Is there some upper limit in the moon size distribution?

Small point, but Ganymede is slightly larger than Titan as moons are measured by their solid surface, though with atmosphere, Titan appears slightly bigger.   The drawing is probably accurate, but slightly misleading.

It's worth adding that Titan is out-gassing so it may have, quite some time ago, been larger than Ganymede.

Source

According to wikipedia, the larger, close to Jupiter and Saturn moons are unlikely to have been captured moons.

https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System#Moons

  Jupiter and Saturn have several large moons, such as Io, Europa,
  Ganymede and Titan, which may have originated from discs around each
  giant planet in much the same way that the planets formed from the
  disc around the Sun.[76] This origin is indicated by the large sizes
  of the moons and their proximity to the planet. These attributes are
  impossible to achieve via capture, while the gaseous nature of the
  primaries also make formation from collision debris unlikely. The
  outer moons of the giant planets tend to be small and have eccentric
  orbits with arbitrary inclinations. These are the characteristics
  expected of captured bodies

and the 3 listed methods of moon formation are:

Co-formation
Impact
Capture.

The largest co-formation ratio is Titan to Saturn, about 1 to 4,000 in terms of mass.   Ganymede to Jupiter, less than 1 to 12,000.   The wiki article points out that ratios of 100 to 1 are likely giant impacts and you're unlikely to see that kind of ratio in co formation, so a conservative estimate, the largest co-formation you mgiht see might be around a few hundred to 1.   If you figure the largest planets are about 12-12.5 Jupiters (more than that, they become brown dwarfs, though a cool brown dwarf orbiting a sun might get called a planet too at greater than 13 Jupiter masses), but lets say, 12 Jupiters.  That's about 4,000 earth masses.    If we look at a co-formation ratio of a few hundred to perhaps over a thousand to 1 as about the maxiumum, there's a possible ballpark answer of a handul of earth masses might be about as big as a co-formation moon is likely to get in a perfect formation situation, And those would probably be pretty rare.

Giant impacts are different.  Pluto-Charon could be an impact generated dwarf planet/moon system and Charon is 1/8th the mass of Pluto.    How likely that kind of impact is to happen on a super earth is hard to say, but theoretically a super-earth with a giant impact could have a large moon too - a bigger earth with a bigger impact and bigger moon, but there's problems here.  Giant planets are likely to form close to the sun and the orbital sphere of influence gets smaller closer to the sun.   Also, giant planets might tend to form thick gaseous atmospheres, retaining all their hydrogen and helium, so too big, it might begin to resemble a gas giant, which aren't as good at creating debris when impacted, so there's a few limiting factors here and we might be limited to, certainly several times larger than our moon, but perhaps not as large as co-formation moons, further from the sun.

Finally, moon capture.    While there's no examples of this in our solar system, I see no reason why a large planet couldn't capture another planet, though you then have to determine whether it's a co-planet orbit, which I think a captured planet should be considered, or whether the captured object is a moon. - personally I think a captured planet that orbits a planet would be a co-planet system, so I don't like the capture as a method for getting as large a moon as possible, though I suppose it could be looked at either way.

We know almost nothing about moons in other solar systems (exomoons).  The observations are just barely possible with current technology, but very difficult.  That's likely going to change in the next few years, but for now we know virtually nothing in other solar systems, so, any present predictions on the largest possible moons would have to be modeled or estimated.

A few articles on exomoons:

http://www.astrobio.net/news-exclusive/new-exomoon-hunting-technique-could-find-solar-system-like-moons/

http://discovermagazine.com/2012/jul-aug/06-hunting-moons-outside-the-solar-system

http://www.space.com/25438-exomoon-around-alien-planet-discovery.html

http://phys.org/news/2014-10-exomoons-abundant-sources-habitability.html

http://blogs.scientificamerican.com/life-unbounded/has-an-exomoon-been-found2/


How did Copernicus get the idea of heliocentric model?

The heliocentric model is historically thought first by Aristarchus. He eventually gave up on the idea due to not being able to give evidence for a concept called stellar parallax: If earth orbits the sun then we should see the closer stars shift in relation to the other stars (strong evidence for this didn't come until Galileo).

Copernicus became to accept the heliocentric model given by Aristarchus due to his study of the Ptolemaic model. His studies showed that planetary motion in a geocentric model was inaccurate.

His model, though, was still based on the fact that "heavenly objects" moved in complete circles. He had to add much complexity to account for this fact which is why the heliocentric model was still not widely accepted. Side note, it was Johannes Kepler, over 100 years later, that claimed that planets move in ellipses rather than circles.


I think there might have been a problem with your compass readings, or "in the north" really means just a little north of east.

Cuba is in the northern hemisphere at around 20 to 23°N (tropic of Cancer) and so the Sun can be directly over head and the Moon with an additional 5 degrees inclination with respect to the ecliptic can go several degrees North. (see this answer for a bit about the Moon's motion).

  it started in the north position around 4 pm.
  By 7pm, it was in the east position,
  and by 11 pm, it was in the south position.

I think there might have been something affecting your compass. The Magnetic declination there is about 6-8 degrees, that might have had a tiny effect, but you might look for problems like

iron or other ferromagnetic material,
mechanical compass from the wrong hemisphere sticking due to dip angle
or if it was an electronic compass, incorrect degaussing/calibration.

The other possibility is that by "in the north" you really mean only "slightly north of east" or "east-north-east".

I wrote a short script in Python for your trip, and the three lines in the plot represent January 15 (lowest, blue), 17, and 19 from Santiago de Cuba, the southernmost area of Cuba. The moon does get to the zenith and perhaps a tiny bit past it (towards North) but I don't think that would be visually noticeable.

The Large black dots represent 7PM local time, and each dot to the west is one more hour.

Some Python:

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Loader, Topos

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads        = 180/pi, pi/180

load  = Loader('~/Documents/fishing/SkyData')  # single instance for big files
ts    = load.timescale()
de421 = load('de421.bsp')
earth = de421['earth']
moon  = de421['moon']

Santiago_de_Cuba = earth + Topos(latitude_degrees  = 20.019833,
                                 longitude_degrees = -75.813917,
                                 elevation_m       = 10.)

hours = np.arange(16, 23.1, 0.5) + 5

days   = (15, 17, 19, 21)

altazs, lines, linez= [], [], []
for day in days:
    times = ts.utc(2019, 1, day, hours)
    alt, az, d = (Santiago_de_Cuba).at(times).observe(moon).apparent().altaz()
    alt, az    = [thing.degrees for thing in (alt, az)]
    alt[alt<0] = np.nan
    altazs.append((alt, az))
    r = (1 - alt/90.)
    theta = rads * az
    lines.append((r, theta))
    x, y = [r*f(theta) for f in (np.sin, np.cos)]
    linez.append((x, y))

if True:
    plt.figure()
    plt.subplot(2, 1, 1)
    for alt, az in altazs:
        plt.plot(hours, alt)
        plt.plot(hours[::2], alt[::2], 'ok')
    plt.subplot(2, 1, 2)
    for alt, az in altazs:
        plt.plot(hours, az)
        plt.plot(hours[::2], az[::2], 'ok')
    plt.show()

th = np.linspace(0, twopi, 201)
cth, sth = [f(th) for f in (np.cos, np.sin)]
if True:
    plt.figure()
    plt.plot(cth, sth, '-k', linewidth=1.5)
    plt.plot([0], [0], 'or', markersize=8)
    plt.plot([-0.1, 0.1], [ 0,   0  ], '-k')
    plt.plot([ 0,   0  ], [-0.1, 0.1], '-k')

    plt.text(-0.05,  0.85, 'N', fontsize=16)
    plt.text( 0.9,  -0.05, 'E', fontsize=16)
    plt.text(-0.95, -0.05, 'W', fontsize=16)
    plt.text(-0.05, -0.90, 'S', fontsize=16)

    for x, y in linez:
        linewidth= 1
        plt.plot(x, y)
        plt.plot(x[:-1:2], y[:-1:2], '.k')
        plt.plot(x[:1], y[:1], 'ok')
    plt.show()


Using a telescope to look at the Earth History in Some Detail

No.

Let's say we want to see Earth as it was 1000 years ago.  Assume that someone has set up a perfect mirror 500 light-years away, so that we can actually see the light that left Earth 1000 years ago. (That's a really big assumption.)

The best telescopes in the world can't see the Apollo landing sites from Earth. We didn't get decent images of the descent stages, which were left on the surface, until the Lunar Reconnaissance Orbiter sent back photos it took from Lunar orbit. See http://www.nasa.gov/mission_pages/LRO/news/apollo-sites.html

1000 light-years is about 20 billion times as far away as the Moon. There's no way we could see people's faces at that distance with current technology -- or with any reasonable future technology. (There are physical limits on the resolution of an optical telescope of a given size.)

And all this assumes that we have that perfect mirror out there. As far as I know, nobody has set up such a mirror for us, and all the light that left Earth 1000 years ago is now 1000 light-years away, badly faded, and beyond our reach.

It's conceivable that we could develop faster-than-light travel (which may or may not be physically possible), go out there, build a telescope with a really big aperture, and point it back at Earth. But that's not likely to happen any time soon, and I wouldn't know how to determine how good an image we might be able to get.

And going back millions of years just makes the problem worse.


Relationship between Periapsis and apoapsis longitude

The periapsis and apoapsis are 180° apart.
According to Kepler's first law,
an orbit is an ellipse with the central body at one focus.
The periapsis (P), central body (F), and apoapsis (A) all lie on the major axis of that ellipse.


how did James Bradley discover light aberration?

That is a horrible value.

That value for aberration of starlight is "horrible" in two ways: One, in that's is very small. And two, in that it's large enough to interfere with parallax measurements.

