(Thought Question) What would a giant far away mirror look like to a telescope?

If there was a theoretically perfect mirror the size of say, our solar system, somewhere out in space...

Great thought experiment so far!

...that had a focal point of literally earth...

Rats! I was hoping for a flat mirror so the answer would have been slightly simpler.
It is true that concave mirrors are "magnifying mirrors" and if we stay closer to the mirror than twice its focal point (see below) we can see a magnified image. I don't believe that it provides better optical resolution than a normal mirror, but since the construction of our eyes is fixed (we can't change the optical system nor "pixel density") we use the mirror to "blow up" the image in the same way that a photographic enlarger blows up the image on a negative without improving the resolution of the image.
 source
In optics image information is encoded in the wavefront at any point whether in focus at that point or not. If we know the mirror's diameter and distance from Earth, we can apply the principle of diffraction in a simple way no matter how complicated the rest of the optical system.
For a hard-edge circular mirror (top-hat shaped apodization) we know that the Airy disk is the right principle to apply, and a simple definition of resolution gives us
$$\theta \approx \text{1.22} \frac{\lambda}{d}$$
for the angular resolution. Let's use 500 nm green light for $\lambda$ and twice Neptune's orbital radius (60 AU) for $d$. That gives us $4 \times 10^{-18}$ radians.
If the mirror were at the distance of Proxima Centauri or only $4 \times 10^{+16}$ meters, then the impact on resolving a wavefront from the perfect mirror based only on diffraction of a circular aperture would be of the order of centimeters!
The focal length of a concave mirror is the parallel-to-point distance, so we would really need the mirror to be at roughly the point-to-point distance of twice the focal length.
If the concave mirror brought our image back to us, say on a sheet of paper, it would be incredibly dim i.e. pretty much no photons except from the Sun itself. But if we ignore that then we would see ourselves blurred by about 1 centimeter.
If the image was kilometers in front of us then we could focus a telescope on that image in space and re-image it at the entrance pupil of an eyepiece or on to a sensor.
If the telescope were 1 meter in diameter, we could see roughly a 1 meter wide swath of Earth.
Of course this doesn't work so simply because the mirror would have to be pointed such that the location of Earth 8.5 years ago would be imaged where we were today.
Conclusions
Yes this is kinda possible from a Gedankenexperiment point of view; a 60 AU mirror out at Proxima Centauri with a focal length of twice that distance could produce an image of Earth nearby Earth (light time considerations in force) and we could look at that location in space with a large diameter telescope and see a patch of Earth 8.5 years ago roughly vignetted by the size of the telescope's aperture.


What effects do CMEs have on planets like Mercury?

Much like Earth, Mercury also has a magnetic field, only much weaker. Because of this, the high velocity charged particles accompanying CMEs distort Mercury's magnetic field at a much greater magnitude than it does earth. Also, because it extends beyond the sun's corona, it has the potential to burn the sun facing side of Mercury's surface even more than usual. Because of the high solar winds accompanying the CMEs it has the potential to actually strip particles off the surface of Mercury to become space dust.
CMEs contain high radiation, large amounts of high velocity charged particles (causing a HUGE magnetic flux/field), and extreme solar winds. (among some other things) so any way that these phenomena will normally affect Mercury will happen during a Coronal Mass Ejection


I don't have a complete answer, but I do know where the $7.6$ factor comes from: It equals
$$(2×0.0167)×(24×60)/(2 \pi).$$

$0.0167$ comes from the eccentricity of the Earth, and the factor of $2$ comes from the above equation. $24×60$ converts from days to minutes, and $2 \pi$ is the number of radians in a complete revolution.

What I still don't completely understand is why these coordinate conversions need to be done, specifically the division by $2 \pi$.

Thanks everyone for your help.

Edit: I now have a somewhat better explanation for the computation in this answer. 2ε sin(M) is the number of radians that the mean Sun is behind the apparent Sun on Noon on a day with mean anomaly M. Dividing the 2ε factor by 2πgives the fraction of the rotation of the Earth (i.e. one day). When the result is multiplied by the number of minutes in a sidereal day (23×60+54) yields the desired coefficient 7.63345.


Looking for a routine to correct for Local Standard of Rest (LSR)

There are several corrections here, and as you suspect, all of them can be done in astropy.  You'll need to do something like this:

Transform your observed velocity to the solar system barycenter (thus correcting for Earth's rotation on its axis, and its orbit around the Sun.  There is an example here in the astropy docs that shows how to do that.  If you follow that example, you'll first need to define your radial velocity as an astropy "quantity" object with units, e.g. something like this:

import astropy.units as u
# Define RV of 30 km/s
rv = 30 * u.km / u.s

You can ignore the discussion in that example of the relativistic correction, and just add the barycentric correction to your velocity, e.g. corrected_rv = rv + vcorr

Then, you can correct for the fact that the Sun isn't moving quite in step with the average of the nearby stars, and convert from the barycentric radial velocity to the local standard of rest, or LSR.  Astropy has some examples here but they don't do anything quite like this, so here's some example code.

Set up an observation that has the coordinates you observed, and the corrected radial velocity from step 1.  (I'm continuing with 30 km/s here, but you would substitute whatever you got in step 1.)
from astropy.coordinates import ICRS, LSR
my_observation = ICRS(ra=17.75*15*u.deg, dec=-29*u.deg, \
     pm_ra_cosdec=0*u.mas/u.yr, pm_dec=0*u.mas/u.yr, \
     radial_velocity=30*u.km/u.s, distance = 1*u.pc)

For specificity, I've chosen coordinates that are toward the Galactic Center. Here we're specifying zero for the proper motion terms since astropy won't do the velocity transformation if it doesn't have all three velocity components; that's also why we need the distance.  (You can experiment and convince yourself that the values don't matter in terms of what you get for the radial velocity.)  Now, you can transform that to the LSR frame and print out the new radial velocity:
new_rv = my_observation.transform_to(LSR()).radial_velocity
print(new_rv)

which gives 41 km/s, since the motion of the Sun relative to the LSR has a component of 11 km/s (the "V" component in the UVW system), for example in this derivation of the solar motion from Gaia data.


Why is there a density dropoff in the stellar winds of magnetic O-type stars?

Conservation of mass?

In a steady-state wind, the mass loss rate through each shell is the same.
$$ \frac{dM}{dt} = - 4\pi r^2 \rho(r) v(r),$$
where $\rho$ and $v$ would be the density and velocity of a spherically symmetric wind, with a fixed mass-loss rate $\dot{M}$.

Thus for a fixed wind speed you expect the density to fall as the square of the radius in spherically symmetric wind - and it looks to me that a little outside the closed field loops then spherical symmetry is approximately recovered.

This alone, would not account for the sharp drop off in density seen, but the other factor is that a radiation driven wind accelerates with radius when close to the star, which exacerbates the decrease in density. Typically, $v$ would increase by a factor of a few between say 1.5 and 5 times the radius. Multiplying this by a $r^{-2}$ factor would give a factor $\sim 100$ decrease in density overall.

The behaviour of the temperature is much more complex, governed by heating and cooling processes and the expansion of the gas.


How do we know the big bang didn't happen in an existing universe?

We don't know what's on the other side of the Big Bang, if anything at all. The Big Bang theory as well as the rest of physics is agnostic on this possibility. During the time period after the Big Bang, the entire universe was basically in a hot, dense plasma state; any physical signals propagating from a period before the Big Bang that could potentially influence the present or provide us with information of a pre-Bang past would be eradicated by the extreme conditions.

So the singularity of the  Big Bang is effectively the start of our universe because it is effectively an impenetrable cosmic firewall disconnecting us from anything that came before. Maybe there's another room on the other side of the wall, or maybe it's just more wall. Until someone comes up with a more refined theory or an experiment finds a way to detect a tiny signal that somehow made it through, it's pointless to speculate.


What is the escape velocity of our Galaxy relative to?

One way to think of the Escape velocity is to imagine it backwards.

Instead of a rocket being fired into space, think of the same rocket, starting at rest relative to the Earth at a great distance (for this thought experiment pretend the sun, doesn't exist and the rest of the universe is empty), then let it fall to the ground. No matter how far away you start, the speed when the rocket hits the Earth will always be 11km/s or less.

So if you start with 11km/s or more at the Earth's surface, you have enough energy to coast forever into space. The escape velocity at the surface is 11km/s, the escape velocity would be different, and lower, if you were already in a high orbit.

We can do the same thought experiment with a galaxy. We know, roughly the distribution of mass in the galaxy (most is in a large blob of dark matter  with the visible disc inside it) Now take a rocket ship a very great distance from the galaxy, and allow it to fall.  As it passes the sun, the rocket will be moving at 537km/s relative to the centre of the galaxy. (there is a black hole there, but it is tiny compared to the galaxy, so has very little effect on the gravity of the galaxy)

If you know how the mass of a galaxy is distributed we can get a good idea of how fast something would be travelling at any point, if allowed to fall. And by the same argument as above, that gives the escape velocity at that point.

The escape velocity around the sun is about 1/600 of the speed of light. The journey to leave the Milky way is at least 30000 light years long, This means it is going to take about 20 million years to reach the edge of the galaxy, since you will be getting slower as the galaxy's gravity pulls you back. Space (as has been observed) is big.


Planets and moons positions in cartesian coordinates?

Horizons Ephemeris Generator

Here is a screen capture where vector option is chosen:


How long New Horizons will remain operational?

The answer is in the other section:

New Horizons Mission Info

What I can say off the top of my head is that the mission has been extended and they have identified another object that the spacecraft can reach. However, after the next rendezvous it will be out of fuel and will not be able to maneuver.
BUT -- the probe is powered by a Plutonium-238 radioisotope thermal generator (RTG), and will likely to continue functioning for the foreseeable future. The relevant question is: how long will the government fund the NASA operations team that is 'talking' to the probe. That is anyone's guess.

New Horizons Spacecraft Components

PS - I'll update this answer later once I talk with some colleagues. I started, and worked for the company (KinetX, Inc) that did the navigation for New Horizons. I no nothing of the science, but my friends did the navigation and know the near-term details for the mission.


Is it possible to predict a supernova with years or decades of advanced notice?

While a whole slew of signals will arrive once the supernova actually occurs, from neutrinos to light of all different energies and wavelengths, the outward, visual appearance of the star will not give any surefire clues that a supernova is imminent. But the nuclear reactions powering the star do change over time, and at just 640 light-years away, Betelgeuse’s neutrinos may give us the early warning signal we need to predict its supernova accurately, after all.

