Can the martian dichotomy (SIH) and the moon's formation (GIH) be related?

Early in the solar system's history, the Sun was surrounded by a protoplanetary disk, full of gas, dust and rock. Eventually, protoplanets began to form. These were small, rocky bodies - smaller than the rocky planets today, but bigger than asteroids. Think of them as dwarf-planet-sized objects. All the terrestrial planets formed from them.

Anyway, as you might know, the object that slammed into the Earth (according to the GIH) was one such protoplanet, lovingly named Theia. The resulting debris coalesced into the Moon. The SIH tells of a similar story: Once again, a protoplanet headed on a collision course, this time towards Mars. It slammed into Mars, creating a large basin. Simple but effective.

There's one thing that can relate these two disaster tales together, and that is the protoplanetary disk. Remember, the protoplanets formed from the same disk. The differences in them stemmed from the materials they formed from. So we could relate the two suicidal protoplanets to each other via that connection. Also, most collisions in the inner solar system were due to the erratic orbits of these protoplanets, so it is likely that the fault for the collisions lies solely on the protoplanets.

I think you want something a bit different from random collisions, though; the previous reason is a bit of a cop-out. Well, there is something more interesting: the rocky planets weren't alone in the early solar system. There were still their big siblings: the gas giants. The gas giants hadn't yet grown to their full sizes, but their cores and early atmospheres were still pretty massive, and commanded a strong presence in the outer solar system. It is possible that the two objects that hit the Earth and Mars formed together farther away from the inner planets, and were gravitationally perturbed by the gas giants, swinging in from an outer orbit towards the inner solar system.

You might have heard of the Nice Model. Basically, this says that the gas giants formed closed to the inner solar system than they are now. They encountered some of these early protoplanets and flung them into the inner solar system, moving outwards in the process. thus, collisions increased in the inner solar system.

It's a "nice" explanation. Trouble is, the majority of this "flinging" happened later on in the solar system's history; it has been suggested that the gas giants did this and thus caused the Late Heavy Bombardment. Still this idea shows that there could have been interaction between the protoplanets and the gas giants.

So how does this relate the two suicidal protoplanets? Well, it means that they could have formed together, further out in the solar system, and have been brought in by a gas giant. They could even have been part of a larger object, which broke apart after various collisions. But if the two objects had formed together (even somewhere in the inner solar system), or been part of a larger object, they should be made of the same materials, right? So if we were to travel to Mars we should be able to find similar materials as we did on the Moon. So far there hasn't been much luck along that avenue, but the search will continue.

To summarize: The early solar system was chaotic. It is likely that the impacts were unrelated, and that the two suicidal protoplanets formed apart in the inner solar system, but they could have formed together further our and been perturbed by one of the gas giants. They may or may not have been formed neary each other, or even been fragments of the same object. Detailed analysis in the future could give us some clues as to their formation.


Constructing a periscope/telescope - trouble with lenses

You're close!  What you built is a back-to-back set of Galilean telescopes (with an overall magnification of 1:1).  The reason you're getting a smaller circle, but with a life-size image, is that the light bundles in the middle are diverging faster, and so you lose the edges if the Galilean assemblies are too far apart.

Military perisopes use optical relays to get the image down the tube without blocking out any of the view.  In simple terms, they have one relatively short-focusing  group of achromatic/apochromatic lenses (called an Objective) to focus the light to a crosshair, like in a rifle scope.  From there, the image is picked up by another long-focusing lens group called a relay.  The relay re-collimates the light (but at low angles).  The light is focused and collimated several times through the tube, until it reaches a final image plane, which is then looked at with an eyepiece (like you might have in a telescope).

Overall, you can think of a periscope as a telescope/monocular/riflescope with relatively low magnification with several pairs of relays in between.  A point of note, you need an even number of image planes (0, 2, etc) to avoid having an
upside-down image.  I can go into a lot more detail or provide some resources, but I think this most directly answers your question.  Please let me know if there's any other questions on this.

Source: I design optics for periscopes for naval applications.

Edit for additional questions:
First, the most comprehensive resource I found in my lazy search is this page: http://taylortechassoc.com/?page_id=97, which is actually written by a previous engineer at my company.  There's a ton of information there, mostly intended for people who already know about optical design, but there's probably a good amount you can get from it as well.

To the question in your comment: in your preliminary design, the positive and negative lenses are taking collimated light input, and outputting collimated light.  However, since they are acting as a telescope with magnification power, the angle of the rays coming out is wider than the angle going in (by a factor equal to your magnification).  The trick to a relay lens design (used in periscopes and some kinds of endoscopes, as well as other applications) is to get the collimated spaces short enough that you don't lose that light before it spreads out.  Or more accurately, that you get the angles of the collimated light to be small enough that you can spread out your lenses quite a bit while still maintaining this condition.

Regarding field of view: you're absolutely not stuck with a small FOV.  That depends entirely on the focal length of your objective.  A short focal length lens as your objective will give you a wider field of view (so long as your light isn't clipped by your top mirror).  You could also have an objective that changes focal length (a zoom lens) to change your field of view as desired.  For your application, an SLR camera lens could work quite well for this.

Regarding tube length: You could vary the spacing in the sections where the light is collimated.  This will impact how much light is being lost from it spreading out too much (typically limited by whichever section is longest).  The result will be a dropoff in the relative illumination (how bright the edge of the FOV is compared to the center).  This is an effect known as vignetting, and can be mitigated, but not without tradeoffs.


Are there known centaurs which orbit between Jupiter and Saturn without crossing their orbits?

The JPL Small Body Database lists 5 asteroids and 5 comets with q > 4.56 au and Q < 9.04 au.
Two of those asteroid orbits have large uncertainties due to few observations and may not actually meet your criteria. These are fairly well determined:

Asteroid 494219 has a highly inclined 14-year orbit which occasionally brings it within 2 au of Jupiter. It was discovered in 2012.
Asteroid 2000 GM137 has a 22-year orbit which occasionally passes within 2 au of Saturn. It was discovered in 2000, unobserved for a few years, and recovered in 2014.
Comet 29P/Schwassmann-Wachmann 1 was discovered in 1927, has a nearly circular 15-year orbit, and is known for frequent outbursts.
Comet 39P/Oterma was discovered in 1943 on an 8-year orbit. A 1963 encounter with Jupiter switched it to its present 20-year orbit, where it became dormant and was lost until 2001. Numerical simulations indicate a prior Jupiter encounter in 1937; it was likely also a Centaur before that.


The so called expansion of the universe is not as trivial as most people think. What is happening, in fact, is that the distance between two points in space (note that I'm not talking about objects with velocities, but just coordinates in space) increases with time in a manner proportional to a given factor (in this case, the Hubble constant - which is actually not a constant in time, but let's ignore that for now).

This means that galaxies are not exactly moving away from a specific point with a particular velocity (i.e., a particular rest frame), but that the space between them is expanding. It is somewhat hard to wrap your head around the difference between these two situations, but this has to do with the geometry of space, and not with what's inside it.

It is widely accepted that the expansion of the universe is homogeneous and isotropic (the Cosmological Principle), which means that there is no "special" position (rest frame) on the universe, and whatever your velocity and your location, the universe will seem to expand the same way.

The best way to imagine this is to reduce to 2 dimensions expanding in a third dimension. For example, suppose that you live in a galaxy embedded on the surface (a 2D universe) of a balloon (completely unaware of the third dimension), with other galaxies distributed homogeneously. If, for some reason, the balloon is expanding isotropically and, every point of the surface will be moving away from every other point, so will every galaxy from each other. Think about this: is it possible to find a special, "still", point (without resorting to the elusive third dimension) where you will see galaxies moving from you in one side and galaxies moving towards you on the other side?

Our universe works similarly, but we're embedded on a three-dimensional space instead of a two-dimensional one.


How many science instruments can be used in parallel with the Hubble Space Telescope?

There are two types of parallel observations: Coordinated Parallel and Pure Parallel.

Coordinated Parallel means part of the same proposal, so it has to have the same scientific justification: you have to come up with a good case for why the different observations would be useful for the same project. (This is case with CANDELS.)

Pure Parallel means something independent of the main observation, which I think is more what you're asking about. The proposal guidelines currently note that these are allowed only when the primary observation is spectroscopy (with the COS or STIS instruments).

One factor is that you need some specification for how you use a given instrument: what filter, what exposure time, what spectroscopic grating, etc. This requires that someone consider a plausible science case for what are effectively random pointings. (And you need to make sure that you're not accidentally going to try taking a long exposure of Vega, which might actually damage an instrument.)

And, yes, I gather there are data-storage and overhead limits, as well as the idea that running the data transmitters at maximum may shorten their lives. From the WFC3 Instrument Handbook (e.g., here):

  The primary restriction on parallel observations, both coordinated and
  pure, is that they must not interfere with the primary observations:
  they may not cause the primary observations to be shortened; and they
  must not cause the stored-command capacity and data-volume limits to
  be exceeded.

and

  In order to prolong the life of the HST transmitters, the number of
  parallels acquired during each proposal cycle is limited. Proposers
  must provide clear and strong justification in order to be granted
  parallel observing time.

(I think the practical answer to your "how many instruments" is two, though I haven't seen an explicit statement to that effect. This page includes specific limitations for individual instruments -- e.g., which ones can be used in Pure Parallel mode and with what other instruments.)


Why was helium hydride (HeH+) the universe's first molecule?