In the late 1600s, scientists had already adopted the heliocentric model, and so were looking to measure the distance to the stars via parallax. Robert Hooke was among the first to attempt it, and probably the first to claim a measurement of the parallax. His observations were not very convincing at that time, but he probably did observe some effects from aberration of starlight (though they didn't know such a thing existed back then).

In the early 1700s, Bradley attempted to make a more precise measurement of parallax, and did obtain significant data, but the variations in the star positions were out of phase with the model. Using the speed of light estimate by Ole Roemer from Hooke's time, he correctly proposed that this was not parallax, but a speed-of-light phenomenon, now known as the aberration of starlight.

More info here:

http://arxiv.org/pdf/1208.2061.pdf

It took another century, until early 1800, for Bessel to measure the first parallax.

Basically, aberration of starlight produces the same variation in the positions of all stars, a periodic elliptic "wobble" that is the same for all stellar objects. Parallax produces a similar kind of wobble, but the amplitude depends on the distance to the star, and the phase is different from the aberration's.

There are also other effects, such as nutation, which introduce further perturbations.

http://cseligman.com/text/history/bradley.htm

Aberration per se is easy to explain via the old analogy to running through the rain. When you're standing, rain falls vertically. When you're running, rain appears to come from a different direction. Same with starlight. Like in this diagram (by Professor Courtney Seligman, linked above):

Now, small as that value is, it's still 20 arcsec. The resolving power of early telescopes (and modern amateur telescopes) is 1 arcsec or better - which is more than enough. What you need is a very solid mount with good setting circles. Lacking that, you could build a telescope with a fixed position, very rigid, forever pointing at the same place in the sky - that's what Hooke did - and then you're only limited by resolving power.

In any case, the precision required was quite doable with 1700s technology. An amateur could do it today, at very low cost - what you need is persistence to follow through with a long term project.

  Could the effect of parallex, proper motion and light aberration of a
  star be separated clearly?

Sure. Aberration is always the same. Parallax depends on the distance, and the phase is different.

EDIT:

  Why is their phase different? They all have a one year period.

Go to the site by Prof. Seligman, link above, and look at the two diagrams showing the aberration and the parallax. In point B (in April), the star is in the middle of the ellipse due to parallax, but it's at the right-hand tip of the ellipse due to aberration.

Broadly speaking, parallax is a position-based effect, whereas aberration is a speed-based effect. It makes sense that they are out of phase by 90 degrees (remember trigonometry?)

Proper motion appears as a long-term drift of stellar positions that remains even after you subtract aberration, parallax, nutation, etc. There's a back-and-forth-back-and-forth from aberration and parallax, whereas proper motion keeps going in one direction only.

You could measure proper motion with relatively short term observations, but it only works for stars that are nearby and are moving fast. If you're willing to wait decades, proper motion is much more easy to measure.

EDIT:

  Generally we should not worry about the light aberration, right?

Yeah, sort of. You know, at any given moment, the size and the direction of it. So you could simply subtract it from the measured position of the star you're watching.

OTOH, it was an additional source of error, until speed of light and the Earth's orbital parameters were very precisely determined. Perhaps it still contributes to the noise added to the data, to some extent - I'm not entirely sure here.


SIMBAD TAP queries seem to indicate there are no distance measurements for a great many nearby stars, including Alpha Centauri or Barnard's Star

Well - not sure why distance and parallax tables are empty, but I missed something super simple, the PLX_VALUE column on the BASIC table.
SELECT * FROM BASIC  WHERE plx_value IS NOT NULL ORDER BY plx_value DESC
OR convert to distance with D = 1000/PLX_VALUE for dist in parsecs


The odds of a supernova being visible to the naked eye is slim. Adams et al. (2013) estimated that there is a 20% chance that we will be able to see such a supernova within the next 50 years. That should give you an idea of how unlikely this is.

The vast majority of supernovae that have been discovered lately have magnitudes ranging from +20 to +16 (check out this page, updated daily). The naked eye can see objects of magnitude +6 or less, meaning that a supernova of magnitude 16 is $10^{10}$ times fainter. Powerful telescopes are needed to observe these events.

Supernovae also aren't that quick. They fade over days or weeks or months or even years. You would not see a bright, quick flash of light. I don't want to say that it's impossible . . . but it's impossible.

If this was a supernova, it's even more unlikely that you could have seen such detail. When I say that a supernova is visible to the naked eye, that just means that you can see the light from it. It doesn't mean that the structure of any gas or dust being expelled can be observed. To make out that level of detail, it would need to be quite close to Earth - and so it would definitely have been observed by astronomers and amateur stargazers in quite a few places.

So no, I don't think this was a supernova. I have no idea what you saw, but it was not what you think it is.


In the 1950's how were radio-astrometric positions with portable dishes so precise they could be assigned to their dim optical counterparts (Quasars)?

In many cases in astronomy, there’s a difference in finding the position of something precisely vs. seeing detailed structure in that object.  In the former case, what you need to do is to find the centroid of the image.  So even with relatively coarse spatial resolution, you can often get very precise positions because you can centroid the images very well.  (EDIT: However, this may not apply to early radio observations because single-dish radio telescopes don’t form images.)
Specifically for 3C273, the first quasar identified, Wikipedia says that its position was refined with lunar occultation. This technique involves observing a source as the Moon approaches it and determining the precise moment when it disappears behind the edge of the Moon. Here you need to measure time precisely and to know the Moon's path very well.
Also see this answer to Has radio astronomy ever been done on objects that appear very close to the Moon? Is this avoided?
The first radio source to have a star-like optical counterpart associated with it was 3C84, and that was done by measuring the radio position precisely using interferometry, i.e. combining signals from multiple telescopes.
So I think that this was not in fact possible with single-dish radio telescopes in the 1950s, and required more advanced techniques that came a bit later.


@Ingolifs' answer is exactly right. I'll just add to it some more plotting using the Python package Skyfield.

The "inner" plotted bodies (from outer to inner) are Mars, (Earth), Moon, Venus, Mercury, The Sun. The data is for 2000 days.

The "outer" plotted bodies are Uranus, Saturn, Jupiter, Earth, and The Sun. The data is for 2000 months.

Distances are in AU.

Click the plot to see larger, or just run the Python script and make your own plot.

class Body(object):
    def __init__(self, name):
        self.name = name

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Topos, Loader, EarthSatellite

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads = 180/pi, pi/180

AU = 149597870.700  # km

load    = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files
data421    = load('de421.bsp')
data405    = load('de405.bsp')
data       = data405   # need the longer time range for the outer planets
ts      = load.timescale()

things  = ('sun', 'mercury', 'venus', 'earth', 'moon', 'mars',
           'jupiter barycenter', 'saturn barycenter', 'uranus barycenter', )

bodies = []
for thing in things:
    name = thing.split()[0]
    body = Body(name)
    bodies.append(body)
    body.obj = data[thing]

sun, mercury, venus, earth, moon, mars, jupiter , saturn, uranus = bodies

days_inner = np.arange(2000)
times_inner = ts.utc(2019, 1, days_inner)

months_outer = np.arange(2000)
times_outer = ts.utc(2019, months_outer, 1)

for body in bodies:
    body.posn_barycentric_inner = body.obj.at(times_inner).position.km
    body.posn_barycentric_outer = body.obj.at(times_outer).position.km
for body in bodies:
    body.posn_geocentric_inner = body.posn_barycentric_inner - earth.posn_barycentric_inner
    body.posn_geocentric_outer = body.posn_barycentric_outer - earth.posn_barycentric_outer
    body.posn_heliocentric_inner = body.posn_barycentric_inner - sun.posn_barycentric_inner
    body.posn_heliocentric_outer = body.posn_barycentric_outer - sun.posn_barycentric_outer

print (jupiter.posn_geocentric_inner.shape)

if True:
    plt.figure()
    lw, fs = 0.7, 12

    plt.subplot(2, 2, 1)
    plt.title('geocentric inner', fontsize=fs)
    for body in (sun, mercury, venus, earth, moon, mars):
        x, y, z = body.posn_geocentric_inner / AU
        plt.plot(x, y, linewidth=lw)

    plt.subplot(2, 2, 3)
    plt.title('heliocentric inner', fontsize=fs)
    for body in (sun, mercury, venus, earth, moon, mars):
        x, y, z = body.posn_heliocentric_inner / AU
        plt.plot(x, y, linewidth=lw)

    plt.subplot(2, 2, 2)
    plt.title('geocentric outer', fontsize=fs)
    for body in (earth, mars, saturn, uranus):
        x, y, z = body.posn_geocentric_outer / AU
        plt.plot(x, y, linewidth=lw)

    plt.subplot(2, 2, 4)
    plt.title('heliocentric outer', fontsize=fs)
    for body in (earth, mars, saturn, uranus):
        x, y, z = body.posn_heliocentric_outer / AU
        plt.plot(x, y, linewidth=lw)

    plt.show()


How is the expansion rate of the universe obtained from Baryon Acoustic Oscillations?

I'm under the impression that the sound horizon is a fixed physical
  distance, so its measured comoving distance at any redshift should
  always be the same. Am I right or wrong in this assumption?

You are both right and wrong at the same time. Right in the sense that, yes, in comoving coordinates, the distance is the same at all redshifts. But when we use galaxy catalogues to measure the BAO distance scale, aka the BAO peak, it usually varies with redshift. This variation is very tiny, of the order of a 1-2 h$^{-1}$Mpc.

  But if the sound horizon is always at the same physical scale, how can
  we use measurements today to determine the rate of expansion? Do the
  correlation functions give you proper separations, and from that with
  different cosmological models we can determine what 0 or ()
  value gives a comoving separation equal to the sound horizon? And is
  this the same as the BAO signal?

Before redshift of galaxies were measured, people were just using the galaxy positions on the sky (Right ascension RA, Declination DEC) to calculate the angular correlation function. You can also see the BAO peak in the angular correlation function. Nowadays, since we have accurate redshift measurements, people calculate the 3d correlation function $\xi(r)$. To get constraints from BAO peaks in a very simple way, we can in brief, do the following.

