, maintain a stable luminosity and compensate by increasing its surface (photospheric) temperature.

Given that luminosity is provided by gravitational contraction and the rate of change of gravitational potential energy, then
$$ L_{\odot} \simeq  -\left(\frac{GM_{\odot}^2}{R^2}\right)\frac{dR}{dt}, \tag*{(1)}$$
where $R$ is the radius.

If the luminosity is constant and given by
$$L_{\odot} = 4\pi R^2 \sigma T^4,$$
where $\sigma$ is the Stefan-Boltzmann constant and $T$ is the effective (photospheric) temperature, then
$$ \frac{dL_{\odot}}{dt} = 4\pi \sigma \left(2R T^4 \frac{dR}{dt} + 4R^2 T^3 \frac{dT}{dt}\right) = 0 \tag*{(2)}$$

Combining equations (1) and (2) we find
$$\frac{dT}{dt} = \frac{T R L_{\odot}}{2 GM_{\odot}^2} = 90\ {\rm K/Myr}$$
i.e. The Sun surface would heat up by 90K every million years.

This gradual warming would change the spectral type of the Sun, moving it towards an F-type star. The spectrum of light it emitted would shift to shorter wavelengths (according to Wien's displacement law, the peak wavelength in the spectrum is inversely proportional to temperature) and a greater fraction of the Sun's light would appear in the ultra-violet part of the spectrum.


By an estimated order of magnitude there are about $10^{22}$ in our observable universe (hundred billion galaxies with hundred billion stars, each, as a rough estimate), and the universe exits since (order of magnitude) $10^{10}$ years. Hence the average formation rate of stars has been about $10^{12}$ per year, meaning more than 10,000 per second, give or take a factor of 100. This calculation ignores stars which don't exist any more, but for an order-of-magnitude estimate numbers should be ok.

Although the star formation rate has been much higher in the younger universe than today. (One of the press releases about star formation in the early universe)


Why was this asteroid (4864 Nimoy) chosen to be named after Leonard Nimoy?

I'm not entirely sure, but I would guess that it's simply because the discoverer suggested it. After a few steps in the process of characterizing an asteroid's orbit, the discoverer gets to suggest a name (within reason) to a committee.  The full guidelines are available at the relevant IAU webpage.  (It's longer than is worth reproducing here.)

Note that there are plenty of "celebrity" asteroids, drawn from a wide range of human endeavours: Adamsmith, Balzac, Beethoven, Megryan, etc. If anything, I'm surprised that an asteroid wasn't named Nimoy sooner. There's already a Mr. Spock!


Error propagation methods for orbit parameters

ProfRob's comment is actually already the answer:

Unless you can linearise the model, there is no analytic error estimate.

I consider error estimation very important, so I would like to show a bit more details here: A way of estimating the error of a model is by Gaussian error propagation, see in particular the part of the Wikipedia entry on non-linear combinations. Let me briefly summarize the idea behind as I find the Wikipedia article not so easy to follow. I will also use different variable names so that they do not overlap with those in your question.
We start with a function $\varphi$ which depends on different variables $x_1, x_2, \ldots$, meaning $\varphi = \varphi(x_1, x_2, \ldots)$. We assume that we can linearize around a certain point $\tilde{\bf x} = (\tilde{x}_1, \tilde{x}_2, \ldots)$, which mainly means writing the function at that point as a Taylor series.
Then, we can determine the Gaussian (maximal) error in the vicinity of ${\bf \tilde{x}}$.
$$\left. \Delta \varphi \right|_{\bf \tilde{x}} = \left. \frac{\partial \varphi}{\partial x_1} \right|_{(\tilde{x}_1,\, \tilde{x}_2,\, \ldots )} \!\!\!\!\cdot \Delta x_1 + \left. \frac{\partial \varphi}{\partial x_2} \right|_{(\tilde{x}_1,\, \tilde{x}_2,\, \ldots )} \!\!\!\!\cdot \Delta x_2+\cdots$$
In this formula you calculate partial derivatives $\frac{\partial \varphi}{\partial x_i}$ with respect to each variable $x_i$ and plug in the values of the variables at the central point ${\bf \tilde{x}}$. The $\Delta x_i$ is the error estimate for each individual variable $x_i$.
How to we apply that to your problem where you have fitted an implicit equation? Since you excluded the Monte-Carlo approach (which is the usual method for this case), I suggest the following recipe:

Formulate explicit functions $a(x,y), b(x,y), \ldots f(x,y)$, e.g.
$$ a = - \frac{bxy + cy^2+ dx+ ey + f}{x^2}$$
Linearize the functions around $x_0,y_0$. The $x_0$ is called $x$ in your table, and respectively $y_0$ is in the $y$-column.
Determine $\Delta a, \ldots, \Delta f$ by partial differentiation.
Plugin in $\Delta x$ and $\Delta y$ as also given in your table.


Do gravitationally bound (e.g. orbiting) objects really resist the expansion of space?

Expansion of space does affect orbits, but the effect is really small.

Rate of space expansion is currently about 2cm/s per light year. Sun orbits Milky Way at about 230 km/s, and space expansion contributes about 0,5km/s. It is going to be difficult to measure other variables, such as stellar masses, with such detail.


Using a differential force in the derivation of hydrostatic equilibrium in a star

The whole exercise illustrates the usefulness of vectors in addressing problems.

$$\vec{F_g} = -\frac{GM(r)\delta m}{r^2}\ \hat{r}$$
$$\vec{F_P} = -\left(\frac{dP}{dr}\right) \delta r \delta A\ \hat{r}$$
$$\vec{F_P} + \vec{F_g} = 0$$
leads to the scalar equation of hydrostatic equilibrium
$$ \frac{dP}{dr}=- \rho g,$$
where $\vec{g}= -g\hat{r}$.