If we want to know what’s going on in the core of a star — our only true indicator of when a supernova is coming — observing the electromagnetic properties of the star won’t give it to us; there is no change in a star’s temperature, brightness, or spectrum that occurs after the transition from carbon-burning to heavier elements.

But the neutrinos tell a vastly different story.

In the lead-up to a supernova, the neutrinos carry away the vast majority of the energy produced in those core fusion reactions. For the carbon burning phase, the neutrinos are emitted with a particular energy signature: a specific luminosity and a specific maximum energy-per-neutrino. As we transition from carbon-burning to neon-burning, oxygen-burning, silicon-burning, and eventually the core-collapse phase, both the energy flux of neutrinos and the energy-per-neutrino increase.

During the silicon-burning phase, neutrinos are produced with higher energies than previously, and as the silicon-burning phase continues, shells of silicon fusion begin forming around the core. In the final few hours of this star’s life, shortly before the core collapses, the neutrinos produced cross a critical energy threshold. Your antineutrinos can then interact with the protons in your detector, producing a unique signature: neutrons and positrons, an unmistakable signal of inverse beta decay.

Under normal circumstances, inverse beta decay events are extreme rarities in neutrino detectors, coming about only when a random neutrino from the Universe strikes our sophisticated neutrino detectors. But if a star were burning silicon in its core and had crossed that critical energy threshold to produce sufficiently energetic antineutrinos, and if it were close enough, we should see a large number of inverse beta decay events that all come from the same direction.


What happens over time as a neutron star cools?

The cooling history of a neutron star can be divided into an extremely rapid neutrino cooling phase, followed by an indefinitely long cooling phase due to the emission of photons from its surface.

The key to understanding neutron star cooling is to realise that degenerate neutrons possess almost no thermal energy, even when extremely hot - because the Pauli Exclusion principle forbids the neutrons from losing kinetic energy and falling into an already-occupied quantum state. Any cooling processes therefore act to reduce the temperature of the neutron star very quickly. Secondly, the interior of a neutron star is almost isothermal, due to the high thermal conductivity of degenerate gases, but the temperature at the surface is smaller by about a factor of 100 or so.

When a neutron star first forms, its internal temperature exceeds 10 billion kelvin, its surface temperature would be 100 million degrees and emit hard X-rays. Nevertheless, the small size of the neutron star limits radiative losses and instead it is neutrinos escaping from the interior that can cool it by a factor of around 100 in a few thousand years.

The neutrino losses proceed by something called the modified URCA process (somewhat faster direct URCA or processes involving pions/kaons maybe important in the first seconds/minutes of more massive neutron stars, but are blocked by complete neutron degeneracy and the comparative paucity of protons in the interior): cycles of neutron beta decay and inverse beta decay on protons produce anti-neutrinos and neutrinos. Bystander particles are required to conserve momentum. The temperature dependence of this process is $\sim T^{8}$, so if the temperature falls by a factor 100, the neutrino cooling rate falls by a factor of $10^{16}$.

Meanwhile, photon cooling from the surface, whilst initially negligible compared to neutrino cooling, falls only as $T^{4}$. After around 10,000-100,000 years, when the neutron star surface has cooled to about a million degrees, photon cooling, through soft X-ray emission dominates.

If nothing else happened to an isolated neutron star, it would continue to cool, such that $T \propto t^{-1/2}$. The vast majority of the estimated 1 billion neutron stars in our Galaxy, are in this state, whereas most of the pulsars we can observe are young and in the neutrino-cooling phase.

How cold can they get? The formula above suggests their temperature halves when their age quadruples. Following this, they could cool to the temperature of the Sun in around a billion years. However, there are reheating processes that can occur.

Isolated neutron stars can accrete material from the interstellar medium. This may be very little, since most neutron stars move fast. The gravitational potential energy is partly released as heat when it impacts the neutron star surface. Then, the low heat capacity of neutrons works in the opposite direction and comparatively little accretion is required to balance the photon losses.

Secondly, neutron stars are born as fast rotators and spin down. Angular momentum must be transferred outwards from the fluid interior to the outer crust. This is not a frictionless process. Thus some of the rotational kinetic energy can keep the star warm.

Thirdly, the neutron star has a huge magnetic field that gradually dissipates. Some of that energy may end up as heat in the neutron star through simple Ohmic dissipation of currents.

The bottom line is that the thermal emission from neutron stars older than about 100,000 years has not been observed, so we simply don't know how cool they might get. Some more information and references can be found in the Physics SE answer where I discuss where old neutron stars might appear on the HR diagram.


If you put all planets between earth and moon what would happen?

Talking about instantly teleporting planets isn't very meaningful if you want to get scientific answers, as PM 2Ring pointed out. But is is almost certainly not physically impossible for them to have trajectories that let them all pass between the earth and moon at the same time. But the earth and moon would have to start out farther apart than they are now because all the planets passing in between would pull them closer together, and we want them to be at the current distance just at the moment the planets pass between.

The biggest issue here is tidal forces. Planets pull harder on the near side of other objects than on the far side, and the effect is much stronger the closer you are. Whichever planet was almost touching Jupiter would be torn apart by this difference in force - Jupiter's gravity would be pulling on the near side faster than the far side could catch up. The only way to avoid this is to have the planet passing by very fast, so it would only be in the area of maximal tidal forces for a very short time.

What exactly happened would depend on the exact trajectories. Obviously there are an infinite number of ways a bunch of things can fly past each other. The reason everything doesn't fall into Jupiter in this scenario is that they are moving fast enough and in the right directions. They will all definitely curve towards it, though. Its gravity would dominate the situation, but the gravity of all the planets would have to be considered if you were a super-powerful civilization trying to make this happen.


Photodetector Question: Does converting an RGB image to grayscale produce the same result as using a grayscale image detector?

Really, all cameras/detectors are greyscale detectors.  Every pixel in a CCD detector puts out a single number for the number of photons it detected. To create an RGB image you need three filters.  If you have 3 detectors, you can put a different filter in front of each and get 3 numbers at once.  If you have 1 detector, then you take 3 measurements with a different filter each time.  To create the color image, one combines these values but you have freedom to weight them as you like.  Typically, one tries to imitate the sensitivity of the eye, but that is a complicated issue since the sensitivity of the eye to different colors changes with total intensity and the filters can not exactly reproduce the eye's bandpasses.  So, there is no right way to do this.  Some people like red more than blue, so they weight the R-band more. Once, you get aR+bG+cB, you loose the information you need to derive R, G, and B separately.

So, to do your experiment properly, you need to first get the response function (response vs wavelength) of the filters + CCD of your camera at each wavelength that you are using.  You do this by imaging without the sample.  Then repeat with the sample and divide "with" by "without", where "with" is the output in the R, G, or B channel that has the most response in the "without" case.


I may not qualify as more knowledgable but here's what I can add to @called2voyage's comment.
In the question Have there been any documented mini-moons since 2006 RH120? I link to  Granvik, Vaubaillon and Jedicke 2012 (The population of natural Earth satellites Icarus, 218 (1) March 2012, pp262-277) available without paywall in ArXiv.

Abstract
We have for the first time calculated the population characteristics of the Earth’s irregular natural satellites (NES) that are temporarily captured from the near-Earth-object (NEO) population. The steady-state NES size-frequency and residence-time distributions were determined under the dynamical influence of all the massive bodies in the solar system (but mainly the Sun, Earth, and Moon) for NEOs of negligible mass. To this end, we compute the NES capture probability from the NEO population as a function of the latter’s heliocentric orbital elements and combine those results with the current best estimates for the NEO size-frequency and orbital distribution. At any given time there should be at least one NES of 1-meter diameter orbiting the Earth. The average temporarily-captured orbiter (TCO; an object that makes at least one revolution around the Earth in a co-rotating coordinate system) completes (2.88 ± 0.82) rev around the Earth during a capture event that lasts (286± 18) d. We find a small preference for capture events starting in either January or July. Our results are consistent with the single known natural TCO, 2006 RH120, a few meter diameter object that was captured for about a year starting in June 2006. We estimate that about 0.1% of all meteors impacting the Earth were TCOs.

The paper describes an exploratory simulation of the probability of objects being captured into at least temporary 3-body orbits in the Earth-Moon system, also taking into account perturbations from the Sun. Usually the orbits last a few months. Longer orbits are less common.
There is certainly the small possibility that an object could spend several years or longer, but the probability of this naturally decreases as the the lifetime increases. You can think of it as the "keyhole" for entering a long term 3-body orbit shrinking as the duration of that orbit increases.
The population of near Earth objects to be captured decreases as the size increases. Thus there could be many temporarily captured objects of micrometeor size, but few large enough to be easily observed. The chaotic nature of these loosely-bound orbits makes them harder to track and analyze; you can't assign Keplerian orbital elements and look for repeat passes.
What is really surprising is the first sentence of the conclusion (Section 4):

We provide the first estimate of the orbit and size-distribution for temporarily captured natural irregular satellites of the Earth. We predict that there is a one-meter-diameter or larger NEO temporarily orbiting the Earth at any given time.

You can also listen to the conference call with Bill Bottke (MP3 plus other documents) about "minimoons" presents an alternative to Asteroid Redirect first got me interested in orbital mechanics.

Here is an example of one theoretical orbit from here or here:


Do orbital resonances always form naturally?

If the question is "if I throw two planets to orbit a star at random direction, would they form an orbital resonance?" -- then in general, no. A resonance is an integral ratio (1/1, 2/1, 3/5, etc.) between the periods of motion of objects -- i.e., the ratio of their periods forms a rational number. Formally speaking the odds of getting a integral ratio (let alone a strong, low-order ratio, since those are the dynamically interesting ones) if you set the system up "randomly" should be infinitesimal, because irrational numbers are (infinitely) more abundant that rationals.

However, if the orbits of one or both of the planets can change over time, then the ratio between their periods changes, and they can end up in a resonance. (Which is maybe answering the title question.) How often this happens depends on whether the planets happen to start near a strong resonance, and on how rapidly the orbits change. (If the orbit of a planet changes slowly, then it won't encounter new resonances very often; on the other hand, rapid orbital change can overwhelm the effect of weak resonances, so that the planet passes through the resonance without being caught.)

For example, it's thought that Neptune and Pluto were originally not in resonance; but the gradual outward migrations of Neptune (due to various gravitational encounters between planetesimals and the giant planets) changed its orbital period and meant that eventually it reached 2/3 resonance with Pluto, and Pluto was "captured" by the resonance, after which it stayed in resonance with Neptune.