At the risk of self-plagiarism: (https://physics.stackexchange.com/a/532568/43351 with a bit added).
Molecular chemistry in the early universe requires species with bound electrons. Helium hydride is the first molecule to form because neutral helium atoms, formed about 120,000 years after the big bang, could combine with plentiful protons; but it was another 260,000 years until significant numbers of neutral hydrogen atoms formed, and it is only once these are present that there is a route to forming H$_2$.
Details:
Helium has a much higher ionisation energy than hydrogen and therefore starts to recombine at higher temperatures (about 7000 K at redshifts of $\sim 2500$, compared with 3000 K and a redshift of 1100 for hydrogen). Thus in the primordial gas of hydrogen and helium, it is the helium that recombines first. There is therefore a period of time in the early universe, $120,000 < \tau < 380,000$ years, in which almost all the hydrogen is ionised, but most of the helium is in the form of atoms.
The two react to form helium hydride
$${\rm He} + {\rm H}^{+} \rightarrow {\rm He H^+}$$
As you might expect, the concentration of this molecule is low, because the temperatures were still high enough to easily radiatively disassociate it - about 1 part in $10^{21}$ at $z \sim 2000$ (Stancil et al. 1998; Galli & Palla 2013).
This is the first molecule to be produced with any important level of abundance. Note, this appears to be a somewhat arbitrary definition, since it is also claimed (e.g. Lepp et al. 2002), that He$_{2}^{+}$ formed the first molecular bond, via He$^+ +$ He, but was too weakly bound to survive in any concentration (the concentration peaks about 100 times lower than HeH$^+$ according to Galli & Palla 2013).
It is also possible to form small quantities of $H_2$ at $z> 2000$ via the reaction of a hydrogen atom with another in an excited state: H + H$^* \rightarrow$ H$_2$; but of course, although the hydrogen molecule is much more strongly bound than HeH$+$ (the dissociation energy of $H_2$ is 4.5 eV, versus about 1.8 eV for HeH$^+$), there is very little atomic hydrogen present and this reaction requires not one, but two hydrogen atoms to get together. It is not until H atoms recombine in quantity some 260,000 years later that hydrogen molecules are formed in various gas phase processes and H$_2$ becomes the dominant molecular species.
The dihydrogen cation, H$_2^{+}$ cannot form directly before either helium hydride or atomic hydrogen are formed as precursors. e.g.
$$ {\rm HeH}^{+} + {\rm H} \rightarrow {\rm H}_2^{+} + {\rm He}$$
$$  {\rm H} + {\rm H}^+ \rightarrow {\rm H}_2^{+}$$
and thus H$_2^{+}$ forms after HeH$^+$.


Catalogs for type K stars and their planetary systems?

There are many star catalogues, and from your question I cannot easily tell which would be the best one for you. You probably need one which contains positions of the star and their spectral type, maybe further properties as well.
Since you mentioned planetary systems, I think you might actually want to look at NASA Exoplanet archive (and filter it for K-type central stars).


How to discover Neptune from the Uranus orbit (by computer simulation)

Here is what I did:

Based on their masses, it is safest to initially consider Jupiter and Saturn as well as Uranus. It might also be fruitful to include the Earth in the analysis, to get relative positions, observation angles, etc. So, I will be considering:

Sun
Earth
Jupiter
Saturn
Uranus
Neptune

Get the standard gravitational parameters (μ) for all of them
Get initial positions and velocities via JPL/HORIZONS for all these planets. I had some data lying around from J2000.5, so I just used the state vectors from the 1st of January, 2000, at noon.
Write an N-body integrator with built-in MATLAB tools. Integrate this incomplete Solar system once without Neptune, and once with Neptune included.
Analyze and compare!

So, here's my data, and N-body integrator:

function [t, yout_noNeptune, yout_withNeptune] = discover_Neptune()

    % Time of integration (in years)
    tspan = [0 97] * 365.25 * 86400;

    % std. gravitational parameters [km/s²/kg]
    mus_noNeptune = [1.32712439940e11; % Sun
                     398600.4415       % Earth
                     1.26686534e8      % Jupiter
                     3.7931187e7       % Saturn
                     5.793939e6];      % Uranus

    mus_withNeptune = [mus_noNeptune
                       6.836529e6]; % Neptune

    % Initial positions [km] and velocities [km/s] on 2000/Jan/1, 00:00
    % These positions describe the barycenter of the associated system,
    % e.g., sJupiter equals the statevector of the Jovian system barycenter.
    % Coordinates are expressed in ICRF, Solar system barycenter
    sSun     = [0 0 0 0 0 0].';
    sEarth   = [-2.519628815461580E+07  1.449304809540383E+08 -6.175201582312584E+02,...
                -2.984033716426881E+01 -5.204660244783900E+00  6.043671763866776E-05].';
    sJupiter = [ 5.989286428194381E+08  4.390950273441353E+08 -1.523283183395675E+07,...
                -7.900977458946710E+00  1.116263478937066E+01  1.306377465321731E-01].';
    sSaturn  = [ 9.587405702749230E+08  9.825345942920649E+08 -5.522129405702555E+07,...
                -7.429660072417541E+00  6.738335806405299E+00  1.781138895399632E-01].';
    sUranus  = [ 2.158728913593440E+09 -2.054869688179662E+09 -3.562250313222718E+07,...
                 4.637622471852293E+00  4.627114800383241E+00 -4.290473194118749E-02].';
    sNeptune = [ 2.514787652167830E+09 -3.738894534538290E+09  1.904284739289832E+07,...
                 4.466005624145428E+00  3.075618250100339E+00 -1.666451179600835E-01].';

    y0_noNeptune   = [sSun; sEarth; sJupiter; sSaturn; sUranus];
    y0_withNeptune = [y0_noNeptune; sNeptune];

    % Integrate the partial Solar system
    % once with Neptune, and once without
    options = odeset('AbsTol', 1e-8,...
                     'RelTol', 1e-10);

    [t, yout_noNeptune]   = ode113(@(t,y) odefcn(t,y,mus_noNeptune)  , tspan, y0_noNeptune  , options);
    [~, yout_withNeptune] = ode113(@(t,y) odefcn(t,y,mus_withNeptune),     t, y0_withNeptune, options);

end

% The differential equation
%
%    dy/dt = d/dt [r₀ v₀ r₁ v₁ r₂ v₂ ... rₙ vₙ]
%          = [v₀ a₀ v₁ a₁ v₂ a₂ ... vₙ aₙ]
%
%  with
%
%    aₓ = Σₘ -G·mₘ/|rₘ-rₓ|² · (rₘ-rₓ) / |rₘ-rₓ|
%       = Σₘ -μₘ·(rₘ-rₓ)/|rₘ-rₓ|³
%
function dydt = odefcn(~, y, mus)

    % Split up position and velocity
    rs = y([1:6:end; 2:6:end; 3:6:end]);
    vs = y([4:6:end; 5:6:end; 6:6:end]);

     % Number of celestial bodies
    N = size(rs,2);

    % Compute interplanetary distances to the power -3/2
    df  = bsxfun(@minus, permute(rs, [1 3 2]), rs);
    D32 = permute(sum(df.^2), [3 2 1]).^(-3/2);
    D32(1:N+1:end) = 0; % (remove infs)

    % Compute all accelerations
    as = -bsxfun(@times, mus.', D32);              % (magnitudes)
    as = bsxfun(@times, df, permute(as, [3 2 1])); % (directions)
    as = reshape(sum(as,2), [],1);                 % (total)

    % Output derivatives of the state vectors
    dydt = y;
    dydt([1:6:end; 2:6:end; 3:6:end]) = vs;
    dydt([4:6:end; 5:6:end; 6:6:end]) = as;

end

Here is the driver script I used to get some nice plots out:

clc
close all

% Get coordinates from N-body simulation
[t, yout_noNeptune, yout_withNeptune] = discover_Neptune();

% For plot titles etc.
bodies = {'Sun'
          'Earth'
          'Jupiter'
          'Saturn'
          'Uranus'
          'Neptune'};

% Extract positions
rs_noNeptune   = yout_noNeptune  (:, [1:6:end; 2:6:end; 3:6:end]);
rs_withNeptune = yout_withNeptune(:, [1:6:end; 2:6:end; 3:6:end]);

% Figure of the whole Solar sysetm, just to check
% whether everything went OK
figure, clf, hold on
for ii = 1:numel(bodies)
    plot3(rs_withNeptune(:,3*(ii-1)+1),...
          rs_withNeptune(:,3*(ii-1)+2),...
          rs_withNeptune(:,3*(ii-1)+3),...
          'color', rand(1,3));
end

axis equal
legend(bodies);
xlabel('X [km]');
ylabel('Y [km]');
title('Just the Solar system, nothing to see here');

% Compare positions of Uranus with and without Neptune
rs_Uranus_noNeptune   = rs_noNeptune  (:, 13:15);
rs_Uranus_withNeptune = rs_withNeptune(:, 13:15);

figure, clf, hold on

plot3(rs_Uranus_noNeptune(:,1),...
      rs_Uranus_noNeptune(:,2),...
      rs_Uranus_noNeptune(:,3),...
      'b.');

plot3(rs_Uranus_withNeptune(:,1),...
      rs_Uranus_withNeptune(:,2),...
      rs_Uranus_withNeptune(:,3),...
      'r.');

axis equal
xlabel('X [km]');
ylabel('Y [km]');
legend('Uranus, no Neptune',...
       'Uranus, with Neptune');

% Norm of the difference over time
figure, clf, hold on

rescaled_t = t/365.25/86400;

dx = sqrt(sum((rs_Uranus_noNeptune - rs_Uranus_withNeptune).^2,2));
plot(rescaled_t,dx);
xlabel('Time [years]');
ylabel('Absolute offset [km]');
title({'Euclidian distance between'
       'the two Uranuses'});

% Angles from Earth
figure, clf, hold on

rs_Earth_noNeptune   = rs_noNeptune  (:, 4:6);
rs_Earth_withNeptune = rs_withNeptune(:, 4:6);

v0 = rs_Uranus_noNeptune   - rs_Earth_noNeptune;
v1 = rs_Uranus_withNeptune - rs_Earth_withNeptune;

nv0 = sqrt(sum(v0.^2,2));
nv1 = sqrt(sum(v1.^2,2));

dPhi = 180/pi * 3600 * acos(min(1,max(0, sum(v0.*v1,2) ./ (nv0.*nv1) )));
plot(rescaled_t, dPhi);

xlabel('Time [years]');
ylabel('Separation [arcsec]')
title({'Angular separation between the two'
       'Uranuses when observed from Earth'});

which I'll describe here step by step.

First, a plot of the Solar system to check that the N-body integrator works as it should:

Nice! Next, I wanted to see the difference between the positions of Uranus with and without the influence of Neptune. So, I extracted just the positions of those two Uranuses, and plotted them:

...that's hardly useful. Even when zooming in greatly and rotating the heck out of it, this is just not a useful plot. So I looked at the evolution of the absolute Euclidian distance between the two Uranuses:

That's starting to look more like it! Roughly 80 years after the start of our analysis, the two Uranuses are almost 6 million km apart!

Large as that may sound, in the grander scale of things this might drown in the noise when we take measurements here on Earth. Plus, it still does not tell the whole story, as we'll see in a moment. So next, let's look at the angular difference between the observation vectors, from Earth towards the two Uranuses to see how large that angle is, and if it can stand out above the observational error thresholds:

...whoa! Well over 300 arcseconds difference, plus all sorts of wobbly bobbley timey wimey rippling going on. That seems well within the observational capabilities of the time (although I can't find a reliable source on this so quickly; anyone?)