To do so, we first need to convert RA, DEC and redshift to X,Y,Z cartesian coordinates. It is at this step, we assume a fiducial cosmological model, as in $\Omega_{M}=0.27, \Omega_{\Lambda}=0.7, h=0.67$ etc. Now, we have 3D information of our galaxies and we can count pairs and get the $\xi(r)$, from which we can calculate the BAO peak.

People can then use theoretical model predictions for $\xi(r)$, given a specific set of cosmological parameters. It is here that cosmologists vary the set of cosmological parameters, which will give us a specific $\xi(r)$.
Lets say, we have 10 variations of $\Omega_{M}$, $\Omega_{\Lambda}$ and $h$. This would give us 30 theoretical $\xi(r)$'s, and we can compute the BAO peak for all of them using an empirical fit model. These peaks can now be compared with the observed peak from the data and we can calculate the $\chi^{2} = \frac{(peak_{obs} - peak_{theory})^{2}}{err(peak_{obs})^{2}}$. This can give us constraints on the different cosmological parameters.

Of course, this is just a basic answer to your question and there are a lot of other processes involved! You can check out this paper which was recently published in arXiv which uses only the BAO peak information to get constraints on cosmological distance measures.


Time does not exist for photons.  Quoting Wikipedia's somewhat unnecessarily technical page:

  In relativity, proper time along a timelike world line is defined as the time as measured by a clock following that line.

The proper time between any two points along any photon's path in empty space is zero.  So the photon "experiences" no passage of time.

It's not very clear to me how this could explain the double-slit experiment. Proper Time is an entirely relativistic concept, whereas the double slit results are not explained by relativity, but are explained by Quantum Mechanics.


You seem to have two types of seasons confused.  Astronomical seasons are relatively stable in their durations, while seasons as defined by the weather, can vary by several weeks.  For example, snow 3 weeks after spring begins or freezing weather before winter begins.  This is what is meant when we say an early summer or a late winter.  It simply means that the weather usually associated with that season is later or earlier than usual.

Generally, the lengths of the seasons ranges from about 88 to 93 days every year.

For 2014 (Northern hemisphere season lengths)
Length of spring = 92d 17h 54m
Length of summer = 93d 15h 37m
Length of autumn = 89d 20h 34m
Length of winter = 88d 23h 42m

The axis of the earth has nothing to do with it unless we are considering a time span extending to many thousands of years.  The Earth's axis hasn't changed enough in recorded history to significantly effect the seasons.

The astronomical seasons are defined by the geocentric position of the sun and have little to do with the weather.

March Equinox Beginning of spring in the northern hemisphere = Beginning of autumn in the southern hemisphere, when the sun reaches declination 0 degrees while crossing the celestial equator northward.

June Solstice Beginning of summer in the northern hemisphere = Beginning of winter in the southern hemisphere, when the sun reaches right ascension 90 degrees (6h).

September Equinox Beginning of autumn in the northern hemisphere = Beginning of spring in the southern hemisphere, when the sun reaches declination 0 degrees while crossing the celestial equator southward.

December Solstice Beginning of winter in the northern hemisphere = Beginning of summer in the southern hemisphere, when the sun reaches right ascension 270 degrees (18h).

These astronomical season lengths vary by only a matter of minutes either way each year, while the seasonal weather can vary considerably and winter-type weather can persist even after astronomical spring has begun (a late or short spring), but it is not yearly changes in the axis of the Earth that causes these annual weather variations.

SEASONAL STATISTICS FOR THE YEAR 2014 @ TZone = UT-05:00 (New York City)
BEGINNING DATES, TIMES TT AND LENGTHS OF THE SEASONS AND SEASONAL YEARS

Equinox or Solstice      Season Length     Since Previous     Until Following
E 2014 Mar 20 11:57:58   92d 17h 54m 23s   365d 05h 54m 51s   365d 05h 48m 15s
S 2014 Jun 21 05:52:21   93d 15h 37m 26s   365d 05h 47m 17s   365d 05h 46m 41s
E 2014 Sep 22 21:29:47   89d 20h 34m 21s   365d 05h 44m 56s   365d 05h 52m 40s
S 2014 Dec 21 18:04:08   88d 23h 42m 05s   365d 05h 52m 01s   365d 05h 44m 56s

Perhaps these links will help:
http://neoprogrammics.com/equinoxes_and_solstices/
http://neoprogrammics.com/equinoxes_and_solstices/seasonal_statistics/


Finding the solar noon based on longitude and latitude?

Taking solar noon as being the time when the sun is at its highest elevation (altitude) in the sky for a given day, then your assumption of the LST (local solar time) is correct, as it is stated in the website you linked as being:

  Twelve noon local solar time (LST) is defined as when the sun is highest in the sky. Local time (LT) usually varies from LST because of the eccentricity of the Earth's orbit, and because of human adjustments such as time zones and daylight saving.

This is something I have, in part, programmed into an app for a separate project (quite unrelated to yours).

You would have to equate the precedent quantities first (Local Standard Time Meridian, Equation of Time and Time Correction factors) as they have done on the website you linked (as these are reliant on the observer's position and time of observation), but to determine at what time solar noon occurred, the final step needs to have local solar time (LST) set to 12:00 (solar noon).

A worked exampled is given within the first 7 pages/slides of Sunlight and its Properties II.


Why does Jupiter atmosphere does not get thinner on edges

Why does Jupiter's atmosphere not get thinner at the edges?
Looking at an image of the atmosphere's edge it is very sharp as if it has a solid surface, which it has not since it’s a gas giant. There should be some thinning of the atmosphere like we see around Earth. Would this be seen on higher resolution images?

When you look at a spherical object from a distance it appears round, sometimes when you view it from close up it doesn't appear quite so smooth (or round). The wavelength that you look at an object with also affects what you are able to see.

$$\scriptsize \text{Figure 6. High resolution SEM image of CN ball bearing inner ring fabricated from 52100 steel showing} \\ \scriptsize \text{fine globular cementite. Presence of fine precipitates of carbides and nitrides uniformly aggregated in the} \\ \scriptsize \text{fine martensitic microstructure, which act as obstacles to the motion of dislocations.}$$
The above scanning electron microscope image of a smoothly ground ball bearing illustrates this, from a relatively far distance it appears smooth; from much closer the particles constituting its makeup become visible. Image Source: "Effect of Carbonitriding on Endurance Life of Ball Bearing Produced from SAE 52100 Bearing Steels", by Karthikeyan Rajan, Vinod Joshi, and Arindam Ghosh.
A similar principle applies to viewing Jupiter from so far away. While its clouds are far greater than 2000 km thick most of the gases are opaque thus not allowing you to see very far into the atmosphere using visible light.

The upper left image shows Jupiter from a distance using visible light. Source: "PJ14 Image 26 - Hot Spot (Enhanced)".

The upper right image  shows the different layers of Jupiter's atmosphere. Source: "How deep is the atmosphere?".

The lower left image from NASA's Juno spacecraft approaching Jupiter on Aug. 27, 2016, the Jovian Infrared Auroral Mapper (JIRAM) instrument captured the planet's glow in infrared light. Source: "Juno Captures Jupiter's Glow in Infrared Light".

The last two images are from: "NASA's Juno Mission Provides Infrared Tour of Jupiter's North Pole", there is also a close-up video flythrough of the atmosphere created from infrared imaging. A second video on that webpage, titled: "Jupiter's Dynamo", shows magnetometer imaging overlaid on visible spectrum imaging. The YouTube links for those videos are: "Low 3-D Flyover of Jupiter’s North Pole in Infrared" and "Jupiter's Dynamo".


In practice, can very high magnetic fields significantly alter the dynamics of an astronomical-scale bodies?

Where such an effect becomes important is during the accretion of matter from a regular (more-or-less) main sequence donor star onto a magnetized neutron star.
Here, the magnetic field at the NS can reach about 10$^{12}$ G and the rate of accretion about 10$^{13}$kg/s. On its way down the gravitation well towards the NS the material, by now mostly a plasma, will reach a point where the Lorentz-force will force it to spiral around and follow the magnetic field lines. So from that point onward its dynamics will be determined by the structure of the magnetic field, and the material will ultimately fall onto the magnetic poles, creating pulsars.
This forcing onto the magnetic field lines happens roughly when the magnetic pressure exceeds the ram pressure of the material. So,
$$
P_{\mathrm{ram}}=P_{\mathrm{B}} \\
\rho(r) \cdot(\dot{\vec{r}})^{2} = \frac{B(r)^{2}}{2 \mu_{0}}
$$
Here B(r) is a dipole field around the NS which drops off as r$^{-3}$ and so P$_\mathrm{B}$ drops off rapidly as r$^{-6}$, as Rob Jeffries  already mention in his comment.
The radius around at which the magnetic field becomes dominant is called the Alfven radius and is in the order of a few 1000 km.

This then looks something like this (Filippus et al., 2008):

I am not sure, however, if that counts as astronomical-scale, but this is the most dramatic effect of magnetic fields I could think of.


With only one shadow in a picture, no, you can not.

In order to get North's position from a shadow you need to know where you are and which solar time and date is it there. These are two extra data and you can not disentangle them, even if the picture is perfect.

e.g. suppose a picture in which you see exact measurements of shadow and people, and they are both exactly of the same size. This means Sun is 45º above Horizon. But that may be local noon at Bourdeaux on March 20th (thus shadow pointing North), or local noon at Christchurch on same date (thus shadow pointing South), or local 3PM solar on Quito on March 20th again (thus shadow pointing East).