The vast majority of objects in the Solar System are not in resonance with anything else, which is perhaps another way of answering your question. (I.e., in practice it doesn't happen very often.)


Did you get this from Wikipedia? If you did, I sympathize; all the entries related to this give no explanation for the phenomenon. If you didn't go there . . . Well, it won't answer your question.

But I think I've found the answer somewhere else. I'll go over some background stuff for anyone out there who doesn't know much about this, to make it a bit clearer (although you should be able to skip all this). Protostars are small pre-pre-main-sequence objects. At first, they aren't more than small clumps of gas and dust in a stellar nursery (a special nebula, perhaps). Soon, though, more matter begins to come towards a clump, and gradually the clump builds up enough matter that it begins a slight gravitational collapse into a definite object. Temperatures build, and eventually the object becomes a pre-main-sequence star.

These pre-main-sequence stars begin to accrete matter (here's where you should stop dozing off) into a circumstellar disk. At the moment, it doesn't matter if the disk will become a protoplanetary disk or something completely different; for now, we can treat it as a circumstellar accretion disk. Matter "falls" into the disk and towards the star; the star absorbs most of it. However, something else comes into play, and that's the star's magnetic field.

A lot of objects in the universe have magnetic fields, and a young pre-main-sequence star is no exception. It's field may be stronger than that of the Earth - so strong, in fact, that some of the matter in the circumstellar accretion disk follows slightly different paths than it normally would - paths along the field lines of the star's magnetic field. While there isn't enough evidence (scarcely any at all, really) to support or, more importantly, disprove this idea, it is though that the magnetic field lines lead the matter towards a point on the star, perhaps at its poles, and then compress it together into an astrophysical jet (or bipolar outflow) that shoots out into space. Such jets are often associated with other cool phenomena, such as Herbig-Haro objects.

So for everyone who considers this post a "Too long; didn't read" answer, here's  a summary:

The truth is that nobody really knows. The current explanations for this phenomenon are that they are caused by matter from accretion disks being concentrated by magnetic field lines onto a single (well, double, one on either end of an axis) point, which then becomes an astrophysical jet.

I hope this helps.

Note: This paper, specifically the second page, was extremely helpful. Other pages that might interest you (or provide more information for those who are curious about this and related topics) are:

Protostar

Herbig Ae/Be star

Herbig-Haro object

Paper #1

Paper #2

Pre-main-sequence star

Circumstellar disk

Astrophysical jet

Bipolar outflow

Softpedia.com

Sci-news.com

IOP


Relationship between metallicity and color? Should Pop. I stars be blue?

I think the misunderstanding (which I've been asked about before) is because Population II stars are redder as a population, even though for a given mass, a main-sequence Pop II (i.e. metal-poor) star would be bluer.  That is, if you compare representative samples of Pop I and Pop II stars, the Pop II stars will on average be redder.

This because Pop II stars are generally old so the blue stars have already died as supernovae or evolved into compact remnants. So looking at the whole population, you'll see more of the longer-lived, lower-mass stars (including red giants), which are redder.

The same argument applies to galaxies that aren't forming stars appearing redder than those galaxies that are still forming stars.  Since new stars are being born, some are blue, but they live fast and die young, so unless the galaxy keeps forming stars, it will become red overall.  Again, this isn't because of the stars being redder for a given property but because the population lacks young, hot, blue stars.


A recent study indicates that Cold Jupiters similar to Saturn and Jupiter greatly outnumber Hot Jupiters. The authors studied 18 years worth of data to find long-period exoplanets, that is planets far from their host star.

Cold Jupiters, being farther from their host star, have longer periods than Hot Jupiters. Therefore, they need to be observed over a longer time frame to see multiple transits. ETs would need to observe the Sun for 12 years before seeing two transits of Jupiter, and that still wouldn't be enough to confirm its presence.

Planets that are further out also induce smaller variations in the velocity of their host star, so the spectrum of the star is shifted by a smaller amount. Therefore higher resolution spectrographs are needed to detect Cold Jupiters using the radial velocity method.

Finding more Warm and Hot Jupiters at the beginning of the era of exoplanetary discovery was an observational bias due to the limits of the instrumentation available at the time, and the amount of time needed to find long-period exoplanets.


How are parachutes usable in other places than Earth?

Parachutes have never been used on the moon, but they are viable for Mars because Mars does have an atmosphere - albiet one much lighter than that of Earth.  For that reason, parachutes cannot be the only means of slowing down on Mars --- for example, in the case of the Curiosity Rover, that's why an elaborate booster and crane combination was used (see figure).  Another approach, taken by the Pathfinder Rover, was to use large airbags to bounce on landing.

One benefit to using parachutes on Mars, is that they can be used at higher velocities --- unlike on Earth where the heavier atmosphere (and thus stronger drag force) would shred a parachute at comparable speeds.


What do we know about the lifecycle of the Milky Way (or any other spiral galaxy)?

If we look at Newtonian physics, and how galaxies will interact, a central black hole should just be considered as a massive, dense object.

The Milky Way does not fall into its own central black hole, it orbits the centre of mass - as physics tells us it should.

When we get closer to Andromeda, the gravitational influence of Andromeda will act more strongly on us, and when we get really close, individual masses within each galaxy will have dramatic effects on each other, but as with any such system, the two black holes will not suck everything in.

If the two galaxies end up coalescing - which is not a given - the orbits of the stars and the black holes will be very complex. For billions of years the 2 black holes will orbit each other, getting closer as they shed energy, but during that time the stars around them will suffer many effects, including:

some will be expelled
some will hit the black holes
some will go nova
and so on

The best way to understand what will happen is to avoid thinking too deeply about black holes being weird, and treat them as dense masses. For most purposes this will help you model galactic collisions.

Have a look at this NASA simulation of the collision between the Milky Way and Andromeda:


Where do we have it from that the Moon is migrating away from Earth?

There are, I think, at least four parts to this argument: the first being the theoretical argument that ties it all together and the remainder being observational evidence for the Moon's orbit increasing in size.
1. The underlying theoretical argument. This, of course, is the idea that tidal braking causes the Earth to slow down in its rotation and the Moon to move further out in its orbit. (Once upon a time this also caused the Moon to slow its rotation, until it became tidally locked, with its rotation period = its orbital period.) This predicts two things: the Moon should be gradually getting further away from the Earth over time; and 2) The Earth should be slowing down in its rotation (days getting longer) over time.
There are at least three sets of evidence supporting this.
1. The increasing distance of the Moon, as derived from lunar laser ranging. The key point here is that it's the average distance that's increasing. Since the Moon's orbit is elliptical, the distance varies over the course of one lunar orbit, and there are additional variations introduced by the gravitational influence of the Sun, other planets, the not-perfectly-spherical shapes of the Earth and Moon, etc. What this means is that if you trace the Moon's distance over time, you will see it increasing and decreasing, but you will also see that the average distance is gradually getting larger. (And this is what has been measured with the laser ranging experiments; e.g., this 1994 article by Dickey et al. which found that semi-major axis of the Moon's orbit was increasing by about 3.8 cm/year.)
2. The slowing down of the Earth's rotation. As required by basic physics, if the Moon is getting further out in its orbit, and thus gaining angular momentum, there has to be a compensating loss of angular momentum, and this happens via the Earth gradually spinning more slowly. This has been measured in at least two ways:
A. Timing of historical lunar and solar eclipses. There are Chinese records of eclipses going back to roughly 400 BC, and Babylonian records back to almost 800 BC, as well as more recent Greek, Arab, and European records. These can be used to estimate changes in the length of the day, in two ways. First, some records include an approximate time of day, so we can run our calculations backwards and figure when an eclipse visible at, say, Babylon in 200 BC should have occurred. And even when we only know the day (and location) of an eclipse, we can predict where it should have been visible. As explained by this article describing a 2016 study by Stephenson et al.: "The oldest event in the catalog, a total solar eclipse that occurred in 720 B.C.E., was observed by astronomers at a site in Babylon (now modern-day Iraq). But, working backward, today’s astronomers would have predicted that the eclipse should have been seen a quarter of a world away, somewhere in the western Atlantic Ocean. The discrepancy means Earth’s rotation has gradually slowed since the 8th century B.C.E."
B. Geological and paleontological measurements of the length of the day.* It turns out there are some geological records which can be used to derive the number of days in a lunar cycle, or the number of days in a year, millions or hundreds of millions of years in the past.
One way is via the analysis of "tidal rhythmites", which record alternating effects of ocean tides. Since tides have cycles of twice per day (high and low tides) due to the Earth's rotation and twice per lunar orbit (spring and neap tides), you can use the combination to work out how many days there were in a lunar orbit. There can also be yearly variations, which allows you to work out how many days there were in a year when the tidal rhythmites were formed.
Another way is by studying certain fossils, where the growth of part of the organism is recorded in both daily and annual variations: for example, the growth of a shell can vary over the course of a day, but also over the course of a year (e.g., more growth in summer, less in winter). Putting this together allows you to figure out how many days there were in a year when the organism was alive. A fascinating recent example is this careful study of a 70-million-year-old fossil of a rudist (an extinct type of mollusc), which reveals that a year had 372 days back then, so that each day was 23.5 hours long instead of 24 hours.


Frames per second capabilities cameras currently installed in Modern Telescopes?

The types of sensors that are used for "Lucky Imaging" can work at frame rates easily capable of sampling changes in the speckle pattern due to turbulence in the atmosphere; 50 Hz is certainly possible.

The detectors used are electron-multiplying CCDs and are available on a number of telescopes around the world.

A video camera can of course be put at the focus of any ground-based telescope.

Such devices are not usually put on space telescopes because they have no need to deal with a turbulent atmosphere. Space telescopes are usually concerned with taking pictures of very faint objects, where minimising the readout noise of the device (to a few electrons) is a priority, and necessitates longer readout times - often tens of seconds.


How to calculate the frequency of a gas giant eclipsing the sun from a moon?