Just for good measure, I also produced that last plot leaving Jupiter and Saturn out of the picture. Although some perturbation theory had been developed in the 17th and 18th centuries, it was not very well developed and I doubt even Le Verrier took Jupiter into consideration (but again, I could be wrong; please correct me if you know more).

So, here is the last plot without Jupiter and Saturn:

Although there are differences, they are minute, and most importantly irrelevant for discovering Neptune.


Whats the farthest distance a spacecraft has been detected (except by radio)

OSIRIS-Rex has been spotted on its approach to Earth, at a distance of approximately 7 million miles (12 million kilometers) away, with a brightness of approximately 25th magnitude. The Large Binocular Telescope was used for this observation, this has a pair of 8.4-meter mirrors.


How would a small TCO (temporarily captured orbiter) or other natural Earth satellite most likely be detected?

Cruithne and J002E3 were detected using ground based astrophotography.
J002E3 is thought to be "the S-IVB third stage of the Apollo 12 Saturn V rocket ... It is thought that J002E3 left Earth orbit in June 2003, and that it may return to orbit the Earth in the mid-2040s.1"
Ground telescope also found 6Q0B44E which orbits beyond the moon.
2002 AA29 was found via automatic sky survey.
Several of these objects are in horseshoe orbits.

Regardless, the discovery of these objects usually comes via conventional astrophotography.


Why do we not send diggers to Mars if we think there's underground water?

Sending anything to Mars is expensive. Sending a machine capable of drilling to the required depths to find subsurface water would be prohibitively expensive with current rocket payload limits.

There are also far more accessible sources of water on the surface - lakes of ice in craters, water rich minerals, and even an equatorial frozen sea .

There is also water at the surface of Mars on the poles, but the conditions here are much harsher than nearer the equator, and there is less sunlight, reducing the capability of any solar powered rover.

The subsurface water is not necessarily that far below the surface, but it is also not a torrent; more likely trapped as ice particles in between the rock or in secondary minerals.


Can an SMBH recycle dark matter into energy?

The accretion of dark matter by a black hole is possible, but difficult. Unless its trajectory takes it into the black hole, then it will simply approach the black hole, speed up and then escape again, since non-interacting dark matter cannot dissipate energy in any way.

Presuming a black hole did accrete the dark matter then the energy possessed by the dark matter would become part of the rest mass energy of the black hole. So yes, the black hole would grow. Not sure what you mean by recycle the energy. Energy is always recycled (conserved). If you mean can the black hole lose its rest mass energy? I suppose it may do via Hawking radiation, but this is vanishingly small for a SMBH. The dark matter will also carry angular momentum into the BH and there are ways that this rotational energy can be extracted (e.g. the Penrose process https://en.m.wikipedia.org/wiki/Penrose_process ).


Smallest body that can be observed from Earth?

The smallest natural object that I know of that has been detected from an Earth-bound telescope is 2011 CQ1, a one-meter asteroid observed while specifically looking for NEOs.

As DeerHunter mentions in a comment, smaller artificial objects are identified and tracked, mostly by NORAD and some other national defence centers.


How long does it typically take to make a telescope yourself for the first time?

Making your own telescope is very do-able given basic working skills. Usually a DOB or Newtonian reflector is a good choice. You can buy a mirror and components all ready to assemble or you can choose to grind your own (which is more difficult). If you have an astronomy club in your area check to see if they can help putting your project together. If not, check online ( Youtube, etc) for guidance. Websites like Edmonds Optics may still be a good choice for components. Once you get all the components together, they can be assembled in a pretty short time. Good Luck!


When calculating absolute magnitudes of distant objects, do you use the light-travel distance or the comoving distance?

Neither. You use the Luminosity Distance.
$$M = m -5\log D_L + 5$$
This assumes bolometric magnitudes. If you are trying to estimate it in some photometric band then you must also calculate and apply a K-correction that will depend upon the intrinsic spectrum of the source.
For $z=11.09$ this cosmology calculator gives a model-dependent luminosity-distance modulus of 50.39 (corresponding to $D_L=1.2\times 10^{11}$ pc).
NB: The distances you quote in the question do not give the absolute magnitudes quoted.


What will the last day of humanity look like?

The last humans in this scenario likely die of starvation and/or drought during a natural short-term climate fluctuation.

The basic issue is that as the sun heats up there will be a gradual reduction of carbon dioxide in the atmosphere (due to plant life but also silicate weathering, which speeds up in warm environments). There may also be a reduction in volcanic outgassing, although I think the timeframe is too slow. This is bad for plants, since they will get less CO$_2$ to photosynthesise from, and it will reduce the climate stabilising feedback so temperatures will start going up as the sun increases in luminosity. Eventually we get to a runaway greenhouse stage where water vapour alone triggers overheating, but this is possibly long after the biosphere is mostly dead.

The most complete model of this I have seen is this paper by Franck, Bounama, & von Bloh. They estimate complex multicellular life to go extinct in 0.8–1.2 Gyr.

Note that this will happen long before the sun becomes even a subgiant.

In any case, if this model is correct we should expect that slowly but surely the future humans would be forced to migrate to cool regions and would be especially limited by the dwindling biomass available. But beside this they will be subject to normal random climate fluctuations that sometimes create decade-long droughts or other misfortunes, and it is likely one of those that actually gets the last individuals.

Of course, the scenario assumed makes some very strong assumptions. In order to remain recognisably humans over millions of years these humans would have to engage in deliberate genetic control, otherwise genetic drift will make them mutate into new species. It assumes that nobody does anything about the long-term trend. And so on.


Would one have to be more than halfway to the Oort Cloud for the Sun to appear as bright as the full moon?

Yes, brightness is inversely proportional to the square of the distance, and the square of 632 is about 400,000. It's probably appropriate to round the value to 1s.f. to give "600AU".  Of course, at that distance the sun appears point-like.

The distance to the Oort cloud is rather hard to define. It's not like you reach 1000 AU and see a Star Wars like "Asteroid field" of comet nuclei. For example Wikipedia gives a distance of 2000AU.


Why aren't there galaxy-sized balls of iron out there?

In short, you can't make a galaxy-mass ball of iron because there's no way to support the ball against the inward crush of its own gravity. First, it would collapse into a neutron star, but even neutron stars can't hold themselves up beyond about 2-3 solar masses.  Your galaxy-sized object would collapse into a black hole.  But we do see these!  All massive galaxies are believed to have supermassive black holes at their centres. These are up to 10 billion solar masses in size, which is comparable to smaller galaxies.

In any star, be it a main-sequence star like the Sun, a giant, or even a white dwarf or neutron star, the inward force of gravity must be balanced by some outward force or pressure.  (In fact, we basically assume this when we construct stellar models, although it's a very good assumption.) In main-sequence stars, this pressure is provided by the nuclear reactions happening inside them but as a star evolves, it gets a bit more complicated. In some situations, the nuclei get so close together that they their electrons begin to share quantum states. And because no two electrons can be in the same state, they have to occupy higher energy states. This in effect exerts a kind of pressure (called degeneracy pressure), because the electrons are forced to move around faster than they otherwise would. This is what supports white dwarfs and the cores of some evolved stars. e.g. low-mass red giants.

Degeneracy pressure is fine up to a limit known as the Chandrasekhar limit, which is about 1.44 solar masses (depending on the composition of the object in question). Anything above this would collapse. First, this collapse would yield a neutron star, where things are a bit different. But a similar principle holds, and there's a degeneracy pressure because the neutrons themselves also occupy one quantum state each.  The details are much more difficult here, but the overall consensus is that neutron degeneracy pressure can support objects up to about 2-3 solar masses. After that, there's nothing left to balance gravity, and the star collapses into a black hole.


Jupiter appearing odd through a 6" reflector

Most likely would be the the main mirror needs collimating. On big things like the moon you won't notice bad collimation as much as something small like a planet or star.


Solar Noon: meridian crossing time versus time of maximum elevation

There are a couple of things you have to think about in order to understand the problem at hand. The first is that there are three coordinate systems at play. The first is the celestial coordinate system, which is a coordinate system based upon the latitude and longitude of the Earth (the celestial equator is the Earth's equator projected out into space). See this post.

The second coordinate system is that of the solar system: The plane of the solar system (which contains the sun and planets, and traces out a line in the sky called the ecliptic). Now because of the tilt of the Earth's axis with respect to the normal vector of the plane of the solar system, the celestial equator is not the same as the ecliptic. What's more, is that the relative position of the ecliptic and the celestial equator change throughout the course of the year.

Lastly, there is the altitude/azimuth coordinate system which locked to the observer. The zenith (the point at which all horizons are equidistant from the point you are looking at) is $+90^{\circ}$ in altitude and is a degenerate point in azimuth, and all horizons are $0^{\circ}$ altitude but lay on a unique azimuthal arc stemming from the zenith and intersecting the horizon.

Below is a diagram showing the relationship between the first two coordinate systems, the celestial coordinate system and the coordinate system of the solar system.

Why does all of this matter for the question you've asked? It's because as you look due East and due West, you're looking at $0^{\circ}$ declination (the celestial equator intersects the horizon at exactly due East and due West) whereas the ecliptic intersects the horizon at different points. These points change throughout the year as the relative position of the ecliptic changes with respect to the celestial equator. Now, the amount the sun travels on the ecliptic over a 24 hour period is $0.9856^{\circ}$ ($360^{\circ}/365.25 \text{ days}$; actually I may want to use a sidereal day here but nonetheless it's about a degree).

Due to this, there is potentially a little bit extra in altitude that the sun may travel throughout the course of a day (since the path of the ecliptic is neither aligned with the celestial coordinate system, nor aligned with the local alt/az coordinate system of the observer; unless you're at the north or south poles in which case your local alt/az coordinate system is the same as the celestial coordinate system, however both are still mis-aligned with the coordinate system of the solar system). I think the word "approximate" needs to be there.

All told, I don't personally know how to quantify the extra bit of altitude the sun may gain from the misalignment of the ecliptic with lines of constant altitude and from the sun's motion along the ecliptic, but I would welcome references and calculations from other people. This little bit extra may not be a really small number. In other words, you might want to see how the rate of change of altitude of the sun at that point changes simply due to the Earth's rotation and how it may compare to the rate of change of altitude due to the motion of the sun along the ecliptic.

EDIT:
I also just looked to see what the ecliptic looks like on the sky (as seen from Philadelphia), and how it compares to the local alt/az coordinate system. Below is a picture with all coordinate systems present to try to illustrate the point I was trying to make.

The green line is the meridian, the orange(ish) lines are alt/az coordiantes, the blue line is the celestial equator, and the red line is the path of the ecliptic. Hopefully you can make this out sufficiently well.