I chose March 20th because it is easier to visualize where the Sun would be (exactly over the Equator), and these cities due to their latitudes:

Bourdeaux: 44º 50' N (almost 45ºN)
Christchurch: 43º 31' S (almost 45ºS)
Quito: 0º 13' N (almost on Equator)


Dynamical modelling of galaxies with counter-rotating components using the Schwarzschild method

Convert the 3D luminosity distribution to mass using a bunch of mass particles.

Well, you probably could do that, but I suspect it would be inaccurate and extremely time-consuming. Most Schwarzschild modeling approaches parameterize the 3D luminosity-density distribution using functions from which the gravitational potential can be relatively easily calculated.

For example, one popular approach is to approximate the observed 2D light distribution using the sum of multiple Gaussians (as you mention), which can be deprojected under certain assumptions into the sum of multiple 3D Gaussians to get the 3D luminosity-density distribution. Then you multiply this distribution by a mass-to-light ratio (usually one of the parameters of the model you're ultimately trying to fit to the data) and derive the gravitational potential from that. No need to approximate things with a bunch of particles.

This also allows you to easily add extra hypothetical "dark" components, like dark-matter halos or central supermassive black holes, which have simple analytical descriptions and don't need to modeled with particles.

  To my understanding, the orbits of the particles in a Schwarzschild model can be counter-rotating.

Correct. In practice, if the system is (assumed to be) axisymmetric, you can calculate a set of orbits, and then easily generate their counter-rotating counterparts by just reversing the velocity vectors.

  Yet, all papers on dynamical modelling using the Schwarzschild method fit models to velocity and velocity dispersion maps instead of a spectral cube.

You're clearly not reading the right papers ;-).
Let me divide this into two parts:

1. Converting spectroscopic data to stellar-kinematic data

Before any modeling takes place, the observed spectroscopic data (long-slit and/or integral-field-unit "spectral cubes") must be analyzed to extract the stellar-kinematic data. This is done by dividing up the data into different spatial bins on the sky (bins along the slit, or 2D bins for a spectral cube) and summing the spectra within each bin, to improve the S/N.

Then you fit each spectrum with a weighted combination of stellar spectra (observed or from models), broadened by a model1 line-of-sight velocity distribution (LOSVD). In very simple cases, the LOSVD is assumed to be just a Gaussian with a specific central velocity $V$ and velocity dispersion $\sigma$. The more common present-day approach is to model the LOSVD with a Gaussian which is multiplied by third- and fourth-order Gauss-Hermite polynomials. The strength of the 3rd-order polynomial is parameterized by $h_{3}$ and measures skew; the 4th-order polynomial is parameterized by $h_{4}$ and measures kurtosis. For galaxies with counter-rotating components, $h_{3}$ and $h_{4}$ provide some information about the non-single-Gaussian nature of the LOSVD, although it's rather fuzzy and poorly constrained.

A more sophisticated approach (requiring high-S/N data) is to "non-parametrically" model the LOSVD with a histogram having twenty, thirty, or more velocity bins. To show how the non-parametric approach works, consider the following stellar-kinematic data for the S0 galaxy NGC 4191:

The first plot, from Fig. B1 of Krajnovic et al. 2011 (MNRAS, 414, 2923), shows $V$ and $\sigma$ maps (from the SAURON IFU). In the velocity field, you can see that the velocity reverses between $r \sim 5$ and 10 arc seconds; at roughly the same radius, $\sigma$ peaks. The paired off-center peaks in $\sigma$ are suggestive of a region where the two counter-rotating components are about equal (as argued by Krajnovic et al. in their paper).

The second plot, based on Fig. 2 of Coccato et al. 2015 (A&A, 581, A65), shows the $\sigma$ map (colored panel, rotated 90 degrees with respect to the figure above) and non-parametric LOSVDs (31 velocity bins) for three of the bins (histograms) from VIRUS-W IFU observations of NGC 4191. You can see how, in the second histogram, the double peaks due to the overlapping, counter-rotating stellar components are clearly represented.

2. Comparing the Schwarzschild model with the kinematic data

After computing the library of stellar orbits for a given galaxy potential and assigning a set of weights to the individual orbits, you "observe" the model by orienting the galaxy to match the observed orientation, dividing it up into projected 2D spatial bins that match the 2D binning used for determining real galaxy's stellar kinematics. Then (for each bin) you compute the weighted LOSVD by integrating along the line of sight through that bin, adding up the radial-velocity contributions from each orbit that intersects the line of sight. If your data consists of $V$, $\sigma$, $h_3$, and $h_4$, then you fit the resulting model LOSVD for that bin with the Gauss-Hermite function to get the model bin's $V$, $\sigma$, $h_3$, and $h_4$. If your data are non-parametric LOSVDs, then you can simply bin the model LOSVD using the same velocity-binning scheme you used for the data.

This set of predicted LOSVD values (either $V$, $\sigma$, $h_3$, and $h_4$ or the non-parametric binned velocity histogram) is what you compare with data's LOSVD values for that spatial bin, repeating for all the spatial bins with kinematic data.

1 "Model" in this case having nothing to do with the Schwarzschild modeling.


Orbiting and landing on non-sperical objects

For bodies small enough to be non-spherical, the mass is relatively low (as an example, Rosetta's bounces took a long time - it touched the surface at 15:34, 17:25 and 17:32 GMT comet time) but the centre of mass is still the point the lander will orbit.

So a chaotic orbit is not a problem here

What will be problems are:

trying to arrange for the lander's velocity to match that of the landing site. For example if the body is spinning rapidly and an outcrop spins round into the path of the lander that could end badly
staying on the surface - Rosetta tried to use harpoons. Another proposed solution was a low powered rocket pointing upwards
while small bodies won't have an atmosphere as such, outgassing and dust can cause problems

I'm assuming 1 is the one you are most interested in here. The solution Rosetta favoured was to pick a site that was on one of the outer surfaces, rather than on the 'neck' of the comet, as this offered the cleanest route to landing. A landing on the neck would have necessitated flying into a channel on a spinning body, and that would make the calculations much harder.


What is the mean separation between a star and its closest neighbor in the Milky Way halo?

I did some rough calculations, and 100 light-years doesn't seem to be a bad guess. If we assume that the average mass of a halo star is $\sim0.3M_{\odot}$, as would be expected for a typical IMF, and that the total stellar halo mass is $\sim10^9M_{\odot}$ (Deason et al. 2019), then we should expect there to be $\sim3.3\times10^9$ halo stars.$^{\dagger}$ The halo extends to somewhere near 100 kpc, so if we take it to be roughly spherica-ishl, this leads to an average number density of
$$n\approx\frac{3.3\times10^9\;\text{stars}}{\frac{4\pi}{3}(100\;\text{kpc})^3}\approx7.9\times10^{-7}\;\text{pc}^{-3}$$
and a mean inter-star distance of $l\approx n^{-1/3}\approx108\;\text{pc}$. So my back-of-the-envelope result differs from yours by only a factor of 3.
Now, the halo has a distinctly non-uniform density. It follows a power law of $r^{-n}$, with an index somewhere near $n\approx3$. Different surveys have yielded values ranging from $2.5\lesssim n\lesssim 3.5$ (ish), depending on the methodology and groups studied. The upshot is that in the outer halo, you'll certainly see much larger separations, whereas closer to the Galactic plane, you could see densities perhaps within a couple orders of magnitude of disk populations.

$^{\dagger}$Two things: First, the halo population's mass function should be different than a typical IMF, because it's composed largely of aging Population II stars - but I think $0.3M_{\odot}$ is still reasonably close. Second, I also checked the number by actually integrating via a Kroupa IMF (normalizing by total halo mass), (and again, maybe not a totally accurate distribution) and found a value within a factor of 2 of 3.3 billion.


How to find the distance between two galaxies from ra and dec and redshift

Since you are presumably using Hubble's law to get the distance to each of the galaxies, the approximate estimate would the "the greater distance ± the smaller distance".
But the geometric solution is useful too. Your strategy appears to be something like:

Calculate angular separation $\Theta$ from RA and DEC.
By trigonometry, the distance is then $d_{a,b} = \sqrt{d_a^2\sin(\Theta)^2 + (d_b - d_a\cos(\Theta))^2}$

The problem is that you can not calculate the angular separation on a sphere by the Pythagorean formula, instead, you need the haversine formula. To keep this post self-contained, that is:
$$\Theta = 2\sin^{-1}\left(\sqrt{\sin^2\left(\frac{\phi_2 - \phi_1}{2}\right) + \cos(\phi_1)\cos(\phi_2)\sin^2\left(\frac{\lambda_2 - \lambda_1}{2}\right)}\right)$$
Where $\phi$ is declination and $\lambda$ is right ascension.
People are also linking this answer from the comments, which deals with the problem by converting everything to Cartesian coordinates first, which may or may not be easier for your use case.


Would it be possible for life found on moon of rogue planet

No one knows for sure, but the idea is plausible. There would not be enough radiation from a rogue planet to provide much help, but we know from the moons of Jupiter and Saturn that tidal heating can keep water liquid. There might have to be multiple moons for it to work. The outer moon keeps making sure the inner moon has an elliptical orbit, and as the inner moon gets farther and closer to the main planet, the tidal forces get weaker and stronger, causing a "squeezing and unsqueezing" that warms the moon.

Of course, at the distance of Jupiter the Sun still provides some warmth, so it would take a lot of tidal heating. But I don't think it's out of the question.


Could the Universe really be 12.5-13 billion years old?

Because some measurements are difficult to make with absolute precision, it often happens that there is some uncertainty about them. The half life of the neutron was once thought to be 12 minutes, but in the 1980s this was amended to 10.5 minutes. In the early days, C14 dating was often substantially out by comparison with dendrochronology and other methods. It depends on the amount of C14 in the atmosphere at any particular time, and when they looked into the matter, scientists found that this varies from one century to another. When they took that factor into account, C14 dates became much more accurate and reliable. Back in the 1970s the age of the universe was thought to be about 15 billion years, but that has been revised to 13.8, and as new data comes in it may need to be revised again. This is why gravitational collapse of the universe has not been entirely ruled out, despite the advent of dark energy, which is thought to be accelerating the expansion when it should be slowing down. It is possible that like some other measurements in science, the dark energy measurement may also need to be revised at a later date.