Let’s start with an easy example and say the orbit of the planet with radius $r_p$ around the sun with radius $r_s$ is circular, the orbit of the moon around the planet is circular, the inclination of the moon's orbit is 90 degrees from the planet's orbital plane, and the orbital period of the moon is 1/360th the orbital period of the planet. Let's call the vector from the sun to the planet $v_1$ (with length $d_1$), the vector from the planet to the moon $v_2$ (with length $d_2$), and the vector from the sun to the moon $v_3$.  Let’s assume the moon is so small we don’t have to worry about orbital passes in which there is a solar eclipse visible from one side of the moon, but not from the other.
Every consecutive time the moon ascends through the planet's orbital plane, the angle $\theta$ between $v_1$ and $v_2$ will increase by 1 degree.  A little trigonometry (does everyone remember SohCahToa?) will give us the angle between $v_1$ and $v_3$ as $\phi = \arctan (d_2 \sin\theta/(d_1+d_2 \cos\theta)$.  So from the perspective of the moon, the angle between the sun and the planet in the sky will be $\alpha=\theta-\phi$.  The planet will be at least partially eclipsing the sun in the lunar sky whenever $\alpha<\beta_p+\beta_s$ where $\beta_p= \arctan(r_p/d_2)$ is the half angle of the planet in the lunar sky, and $\beta_s= \arctan(r_s/\sqrt{(d_1+d_2 \cos\theta)^2+(d_2 \sin\theta)^2}$ is the half angle of the sun in the lunar sky.
Here is a diagram looking at the planet's orbital plane when the moon passes through the plane:

Clear as mud?  Let’s do an example with Jupiter (which we will now assume has a circular orbit) and our Sun!  Jupiter’s orbit is 4333 days, so our moon’s orbital period should be 4333/360 =~12 days.  Kepler’s third law says the ratio of the cube of the SMA (semi-major axis) to the square of the orbital period is constant for moons about a body, and we know Jupiter’s moon Callisto has a SMA of 1.883e6km and an orbital period of 16.69 days, so we have the equation $d_2^3/12^2=(1.882e6)^3/16.69^2$.  Solving yields $d_2=1.51e6km$ is how far our moon is from Jupiter.  Jupiter has a radius of $r_p=6.9911e5$km so the half solid angle of Jupiter from our moon is $\beta_p = \arctan(6.9911e5/1.51e6)=0.5radians$ or 28.6 degrees.  The solid angle of the sun from our moon will average about $\beta_r = \arctan(6.957e6/7.8e8)=0.00892radians$ or 0.551 degrees, but it varies a bit in the third significant digit.
Plug it all into the above equations and you get an orbit without an eclipse, followed by 59 orbits with an eclipse and then another orbit without an eclipse.  Here is the matlab code for posterity:
theta = 0:1:359;
d1=7.8e8;
rs=6.957e6;
d2=1.51e6;
rp = 6.9911e5;
phi = atan(d2*sin((pi/180)*theta)./(d1+d2*cos((pi/180)*theta)));
bp = atan(rp/d2);
bs = atan(rs./sqrt((d1+d2*cos(pi/180*theta)).^2+(d2*sin(pi/180*theta)).^2));
alpha =pi/180*theta-phi;
sum(alpha<bp+bs)*2-1

Notes:

This was an easy example because I got to choose all the orbital geometries.  Want to make it harder? Choose a planetary orbital period that isn’t an integer multiple of the lunar orbital period. Want to make it harder? Choose an inclination less than 90 degrees so that some eclipses occur when the moon is outside the planetary orbital plane. Want to make it harder?  Choose highly elliptical orbits for both the planet and the moon, so that $d_1$ and $d_2$ both vary as separate functions of time.  Want to make it harder?  Specify a large moon so that some eclipses can only be seen from some points on the moon, but not others.

I provided the solution for partial eclipses.  If you want the solution only for full eclipses, use this equation: $\alpha<\beta_p-\beta_s$ instead of this equation: $\alpha<\beta_p+\beta_s$ above.

I did not provide a closed analytical solution for the general problem of counting consecutive solar eclipses from a moon given the orbital elements for the planet and moon and the radii of the sun and planet.  However, in describing the solution for a particular class of orbits, I think I've clearly detailed the calculation method that could extend to any particular orbit. The key is in understanding that even though a moon's orbital plane is fixed in an PCI (Planet Centered Inertial) coordinate system, it appears to precess in a sun-centered, planet-fixed coordinate system. A similar example of this phenomenon is the four seasons here on earth due to the Earth's axial tilt.  The Earth's axial tilt is nearly constant, but from our perspective here on Earth, it appears to precess.

Classical orbital mechanics is disgustingly hard and unforgivingly unintuitive. It's much easier to use Satellite Toolkit or Universe Sandbox to plug in your initial orbits, propagate them, and count the eclipses as others suggest.


When would we detect a tiny meter size natural satellite in a geostationary orbits?

tl;dr: At distances far enough from Earth that the motion with respect to the stars was slow, a serendipitous survey photographic plate from a large enough telescope might catch a trail, and in a doubly serendipitous situation it might have been a short exposure, duplicated on the next night, an Earth orbit suspected, and a hunt for a second Earth satellite begun.
However starting in the 1960's and 1970's radar and visual scans for artificial satellites in Earth orbit would have found this natural satellite in Earth orbit if it were low enough.

I'll start with @CarlWitthoft's 5 meter asteroid refer to this answer and especially this answer. Two equivalent equations to get the absolute magnitude of an asteroid are:
$$ H = C - 5 \log_{10} D - 2.5 \log_{10} p_V$$
where $H$ is absolute magnitude, $p_V$ is albedo, D is in km, and $C$ = 15.618, and
$$M_{Abs} = 5 \left(\log_{10}(1329) -\frac{1}{2}\log_{10}(\text{albedo}) -\log_{10}(D_{km})\right).$$
A 5 meter diameter asteroid with an albedo of 0.1 has an absolute magnitude of +29.6.
The apparent magnitude from this answer:

Knowing the absolute magnitude of an object, you calculate the apparent magnitude $m$ using:
$$ m = M_{Abs} + 5 \log_{10}\left(\frac{d_{SR} \ d_{RE}}{1 \ \text{AU}^2 O(1)}\right), $$
where $d_{SR}$ and $d_{RE}$ are the Sun-Roadster and Roadster-Earth Sun-satellite and satellite-Earth distances, each normalized by 1 AU, and the factor $O(1)$ is the phase integral, of order unity, taking into account the angular difference between the direction of illumination and the direction of viewing. In an order of magnitude calculation, this only becomes really significant when the body moves between the Sun and the viewer. See https://en.wikipedia.org/wiki/Absolute_magnitude#Solar_System_bodies_(H).

Let's pick two distances. One is the geostationary distance where the satellite would appear to hover above the observer, probably drifting up and down in a roughly annelema shape because the Earth's oblanteness will eventually tilt the orbit. See Geostationary orbit; Orbital stability. It will have a distance to Earth as close as 36,000 km.
The other is a low Earth orbit, but high enough that it will not decay due to drag too soon. Call it 1000 km altitude, or a circular orbit with a 7378 km semi major axis.
Plugging all of this into the equation above, I get:
     orbit                   closest distance      visual magnitude
Geosynchronous altitude         36,000 km                +20.6
Low Earth Orbit                  1,000 km                +16.7

In low Earth orbit the apparent magnitude is almost as bright as Pluto, but it's going to be moving pretty fast. $\sqrt(GM/a)$ gives 7350 m/s, at a distance of 1000 km that's about 0.4 degrees per second. Any large telescope that's being used in astronomy will be tracking the motion of stars, or close to that, so it will be fast +17 magnitude track rather than a dot, and last only a fraction of a second. That probably wouldn't expose a photographic plate or if it did it would be dismissed as an artifact, meteor, or scratch. Visually it wouldn't be noticed.
At GEO type distances and +20.6 magnitude, the object would move about 0.25 degree in a minute, so it might also be captured on a photograph, but by the time the plate was developed it would be impossible to know when it appeared in a long exposure. However if the exposure (say at the Hale 200 inch telescope) were short, it really is possible that at a short-term trajectory on the celestial sphere could be considered. The problem is that nobody would suspect that it was in Earth orbit, and they would extrapolate to a heliocentric orbit and never find it again.
If the plate happened to be a series and there was another exposure of the same patch of sky the next night, then they would see it again and get pretty suspicious that it was in Earth orbit.
However, in a post-Sputnik cold war era radar and optical searches of the sky for objects in Earth orbit became particularly interesting.
So I would say that satellite surveys (both optical and radar) in the 1960's and 1970's would be the first likely candidates to find this 5 meter, 0.1 albedo satellite.

For some insight on optical tracking, see the two videos linked in Are commercial communications satellites in GEO being constantly monitored by telescopes?. Currently these links will take you to a new tab with the YouTube video:
https://www.youtube.com/watch?v=8ebIAUjFfZM
https://www.youtube.com/watch?v=4FXX1kSNljU
If you would like to see them here, then leave a comment or answer or vote at Interest in looking into adding the YouTube viewer?.


What implications do younger Earth and Moon have on Late Heavy Bombardment genesis hypotheses?

From the article: "In fact, its age may be 4.36 billion years old." This is still older than the estimated age of the Late Heavy Bombardment, so it wouldn't affect that hypothesis much. In fact, it may actually help the hypothesis, as it means the Moon would have more of its interior molten (though probably not by much). It's suggested that the magma from underneath the crust filled the lunar mare after the giant impacts through fissures in the crust. (Wikipedia)


Where can I find source data for the graph of metric expansion over time, as shown in this ESA web page?

I was hoping a physicist might answer this, as I'm just an enthusiastic armchair scientist whose undergrad calculus was 30 years ago. But as no one else has had a go, let me try an answer.

The graph in the web page is a simplistic and somewhat exaggerated representation of the evolution of the cosmic scale factor a(t) over time. You can derive the "source data" yourself using the following metrics for the three distinct periods in the expansion of the Universe, since the only variable is t which in this case is the age of the Universe. The scale factor a(t) is dimensionless.

For the earliest period of the Universe up to about 47,000 years after the Big Bang, the Universe was dominated by radiation. The evolution of the scale factor in the FLRW metric for this period is given by: $$a(t) \propto \sqrt t$$

After that very brief time (in cosmological terms), matter took over as the dominant factor in the Universe for the next 9.8 billion years, and the evolution of the scale factor is given by: $$a(t) \propto t^{2/3}$$

Finally, the Universe in the current period is dominated by dark energy, and the evolution of the scale factor is given by: $$a(t) \propto exp (Ht)$$
where H is the Hubble constant.

In practical terms, if you'd like to generate a graph similar to that in the web page, you can set $t_1$ = 13.79 Gyr (i.e. the current age of the Universe). I found that the first era of radiation domination is so short that it's not worth including in the graph.

Here's my simplistic rendition:


Is the Alpha Centauri star system moving closer to us?

Yes, it's moving closer.

Per Wikipedia.

Source.