As viewed from Mars, what are Jupiter's and Saturn's maximum brightness in apparent magnitude?

The inverse-fourth-power law you're referring to is valid for light emitted from a source, reflected non-specularly  — i.e. in all directions — from a reflector, and detected by the original emitter. If the reflector is a mirror, the observed flux just follows the normal inverse-square law with the nominator equal to $(2d)^2$ instead of $d^2$, since the light has to go back and forth. But if the reflector scatters light in all directions — i.e. into a $2\pi$ hemisphere — then the detected flux is $\sim r^2/d^4$, where $r$ is the radius of the reflector (see this answer for a more throrough explanation).

An example of this is a radar. But in our case, it isn't us that emit the light, it's the Sun. The amount of light reflected from Jupiter and Saturn depends on their distance to the Sun, and that distance doesn't change if you move to Mars. The relevant distances (that I got from NASA's Planetary Fact Sheet) are:

Earth semi-major axis    $d_\mathrm{E} = 1.00\,\mathrm{AU}$
Mars aphelion$^\dagger$  $d_\mathrm{M} = 1.64\,\mathrm{AU}$
Jupiter semi-major axis  $d_\mathrm{J} = 5.20\,\mathrm{AU}$
Saturn semi-major axis   $d_\mathrm{S} = 9.58\,\mathrm{AU}$

Now the differences between them:

Earth to Mars    $d_\mathrm{M-E} = 0.64\,\mathrm{AU}$
Earth to Jupiter $d_\mathrm{J-E} = 4.20 \,\mathrm{AU}$
Earth to Saturn  $d_\mathrm{S-E} = 8.58\,\mathrm{AU}$
Mars  to Jupiter $d_\mathrm{J-M} = 3.56 \,\mathrm{AU}$
Mars  to Saturn  $d_\mathrm{S-M} = 7.94 \,\mathrm{AU}$

Hence, from Mars the distance to Jupiter is $d_\mathrm{M-J} = 0.85 d_\mathrm{J-E}$, and the received flux  is thus $1/0.85^2 = 1.4$ times that on Earth. The change in apparent magnitude is then
$$
\Delta m = -2.5 \log\left( \frac{0.85^2}{1}\right) = -0.36,
$$
i.e. Jupiter would be $m = -2.94 - 0.36 = -3.30$ as viewed from Mars (assuming the values you provide are correct; I didn't check this).

Following the same approach for Saturn, I get $m = -0.41$.

$^\dagger$The reason I used Mars' aphelion instead of Mars' semi-major axis is because Mars has a rather eccentric orbit ($e\sim0.09$). Jupiter and Saturn are somewhat closer to circular orbits, though ($e\sim0.05$). This of course is still an approximation. If you wish to take into account the eccentricities of all orbits, you also need to know the angle between their semi-major axes. This also doesn't take into account the inclination of the orbit; however, these are only 1º-2º.
And of course this minimum distance will not occur every Martian year.


How big would a disc have to be to blot out the sun covering 5% of the land mass of earth?

The angular diameter of the sun is approximately 30 arcminutes (roughly .0087 radians), or half of a degree.

The formula for angular diameter of a round object in radians is:
$$
\delta = 2arctang\frac{diameter}{(2)distance}
$$

Given this, we should also think about position. Assuming we want to continually block the same spot on Earth we need to the orbital period to one year. (I'm not sure if there is a technical difference between orbiting the Earth once per year such that you're always between the Sun and Earth, and just plain orbiting the Sun)

According to this calculator, our sun blocker would need at a height of 2,152,050km to orbit Earth once per year. So, given that we can set up our formula with...

$$
.0087 = 2arctang\frac{diameter}{(2)2152050}
$$

...and the needed diameter of an object should come out to be about 18,722km in diameter. Given that, Mr. Burns's terrestrial solution is far more practical, though certainly a nice orbital one is far more devious. Being evil isn't cheap.

Edit:

This answer is a bit rough as it doesn't calculate for the 5% of the Earth's land mass stipulation. The answer which would meet this would depend on whether only an umbral shadow qualifies, or if a penumbral shadow would also suffice. I suspect probably the prior. The shadow would certainly reach the Earth, but probably not cover 5% of it, which is a shame considering its diameter is actually larger than Earth's. Regardless, it's not a very elegant solution.


How long does lunar opposition surge last? Are there measurements of the full Moon getting suddenly brighter?

The lunar opposition surge has been well studied, likely because we can study it in detail, we have surface samples, and so it serves as a baseline for other bodies in the solar system (as it does for many other kinds of surface studies).  It is quite substantial in visible light.
Probably Clementine data are the oldest-modern reference, and among others, Buratti et al. showed that from a baseline normalized reflectance of ~0.6 at a phase angle of 10°, it increased to 1.0 at a phase angle near 0° (hence, normalized reflectance).  Their Fig. 6 shows that this increase happens over just the last few degrees, which is typical of opposition surges.  Shkuratov et al. (1999) also published based on Clementine data and their Fig. 5 again shows a ~10s% increase over the last ~10° of phase angle.
What you really want is an expanded plot of just the last few degrees of phase angle, which they include in their Fig. 10 (I'm not reproducing the images here due to © concerns).  They show that the Buratti et al. (1996) results increase by ~20% in brightness when going from just a phase angle of 0.5° to 0°.  The paper compares this to numerous other materials.  That's all in UV to visible light.
Velikodsky et al. (2016) also looked at this from LROC-WAC data.  For the UV (415 nm) and red (689 nm) bands, they show (Fig. 6) that from a phase angle of 1° to 0°, the increase can be around 50%.  Going from 5° to 0° in a different part of their paper (Fig. 9), the increase can be a factor of ~35–45%.  This is again wavelength- and material-dependent.  I would say that that paper is probably going to be the best quality plots, or at least a paper that uses LROC-WAC data.
As the Shkuratov et al. (1999) paper notes: "Below about 10° the brightness of the Moon grows more sharply than at large phase angles. The lunar opposition effect is about 1.25 at 1°–10°."  So it could be considered that the "effect" starts <10°.
Those all seem to tell a reasonably consistent story that, within those last few degrees, the phase function increases the visible light returned by up to about 50%.


Searches for dipole form of anisotropies (on luminosity distances) in the universe with type Ia supernovae

Isn't the point of using type Ia supernovae that they have an estimated distance that is independent of redshift?

${\bf \hat{n}}$ here is not the direction of the CMB dipole, it is a free parameter in their model.

If distances were being estimated using redshift I guess you would have a point.


How much does a linear lightyear of space expand in a year?

A recent estimation of the Hubble constant is 71.9 (km/s)/Mpc.

This means there is an expansion of

$$\left(71.9\ \mathrm{\frac{km}{s\ Mpc}}\right)\left(\mathrm{\frac{60\ s}{min}}\right)\left(\mathrm{\frac{60\ min}{hr}}\right)\left(\mathrm{\frac{24\ hr}{day}}\right)\left(\mathrm{\frac{365\ day}{yr}}\right) = 2,267,438,400\ \mathrm{\frac{km}{yr\ Mpc}}$$

To apply this expansion over a light year:

$$
\frac{2,267,438,400\ \mathrm{(km/yr)/Mpc}}{3,261,564\ \mathrm{lyr/Mpc}} = \boxed{695.2\ \mathrm{\frac{km/yr}{lyr}}}
$$

which is about $7.3483\times10^{-11}\ \%$ per year or $432\ \mathrm{miles/year}$.

A linear light year of space expands by 695.2 km/yr with a given Hubble constant of 71.9 (km/s)/Mpc.

Edit:
A better measurement released on 2018-02-22 using parallax found the Hubble constant to be 73.45 ± 1.66 (km/s)/Mpc.


How do we know what the biggest star is?

As @uhoh stated, there is a crucial distinction between what is in principle "observable" and what we have to-date "observed."
Also, you have to specify whether you're talking about biggest in radius or biggest in mass. For stars with very large radii, like the St2-18 system you've referenced, they often can't even constrain the mass! And when they are able to, they end up having pretty small masses, i.e. UY Scuti. Stars, especially high-mass stars, generally increase in radius as they age, but they lose mass as they age. So it's not surprising to find red supergiants with very large radii (because they're at a late stage in their stellar evolution) and with small masses. But the theory behind supergiant stars is very complicated and uncertain still, making mass measurements very very difficult (And thus are not usually attempted) since mass measurements require some kind of binary interaction or a well-understood model of the stellar atmosphere (which is lacking for supergiants as I understand).
Regarding what's "observable," keep in mind that this is a fast-changing landscape: for example, gravitational wave detections of intermediate-mass black holes suggest that they possibly form from core-collapse of super-massive stars (among other possibilities). And the link to the question "what is the size limit theoretically of a star" that you provided has many answers which explain how uncertain such a theoretical limit is for mass and for radius.
Regarding "observed" systems, you must take into consideration the error bars on mass measurements of stars. The largest observed star with a known mass that has respectable error bars is the Wolf-Rayet star R136a1. See here for a comprehensive list.
With that said, you asked,

How sure are we (in percent) that St2-18 is right now, in 2020, still the largest observed star?

Of course there is always the possibility that with a new survey with better resolution, larger field of view, etc... even larger stars may be in the universe. It's a zoo! We were surprised to find stars as big as St2-18, so why would the universe stop there, eh? There COULD BE SOME reason in principle that disallows larger stars, but the universe keeps providing bigger and bigger stars for us to ponder, so there's no reason a priori to assume so. Regarding the survey you're asking about in particular, the wiki article has a nice discussion explaining the great uncertainty in determining the radius of St2-18, i.e. do we rightly consider it as part of the cluster or not in order to make assumptions about its distance, etc...:

A calculation for finding the bolometric luminosity by fitting the Spectral Energy Distribution (SED) gives the star a luminosity of nearly 440,000 L☉, with an effective temperature of 3,200 K, which corresponds to a very large radius of 2,150 R☉ (1.50×109 km; 10.0 au; 930,000,000 mi),[a] which would be considerably larger and more luminous than theoretical models of the largest, and most luminous red supergiants possible (roughly 1,500 R☉ and 320,000 L☉ respectively).[11][6] An alternate but older calculation from 2010, still assuming membership of the Stephenson 2 cluster at 5.5 kpc but based on 12 and 25 μm fluxes, gives a much lower and relatively modest luminosity of 90,000 L☉.[7] A newer calculation, based on SED integration and assuming a distance of 5.8 kpc, gives a bolometric luminosity of 630,000 L☉ although the authors are doubtful that the star is actually a member of the cluster and at that distance.