Lightning in Jupiter's "string of pearls?"

According to the wikipedia page, there are two preconditions necessary for lightning:

Firstly, a sufficiently high potential difference between two regions
of space must exist, and secondly, a high-resistance medium must
obstruct the free, unimpeded equalization of the opposite charges. The
atmosphere provides the electrical insulation, or barrier, that
prevents free equalization between charged regions of opposite
polarity

Since the "string of pearls" are massive counterclockwise rotating storms, one would presume that high potential differences would exist in these turbulent regions. So the only remaining reason for not having lightning would be if there was an electrically conductive atmosphere.  But there are numerous examples of lightning strikes in storms in Jupiter, so, unless the atmosphere at the Pearls is somehow different than the rest of Jupiter, one would expect lightning there as well. Indeed, "storms on Jupiter are always associated with lightning".
From NASA's website, here is a diagram of one of the mechanisms thought to be responsible for lightning on Jupiter:

In conclusion, if you want a reason not to have lightning at the string of pearls, there would need to be little or no turbulence or an an electrically conductive atmosphere.  Since neither of these conditions is the case, one could expect lightning at these storms.


What is there in the intergalactic space?

There can be stars (and small star systems). Stars need not be found only in galaxies.

There can be gas clouds.

Most nebulae are in intergalactic space. Indeed, for a while, both galaxies and nebulae were termed as "nebulae" until the differing nature of galaxies was discovered.

Almost any (small) structure that can be present in a galaxy can be present in intergalactic space, for that matter. Black holes (neutron stars, white dwarfs, brown dwarfs, ... ), star clusters, star systems, etc can all be present.

Finally, there's a very rare (rare in the density sense) plasma, mainly of hydrogen, that makes up almost all of the intergalactic medium. Zero pressure is an unattainable ideal case.


Magnitude is defined relative to a standard flux of a magnitude 0 object $F_0$.

$$M = -2.5\log(F/F_0)$$

if the error is $F_E$, then $F =F_0\pm F_E$, so the magnitude M is

$$ -2.5\log((F_0+F_E)/F_0) < M <-2.5\log((F_0-F_E)/F_0)$$

The errors here are tiny in comparison to the measured flux, only one part in 10000, so the range of magnitudes is similarly tiny +-0.0001


How close would quasars have been to each other?

That's an image of two quasars, very close to each other, "separated on the sky by only about 70 thousand light years".

According to this paper the peak density should have been about $10^{-6}$ quasars per Mpc, resulting in a peak mean distance between two quasars of about 100 Mpc. (Density of $10^{-6}$ quasars per Mpc results in one quasar per cube of $10^6 Mpc^3$, hence on average one quasar each $\sqrt[3]{10^6 \mbox{Mpc}^3}=100 \mbox{ Mpc}$)

With about 1 million known quasars in the observable universe I get a mean distance of about half that distance. (With a considered light-travel distance of about $4 \mbox{ Gpc} = 4\cdot 3.26 \mbox{ Gly} = 13.04\mbox{ Gly},~~$  a little less than the 13.81 Gly light travel radius of the observable universe, the volume of a sphere with this radius is about $\frac{4}{3}\pi(4\cdot 10^3 \mbox{Mpc})^3=2.68\cdot 10^{11}\mbox{Mpc}^3$. With 1 million quasars, it's about one quasar per $268000 \mbox{ Mpc}^3$ or one quasar per $\sqrt[3]{268000 \mbox{ Mpc}^3}=64\mbox{ Mpc}$ in one dimension. Due to non-constant quasar frequency over cosmic time, the minimum mean distance should have been below this overall average distance. Using the light-travel distance considers the higher density of the earlier universe better than the comoving distance of 13.8 billion years after big bang. 4 Gpc are used as light travel radius to restrict roughly to redshift z < 7, the range where quasars have been observed. A redshift calculator can be found e.g. here, take $H_0=70.4$, $\Omega_m = 0.265$, $\Omega_{\Lambda}=0.728$, and $z=7$.)

The peak number density seems to be somewhere between a redshift of 2.0 and 2.5 (interpreting the paper). If it will turn out by future observations, that the peak number density has been earlier in the universe, we get much shorter peak mean distances.

So you have at least an order of magnitude: Somewhere between 150 and 300 million lightyears (between 50 and 100 Mpc, 1 parsec equals 3.26 lightyears) should have been the peak average distance between two quasars. Might be this estimate will shrink over time, as more (distant) quasars get known.


What leads an observer to choose a narrow slit over a wide one (and vice-versa) when observing?

You seem to have all the ingredients apart from the variables of what size your detector pixels are (either physically or binned in software/hardware) and the angular extent of the object you are taking a spectrum of.
The basic trade-off, as you say, is between flux and spectral resolution, but there are limits to that trade off.
You should not reduce your slit width such that its projected size at detection is less than 2 pixels. This would be undersampling and you will not get the gain in spectral resolution you hoped for.
There is no point widening your slit in the hope of more flux if it already encloses most of the light from the object. In other words, if observing a point source, you don't usually open up the slit much beyond the "seeing" width.
If you do, this can actually damage your data because the spectrum you get will start to vary depending on exactly where the star was positioned in the slit. Definitely bad news for accurate radial velocities.
The exception might be if trying spectrophotometry where you really want to get most of the flux and not be affected by seeing variability or guiding errors.
So your extremes would be: Use a narrow slit to achieve maximum (2-pixel) resolution on bright objects where radial velocity precision is important and the seeing is good enough so that your narrow slit still accumulates enough flux. Use a broad slit if the seeing is really bad but you don't care too much what your spectral resolution is. Or open the slit wide if you want to do flux-calibrated spectrophotometry and avoid flux variation due to seeing changes or guiding errors.


How to improve the precision in a star's distance

You would have to pay a lot more than buying time on existing instruments.  You would have to pay for the construction of new and very expensive instruments
You could pay to have copies of the Gaia spacecraft built and sent into orbit around the giant planets in our solar system and/or at their L4 and L5 positions.  Thus two or more such observatories could study the same star at the same time with a baseline between them tens of times greater than the baseline resulting from the orbit of the Earth around the Sun.  Thus the parallaxes and distances should be tens of times as precise.
And if you pay for the development of more precise instruments than Gaia has, you can increase the accuracy and precision of the parallaxes and distances even more.
And you could pay to have several such observatories sent out of the solar system in various directions so that they will eventually be separated by even greater distances than the outer planets are, thus making their parallax and distance measurements even more accurate.


If the inertial mass is not equal to the gravitational mass, it would be equivalent to the gravitational constant, G, being different, or perhaps non-constant, for dark matter. If that were so, the manner in which dark matter would orbit the milky way would be different, leading to a different distribution of dark matter

We know from the rotation curve of the milky way roughly how dark matter is distributed in the milky way. Our models are consistent with the inertial mass of dark matter being equal to the gravitational mass.

This is, of course, far from "proof". As we don't even know what dark matter is, it is impossible to be certain of any of its properties. And the equivalence of inertial and graviational mass is not "proved" even for normal matter (but no experiment has ever detected a difference) But it would be exceedingly surprising if gravity affected dark matter differently, and there would need to be strong evidence for that. As it stands, there is none.


What came first: the Sun shining or the existence of helium?

The Big Bang theory predicts that (depending what assumptions you choose) the initial elements were formed from 10 seconds to 20 minutes after the Big Bang.

The initial elements were

mostly hydrogen
some deuterium
some helium-3 and helium-4
a little lithium-7
a couple of unstable isotopes that decayed to lithium-7 or helium-3.

As the linked Wikipedia article says

  Essentially all of the elements that are heavier than lithium and beryllium were created much later, by stellar nucleosynthesis in evolving and exploding stars.

So the answer to your question is: the Helium came first.


Is there a maximum temperature a planet can have?

A planet in a stable orbit can only receive a finite maximum power from it's star and likewise can only generate a finite energy from internal heat (like our cooling core, which is a tiny power output compared to what we get from the Sun).  There's a finite limit to available power.

If that power is not radiated out (which is thermal radiation) the planet's temperate would rise without limit.  This never happens because hot things radiate heat (as EM radiation) to try and reach thermal equilibrium with their environment (which in the case of a planet is the very cold space around it).

So there must be a finite limit to the temperature any planet in a stable orbit can reach.

As @Cody indicated the actual point of maximum temperature may not be at periapsis.

Atmospheres can act in complex ways to raise the temperature of the surface significantly.  Venus is a prime example as this page demonstrates.  Without it's pressure cooker-like atmosphere we'd expect a temperature of around 240 K, whereas it's actually a rather staggering 740 K.  But all of that is solar powered and can't rise without limit.  But it illustrates the importance of atmospheric composition (and indeed the Greenhouse effect) on planets.

Update : I came across another page on the basic theory of planetary temperature while looking into something else. It may be of use so I've added it in this edit.


It's not possible to see photons from "beyond" the cosmic microwave background (CMB), because (assuming our cosmology is generally correct) the CMB is the result of recombination of electrons with protons in the early universe. Photons from before that era were absorbed in the plasma.

Essentially then, the earliest matter we can see (in photonic terms) is this glowing hot hydrogen, but massively red-shifted so it looks like hydrogen at 3K.

We see the CMB in every direction.


When will we be able to detect Earth-sized exoplanets?

In 2013, the smallest detected exoplanet was Kepler-37-b, which is actually smaller in mass and size than Earth, so we already had a limited ability to detect these size exoplanets.

It is worth noting that there wasn't any new technology that allowed this advance, the paper linked to by NASA indicates that the same methods were used as they would usually use. Making it remarkable that they were even able to discover something this small.