In about 30,000 years Alpha Centauri will be 3 (and change) light years away, then it will start moving further away.

If anyone wants to provide a more detailed explanation on stellar movement and charts, please feel free.


Probably the most reliable guide to vsop2000-p11.dat is the reference implementation vsop2000.for.
The input line in question is the first of 5399 k=0 records contributing to $x$.
Those are followed by 5842 k=1 records, 610 k=2 records, and so on up to k=8.
Then $y$ (k=0..8) and $z$ (k=0..7) follow the same pattern.

If I read the Fortran code correctly, the fixed-length (not blank-delimited) record format is:

Columns 1-7 are an index $i$
Columns 9-63 are 17 per-body coefficients $b_{kij}$
Columns 64-72 are three zeroes, ignored
Columns 73-126 are coefficients $s_{ki}, c_{ki}$ in astronomical units

To compute a position for time $t$ in Julian years since J2000:

For each body $j$, compute $\theta_j(t) = \dot{\theta}_j t + \theta_j(0)$, where $\dot{\theta}_j$ is the mean motion and $\theta_j(0)$ is the J2000 mean longitude from vsop2000-exp.txt
For each record $x_{ki}$, compute $\phi_{ki} = \sum\limits_{j=1}^{17} b_{kij} \theta_j(t)$
$x = \sum\limits_{k=0}^8 t^k \sum\limits_{i=1}^{n_k} (s_{ki} \sin \phi_{ki} + c_{ki} \cos \phi_{ki})$
For $y$ and $z$, repeat steps 2 and 3 using the respective sets of records.

This should give you a geocentric Moon position in astronomical units.


Are all planets/galaxies moving away from *us*?

No, not all of them. Only the ones very far away.

When you walk your dog, the dog is tied to you via the leash. It does not recede into the distance.

Planets in the Solar System are all tied up to the Sun via gravity. They don't recede into the distance.

All stars in our galaxy are bound together in a big clump (the galaxy) and so they stay together.

Finally, all galaxies in our neighborhood are also bound by gravity, so they stay together.

Beyond that range, things are not bound by gravity, so they keep running away from each other.

Gravity is like a leash. If things are close enough, they stay together.

Keep in mind that the universe does not have a center. It's infinite (as far as we can tell), and infinite things don't have centers. Rather, if you take any two distant galaxies, they all recede from each other. But the expansion does not go back in time to a common center.

Here's a comparison. Imagine you're a bi-dimensional creature; instead of having 3 dimensions, you only have 2. Like a stick figure drawn on a paper. You only exist in the 2 dimensions of the paper, and can't even imagine the 3rd dimension.

Now imagine a balloon. You draw a bunch of stick figures on it. Now blow air into the balloon. All stick figures recede from each other, but for them there's no center for that expansion - because all they know is the surface of the balloon. They don't know anything that's not on the surface. So they all see each other running away, but there doesn't seem to be a center for that expansion.

It's just an analogy, because the universe is not a balloon, but it will get you started.

Finally, red-shifted galaxies don't emit only red light. They emit the usual radiation like everything else. But because us and them are running away from each other, radiation from one side reaches the other side a little bit stretched out. For the visible part of the spectrum, it's like taking the spectrum and shifting it towards the red part.

That's all there is to it. It's still the usual radiation, but all wavelengths are a little bit blown up.


What is the difference between Terrestrial Time (TT) and International Atomic Time (TAI)?

It's historical. TT is a successor to The International Astronomy Union's ephemeris time, and the offset between TT and TAI is a result of getting TT to match ET when they switched.

Excerpts from https://en.wikipedia.org/wiki/Terrestrial_Time:

  Terrestrial Time (TT) is a modern astronomical time standard defined by the International Astronomical Union, primarily for time-measurements of astronomical observations made from the surface of Earth.[1] For example, the Astronomical Almanac uses TT for its tables of positions (ephemerides) of the Sun, Moon and planets as seen from Earth. In this role, TT continues Terrestrial Dynamical Time (TDT or TD),[2] which in turn succeeded ephemeris time (ET). TT shares the original purpose for which ET was designed, to be free of the irregularities in the rotation of Earth.

  The present definition of TT is a linear scaling of Geocentric Coordinate Time (TCG), which is the proper time of a notional observer who is infinitely far away (so not affected by gravitational time dilation) and at rest relative to Earth. TCG is used so far mainly for theoretical purposes in astronomy. From the point of view of an observer on Earth's surface the second of TCG passes in slightly less than the observer's SI second. The comparison of the observer's clock against TT depends on the observer's altitude: they will match on the geoid, and clocks at higher altitude tick slightly faster.

  Time coordinates on the TT and TCG scales are conventionally specified using traditional means of specifying days, carried over from non-uniform time standards based on the rotation of Earth. Specifically, both Julian Dates and the Gregorian calendar are used. For continuity with their predecessor Ephemeris Time (ET), TT and TCG were set to match ET at around Julian Date 2443144.5 (1977-01-01T00Z). More precisely, it was defined that TT instant 1977-01-01T00:00:32.184 exactly and TCG instant 1977-01-01T00:00:32.184 exactly correspond to the International Atomic Time (TAI) instant 1977-01-01T00:00:00.000 exactly. This is also the instant at which TAI introduced corrections for gravitational time dilation.

(Bold emphasis mine).

For more on International Atomic time see https://en.wikipedia.org/wiki/International_Atomic_Time . Basically, it's a weighted average of multiple atomic clocks - and the altitude effect mentioned in the middle paragraph above is why they added the gravitational time dilation corrections. From the Wikipedia link above:

  n the 1970s, it became clear that the clocks participating in TAI were ticking at different rates due to gravitational time dilation, and the combined TAI scale therefore corresponded to an average of the altitudes of the various clocks. Starting from Julian Date 2443144.5 (1 January 1977 00:00:00), corrections were applied to the output of all participating clocks, so that TAI would correspond to proper time at mean sea level (the geoid). Because the clocks had been on average well above sea level, this meant that TAI slowed down, by about one part in a trillion. The former uncorrected time scale continues to be published, under the name EAL (Echelle Atomique Libre, meaning Free Atomic Scale).[7]:215


For there to be a significant greenhouse effect there has to be a lot of greenhouse gas in the atmosphere. Not just a large percentage, but a large amount of gas

There is a greenhouse effect on Mars, however, as its atmosphere is very thin. The surface pressure on Mars is only 0.6% of the pressure on Earth. As such there very little of anything to trap heat. Even though a large percentage of the atmosphere is CO2, the greenhouse effect only raises temperatures by about 5 Kelvin. http://nova.stanford.edu/projects/mod-x/id-green.html

It wasn't always this way. During a dust storm, more heat can be trapped, and temporarily raise temperatures. Early in Mars's past the atmosphere was thicker at temperatures were higher. It is even possible that at one point, a runaway greenhouse effect heated Mars, until the atmosphere thinned and the planet cooled http://www.astrobio.net/mars/did-ancient-mars-have-a-runaway-greenhouse/

The only requirement for a planet to be warmed by the greenhouse effect is for its atmosphere to contain greenhouse gases. Mars does have a greenhouse effect, as other bodies with atmospheres: Earth, Titan and Venus. https://www.nasa.gov/home/hqnews/1991/91-143.txt Other rocky bodies don't have a thick enough atmosphere for a greenhouse effect to be measured.


Does the age of the universe take into account General Relativity / Special Relativity?

This is a confusing question - your title mentions GR, but of course the age of the universe is entirely derived as a result of using GR to solve for the dynamics of the expanding universe.

The text of your question is talking about time dilation and the effects of special relativity (a subset of GR). Here there is a point to be addressed. It is not a good way to think about the explansion of the universe in terms of things receding from each other at ever increasing velocities. It is better to think about space expanding. It is also not possible to apply special relativity except in local inertial frames.

It can and has been shown that the redshift-distance relationship cannot be simply interpreted as the doppler shift due to things moving away from us at ever increasing speeds - since this of course predicts a maximum recession velocity that approaches the speed of light. In fact, beyond redshifts of $z=1.5$, the "recession velocity" in all viable models exceeds the speed of light. What actually happens is that light emitted at some earlier epoch travels across expanding space to get to us and this expansion increases the wavelength of the light.

Essential reading: Davis & Lineweaver (2003).


Can we get the gas compositions from a gas planets?

Yes.  It's certainly feasible to obtain a sample from Jupiter or Saturn, effectively getting gas from the planet, but it's not easy.

Massive planets, as a byproduct of their high mass, create large gravity wells which require a lot of fuel to fly close enough to get a sample from and return with it to where it's needed.  It would take perhaps a ton of conventional fuel to return from Jupiter with a few pounds of gas, which isn't a good return on investment.

Collecting hydrogen from Jupiter would be expensive almost certainly not worth the cost.   It would be much easier to obtain hydrogen in space from frozen water ice, frozen methane or frozen ammonia, where hydrogen is abundant.     This can be obtained in icy comets and there's even some water-ice on the moon in craters, which has a much smaller gravity well.   Also on Mars' ice caps and inside Ceres, it's believed there's about an entire oceans worth of water, perhaps more water inside Ceres than on Earth.

That's a lot of hydrogen that would be far easier to harvest than hydrogen from Jupiter.   And while it would need to be separated from Oxygen, chemical separation would require many times less energy than collecting hydrogen from a gas planet's gravity well.   You'd also have the advantage of not just harvesting hydrogen but oxygen as well.

It's not impossible to harvest hydrogen from Jupiter or Saturn, it's just difficult to imagine it being worth the trouble.


A map of all galaxies gets kind of unwieldy, like a map of all stars in the milky way or a map of every house in the country, or every grain of sand on a beach . . . you get the idea.

Start here - Local Group

Source

Then Virgo Supercluster

Source

Then local superclusters

Source

And an article, even if it's a summary it's very much worth reading, with a more recent map with a "ginormous" supercluster that includes the milky way.

and here's another, more info here.

The largest of these maps is some 520 million light years across, so this is just a tiny part of the entire known universe, which (depending on how you measure) is either 27.5 billion light years across or 84 billion light years across.

It's worth pointing out, I haven't actually answered your question, and that was deliberate, cause I think the nearest 520 million light years is enough and it's all we have a really good picture of anyway (as far as I know).  The farther out you go the more holes and inaccuracies there will be in the map.

There's also some closer stuff that's blocked by the Milky way so we can't see it, like the great attractor.  No worries about superclusters eating our galaxy one day because dark energy expansion keeps these super-clusters from merging most of the time.  Andromeda will merge with the Milky way in about 4 billion years though but we're not expected to ever merge with the great attractor, even as it pulls our galaxy towards it, the distance between us is growing.