Lastly, you asked

How are we sure that St2-18 was indeed the largest object within the given survey? Technically speaking, one would have to calculate at least the distance or lumnosity for each object of the survey?

I quickly looked at the survey report paper, and it seems that this is actually unclear. Keep in mind that they were surveying a few different clusters, the Stephenson#2 being only one of them. The tables at the end of their paper make that clear. However, they only list the technical properties, and do not even attempt to constrain the radii for all of these stars. So I'd take it with a grain of salt.


Could a habitable satellite of a gas giant have a stable subsatellite?

This is a fun question, so I spend some time thinking about it. This is what I came up with.

Is this possible?

Yes, this is certainly possible. Just consider the Solar System itself. We have a massive central body, the Sun, and several tiny subsystems (planets with their moons) orbiting it.

Is it stable?

Again, from the simple observation that the Solar System exists in its current form, we know that such a system can remain more or less stable over a significant portion of the stellar lifetime.

A more difficult problem, I think, is if can it be formed in the first place.

Could it be habitable?

This is a tricky question as it critically depends distances between the several orbiting bodies. Lets consider the orbital mechanism and derive from what we know of the Solar System.

In the following I've assumed the gas planet is Jupiter and the fictional planetary system consists of a Sun', Jupiter', Earth' and Moon' such that the apostrophe denotes the fake object. These fake objects are identical to the real equivalents, except for their orbital configuration and relative distances. If you want a different system you can follow the same line of reasoning, but plug in different numbers for the fake objects.

Now, we know that the Earth-Moon system exists in a stable orbit around the Sun, which means that at 1 AU the gravitational force of the Sun is not strong enough to destabilize the Moon orbit. From this we can find a lower limit on the gravitational force that Jupiter' may exert on the Moon'
\begin{aligned}
  F_{J'M'} &= F_{SM} \\
  \frac{ M_{J'} M_{M'}}{r_{J'M'}^2} &= \frac{ M_{S} M_{M}}{r_{SM}^2}
\end{aligned}
Solving this equation we get
\begin{aligned}
 r_{J'M'} &= r_{SM} \left( \frac{ M_{J'} M_{M'} }{ M_{S} M_{M} } \right)^{1/2} \\
   &= r_{SM} \left( \frac{ M_{J'} }{ M_{S} } \right)^{1/2}
\end{aligned}
which evaluates to about 0.03 AU. So the Earth'-Moon' system should have an orbital radius around Jupiter' of at least 0.03 AU.

Now the question is if the Earth-Moon system can remain bound in its orbit around Jupiter. To answer this we can safely ignore the Moon' and assume it remains tightly bound to the Earth'. The potentially destabilizing force on the Jupiter'-Earth' system is again the gravitational pull of the Sun', but now on the Earth'. We can do a similar trick as before, but now we need to account for a different mass ratio. So lets require that the fraction of forces be equal in both cases:
$$
  \frac{F_{SM}}{F_{EM}} = \frac{F_{S'E'}}{F_{J'E'}}
$$
with some algebra this gives
$$
 r_{S'J'} = r_{J'E'} \frac{r_{SM}}{r_{EM}} \left(\frac{M_E}{M_J}\right)^{1/2}
$$
which for $r_{J'E'} = $0.03 AU evaluates to a minimum orbital radius of about 0.7 AU.

This greatly surprised me as it means you can comfortably fit the entire Jupiter'-Earth'-Moon' system inside the habitable zone, meaning that in principle life should be possible.

Of course the seasons on such a planet would be rather extreme.


According to this (https://spotthestation.nasa.gov/sightings/view.cfm?country=Bangladesh&region=None&city=Dhaka#.WcATdWdSz0o) the ISS was around at that time of morning. The sunrise was 5:45, so it could have been illuminated by the sun before it rose.


What will it finally take to accurately measure the distance to Betelgeuse?

The problem is that the apparent diameter of Betelguese is about 50 mas (milli arc seconds --- 1 mas  is about 5 nano-radians) while its parallax is about 5 mas and its shape and surface brightness are both irregular and variable.
Given that, the current measurement are amazingly accurate. So I can identify about three approaches to doing this measurement, and I don't know enough to know which one(s) might really work, or which might be expected to work first.

Measure the distance without using parallax. This most likely means getting an accurate estimate of the absolute brightness or all, or part of, the disk of the star and comparing that with the apparent brightness. Spectroscopy might give an accurate idea of the temperature, which should in turn allow the brightness to be calculated and interferometry might allow the measurement to be restricted to a small area  near the centre of the disk which was not obviously a star-spot or anything.

Simultaneously image the disk and measure its position relative to background stars, allowing the position of the centre of mass to be estimated accurately.

Increase the baseline of the parallax measurement. A pair of telescopes on opposite sides of Saturn's orbit, for instance would have 10 times the baseline, and more or less ten times the accuracy.

Update: It seems the Gaia consortium now expects to work around the brightness constraints and get data for all stars. From the wikipedia page

Although it was originally planned to limit Gaia's observations to stars fainter than magnitude 5.7, tests carried out during the commissioning phase indicated that Gaia could autonomously identify stars as bright as magnitude 3. When Gaia entered regular scientific operations in July 2014, it was configured to routinely process stars in the magnitude range 3 – 20.[57] Beyond that limit, special procedures are used to download raw scanning data for the remaining 230 stars brighter than magnitude 3; methods to reduce and analyse these data are being developed; and it is expected that there will be "complete sky coverage at the bright end" with standard errors of "a few dozen µas".[58]

I don't know how much that will help with Betelgeuse though, due to its large apparent diameter.


Could Venus be a source of Earth's apparent overabundance of water?

No idea is stupid per se.  But in order to place an answer to your question, please consider Venus as a comet. It once had, in that image, a coma (a tail) pointing outwards from the sun, made of evaporated water pushed away by solar wind. How frequently is the Earth exactly inside that coma?

As a comparison, when the Earth strikes a real comet coma we see meteor showers that last no more than one day. So if we consider that Earth may go through Venus imaginary coma once a year, we have an upper limit of 1/365 of Venus' water captured by Earth. This, without taking into effect that Venus had not have a real coma, that evaporation and expelletion of water from Venus was not a so directional process, and surely other factors I can not think about just now.

Besides, do not consider comets the main source of water on Earth. You need to count also with the water vapour generated on volcanoes and that generated on acid-base reactions among Earth's rocks.


Could the Sun's magnetosphere block extraterrestrial contact?

The Sun's magnetosphere is too weak to have a significant effect on extra solar radio waves. If it did, we wouldn't have any radio telescopes as it would block the signals they detect.


What is happening with these solar particles detected near the Sun that is so newsworthy?

If I understand your question correctly, it is about EPI-Lo which a part of the Integrated Science Investigation of the Sun (ISʘIS) . The picture is also taken from there:

EPI-Lo stands for Energetic Particle Instrument-Low energy which indeed sounds confusing at first sight. The essential definition can be found in the instrument paper:

EPI-Lo measures ions and ion composition from ∼20 keV/nucleon–15 MeV total energy and electrons from ∼25–1000 keV. EPI-Hi measures ions from ∼1–200 MeV/nucleon and electrons from ∼0.5–6 MeV.

The paper also addresses the scientific background of the measurements of SEPs, which are defined to have the following energy range:

They consist of protons, electrons and HZE ions with energy ranging from a few tens of keV to many GeV.

In short: The EPI-Lo measures the lower part of the SEP events.
Concerning the newsworthyness, I can only guess. Why SEPs are worth studying is nicely stated at Wikipedia:

They are of particular interest and importance because they can endanger life in outer space (especially particles above 40 MeV).
Solar energetic particles can originate either from a solar-flare site or by shock waves associated with coronal mass ejections (CMEs). However, only about 1% of CMEs produce strong SEP events.

I hope that my remarks answer (at least partially) your questions.


Why are Saturn bands much fainter than Jupiter's?

I'll give this one a shot. Correction is welcome.

Upper atmosphere temperature.

It's not just elements that give a planet color, but the temperature of elements. When we examine what a planet looks like, we're basically talking about reflected sunlight from the planet's surface or atmosphere. With Earth, its atmosphere is transparent enough that its surface can be seen from space. However, it is the only planet in the Solar System where that is the case. The other planets have thick atmospheres, so all we see is basically just high altitude clouds.

Also, as I understand it, the pictures of planets are often enhanced to make any distinctions more apparent. So if you actually flew above Jupiter, its lines might not be quite so distinct as the pretty pictures you'll usually find on the web. Regardless, that doesn't change your question; its lines are still far, far more distinct than Saturn's are and Galileo was able to see its Red Spot, so I think it's safe to say its lines are pronounced, even if the pictures we get are doctored a bit.

So anyway, Jupiter is at the right distance from the Sun to undergo atmospheric transition from ice to gas. As Wikipedia puts it,

  The upper ammonia clouds visible at Jupiter's surface are organized in
  a dozen zonal bands parallel to the equator and are bounded by
  powerful zonal atmospheric flows (winds) known as jets. The bands
  alternate in color: the dark bands are called belts, while light ones
  are called zones. Zones, which are colder than belts, correspond to
  upwellings, while belts mark descending air. The zones' lighter
  color is believed to result from ammonia ice; what gives the belts
  their darker colors is not known with certainty

Ammonia ice, like all ice, is very reflective, so the colder bands with ice are lighter. Ammonia gas is transparent, but anyone who has ever flown over the ocean knows that if you have enough of a transparent thing (water), it has a distinct color. While the quote from Wikipedia above says the reason for the darker colors is uncertain, less ice in its uppermost atmosphere means less reflected light and darker color.

Saturn, Neptune and Uranus are far enough away from the Sun where they always have ice in their upper most atmospheres, so they have less variation in color.  Earth's clouds are also primarily ice (not water vapor), so that is basically the answer. Jupiter is the right distance from the Sun for its upper atmosphere to transition and have distinct bands, some with ice, some without.

"But why are the bands in straight lines?"

This is due to the Coriolis effect.  Viewed from above, the Coriolis effect creates bands that line up with the equator.  Jupiter and Saturn both rotate quite quickly (9.5 and 10.8 hours respectively), so both have strong Coriolis effects.

Jupiter's atmosphere isn't just moving along those visible lines; it's circulating up from the warmer lower parts of the atmosphere to higher parts by convection. This is because there's a lot of heat to be transferred. Jupiter (along with Saturn, Uranus and Neptune) are all radiating more heat into space than they are receiving from the Sun, so while the heat from the Sun plays a role in not freezing Jupiter's upper atmosphere over warmer bands, it is the Coriolis effect that creates the bands.