Now, even more Earth-sized planets have been detected. One method used to aid in their detection is described as follows:

  We account for Kepler's imperfect detectability of such planets by injecting synthetic planet-caused dimmings into the Kepler brightness measurements and recording the fraction detected.

Sources:

Kepler-37 - NASA
Prevalence of Earth-size planets orbiting Sun-like stars - Petigura, Howard, and Marcy


What is the light that we see from a neutron star come from?

The visible "surface" of the neutron star contains no free neutrons. It consists of completely ionised atomic nuclei and free electrons. The exact composition is uncertain - either at or just below the surface will consist of iron-peak elements, becoming increasingly neutron-rich with depth, however the outer few cm (which is where any electromagnetic radiation arises) could consist of ionised hydrogen and helium that has been accreted.

Neutron stars also have magnetospheres and these are also a source of radiation. Charged particles are accelerated away from the neutron star magnetic poles, spiral in the magnetic fields and emit (mainly the electrons, which undergo more extreme acceleration) synchrotron and curvature radiation.

The high density "neutron fluid", containing mostly free neutrons, with a small proportion of protons and electrons, is not reached until about 1-2 km below the neutron star surface.


How can a gas planet become tidally locked with a star?

Gaseous planets would tidally lock, just as binary stars do.  That would occur in addition to convection-- tidal effects are weak and operate on very long timescales, convection is strong and operates on much shorter timescales.  It is true that a gaseous planet conforms more easily to the tidal potential even if it is rotating, so there is less of a turning of the tidal bulges, but the tidal bulges are larger because they fill the equipotential, whereas solid surfaces don't.  So the rapid response of gas could make the locking process either faster or slower, depending on which effect dominates.  But you are probably right to focus on energy dissipation as the key requirement, and gas is probably less dissipative, so I would guess the locking time is longer for gas.  Still, there is energy dissipation, there is a delay as the gas flows, and there is a tidal locking effect, all going on on timescales much slower than convection.


Light year is a measure of distance. Light in vacuum travels at a (large) finite value, $3 \cdot 10^5$ kilometers per second to be almost exact. We can exploit this speed as a function of time. If an object is 5 light years away, it means light must travel at that large speed for 5 years to reach Earth from where it left.

The Sun is about 8 light minutes away. It takes light 8 minutes to reach Earth from the Sun. The sunlight you see is from 8 minutes ago, as it was when it left its source — the Sun.

If there were a telescope powerful enough to clearly see objects hundreds of millions of light years away, and if some hypothetical advanced alien race used such a telescope right now from hundreds of millions of light years away, they would see dinosaurs on Earth since that light had enough time to reach its destination. We aren't visible to them yet because our light is still traveling on its way to destination.


Stars in constellation patterns by catalogue number?

If you visit https://github.com/Stellarium/stellarium/tree/master/skycultures you see several sky cultures that include various constellation shapes (you also get this directory when you download stellarium).

As an example, lets look at https://github.com/Stellarium/stellarium/tree/master/skycultures/western_rey, in particular the file https://github.com/Stellarium/stellarium/blob/master/skycultures/western_rey/constellationship.fab and line number 127 which reads:

Aql 10 98036 97649 97649 97278 97278 96229 96229 95501 95501 97804 99473 97804 95501 93747 93747 93244 95501 93805 93805 93429

The "Aql" stands for the for the constellation Aquila, the Eagle (https://github.com/Stellarium/stellarium/blob/master/skycultures/western_rey/constellation_names.eng.fab verifies this)

The "10" means there are 10 lines that make up the markings for Aquila. Note that these lines are specific to the H.A. Rey skyculture. Different cultures will have different lines.

The next 20 number represent the lines themselves. For example "98036 97649" is a line from star 98036 to star 97649, and "97649 97278" is a line from star 97649 to star 97278. Since these two lines share star 97649, they are connected, but that's not always the case with adjacent lines. For example "95501 97804 99473 97804" represents a line from star 95501 to star 97804 and another unconnected line from star 99473 to star 97804.

The next question: what exactly IS star 98036?

If you look at https://github.com/Stellarium/stellarium/blob/master/stars/default/name.fab you'll see it's Beta Aquilae, but not all stars have names in this file, and the name isn't particularly helpful in finding the star itself.

Stellarium itself stores this data in the "catalog files" at https://github.com/Stellarium/stellarium/tree/master/stars/default with the format described at http://stellarium.sourceforge.net/wiki/index.php/Star_Catalogue_Format

However, this can be ugly. Since Stellarium uses the standard Hipparcos catalog, you can visit ftp://cdsarc.u-strasbg.fr/pub/cats/I/239/ and download "hip_main.dat.gz". The format is described at http://cdsarc.u-strasbg.fr/viz-bin/Cat?I/239

The line relating to star 98036 starts:

H|       98036| |19 55 18.77|+06 24 28.6| 3.71|...

which tells us star 98036 is at J2000 right ascension 19h55m18.77s, declination +06 degrees, 24'28.6" declination, and has a magnitude of 3.71. This agrees almost perfectly with https://en.wikipedia.org/wiki/Beta_Aquilae (the declination is slightly off, but this may be because Beta Aquilae is a double star, and Hipparcos references the brighter star and Wikipedia references the midpoint of the double star-- or something).

There are other files in the stellarium github (eg, culture-specific star names) that also use the Hipparcos numbering designation.


Here's one way to arrive at ProfRob's answer step by step:
$$t = \dfrac{5c^5r^4}{256G^3(m_1m_2)(m_1+m_2)}$$
If we lump all the constant terms together:
$$C = \dfrac{5c^5}{256G^3(m_1m_2)(m_1+m_2)}$$
then the problem suddenly looks a lot simpler:
$$t = r^4 C$$
$$\frac{dt}{dr} = 4r^3 C$$
$$\frac{dr}{dt} = \frac{1}{4r^3} \frac{1}{C}$$
$dt$ is how much longer the system will live if the starting radius is increased by $dr$. We know that if we watch it decay, the rate of decrease per unit time $dr/dt$ will be negative, so we'll add a minus sign.
Now put it back together:
$$\frac{dr}{dt} = -\frac{1}{4r^3} \frac{1}{C}$$
$$\frac{dr}{dt} = -\frac{1}{4r^3} \dfrac{256G^3(m_1m_2)(m_1+m_2)}{5c^5} $$
$$\frac{dr}{dt} = -\frac{1}{r^3} \dfrac{64G^3(m_1m_2)(m_1+m_2)}{5c^5} $$
$$\frac{dr}{dt} = -\dfrac{64G^3(m_1m_2)(m_1+m_2)}{5c^5r^3} $$
With gravitational constant $G = 6.674 \times 10^{-11} \text{m}^{3} \text{kg}^{-1} \text{s}^{-2}$ the speed of light $c=2.997 \times 10^{8} \text{m/s}$ each body with a solar mass $M_{\text{Sol}} = 1.9885 \times 10^{30} \text{kg}$ and an almost-touching separation of a million kilometers $r=1\times 10^{9} \text{m}$, then $C = 1.0102 \times 10^{-20}$ and the rate $dr/dt = -2.4748 \times 10^{-8} \text{s}^{-1}$ or -78 centimeters per year.


The CCD has no way of recording the direction, the point in the sky, from which a photon is coming.

Say you point your mirror-less telescope at the Moon.  Every point on the moon's surface would be reflecting photons onto every part of the CCD at the same time.

You've just created an expensive, sensitive, ambient light meter. There would be no image information whatsoever.


Draw the celestial sphere for latitude, showing the star Sirius, Estimate the approximate date when Sirius rises with the Sun

The figure doesn't show the -17° parallel of declination.
If it were shown, it would be parallel to the equator as you expected, as would the parallel of the Sun's declination.
PAQ and PBQ are semicircles of right ascension, intersecting at the celestial poles P and Q.

For an interactive aid, try Stellarium.
You can highlight the meridian, equator, ecliptic, etc. in the View > Markings dialog.
Under View > Sky, orthographic projection renders the sky as a celestial sphere, mirror-reversed from the book figure.
To reproduce the problem scenario, set a location at 30° N and a date of...July 24.
The Sun's right ascension on that date is closer to 8h10m.

The authors don't explain how they determined an RA of 8h40m for the Sun, and they're off by half an hour; maybe they eyeballed it as 30° east of Sirius in the diagram.
The Sun's RA is exactly 0h at the March equinox, 6h at the June solstice, 12h at the September equinox, and 18h at the December solstice.
Between those milestones, the Earth's orbital motion makes the Sun's RA increase about 4m per day or 1h in 15 days.


What magnification is required to see detail on Mars / other local bodies?

You have several questions here:

Mars is relatively close to Earth right now (October 2018) - about 90 million km.
Is your telescope powerful enough for you to see detail on Mars?  From what you have said, it's difficult to tell.  You say you have magnification of 6x.  This is almost certainly wrong.  If you state the focal lengths of your telescope and its eyepieces we can work out the correct magnification available.  To see much detail on Mars you would need in excess of 100x magnification (same for Saturn, less for Jupiter), and ideally a lot more.  Your scope should be able to manage 100x.  However the Jones Bird design is known to be difficult to align (collimate) properly, so even with that magnification, you may just see a blur.  How much detail you can see will also be heavily affected by the state of the atmosphere - ideally you need a night when the air is calm. Sometimes a slight mist can indicate a steady atmosphere.
Would the CPC 1100 be overkill etc?  This is much more opinion based, but in short, yes - for a beginner anyway.  For an experienced observer, the answer is much more nuanced, and I'm not going to go into any of that.  The scope is generally regarded as pretty good for what it is.


What are the dark spots in the "face of Mars" picture?

What are the dark spots in the “face of Mars” picture?