Will Sirius B start accreting from A and become a supernova type Ia?

Will Sirius B start accreting? Yes, it is doing so now. Sirius A will have a wind and some of that wind will be captured by the white dwarf.

The effectiveness of wind capture is a strong function of relative wind speed. An analytic approximation to the accretion rate, known as Bondi-Hoyle accretion, goes as the inverse cube of the relative speed. In its present evolutionary state, the mass loss from Sirius A will be relatively weak (like the Sun) and relatively fast (like the Sun). This disfavours any significant accretion by the white dwarf.

However, in the later stages of its life, Sirius A will swell to become an asymptotic giant star. The envelopes of such stars are gradually (on timescales of millions of years) blown away quite slowly by a dusty wind. If Sirius A is around 2 solar masses now, it will lose about 1.4 solar masses during this phase at speeds of only 10-20 km/s.

Only a fraction of this mass may be accreted by the white dwarf, because the separation between the stars is still large at 25 AU (and will become larger as mass is lost from the system) compared with the likely terminal size of Sirius A (probably of order 2 au). If you look at the likely Roche lobe size, then Roche lobe overflow would require A to reach about 40% of the separation, which isn't going to happen. Exactly what fraction is captured by the less efficient wind accretion process (the majority will likely disappear into space and widen the orbit) depends strongly on the wind speed, which is hard to predict.

Even if Sirius B could accrete the 0.35-0.4 solar masses (I think that is unlikely, but lack the wherewithal to do the hydrodynamic simulation) it needs to become unstable, it is not clear if that mass will "stick". A build up of hydrogen-rich material can ignite and explode in a nova (not supernova) on the surface of a white dwarf, causing mass loss!

Finally, when will this happen? Well Sirius is probably about 300 million years old now and has another perhaps 500 million years before it starts to evolve in the way I described. It will be nowhere near the Sun then.


What is the 2D elliptical projection of the Celestial Sphere called, and how can I make one?

I want to say that's a Mollweide projection. I know that they're pretty common in astronomy; many images of the CMB use them. I was actually working with one recently. Given latitude $\varphi$ and longitude $\lambda$, the $x$ and $y$ coordinates of an object are
$$x=R\frac{2\sqrt{2}}{\pi}(\lambda-\lambda_0)\cos\theta,\quad y=R\sqrt{2}\sin\theta$$
for projecting from a sphere of radius $R$. $\theta$ is the solution to
$$2\theta+\sin2\theta=\pi\sin\varphi$$
Similar images can be found here (labeled as Mollweide projections), which are of the same shape and proportions:

Matplotlib will let you do Mollweide projections; a nice tutorial can be found here. Basemap is a possibility, although I haven't used it before. It has quite a few different built-in projections. There are some interesting features available in the different matplotlib implementations, and using the ephem package, you can use astronomical data. That said, I've found that the transformations aren't that hard to implement by hand; I was able to get some basic projections done from scratch.


AR Scorpii1 appears to be an example. It is a binary system, consisting of a white dwarf and an M-type red dwarf. It was apparently originally classified as a $\delta$ Scuti variable star, but it was later discovered to be a true close binary, with a period of 3.56 hours.

The white dwarf emits pulses, as discovered by Marsh et al. (2016), with a period of about two minutes. These are visible in wavelengths from ultraviolet to radio (see Fig. 2):

The most noticeable patterns are in the ultraviolet, and the weakest are in the radio band, but the signals are certainly detectable.

The rational for determining that the compact object is a white dwarf follows from its luminosity, its relatively long spin period (longer than any neutron star pulsar yet discovered), and the fact that the pairing with a red dwarf is "more natural" than the odd alternative involving a low-mass neutron star and high-mass red dwarf.

The system may form what is known as an intermediate polar, involving accretion onto a white dwarf influenced by the object's strong magnetic field. The other notable example is AE Aquarii, as Rob Jeffries mentioned. However, its radio emissions do not involve pulsations. Why it doesn't emit pulsating signals at radio wavelengths is another question entirely, and not one I can answer right now.

More recent results (Buckley et al. (2017)) indicate that the radio emissions from the system are polarized, similar to those of neutron stars. It seems that AR Sco shares even more characteristics with pulsars than was previously thought.

1 The Wikipedia page, I just discovered, is terrible.


Is normal matter always accompanied by dark matter and vice versa?

We lack the precision to say that there aren't regions where there is matter without dark matter or vice-versa. But what is clear is that the ratio of dark matter to normal matter, which is (or needs to be) around 5 on average to explain the flatness of the universe, varies by orders of magnitude from place to place.

The reason for this is that matter interacts with matter in a completely different way to that in which dark matter interacts (weakly) with normal matter or itself by any other means than gravity. Gravitational interactions between normal/dark matter are not dissipative.

By that I mean that the sum of gravitational and kinetic energy is preserved. When normal matter interacts with itself there are normally dissipative consequences. For example, interacting matter may become hot and radiate away energy.

That is why the normal matter in our Milky Way galaxy is concentrated into a plane. Dissipative interactions during the gravitational collapse of our Galaxy, combined with the requirement for conservation of angular momentum result in disc formation. The same is not true for the dark matter associated with the Milky Way, which is expected to have a much more spherical distribution.

Likewise, when we look at the required density profiles of normal and dark matter in order to explain the kinematics of stars and gas in our Milky Way, we see that the normal matter is much more centrally concentrated than is the dark matter - hence we talk about a "dark halo". Again, dissipative interactions in normal matter are the reason.

Another famous example of the separation between dark and normal matter is the bullet cluster. The image below shows where the hot gas is (normal matter, observed by an X-ray telescope) compared to where it is inferred that the dark matter is (shown by contours) on the basis of gravitational lensing. This pair of galaxy clusters have recently interacted, passing through each other. The hot, intracluster gas has been stripped out of the pair and shock-heated, so that it appears concentrated between the two galaxy clusters. The dark matter content of each cluster has sailed on oblivious and remains concentrated around the visible cluster galaxies.

Yet another example would be dwarf spheroidal galaxies, these are relatively small galaxies that are nevertheless almost totally dominated by dark matter - they have mass-to-light ratios of $\sim$1000, compared with a "normal" galaxy, where this ratio might be $\sim$ 10.

There are also examples of dark galaxies. These may not be galaxies made of non-baryonic dark matter, but dark matter in the sense that for whatever reason, these galaxies do not contain many bright stars. Thus although they may be massive, they do not emit much light and contain a lot of "inert", non-starforming gas.


Deriving the first moment of Collisionless Boltzmann Equation in Spherical Polar Coordinates

After correspondence with the author of those notes, I arrived at the following conclusion.
When we do $\frac{\partial f}{\partial r}$, we require that all other (other than $r$) variables which f is dependent on, are held constant. That is: $\theta, \phi, p_r, p_{\theta}, p_{\phi}$ are constants.
Note that:
$(p_r, p_{\theta}, p_{\phi})=(\dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta})$
Keep in mind that:

$p_r$, ie $\dot{r}$ is held constant: it does not mean $r$ is held constant (in fact, we are differentiating with respect to $r$).

$p_{\theta}$, ie $r^2\sin^2\theta\dot{\phi}$ is held constant. This does not mean that either $r$, $\theta$ or $\dot{\phi}$ is held constant.

$p_{\phi}$, ie $r^2\dot{\theta}$ is held constant. We are not claiming this about $r$ or $\dot{\theta}$.

Writing the steps emphasizing this statement above:
$$\int d\vec{p} p_r \frac{\partial f}{\partial r}  p_r \Bigr|_{\theta, \phi, p_r, p_{\theta}, p_{\phi}}$$
Writing out explicitly what we mean by $d\vec{p}$:
$$=\int \mathrm{d}(p_r,p_{\theta},p_{\phi}) p_r \frac{\partial f}{\partial r}  p_r \Bigr|_{\theta, \phi, p_r, p_{\theta}, p_{\phi}}$$
We can move out the partial differentiation because all other term than $\frac{\partial f}{\partial r}$ is held constant: [THIS IS THE MAIN STEP]
$$=\frac{\partial}{\partial r}\int \mathrm{d}(p_r,p_{\theta},p_{\phi}) p_r^2 f \Bigr|_{\theta, \phi, p_r, p_{\theta}, p_{\phi}}$$
Changing from the $p$-nomenclature to the variables they represent:
$$\frac{\partial}{\partial r} \int \mathrm{d}(\dot{r}, r^2\dot{\theta}, r^2\sin^2\theta\dot{\phi})  v_r^2 f \Bigr|_{\theta, \phi, \dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta}}$$
ie
$$\frac{\partial}{\partial r} \int \mathrm{d}(v_r, r \sin\theta v_{\phi},  r v_{\theta}) v_r^2 f \Bigr|_{\theta, \phi, \dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta}}$$
This step I am unsure about (ie why I don't have a $dr$ and $d\theta$ term, for example), but we say:
$$\mathrm{d}(v_r, r \sin\theta v_{\phi},  r v_{\theta}) = r^2 \sin\theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi}$$
Rewrite integral
$$\frac{\partial}{\partial r} \int r^2 \sin\theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi} v_r^2 f$$
$$=\frac{\partial}{\partial r} r^2 \sin\theta \int \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi} v_r^2 f$$
$$=\frac{\partial}{\partial r} \left( r^2 \sin\theta \nu \bar{v_r^2} \right)$$
As claimed by the notes.

Same answer on Math.SE & Physics.SE.


How do rogue planets orbit around stars in other planetary systems?

Capture of any object is never common. A rogue planet that passes a star will be accelerated by the star's gravity, and provided that it doesn't hit anything will pass the star in a hyperbolic path.

For capture to occur, the rogue planet has to lose momentum, and there are a few ways in which this can happen. The most general way is for the rogue planet to interact with one of the stars existing planets. If the rogue planet passes close one of the star's orbiting planets, it can interact gravitationally, and transfer momentum to the planet, and slow down in the process. This is rather like the reverse of a gravitational slingshot.

It is also possible that a double planet can be captured, if one of the bodies transfers momentum to the other. One of the double planet will escape, the other is captured by the star's gravity.

Capture of rogue planets is very rare. There is no evidence that it has occurred in our solar system (never in 4.7 billion years is a big Never) though some have speculated that Sedna might be a captured object. But we know that captures like this are possible, as Jupiter and Saturn have a collection of moons, some of which seem to have been captured from the asteroid belt.