Saturn has bands too (as you know). They are not as visible because both the warm bands and cold bands of Saturn are icy. See articles here and here and article with false color picture here, crediting Saturn's colors to ammonia ice.

So, for bands to form, all you need is relatively fast rotation. But for highly visible bands, the gas giant planet needs to be the right distance from the Sun and/or have the right amount of internal heat. It's all about the temperature.

I feel that I have to punctuate this with a "probably", because I can't swear that Saturn has ammonia ice all around its upper atmosphere, but I'm pretty sure that is the primary difference between Saturn's more uniform look (Uranus and Neptune too). It is primarily the ammonia that gives these planets their color, even though both Saturn and Jupiter are 99% hydrogen and Helium. Straight gas molecules (O2, N2, H2) and noble gases tend to have very little interaction with visible light.

Here's a fun article on gas giant planet's color. The science of what gives an atmosphere a specific color, however, is quite complex and above my pay-grade. Also, I also found this article interesting, though I can't swear to the accuracy of it.  Saturn radiates a surprising amount of heat given that it's 1/3rd the mass of Jupiter.

Lastly, this one may provide a more accurate but less sexy picture of Saturn than we're used to seeing.

Final note, on Jupiter's dark spots and dark bands. Wikipedia said the reason for the darkness is unknown (as quoted above). While I think the lack of atmospheric ammonia ice is very much a part of the answer, here's an alternate explanation for Jupiter's dark red spot and (perhaps) to some extent, its dark bands. Jupiter gets, on average, about 3.4 times the solar radiation per square meter than Saturn gets. To add to that, it might have more extensive atmospheric mixing going on than Saturn and more organic compounds in its upper atmosphere, which, combined with UV rays, could well play a role in Jupiter's darker lines too.


Is there a website out there where I can find/compare data on stellar parameters for the brightest stars?

As you’ve discovered, there are a lot of different catalogs out there, and many ways to search them.  It would help if you could narrow your question a bit - what kind of data are you looking for?   You seem to want stellar masses at least - what else?  As noted above, there are few stars for which mass is measured directly, but there are robust ways of estimating it from other information, just as effective temperature can be determined from spectral type or photometry.
One catalog that might meet your needs is the TESS Input Catalog (TIC) or its brighter subset, the TESS Candidate Target List.  It compiles information from a lot of other sources, and includes magnitudes from optical through infrared, as well as estimates of stellar mass, radius, temperature, metallicity, luminosity, and surface gravity.  It also includes stellar parallax (i.e., distance) and proper motion from Gaia.
In addition to the ways of searching given in the above link, you can search it through Vizier here - I find the Vizier interface a little easier to navigate than MAST, though both have a lot of options.
Now this catalog may be overkill for what you want: it has 471 million stars, though “only” 27 million of those have mass estimates, and “only” 2 million have distances.    :-)  So knowing more about what you want could help narrow things down.
Edit: The above numbers refer to an earlier release of the catalog, prior to Gaia DR2.  The updated version (which is what I linked to from Vizier) has about 1.7 billion stars, of which 1.27 billion have distances and 455 million have mass estimates.
Edit 2:  After clarification in the comments that the goal is finding information only for very bright stars, this isn’t the best catalog to use.  While it does include entries for some bright stars (e.g. Vega, Betelgeuse), the information included is very limited, presumably because they are saturated (overexposed) in most surveys.
In general it is much harder to get homogeneous data for very bright stars because of this.  There is a wealth of data on them individually, but because they are saturated in systematic surveys, there aren’t uniform sources of data on them to which known relations can be applied in bulk.
The Yale Bright Star Catalogue (searchable here) is a good source of basic data (photometry, spectral type, distance, multiplicity), but for derived quantities like mass or radius you would have to apply known relationships yourself (if you need the data in bulk) or go to individual references for particular stars if you don’t need a lot of them. Honestly Wikipedia isn’t a bad starting point in the latter case - it is generally clear about what references the numbers come from.


Is there a comfortable part of Venus's upper atmosphere?

There is a layer of Venus's atmosphere that has bright white clouds, a white sky, no acid rain, 1 atm pressure, a warm tropical climate and 0.9 gravities...

It is the cloud layers between 50 and 70km altitude above the ground.


Sensitivity of calculated orbital elements to observational errors

Kepler used data from catalogs made by Tycho Brahe which had a varying level of precision and accuracy down to about 0.5 arcminutes for a measurement. However, since Kepler thought all systems should operate in the same way, he could use a larger sample verify his laws. Once you know what the relation should be, you can then map the observations to what you would expect. For an individual case, the precision of measuring orbital parameters depeneds on how much of the orbit you have actually observed.


LocalFluff's comment is spot-on. You need mass to have gas, and moons just don't have enough.

It is thought that gas giants gathered the gas they have today by accreting large amounts of it from the protoplanetary disk in the early days of the Solar System. At first, they were only gas-less cores (not rocky, exactly, but not gaseous), but they quickly became the most massive objects in their immediate vicinities, and thus gobbled up all the gas nearby them.

Now, less massive bodies, like Earth, can still accrete gas. However, they have difficulty retaining it. Lammer et al. (2014) calculated that under conditions like those experienced in the young Solar System, planets of one Earth mass or less could retain captured hydrogen envelopes for no more than ~100 million years - a long time for us humans, but shorter for astronomical objects.

The hydrogen escapes via atmospheric escape, which happens when particles in the atmosphere have velocities higher than escape velocity. This phenomenon is known as Jeans escape. Atmospheric escape happens more often with lighter gases, such as hydrogen. It is also influenced by stellar winds, which can cause even gas giants (generally close to the star, such as hot Jupiters) to lose some or all of their atmospheres.

I wouldn't call gas moons impossible, but certainly very unlikely.


Dark matter, is just a name for something we know nothing of. It was named to account for an extra gravity source for which there have been indirect observations, but yet we cannot explain.

The force of gravity exerted by light is negligibly small yet we have measured the gravitational pull of Dark Matter to be big enough to affect whole galaxies; it is what binds galaxies together.

Furthermore, we have included everything we can observe (all ordinary matter including photons) when we do the calculations for the amount of gravity there should be. So light is already there. 'Dark matter' is that extra gravity which we cannot account for.


From the mysterious site Vaporia, which is a fantastic index for astrophysics terms (and other things I'm seeing now):

  The Bondi radius is analogous to the Hill radius, both giving an idea
  of how far from an object, a medium (gas, dust) is likely to be drawn
  in and accreted. The Bondi radius takes into account the (relative)
  speed of the object through the medium, and the medium's density and
  sound-speed. Accretion due to material falling within the Bondi radius
  is called Bondi accretion.

Try to picture a heavy object, perhaps a planet, travelling through a gaseous or dusty medium. Now, some of the gas is eventually going to be the victim of accretion to the planet. The planet is going to pull gas/dust from the medium and the Bondi radius tells you how far from the planet gas/dust is likely to be accreted to the planet.

Hopefully this is more clear!


Data reduction and photometry without IRAF?

I suspect that everything you want and more is available and written in python or has python wrappers.

Astropy
ccdproc
photutils


Approximate distance from Earth to Mars at a given moment with reasonable (+/-100km) accuracy?

Do not underestimate the astronomers! 100 km? That's absurd. Saturn('s center of mass) is located to within one mile's distance from the Sun. Mars is of course even much better located.

Here's a site I googled that seems to give current distance to Mars to within single kilometers. I don't know how that site makes up its numbers, but it is certainly possible to measure it like that. With mutual radio communication, there are really good opportunities for precis distance measuring. 135,834,832 km just now! Pretty close and that's why the ExoMars space probe was launched towards Mars a few days ago. And here is a table with the conunction distances just past and soon coming up.


Gas giants are believed to have a solid core. They first formed as icy planets, and were heavy enough to accrete hydrogen and helium from the protoplanetary cloud they were in.

Saturn, for instance, is thought to have a central dense core of 10  to  25  Earth  masses  that  was  probably  the  seed  of  the  formation  of  the  planet,  before  it  accreted  its  hydrogen-helium  envelope  from  the  protosolar  disk. (see link)


How did pre-Earth's mass and orbit change at the Moon formation event?

Does this mean that pre-Earth was roughly 9 Moon masses (~10%) lighter
  than today's Earth? Or could much mass have been ejected from the
  Earth-Moon system?

I think that's likely about right.  In a nutshell, there's two basic questions here.   How much mass was lost as a result of the giant impact and how much mass was gained after the giant impact during the late heavy bombardment and following collisions.

We know that smaller bombardments on Mars has resulted in debris landing on Earth, so it seems highly likely that some debris was lost when Theia and Earth collided, but estimating how much is tricky and you'd really need to be well versed in the models to have a pretty good guess.

One theory is that Theia formed in an Earth Trojan point so it shared the same orbit around the sun that the Earth did.   Source   This would cause a slower impact than a more oblique approach where the velocity of impact wouldn't be much greater than the Escape Velocity of Earth.   If the impact velocity is only a bit above the necessary escape velocity and if a fair bit of energy is transferred into orbital momentum and heat, it's unlikely that a high percentage of debris would reach escape.

Now if it was a faster impact from a more oblique orbital approach or a head on collision that would create greater compession and rebound, those two scenarios would likely lead to more material escaping.   I can't give you precise numbers, this is more of general thinking about what might happen.

Again, according to Wikipedia about 20% of the mass of Theia entered the earth's orbit after the impact.    Orbital Velocity has to be at least Escape Velocity over the square root of 2 and less than escape velocity.   Source

So if we use 11 KM/s as an escape velocity estimate, anything that flew off the impact at less than 7.5 KM/s would have returned to Earth pretty quickly.   A bit above 7.5 and maybe about 9 or so KM/s would be low Earth orbit and about 10 KM/s high earth, with 11 KM/s, likely escape and enter in a near earth orbit around the sun.     You'd need a bit faster than that to have a truly outside of earth orbit.

What percentage of Theia flew off at those velocities is hard to say, but I think it's likely that much of the planet slowed down upon impact and not too much escaped the Earth's Hill sphere, but I'd be guessing if I tried to put percentages to that.

  Did the impact substantially and immediately change Earth's orbit,
  such as its average distance from the Sun?

Because it was a (I assume) a Trojan object, the 2 young planets would have shared the same orbit and so there probably wouldn't have been a big change.  Orbital momentum should be largely be conserved though some orbital velocity might be converted to rotational momentum, I don't think there would be a huge change.   Similarly, if any debris from the impact flew off in one particular direction, the remaining material would react with similar momentum in the opposite direction, but having much greater mass, the effect would be small.

Certainly some orbital change probably happened, but compared to the entire Earth, it was probably small.