They are called "bit errors" in the NASA JPL image catalog for PIA01141. The explanation seems a bit handy-wavy to me; so I think we'll need to dig deeper to see if they are received noise on Earth or some electrical or imaging problem on the spacecraft.
Read more about the Voyagers' cameras in What is the brown band in which we find the pale blue dot that is Earth?
By the 1980's NASA spacecraft used solid state CCD detectors and end-to-end digital transmission. See Galileo (spacecraft) Solid State Imager
From PIA01141: Geologic 'Face on Mars' Formation

Original Caption Released with Image:
NASA's Viking 1 Orbiter spacecraft photographed this region in the northern latitudes of Mars on July 25, 1976 while searching for a landing site for the Viking 2 Lander. The speckled appearance of the image is due to missing data, called bit errors, caused by problems in transmission of the photographic data from Mars to Earth. Bit errors comprise part of one of the 'eyes' and 'nostrils' on the eroded rock that resembles a human face near the center of the image. Shadows in the rock formation give the illusion of a nose and mouth. Planetary geologists attribute the origin of the formation to purely natural processes. The feature is 1.5 kilometers (one mile) across, with the sun angle at approximately 20 degrees. The picture was taken from a range of 1,873 kilometers (1,162 miles).

Below is a closer-to-raw-data image from Views of the Solar System: Raw Viking Orbiter Image 035a72 of the Face. Notice that the dots are bright, not dark in the raw data. Voyager's cameras are 1970's technology; they are based on vidicon tubes not solid state detectors.

Prior to the design and construction of the Galileo probe to Jupiter in the late 1970s to early 1980s, NASA used vidicon cameras on nearly all the unmanned deep space probes equipped with the remote sensing ability. Vidicon tubes were also used aboard the first three Landsat earth imaging satellites launched in 1972, as part of each spacecraft's Return Beam Vidicon (RBV) imaging system. The Uvicon, a UV-variant Vidicon was also used by NASA for UV duties

If you look closely you can notice both white and black dots. @Planetmaker's answer mentions this looks like salt-and-pepper noise.
From Arizona State University's Space Exploration Resources Control Network:

The majority of Mariner 10 high resolution images were sent back in real time (117.6 Kbits/s) with an attendant increase in bit errors during transmission [Danielson et al., 1975; Soha et al., 1975]. This resulted in severe salt and pepper random noise [Soha et al., 1975] that hampers the automatic identification of reseaux, especially in areas with low signal such as terminator regions and space. Due to the generally noisy nature of the data [Danielson et al., 1975; Soha et al., 1975] tolerances were set high (0.90 correlation) to help avoid spurious reseau identifications. Failure to find an actual reseau at a given position resulted in default to the nominal position for a given camera/encounter combination.

Danielson, G.E., K.P. Klaasen, and J.L. Anderson, Acquisition and description of Mariner 10 television science data at Mercury, J. Geophys. Res., 80, 2357-2393, 1975.
Soha et al., 1975  Soha, J.M., D.J. Lynn, J.J. Lorre, J.A. Mosher, N.N. Thayer, D.A. Elliot, W.D. Benton, and R.E. Dewar, IPL Processing of the Mariner images of Mercury, J. Geophys. Res., 80, 2394-2414, 1975.

From Digital Elevation Model of the Discovery Region on Mercury(Lunar and Planetary Science XXIX, 1849.pdf):

Excessive salt and pepper noise in vidicon images, can cause the mis-identification of reseaus necessary for accurate image rectification.

From book.google.com Yap et al. (2010) Adaptive Image Processing: A Computational Intelligence Perspective, Second Edition (also here):

9.4.2 Models of Impulse Noise
When an image is coded and transmitted over a noisy channel, or degraded by electrical sensor noise, as in a vidicon TV camera, degradation appears as salt-and-pepper noise (i.e. positive and negative impulses).

Emily Lakdawalla discusses the Voyagers' vidicons and problems with the old "treasure trove" of images


Could a Jovian or Saturnian moon keep its atmosphere if the system was within the habitable zone?

To your first question... from this paper, we can see that there are 3 types of atmospheric escape:

Jeans Escape: Temperature and escape velocity factors determine the gasses and the amount that's lost. Using the chart provided in the comments, and moving Titan along the x-axis to around below Earth's position, we can see Titan will lose most gasses except Xenon.
Charge exchange: Solar radiation creates electrons and positively charged ions in upper atmospheres by tearing electrons off atoms or molecules. Titan is protected ~95% of the time by Saturn's magnetic field, and the other 5% of the time Titan's atmosphere is magnetized, creating an artificially induced magnetic field for a short period of time. This prevents a massive atmospheric loss. Thus, if Titan is placed in an orbit similar to Earth's Titan will lose a lot of atmosphere due to the solar wind.
Vertical atmospheric escape/impact erosion: Energetic objects that strike a planet erodes its atmosphere by creating a plume of heated gas. If Titan is struck, if will lose a considerable amount of its atmosphere. However, this has little to do with your scenario, so let's not consider this.

So we can see if Titan is moved into the Habitable zone, it will lose a considerable portion, or possibly almost all, of its atmosphere.

Regarding your second question, the major factor to atmospheric loss in that scenario, for a small(relatively speaking) celestial body in the Sun's habitable zone is the solar wind. If we look at the inner planets, we can see that they all have some form of magnetic field. Mercury has a relatively strong one; Venus has an weak, induced one; Earth has a strong magnetic field; Mars has remnants of its dissipated one. The magnetic field provides protection to the planet's atmosphere.

Titan does not have a magnetic field. If placed in the Sun's habitable zone, a celestial body would have a hard time satisfying your criteria. So then, let's suppose the moon we put in the Sun's habitable zone has a magnetic field likewise to that of Earth's.

In the habitable zone, we can say that the average temperature of the body would be around 300 K, similar to Earth's. Then, let's define that an Earth-like atmosphere would be comprised of at least nitrogen, oxygen, H2O vapor, and CO2. Using the same chart, we can start by moving Titan to below Earth, and then moving it up to where H2O vapor can be retained. We find that a celestial body with an Earth-like magnetic field can retain an Earth-like atmosphere if its escape velocity is around 6~8 km/s.


Textbook to explain measurements and calculations for heliocentric solar system

It's possible the book you seek is by Christopher Graney.  I cannot find them now, but I believe he had put together instructions on how to build a telescope similar to what Galileo used and how it could be used to argue for and against a heliocentric universe.  But his overall conclusion is that it's poorly suited for the task, and scientists of the time had good reason to not accept it.  (E.g. the scope had a very small aperature, and thus a horrible diffraction limit (which they didn't understand at the time).  It could measure the planets just fine, but through it, stars also appeared quite large too.  Since stars exhibited no detectable parallax, they'd have to be comically huge for the measurements to make sense.)

That said, if you want to actually do it, you don't need any fancy equipment.  Almost any telescope off the shelf today will do.  You'll need a camera with removable lenses (e.g. a DSLR), but not anything "high res" by today's standards, the oldest used one you can find would be plenty.  You'll also need an eyepiece adapter for it (just search for your camera type, e.g. "Canon eyepiece adapter").  You won't need a motorized mount just for the planets.

With that you can take photos of the planets at different times over months, but most likely a least a year or two to get convincing results.  Mars will generally show the most dramatic results, and if you search on Google, you'll find plenty of people who've done it.

You can also prove Venus orbits the Sun just by watching its phases.  In order to go from full, to half full, to crescent, it has to be on different sides of the Sun than we are during some times.  As well as the size change.  You'll also find a lot of people have done that if you search for that.

While not showing evidence of a heliocentric universe, one solid piece of evidence Galileo provided in favor of at least a non-geocentric universe was the moons of Jupiter.  Even with the limits of his telescope, he was able to show that they orbited Jupiter, so the Earth was not the center of everything.  Any telescope of today will be able to see them.  Even most binoculars, or even the scope off a rifle.


Semi-major axis, vs aphelion, vs mean orbital distance?

I think your confusion is just an incorrect (though not uncommon!) visualization of a planetary orbit. The parent star (usually) doesn't lie at the center of the ellipse, but at one of the foci; only for circular orbits are the foci at the same location as the center of the ellipse. Therefore, for non-circular orbits, the aphelion distance is greater than the semi-major axis, and the average of the semi-major and semi-minor axes is the average distance to the center of the ellipse - not the star.

Mathematically, note that the perihelion and aphelion distances are related to the semi-major axis $a$ and eccentricity $e$ by
$$R_p=a(1-e),\quad R_a=a(1+e)$$
As a quick check, adding these together and dividing by 2 yields $a$. In the case of a circular orbit, where $e=0$, we have $R_p=R_a=a$.


The specific radius of $8 h^{−1} \text{Mpc}$ is used because the value of $\sigma_8$ turned out to be close to unity. To quote e.g. from Amendola & Tsujikawa (2010), p. 39:

  If the cells have a radius of $8 h^{−1} \text{Mpc}$, it turns out that $\sigma_R$ is close to unity. Conventionally the normalization of the power spectrum is therefore given by quoting $\sigma_8$.


Absolute magnitudes with dust extinction

Check this lecture note. I believe there is everything you need to understand how to proceed. In brief, you should have extra information about the relative absorption, shown in the table on p.18. Given the correction in one band (which you already have $A_V = 1)$, you can find the correction in other bands using the table.

Note that the relative absorption is different for different scenario, e.g., different RADEC, different host, different type of stars.

Also note that there are two types of extinction: galactic and extragalactic.


Why did the moon abruptly change positions in the sky?

Compared to the planets the moon changes it's rise and set times very quickly.  Using the calculator on this page it can be seen that if one were looking at the sky on May 1st, 2014 from Irvine, California the moon set at 10:08PM (which at 8PM would have had the moon most of the way across the sky as noted in the question.)

The moon rises and sets approximately one hour later each day.  So, by May 14th the moon did not even rise until nearly 8PM - which also correlates with the observations noted above.