In planetary and astronomical science, what exactly is, or is not, a tidal force?

The "tidal force" is simply a manifestation of good old inverse-square-law gravity.  It doesn't need a rigid body to be present to manifest.

Consider a giant circular ring of dozens of space stations established in interstellar space. They're far away from any star and just sit there happily doing whatever it was they were set up to do.  Suddenly, a star sweeps in from stage left (perhaps a fast-moving star created by a supernova-disrupted close binary system).  What happens to the ring of stations?

They are attracted towards the rogue star, of course and start moving noticeably towards it when it gets close enough.  They accelerate towards it pulled by the force of the star's gravity, which follows the inverse square law. But they don't all accelerate at the same rate. The station on the edge of the circle closest to the star accelerates fastest, stations half-way around the circle accelerate, but not as much, and the station on the far side of the circle accelerates least.

Soon the circle becomes an oval with its long axis pointing towards the star.  This is the tidal effect.  Let's say that the rogue star was dark -- a neutron star, perhaps, which had been stripped of its accretion disk by the explosion.  What would the people on the stations observe? They wouldn't notice the acceleration per se because they're in free fall.  Unless they were doing doppler spectroscopy of distant stars, they wouldn't detect their velocity (at least not at first), but they would see their neat ring of stations turning into an oval. From their point of view, some force was tugging on them and dragging them out of alignment. That's the tidal force.

From the point of view of the omniscient observer (us!) there are no tides, there's just simple gravity.  But from the point of vie of the freely falling observers in the gravity field, most of the effects of gravity disappears and what's left looks like a force.

Wikipedia has an article on the tidal force which gives a mathematical treatment. (Personally, I don't think the article is terribly clear, though it appears to be correct.)


Vectorial construction of tidal forces - or why is it centripetal at low tide

The short answer which may or may not be an "Aha!" answer is that what is plotted is what's left over after a much larger, uniform force is subtracted.

The uniform force is the Force from the Moon evaluated at the center of the Earth, and the arrows show the deviation of the actual force from that average.

Why do we do it that way? When looking at the tidal forces on the oceans we treat the Earth as a rigid body with spherical symmetry. With that, we can use a variation of Newton's Shell theorem to say that the extended Earth will move the same way as if it were a point mass at its center.

Now the oceans are fluid (the opposite of rigid) and each bit responds to the Moon's force locally.

That force is (skipping constants unnecessary to make the cartoon plot)

$$F = -\frac{\mathbf{\hat{r}}}{|r|^2}  = -\frac{\mathbf{r}}{|r|^3}$$

where the vector $\mathbf{r}$ is drawn from the Moon to some point on the Earth and $\mathbf{\hat{r}}$ is its unit vector. If the Earth's center is at $\mathbf{\hat{x}} R$ ($R$ is the Moon-Earth distance) and you subtract $-\mathbf{\hat{x}}/R^2$ you'll get that image.

In the plot below I've chosen the Earth-Moon distance to be only 10 Earth radii to highlight the slight left-right asymmetry. The tidal force is stronger on the side closer to the Moon.

import numpy as np
import matplotlib.pyplot as plt

R = 10.0
r_moon = np.array([R, 0], dtype=float)[:, None]
earth  = np.zeros(2)[:, None]
theta = np.linspace(0, 2*np.pi, 49)
positions = earth + np.array([f(theta) for f in (np.cos, np.sin)])

r = positions-r_moon
F = -r * ((r**2).sum(axis=0))**-1.5

r = earth-r_moon
Fmean = -r * ((r**2).sum(axis=0))**-1.5

Ftide = F - Fmean

if True:
    plt.figure()
    plt.subplot(2, 1, 1)
    (x, y), (Fx, Fy) = positions, 50.*Ftide
    plt.quiver(x, y, Fx, Fy, width=0.005)
    plt.plot(x, y, '-b')
    plt.xlim(-2, 2)
    plt.ylim(-1.5, 1.5)
    plt.gca().set_aspect('equal')
    plt.subplot(4, 1, 3)
    for thing in F:
        plt.plot(thing)
    plt.subplot(4, 1, 4)
    for thing in Ftide:
        plt.plot(thing)
    plt.show()


Is there a pattern between the mass of a body and the mass of orbiting objects around it?

No, there is no pattern on 1%. There are cases of double stars and (allegedly) of double planets.

In our Solar System, Charon has a mass of 11.6% of that of Pluto.


Reshaping black holes to "push" the event horizon

The only places around one black hole where gravity would be strong enough to pull an object out from inside the horizon of another black hole are all inside the horizon of the first hole.

Remember that at minimum an object just inside the horizon of a black hole is traveling toward the center at the speed of light (as its speed would be measured from outside the hole, if we could see it) and getting faster. For a second black hole to catch up to the object to pull it out, it would itself have to be moving faster than the object - faster than light - which is obviously impossible but anyway would simply merge the two black holes, without rescuing anything.


The coordinate system for an object in the sky measured relative to the horizon is known as altitude-azimuth. The altitude is the angle above the horizon (10 degrees in your case), and the azimuth is the angle around the horizon, usually measured from North. (Due east would be 90 degrees azimuth in your example.)

See Horizontal Coordinate System on Wikipedia


It is the luminosity in a straight strip like this:

(Credit: West 29 / CC BY-SA from here)
where the center of the strip passes a distance $x$ from the center of the circle at closest approach.  So starting from the center and moving out, each successive strip would have a progressively larger value of $x$, but $x$ is a constant for a given strip (with $x = 0$ for the center strip).
The only differences from the picture are that (a) the strips have an infinitesimal width $dx$, and (b) they are infinite in length rather than the finite circle shown here.


Best locations for large telescopes at high northern latitudes

You can (probably) do much better than Mount Washington, both in terms of latitude and air quality. For one, you're right about the weather. According to Wikipedia, it is a convergence point for several storm tracks and next to a breeding ground for low-pressure areas. This not only means a lot of clouds, but very turbulent air, which also lowers the quality of the sky. Also, at just a little shy of $2~km$ in height, you're going to find yourself in the middle of all these clouds, which won't be pleasant. Oh, and you should also consider the difficulties of actually building an observatory in such weather.

That being said, building a telescope at very high latitudes is certainly not a bad idea. On the other side of the globe, there is Ridge A on Antarctica, which is agreed to be "the best suited location on the surface of Earth for astronomical research".

Yes, being located on low altitudes allows you to see more of the sky, but you can never see all of the sky at once. Observatories near the equator are great for studying a vast number of different objects over the course of the year. If you're doing a more specific study though (for example, if you want to observe a certain object every night during an extended period of time), you'll have to find yourself a more northern (or southern) observatory. On these higher latitudes, there is a bigger portion of the night sky that is visible throughout the year (constellations in these areas are called 'circumpolar'). This may be desirable for some specific projects.

Looking for good northern spots to build a telescope, we can indeed start by looking at mountains in the northern part of the world. If you take any topographic map of the northern hemisphere, you should immediately notice Greenland sticking out. The big issue with Greenland is that it's covered in a giant layer of ice. You can't build on ice; you'll get in trouble when it melts. We'll disregard that for now though, and we're interested in its highest peak: Gunnbjorn Fjeld. We're in luck here; Gunnbjorn Fjeld is not covered in ice, but is a rocky peak protruding through the ice. The major difficulty would be actually building there, but I'm going to take a wild guess here and claim that modern building techniques and machines could do it. So this is probably what you're looking for. I'm not an expert on the weather in Greenland though, so I can't say anything about that (I do know that it's cold, which is good, because that means the air is stable).

Alternative locations at northern latitudes would be Iceland (but that's probably a bad idea given its geologic activity) or Siberia. The highest point of Siberia is Klyuchevskaya Sopka (which is also a bad idea given that it's an active volcano). Apart from that, googling mountains in Siberia yields lots of peaks which may or may not be well-suited for an observatory. You could probably find a mountain in the Urals as well, but they are not as high as one would ideally want.


Can I have a career in observational astronomy if I study astrophysics or does it have to be particularly astronomy?

I don't think there really is a difference between "astronomer" and "astrophysicist". Both are physicists using their knowledge to learn about the Universe. Their is perhaps a tendency to call yourself an astronomer if you're doing observational stuff, and an astrophysicist if you're more theoretical/numerical. If people ask me what I do, I use "astronomer" if I want to sound romantic, and "astrophysicist" if I want to sound clever. :)

So, you are interested in doing observations, rather than doing simulations or pure theory. But in all cases, you start by gaining basic knowledge of both observational astronomy, "normal" physics, theory, and coding. Depending on which country you study in, at some point you will probably have the opportunity to have more influence on the courses you take, so that you could skip courses of, say, particle physics and general relativity, in favor of data analysis and, even better, observing summer schools (which I would recommend even for students planning to do theory, because it's a fantastic experience). Finally, your Master's thesis (to a small degree) and eventually your Ph.D. (to a large degree) will have an impact on the path you choose for your career. But many astro*s have been known to switch.


I think it is Mars.  Although it's difficult to tell, just looking at the image, but if it were Saturn then you would also be able to see Mars.  Mars is much brighter than Saturn at the moment, so if Saturn were visible, Mars would be even brighter.  The fact that there is only one bright object (other than the moon) is, I think, conclusive evidence that this must be Mars.

Looking over the city, the large brick building on the Rue St Gille is visible, and the object is visible directly in line with the phone mast on that building. That gives a direction just East of South. Checking with Stellarium at the given time shows Mars at exactly that position, at an azimuth of 171.


Identifying cataclysmic stellar phenomena

Any of these phenomena would be visible to the naked eye if they occured in the Milky Way, or a neighbouring galaxy (such as the Magellanic clouds) and was not obscured by dust. At the distance of stars, all of these would appear star-like.

Supernovae have been seen, most recently in 1987, when there was a supernova in the Large Magellanic cloud. It had a peak magnitude of 2.9, (about as bright as gamma ursa minor, the star at the bottom of the Little bear constellation, furthest from the polestar). It was visible, but not spectacularly bright. Supernovae that have occurred in the Milky way have been very bright, outshining all the other stars. A supernova in 1006 may have been brighter than Venus. If it is close enough a supernova could be seen during the day.

All of these are highly energetic, and if you were close enough to see any detail (such as a nebula) with the naked eye, you would be toast. The exposive events (Supernovae, Kilonovae, Gamma ray bursts) would fade over time. Magnetars and Quasars are stable and would continue to shine. Remnant nebulae from Supernovae are visible in telescopes, the most well known is the Crab Nebula, M1.