Now, if the impact was different than theorized, that might change the answers quite a bit.    (My 2 cents anyway)


Yes.  Currently it is at least plausible that the universe is spatially flat.  If that is the case then, barring global topological weirdness, it is also spatially infinite.  That would mean it contains an infinite amount of matter.
Disclaimer. I don't know how large the uncertainties on spatial flatness currently are: if Wikipedia is to be trusted $\Omega = 1.00 \pm 0.02$ ($1$ being flat) based on data from WMAP and Planck.


Is the passage of three asteroids near Earth today just coincidental?

It is just a coincidence, and isn't that uncommon of an occurrence. You can see a list of all such occurrences on this page. Looking at the specific orbital parameters, the orbits are very different. They range from an eccentricity of .2-.5, periods of 349-725 days, and otherwise are just plain different. There are 7 listed in the next week of a similar closeness, so having 3 on a single day isn't particularly uncommon. Much further then a week doesn't really matter because these were only spotted recently, they are all 3 quite small objects. The oldest detection of the 3 was only September 30, and one was only spotted on October 10th.


Collimator for Dobsonian, same as Newtonian?

Yes, of course it will work. A Dobsonian is merely a Newtonian telescope on a special mount.


They would if we could see them far enough away, but cepheid variables are only visible up to about 20 million light years distant. That may sound like a long way, but in astronomical terms it is nearby.


Telescope magnification compared to distance from an object

Just divide the actual distance by the magnification.  In your example that would give 750 million /50, which is 15 million miles.  (Actually saturn is currently over 800 million miles from earth).

The reason this works is that a magnification of M means the object is M times its apparent linear size, which requires being M times closer.  This is a slight approximation, but is fine for distant objects.


Are the Trappist-1 planets in stable orbits?

We don't know.
Much of what we do know - or at least think we know - about the planets' orbital parameters comes from simulations the team ran via $n$-body methods[1]. Some methods of integration led to short-term disruptions, on the order of less than $\sim10^6$ years. That said, the system is at least $5\times10^8$ years, and it would be odd if the observations came just before the final instability, in the authors' opinions.
Fortunately, using a different (statistical) method, the team got different results. The found that the system has

a 25% chance of suffering an instability over 1 Myr, and an 8.1% chance of surviving for 1 billion years

This does not seem great for long- or short- term instabilities. However, the authors were extremely cautious, emphasizing that there is plenty of uncertainty. There are many other factors that need to be taken into account:

Tidal interactions between the planets and the star and between each other
Possible other planets in the system (although giant planets may be ruled out)
The masses and orbits are not known to great accuracy, and so interactions can not be modeled as ideally as one would like.
Resonances could be important, and the planets in the system fall into near-integer resonances, which is interesting.

Whether or not the planets are still changing their orbits is an interesting question. The authors believe that the planets moved inwards via gas disk migration, which would require interactions with the initial protoplanetary disk. However, as the disk has clearly dissipated, it would seem that the migration has stopped, and has not happened for some time.


What will be the temperature on Earth when Sun finishes its main sequence?

The difference is that your analysis is assuming that the albedo stays fixed, so the surface temperature simply scales like luminosity to the 1/4 power.  The Wiki entry is including feedback from the greenhouse effect, which will tend to further increase the surface temperature.  Note that an analysis that just looks at solar irradiation would get way too low a surface temperature for Venus, for example.  I can't speak to the accuracy of the feedback included-- as I understand it, that is far from a simple effect to include.


Printed Circuit Board manufacturing file for Radio Jove?

As @JamesK already mentioned in a comment, the RCV manual contains a circuit diagram on page 14:

From your question it is not clear to me how much experience you have with PCB design and which software you are using. Back in the days, I had contact with Autodesk's EAGLE, but there are other free PCB alternatives like librepcb. In worst case, you do in fact have to redraw your design.
For prototyping, so-called breadboards do a pretty good job, which was mentioned in the comments as well by @uhoh.
References

On electronics.SE: Is there a directory of open source hardware projects?


Answering in the spirit of the question, I think he's asking if there is ever a chaotic 3 body system that's long term stable, or, to put it another way, a 3 body system without a standard hierarchy where it's stable.

The answer is no.   3 body systems without Heirarchy are never stable for very long.   They can certainly exist for a while, but they aren't long term stable.

What do we mean by Hierarchy

The Sun-Earth-Moon is an example of a long-term-stable 3 body system in heirarchy.  It will remain stable for billions of years.    The Sun is pretty much in the center.   The Earth-Moon system orbits the sun and the Moon also orbits the Earth.   In a sense, the Moon orbits both the Sun and the Earth.    The Moon can orbit the Earth because it's inside the stable part of Earth's Hill Sphere.

Hierarchies beyond 3 might exist, such as Sun, Planet, Moon, Moon's satellite but there are no natural satellites of Moons in our solar system.

Multiple planets

Obviously, multiple planets around a comparatively massive Central star can be long term stable, and just for clarity, lets call a system stable if it lasts a billion years.    Mercury, Venus, Earth, Mars, Juptier, Saturn, Uranus, Neptune are obviously stable and mathematical models say that they should remain stable for at least the next 3.5 billion years, after which, there's a chance Mercury could become chaotic and fly out of it's orbit, flying into the sun, crashing into another planet or perhaps leaving the solar system.

Any of these orbital systems, for any kind of long term prediction of where the planets will be in a billion years, requires n-body calculations which are very complex and inexact, but they can be calculated within a range of error, which makes it possible to say that a system is stable, for a couple billion years anyway.   You can't, for example, ignore the other planets and calculate where the Earth will be in 50 million years.  The planets affect Earth's orbit but they don't destabilize it.   This is called perturbation theory.

And the final example, binary star system with a planet orbiting the two stars.

This XKCD discusses the binary star system, and from same:

  A planet orbiting two stars can't get too close to them or its orbit
  becomes unstable. If it gets too close, the irregular tugging from the
  gravity of the two stars as they orbit will eventually cause the
  planet to crash into one of them or get flung out of the system.

  For a system with two similar-sized stars, this "critical radius" is
  around six times the distance between the two stars.

What this means is that any 3 body system has to follow the rules if it's going to be stable.   You can have several planets orbiting a central star, and those planets will perturb each other, sometimes casting a planet out but you can also have stability.

You can have Sun, Planet, Moon, but the Moon must be inside the stable region of the Hill Sphere.

And you can have one object orbiting two central objects, like Tatooine, or, like the 4 satellites that orbit Pluto-Charon.

But anything outside those rules is an unstable 3 body or n-body system and it may take some time, but it's ultimately unstable and unlikely to last anywhere close to the billion(s) of years or so that stable orbits tend to last.

So a star with two planets who's orbits cross each other (note, Pluto and Neptune don't actually cross, they just appear to cross when drawn in 2 dimensions).  Their orbits don't come close to crossing.

Horeshoe orbits are 3 body orbits that might last a fairly long time, but I don't believe they are truly long-term stable (if anyone can verify that, please do), but intuitively, I don't see how horeshoe orbits could last very long.

When you have a 3 body system where there's no structured orbit and all 3 bodies kind of dance around each other.   That system isn't stable.

If properly balanced, a 3 body system outside the rules of stability might last for hundreds or thousands or orbits, but in celestial mechanical time that's not stable.

Mathematically some 3 body orbits have been worked out, but they are stable in the same way that a pencil is stable if you balance it perfectly on it's point in a room with no wind.   While this article says that they found 13 solutions, none of these solutions would actually work because space isn't perfectly smooth.  It's full of objects with gravity.   So, like the balanced pencil that gets knocked over by a slight gust of wind, a passing rogue planet would throw any of these "solutions" out of whack and they would progressively destabilize.  Where as a stable system (sun/earth/moon) can hande some perterbations and remain stable.

This diagram below is an example of a perfectly stable 3 body orbit in a computer program, but like the pencil, a small push on any one of the 3 objects would send it towards instability.

Hope that's not too long.  I can clean up if needed.


As far as I know, all known old multiple star systems with stable orbits have highly hierarchical structures.
Binary star pairs in those systems orbit much closer to each other than they do to other binary pairs or single stars within the system.  In such a system it would be very unlikely for a star which is part of one binary to be temporarily captured by another binary and then return to its original binary.
Exceptions are multiple systems of very young stars which have formed recently.  The orbits in those systems are thus very new and have not yet been regularized.  It is believed that the stars in those systems will eventually move into hierarchical orbits or be ejected from those systems.
So I think that your system seems very improbable.  However, you won't know for certain unless you can run highly accurate simulations covering billions of years of evolution of the orbits in your system.


Will the E-ELT use Adaptive Optics at visible wavelengths?

From looking at the E-ELT website, it appears that at first light, the AO will only work for near-IR. Specifically, the instrument that can use AO is the MICADO instrument. The description page for this instrument states

MICADO, or the Multi-Adaptive Optics Imaging Camera for Deep
Observations, is one of the first-light instruments for the European
Extremely Large Telescope (E-ELT) instrument and takes the Adaptive
Optics technique to the next level. It will be the first dedicated
imaging camera for the E-ELT and works with the multi-conjugate
adaptive optics module, MAORY.
MICADO will equip the E-ELT with a first light capability for
diffraction limited imaging at near-infrared wavelengths.

The MAORY instrument will be the specific AO instrument which, at first light will operate at

wavelengths from 0.8–2.4µm

So in short, the answer appears to be that the E-ELT will only have AO capabilities for near-IR at first light.
As to whether or not there are plans for visible AO capabilities, I couldn't find anything specific to that question. Given that first light for this project is 8 years away at best, I think that plans for future upgrades are tentative at best, and likely not easily accessible to those not on the project. I will say that while the MAORY instrument appears to be the intended AO system at first light, there is listed a second AO instrument called ATLAS that will eventually make its way onto the telescope. While I can find no specific details of this particular instrument, its possible that this will provide for visible AO.
On a tangentially related note, this telescope will also have active optics that will help to improve their imaging capabilities.


If space was n>3 dimensional, would clouds of particles still turn into 2D disks due to gravity?

It just so happens that this problem has been rigorously analyzed.  Max Tegmark wrote an interesting paper on it which can be gotten from the arXiv.  It is not super-technical, so I recommend you look at it.  Basically, with more than three space dimensions there are no stable orbits. Particles either disperse to infinity or collapse to a point (presumably a black hole).  (Tegmark also analyzes the effect of extra time dimensions -- weird.)

Added: A comment asks if a cloud of particles behaves differently and might still be stable. The answer is "no".  Here's why.