A supernova in Andromeda would likely be naked eye visible, but not impressively bright, though it depends on the acutal power of the explosion, and other factors, like dust in Andromeda.

Supernovae are visible for months. Kilonovae fade faster. Gamma ray bursts may only last for a few seconds (though may be associated with a supernova that lasts longer). The visible element of a GRB might be visible, there was one that might have reached naked eye visibility, though there is no evidence that anyone actually saw it (it would only have been marginally visible in ideal conditions). Quasars are stable objects that don't fade over time, they are active supermassive black holes.

There are lots of other, more local things that can produce a "star" that brightens then fades: Planes have headlights, and as they turn the appear like a star that brightens then fades over a few seconds. If they are distant enough, the other lights might not be visible.


What of the following moon-situations would be better?

Both would be stable.   A single moon orbiting a planet is generally stable provided it's comfortably inside the true region of stability, in the Hill Sphere and provided that it's not so large that tidal forces become a stability issue or not so close that it dips inside the roche limit and might break apart.

The only way you might run into instability in your scenarios, is if you have the smaller moon very close, inside the Geosynchronous orbital distance, where the Moon would move ahead of the tidal bulge it creates and that would cause the Moon to be drawn towards the planet and in time, crash into it, similar to how Phobos is expected to crash into/break into a ring system around Mars.

As the planet moves away from the sun it's stable region expands.  The geosynchronous orbit is unrelated to the sun, that's between the masses of the planet and moon only, but as long as your moon is orbiting between geosynchronous orbit and inside the stable region of the hill sphere, it should be fine.   (unless you give the planet or moons a retrograde rotation, then you need to re-examine the long term tidal effects).    For most orbits, the planet rotates ahead of the Moon so the tidal bulge pushes the Moon slowly outwards, but this effect gets weaker as the Moon gets further, so it's generally long term stable over billions of years, unless the planet is very close to the sun and it has a tiny hill-sphere.

Only two well known moons in our solar-system orbit ahead of their tidal bulges that leads to orbital decay (a handful of tiny captured moons do too, but those are so small and their tidal forces are largely irrelevant).   Only Phobos and Neptune's wrong way moon Triton orbit ahead of their tidal bulges, resulting in them being pulled towards their planets, not pushed away.   Triton is sufficiently distant that it's timing to crash into/break into a ring system around Neptune is billions of years off.   All the other moons, for tidal reasons, should be moving away from their planets, mostly very very gradually.

The 4 Jovian moons, however, are more complicated as Jupiter has a gonzo magnetic field, a magnetosphere that's almost wind like and it's moons are tidally locked to each other and it's innermost moon is losing mass to Jupiter's magnetosphere - so . . . all bets are off when it comes to Jupiter's 4 inner moons.

2 or more large moons can sometimes interact with each other, creating some instability.  One moon systems are, more often than not, long term stable.

(too long?)


What was the size of the universe around 400 million years after the Big Bang?

There are two issues: what counts as "the size of the universe", and how to rescale this. The second issue is easier.
If you have something at distance $d_1$ at time $t_1$ it will be at distance $d_2 = d_1 a(t_2)/a(t_1)$ at time $t_2$, where $a(t)$ is the scale factor at the relevant time. To further make things nice, $a(t)=1/(1+z)$ where $z$ is the redshift at the time.
For $\approx 400$ myr and a flat universe, $z=11.35$ and $a=0.081$ (from this calculator).
Note the word "flat" above: flat and open universes are by default unbounded. So in a literal sense they are always infinitely large. However, it is entirely possible that they have a nontrivial topology (imagine a universe that repeats like a hall of mirrors; various cases are possible), except that there is at present no evidence for this possibility - if there is something like going on it is likely larger than our current cosmological horizon.
The observable universe on the other hand is finite, and grows both because light from more and more distant objects arrive and because of the expansion makes their distance larger. This is why the size is not just $ct$, it depends on the expansion. The calculator gives 367.92 Mpc as the distance to the horizon at $z=11.35$, or 1.2 giga-lightyear.


How do I obtain the brightness of a star in the night sky?

I don't know about the telescope, but I know that you will need the telescope that has CCD technology to be able to convert from counts/second (as brightness) that CCD detects to quantitative values easily.

For the observational procedure, here is the standard regardless of what kind telescope you use:

Take a picture of your object (let's name it ImObj).
Take a picture of a standard star (let's name it ImStd) or stars (i.e., stars that have real brightness in a database, let's call it MagStdReal).
Take pictures for calibration (let's name it ImCal). If CCD telescope, these images include dark and flat images.
Use ImCal to calibrate both ImObj and ImStd (i.e., removing noise and etc.).
Convert image brightness of calibrated ImStd and ImObj to quantitative values (let's call it MagStdFake and MagObjFake). Note that you can use any arbitrary zero reference point for MagStdFake and MagObjFake, but they use the same reference.
Find conversion equation to MagStdReal = f(MagStdFake). The simplest form of f() is just linear shift: MagStdReal = a + MagStdFake.
Apply MagObjReal = f(MagObjFake). And now you get to do some science.

Also note that, a telescope filters certain wavelengths with certain response function. So, you have to observe everything in one setting of telescope. You cannot mix images or products of different setting of a telescope. Only MagObjReal or MagStdReal those are telescope insensitive (ideally).

I provide very short and conceptual procedure to give you the idea. There are a bunch of details that are important. If you use keywords in the list procedure here and search in Google, you should get more relevant hits to each topic.


How close can planets form to one another?

If you're thinking about how close planets can be, you should probably consider each planet's Hill sphere, the region in which it can retain satellites. Fang & Margot (2013) did an analysis of Kepler data and found that planets had mean values of $\Delta = 21.7$, where $\Delta$ is a parameter given for two adjacent planets by
$$\Delta=\frac{a_2-a_1}{R_{H1,2}}$$
where the $a$s are the semi-major axes and $R_{H1,2}$ is the mutual Hill radius.

One system the authors consider is Kepler-11, which has 6 planets, all with semi-major axes $\leq0.466\text{ AU}$ and with only one semi-major axis greater than $0.25\text{ AU}$. The smallest $\Delta$ there is approximately $5.7$, although all the other $\Delta$s are quite small. Kepler-36, with only two planets, still has a $\Delta$ of $4.7$.

According to the Nature paper about TRAPPIST-1, all seven planets have semi-major axes within $\sim0.063\text{ AU}$. They have mean $\Delta$s of $10.5\pm1.9$ - not much different from the Kepler-11 planets, because they have smaller Hill spheres. They may be closer together, but they can be much closer together without having stability problems. Additionally, they are in a "near-resonant" configuration.

How close planets can be depends strongly on their masses, then, which in turn determines their mutual Hill radii, which determines stability.

All that said, the authors believe that the TRAPPIST-1 planets may have migrated in from further out, thus entering the resonances. Without more information, we can't know whether this is the case, but if so, it is not, then, an example of planets forming close to each other.


What is the radius of observable universe- 46 billion LY or 200 yottameters?

This radius of the observable universe is measured by the comoving distance which is the same in every direction. Using CosmoCalc (a common tool for astronomers measuring cosmological parameters) you get 46 Gly for the comoving radial distance.


Has the conjunction between Venus, Jupiter, and Regulus only occurred twice in 2,000 years?


There is one known earth Trojan - http://en.wikipedia.org/wiki/Earth_trojan

The reason L1, L2 and L3 aren't stable is covered here:   https://physics.stackexchange.com/questions/36092/why-are-l4-and-l5-lagrangian-points-stable

The reason the earth has so few L4 and L5 asteroids is cause other planets like Jupiter and Venus tend to destabilize the orbits - same reason why  Jupiter has so many.


How can we tell the difference between matter and antimatter by observation in space?

Tl;DR

Detection via polarized light - Antimatter interaction with polarized light could be detected by vector rotation;
We're mostly sure, because absence of gamma rays and characteristic Faraday polarization indicates absence of observable antimatter in meaningful amounts.

Long answer

I do believe @userLTK to be correct on his comment.

To my limited knowledge, the absence of gamma-ray bursts that characterizes matter-antimatter interactions hints of baryonic matter prevalence throughout the visible universe; an anti-galaxy near a baryonic galaxy would show a healthy amount of gamma rays originating from boundary particle collisions.

This effect would be observable between structures of any scale - e.g. star/anti-star, galaxy/anti-galaxy or supercluster/anti-supercluster.

"Light from the Depths of Time", by Rudolph Kippenhahn, mentions that

  [...] Even when, like a star, it is radiation, its spectrum remains
  exactly the same, quite independent of whether atoms or anti-atoms are
  responsible for the light.

So the light originating from anti-galaxies would look the same as ordinary galaxies. But what about gravity?

A quote from "Isodual Theory of Antimatter: with applications to Antigravity, Grand Unification and Cosmology" may help us further:

  [...] However, the photons is invariant under charge conjugation and
  travel at the maximal causal speed in vacuum, c. Therefore, the
  photon could well result to be a superposition of positive and
  negative energies, perhaps as a condition to travel at the speed c,
  in which case the photon would be an isoselfdual state, thus
  experiencing attraction in both fields of matter and antimatter.

Meaning that antimatter would also cause gravitational effects, such as lensing.

That would make remote detection of anti-matter quite hard - it would emit and bend light in exactly the same fashion as common matter.

Faraday polarization rotation could give us some hope:

  Polarized light, e.g., from non-thermal synchrotron sources, that
  passes through gas with a non-zero magnetic field will have its
  polarization vector rotated by the process of Faraday rotation. [...]
  Note that regions dominated by antimatter (positrons) cause a rotation
  opposite to that caused by regions dominated by matter (electrons).

(SLAC Summer Institute on Particle Physics (SSI04), Aug. 2-13, 2004)

But the same source mentions that

  The effect has been measured many times and the amount of rotation,
  usually expressed in terms of the so-called “rotation measure” is
  given by the line-of-sight integral [...] The fact that we observe an
  effect at all means that on average, we can’t have equal amounts of
  antimatter and matter along our various lines of sight.

So it seems that the visible universe is missing some antimatter after all.


How many degrees does the sidereal month orbit?

The Moon does move 3.1° in the 0.23 day difference between those month lengths, but it goes 360° in a sidereal month and 363.1° in an anomalistic month.

The anomalistic month, from perigee to perigee, is slightly longer because the perigee moves.
The lunar apsides take 8.85 years to go all the way around the circle.

Similiarly, the draconic month is slightly shorter because the nodes move the other way in a cycle of 18.6 years.