There are two cases here "gas" and "dust". The difference is that gasses have internal pressure which does affect the cloud's dynamics, while the particles in dust are big enough that collisions between them become infrequent and can be ignored for their effect on the dynamics. (Note: the only important difference between "gas" and "dust" is if the particles interact often enough to produce enough pressure to significantly affect dynamics.  It's not their composition. The terms "gas" and "dust" are simply terms of art used to describe the two cases.)

The dust case is trivial: The dust particles spend nearly all of the time in their individual orbits without colliding and if the orbits are not stable, the dust cloud isn't either.

Gas is more complicated, but it's probably easiest to look at what happens in three dimensions.  A non-rotating gas cloud collapses -- that's how galaxies and stars form!  A gas cloud that's rotating enough to support itself against collapse is relying on the centrifugal force to support it against gravity -- which doesn't work in four or more dimensions.


We never saw a planet explosion with Hubble.

To make something explode it has to produce a lot of energy in a very short time. Gravitational collapse and thermonuclear fusion in degenerate matter are two possible ways of producing enough energy to produce a supernova.

These mechanisms can not cause a planet to explode. Planets are solid and supported by their own matter, so can't collapse, and they don't contain large amounts of hot degenerate matter, so don't have any kind of fusion.

Planets don't produce much of their own energy, and there is no mechanism known that could cause a planet to explode.

The closest you may get is planetary collisions, such as that which likely formed the moon. But these have never been directly observed.


How can astrophysicists discriminate between pp-chain solar neutrinos and CNO-cycle ones?

Nuclear reactions can have particular energy spectra, which arise based on how energy and momentum are distributed among the product particles. This holds true for neutrinos produced in the many different stages of the CNO cycle, as well as the pp chain. The authors began the paper with a figure showing the theoretical neutrino fluxes for the various components of the CNO cycle:

The dominant CNO neutrino producers are the steps involving $^{13}\mathrm{N}$ and $^{15}\mathrm{O}$. The peak of the nitrogen spectrum lies in (or at the edge of) a region where the ratio of CNO neutrinos to background is the highest:

Fitting the observed spectra with models of CNO neutrino emission indicated that such neutrinos were indeed present. I'd draw an analogy with making electromagnetic observations: if your statistics are significantly better when, say, you add a black body component to a non-thermal power law fit, then you have evidence of black body emission. It's basically the same thing here - adding another component to a spectral fit.


The main mirror of a telescope forms an image at the prime focus.

You see the lines being brought to a single point, but that doesn't mean all the light entering the telescope is focused to a single point. It means that all the light from a single point in space is focussed at a single point. Light from different sources is focussed at different points, these points form a plane in the body of the telescope, the "focal plane".

This image can be seen. You can project an image using a single lens, or if you hold up a lens at arm's length and look through it, you will see a small inverted image. The purpose of the eyepiece is to act as a magnifying glass and make this small inverted image appear bigger.

The image is in focus only at the focal plane. If a sensor (without any further lens) is placed at any other point, no image will be seen, or at best a blurry image. We can see the image when we look through the eyepiece only because we have another lens (the cornea of our eye) bringing the light to a second focus on our retina.


How did New Horizons take such well-lit pictures of Pluto?

It's brighter on Pluto than you think.

NASA developed a tool called Pluto time, which tells you when at your place the ambient light conditions are similar to the ones on Pluto.  This occurs when the Sun is only 2° below the horizon!  That's quite shortly after sunset, and considerably before the end of civil twilight, which is when it's 6° below.

All of these photos were taken at local "Pluto time":

Pluto time, according to NASA.  Source: NASA

To answer your question: all it takes is a slightly longer exposure time / larger aperture / higher gain (what photographers may call ISO) than taking photos closer to the Sun.  It's easily bright enough for outdoor activities (except perhaps on a cloudy evening and/or in a dense forest, but neither should be common beyond the orbit of Neptune), so have fun glacier hiking on Pluto!


Why is right ascension measured on a 24 hour scale rather than a 23 hours and 56 minutes scale?

Right ascension is a historical oddity.

To specify a point in the sky you need a coordinate system, the one which we have come to use has it's origin at the Point of Aries on the Equator and the Ecliptic (a reasonable choice), It uses the Equator for one axis, and the meridian through the point of Aries for the other, again these are convenient and reasonable.

The coordinate system also requires a unit to be chosen for measuring angular distance. For measuring declension, we use degrees. The more mathematically pure may prefer radians, but degrees are a commonly used unit for measuring angles.

For measuring the angular distance around the equator we have come to use "hours", by dividing the full turn into 24 parts. This is a historical oddity. It has a few conveniences: The sidereal day is only slightly shorter than 24 hours, so a star with RA of 1hr will be due south about 1 hour after a star with a RA of 0hr (not exactly but close enough for a rule of thumb). A telescope with an equatorial mount that is set to make a full turn in 24 hours will track stars well enough for a human observer.

Dividing the equator into (23+56/60) parts would be inconvenient since its not a whole number. For these reasons Flamsteed used Hours for his catalogue in 1712, and we have followed his tradition.

So the use of "hours" may be roughly related to the apparent motion of the stars, but it is really just a way of specifying an angle, and using 24 is easier than (23+56/60)


It is very difficult to predict since the Milky Way galaxy and the Andromeda galaxy are going to collide in about 5 billion years. Since we cannot predict the gravity influences so far in advance, it is impossible to say.

There are 3 outcomes of this collision. The solar system falls into the black hole, the solar system goes into orbit in the new galaxy, or the solar system get propelled away from the new galaxy.


Has celestial navigation been materially impacted by the imperfect nature of celestial reference frames over time?

The best inertial frame is that which is comoving with the cosmic microwave background. There is a notable dipole in the microwave background, the result of the Solar system's motion relative to it. The invariance of this dipole provides evidence that the frame of the barycentre of the solar system is also a very good inertial frame.

This is of no significance to navigation, either by the stars, or among the planets.

Within the solar system, the sun's gravitation influence is dominant. While the sun is orbiting with the galaxy, a space probe is also orbiting the galaxy, and the variation in gravitation due to the galaxy is extremely small. Galactic tides are several orders less significant than photon pressure, various quadrupole effects or the influence of other bodies.


This is because there are many more small NEOs than there are large ones and the larger ones are easier to find and so were found first. The numbers of NEOs follow what's know as a power law distribution with an exponent of ~1.75. This means that given the roughly factor 7 difference between 140m and 1000m NEOs, there will be 7^1.75 = ~31 times as many 140m NEOs as 1000m NEOs. This is shown in this plot below.  We estimate there are about 950 NEOs 1km or larger and about 25,000 NEOs larger than 140m. We have found over 93% of the 1+km NEOs and so there aren't many left to find. In comparison, over 2/3rds of the 140-1000m NEOs still remain to be found, most in the 140-300m size range. However this improved considerably since even 2010 when about 82% of the population was still to be found.


How long before Phobos reaches the Roche limit of Mars?

Sometime before now and "tens of millions of years".
Phobos currently orbits at a radius of about 9400km
The theoretical Roche limit is different for rigid and fluid bodies.  If Phobos were a fluid body, it would already have passed the Roche limit at 10500km.  This is because fluid bodies will deform into an ellipse pointing towards the planet, and this amplifies the tidal forces and so causes break up a larger radius.
If Phobos is fully rigid body, then the Roche limit is half that at about 5400km. In fact a rubble-pile-with-crust like Phobos is neither fully fluid, nor rigid. Wikipedia gives an estimate of "30-50 million years" before it is fully disrupted.
So Phobos is already a body under stress, and (as anybody who knows a little about earthquakes knows) rock under stress is often unpredictable in how it responds: It can creep, fracture, or hold rigid while stress increases and then fail catastrophically.


There's a lot of loosely connected questions here, so let's go through them one by one.

  Which rotational frame does the universe prefer?

The universe doesn't prefer any frame. There is no "universal" reference frame. This is one of the principle axioms of Special Relativity. You cannot pick a frame and say this is the "correct" frame. The reason being that you have no means of establishing what "correct" even means. One person may claim their chosen frame is correct, but another person might choose something different and claim theirs is correct. Neither person would be able to show their frame is better or more "correct" than the other's.

That being said, I think its important to point out that while no true universal reference frame exists, scientists sometimes use a de facto "psuedo-universal" reference frame. That frame is the one which is at rest with respect to the Cosmic Microwave Background (CMB) radiation. It is not a true universal reference frame, but because of the importance of the CMB, the frame at rest with respect to the CMB has been given more (arbitrary) importance by scientists, hence the term psuedo-universal. You can read more about this at this great physics Stack Exchange question.

  We say that the earth rotates once per 24 hours. But is this the master frame?

This is a good question. Anytime someone mentions motion, you should always ask, motion with respect to what? If I'm inside a car, I would say the car isn't moving at all because my motion, with respect to the car is zero. A person standing on the side of the road would say the car is traveling at 25 mph. So you can see that defining motion requires defining a frame to measure that motion in.

When it comes to talking about the rotation of the Earth, there are actually a few different frames that scientists use. They're all very similar and produce similar answers (all about 24 hours). I'll restrict myself to talking about the frame which defines the Solar Day (another example would be the Sidereal Day).

The Solar Day is the time it takes the Sun to return to the same position in the sky as from the previous day. As you can hopefully see, this measure of a day is fixed to a frame with respect to the Sun. It would be the same regardless of the motion of our solar system through the universe and thus wouldn't be considered a "master" frame (not that there is one).

  What is the solution to Newtons Bucket Problem?

This, I would say, is really outside the realm of the Astronomy Stack Exchange and really should be its own question, preferably on the Physics Stack Exchange. For that reason, I won't answer this question here, but I will point you to this great source which talks about the Bucket Problem is detail.


What causes the duration of moon phases (the time the moon is visible above the horizon) to vary for an observer in the northern hemisphere?

The interval between moonrise and moonset varies for the same reason that summer days are longer than winter days.
How long a celestial object appears above the horizon depends on the observer's latitude and the object's declination (angle north or south of the celestial equator).
At full moon on 2019-01-21, the Moon's declination was +20$^\circ$, and there were 15 hours between moonrise and moonset at 40$^\circ$N latitude.
At new moon on 2019-02-04, the Moon's declination was -18$^\circ$, and moonrise and moonset at 40$^\circ$N were only 10h10m apart.

Another way to think of this is that when the Moon is in the first quarter, it's within about 5 degrees of where the Sun would be 3 months later; last quarter, the Moon is where the Sun would be 3 months earlier, etc.
A full Moon in June is up about as long as the Sun in December; a first quarter Moon in March is up about as long as the Sun in June.