Contrasting that with the other context point made in the OP, using this page we can see that on May 1st Saturn rose at 8:07PM and on May 14th rose at 7:11PM.

So while the moon changes its rise a set times by almost an hour each day, the planets change much more slowly by comparison, almost an hour over two weeks.

If one weren't continually watching the moon's progression it would appear to abruptly change positions when comparing it to the planets' movements.


What is a probability of a particle launched from earth hitting any object in the universe?

I can't give you an exact number, but the chances your photon will hit anything within the next several billion years is exceedingly small. Nay, infinitesimally small. Almost no chance that it will happen.

What you're really asking, stated as a physics question is, what is the mean free path of a photon in the present universe?

If we consider an object (be it an electron, photon, or baseball) traveling through some medium (be it the Sun, space, or Earth's atmosphere), we can calculate how far, on average, it can travel before hitting something. This distance is known as the mean free path. You want to know how far a photon can travel through our universe before it hits something, in other words, what is the photon's mean free path?

The answer is that a photon's mean free path in our universe is larger than the observable universe. What that means is your photon can travel from Earth to the edge of the observable universe and still not hit anything (on average).

To support this fact, I point you to the Cosmic Microwave Background (CMB) radiation. This is radiation which was produced more than 13 billion years ago in the early universe and has been traveling through the universe since, almost completely unimpeded. In fact, if you tally up all the photons in the universe, you'll find that a very large fraction are CMB photons. This tells you that these photons, despite being produced 13 billion years ago and traveling through the universe since then, have never hit anything since and most are still traveling through space, unimpeded.

You are correct when you say that its the projected area of objects on our sky that matters, but you have to take into account the extremely minuscule projected area the matter in the rest of the universe has, primarily due to how far away it is.


How is a non-rotating black hole created

All stars have some angular momentum. Conservation of angular momentum when the core of a star collapses means that it is very likely that most black holes are rapidly rotating, just as newly-born neutron stars are observed to be extremely fast-rotating.

Even if one were to have a star with a non-rotating core as it collapsed, the subsequent accretion of material with angular momentum would give the black hole that angular momentum.

Non-spinning black holes are a theoretical ideal that is unlikely to be found in nature.

The details of the accretion of material onto a black hole are complex and still the subject of theoretical and observational research. But yes, the details of accretion in spacetime governed by the Kerr metric appropriate for rotating black holes will differ from those of a non-rotating Schwarzschild black hole. Kerr black holes have an "ergosphere", outside their event horizons, where material is force to co-rotate with the black hole.


Assuming a hypothetical system without gas but only solid rocks, how big of a planet can form through the coalescing of these rocks?

That very much depends on the structure of the solids, i.e. whether they exist in the form of small dust, or ready-to-smash planetesimals.
In any case, the available median dust mass for planet formation is about 158 $\rm m_{earth}$ (see Tychoniec et al., (2020)). If you leave all this mass as dust, without any gas interactions, the dust will not coalesce into planetesimals, as a hydrodynamic instability is required to jump over the meter-sized barrier (Johansen et al., (2014)).
However, if you somehow allow all this dust to be converted into planetesimals, then the size of the planet you can form will be given by how narrow you can pack the planetesimals. The absolute upper mass of the formed planet will be given by the 158 $\rm m_{earth}$, but realistically , that is going to be lower, as planetesimals and collision ejecta are lost during the smashing phase of planet formation.
As you were asking about the size of this hypothetical planet, is we assume no compressional effects and hence the same mean density as earth, you would get a solid ball of the size of $\rm 158^{1/3} \; r_{earth} \approx 5.4\; r_{earth}$.


Does the Moon have any oxygen in its atmosphere?

The Moon's atmosphere is very thin compared to the Earth, so thin that it is usually said to have no atmosphere.  The Moon's gravity is not strong enough to retain lighter elements, so they escape into space.

Apollo 17 carried an instrument called the Lunar Atmosphere Composition Experiment (LACE).  Oxygen is not listed as one of the elements it found on this NASA web page.  The principle gases found were neon, helium and hydrogen.  Others included methane, carbon dioxide, ammonia, and water.  Additionally, ground instruments have detected sodium and potassium in the lunar atmosphere.

The recently launched Lunar Atmosphere and Dust Environment Experiment (LADEE - pronounced laddie, not lady) should be able to tell us more soon.

Update:
According to this article at Spaceflight Now, LADEE did detect oxygen in the Moon's atmosphere.


Can a star fall in a super massive black hole without getting destroyed?

Yes, it is easily possible, but it depends critically on how massive the black hole is. A simple calculation will suffice.

If we take a Newtonian approximation for the tidal acceleration across a star of radius $R_*$ as it reaches the event horizon
$$ a_{\rm tidal} \sim 2\frac{GM_{\rm BH}R_*}{r_s^3}\ ,$$
where $r_s = 2GM_{\rm BH}/c^2$ is the Schwarzschild radius.

The gravitational acceleration at the surface of the star, due to its own mass is just
$$a_* = \frac{GM_*}{R_*^2}\ .$$

If we take the ratio of the two, we can assume the star will survive being ripped up before it crosses the event horizon if this ratio is less than $\sim 1$. i.e. when
$$\frac{a_{\rm tidal}}{a_*}= \frac{c^6 R_*^3}{4G^3 M_{\rm BH}^2 M_*} <1$$

The tidal acceleration reduces markedly for high mass black holes. We can rearrange this inequality to give us the minimum black hole mass for survival:
$$ M_{\rm BH} > \left(\frac{c^6 R_*^3}{4G^3 M_*}\right)^{0.5} = 1.5\times 10^{8} \left(\frac{M_*}{M_{\odot}}\right)^{-1/2} \left(\frac{R_*}{R_{\odot}}\right)^{3/2}\ M_{\odot}$$

To first order, this is almost independent of the type of star swallowed. It depends on the inverse square root of the average stellar density, which doesn't vary a lot.

This means that a star could fall into a black hole like that in M87 ($M_{\rm BH}\simeq 6\times 10^9 M_{\odot}$), but  would not survive to the event horizon of Sgr A* ($M_{\rm BH}\simeq 4\times 10^6 M_{\odot}$).

I think in practice this calculation can only be accurate to factors of a few. I have not taken into account shearing forces that would be present for a rotating black hole, or the compression due to tangential tidal forces, or any heating effects from hot gas near the black hole, or interaction with an accretion disk. A full hydrodynamic simulation is needed (and perhaps has been done?).

Also note, this calculation is just for getting past the event horizon. The tidal forces grow as $r^{-3}$, so any normal star will be ripped up shortly after this, even for the most massive black holes we know of.

Please note: This answer is written from the point of view of the falling star. A distant observer will not see the star (or anything else) cross the event horizon.


Which of Jupiter's moons did I most likely see?

The galilean moons are visible in binoculars, indeed. Yesterday evening (assuming you're in the western hemisphere), 3 of them were visible as separate dots. See the chart here:

http://www.wolframalpha.com/input/?i=jupiter+galilean+moons+on+march+6+2016+at+9pm+pst

You can adjust the query to reflect your time zone, and the actual date/time of the observation.

There are also smartphone apps that allow you to determine the position of the moons at any given time. On Android I use the Moons of Jupiter app.


Can radioactive decay rates be confirmed by looking at supernova?

I think I understand this now. I was trying to imagine how you could take a snapshot of a supernova and infer from it the half-life some some element. In fact, I think the measurement comes from watching the radioactive decay happen over time, in the aftermath of a supernova.
Openstax Astronomy 23.3 "Supernova Observations" shows a graph entitled "Figure 23.13 Change in the Brightness of SN 1987A over Time." It shows brightness increasing during the days right after the supernova, and the rest of the curve not following a consistent fading pattern. It explains:

One of the elements formed in a supernova explosion is radioactive nickel, with an atomic mass of 56 (that is, the total number of protons plus neutrons in its nucleus is 56). Nickel-56 is unstable and changes spontaneously (with a half-life of about 6 days) to cobalt-56. (Recall that a half-life is the time it takes for half the nuclei in a sample to undergo radioactive decay.) Cobalt-56 in turn decays with a half-life of about 77 days to iron-56, which is stable. Energetic gamma rays are emitted when these radioactive nuclei decay. Those gamma rays then serve as a new source of energy for the expanding layers of the supernova. The gamma rays are absorbed in the overlying gas and re-emitted at visible wavelengths, keeping the remains of the star bright.
As you can see in Figure 23.13, astronomers did observe brightening due to radioactive nuclei in the first few months following the supernova’s outburst and then saw the extra light die away as more and more of the radioactive nuclei decayed to stable iron. The gamma-ray heating was responsible for virtually all of the radiation detected from SN 1987A after day 40. Some gamma rays also escaped directly without being absorbed. These were detected by Earth-orbiting telescopes at the wavelengths expected for the decay of radioactive nickel and cobalt, clearly confirming our understanding that new elements were indeed formed in the crucible of the supernova.

So if I understand correctly, the logic would be:

SN 1987A occurred 168,000 light years from earth
Therefore, humans observed it 168,000 years after it happened
As they watched the supernova unfold, astronomers saw radiation that appeared to be generated by the decay of known, short-lived isotopes, and that decay happened at the same rates we observe today
Therefore, the decay rates of these isotopes were the same 168,000 years ago as they are today


Does the dimming of Betelgeuse present any observational opportunities?

In terms of "easier" - not really. There isn't anything made particularly easier or more optimal to measure. Theoretically, as you mentioned, "space stuff" like tiny planetoids or dust that are very close to the star itself might become more visible, but that's about it.

However, the reason we would want to get measurements now is to compare them to data collected when the star was at normal brightness and see differences, perhaps in spectroscopic lines, for instance. As well as that, if the star also returns to normal brightness, then we can observe changes that occurred side-by-side with the dimming or maybe even as a result of it.

Side note: Regarding the composition of the star, which you provided as an example - the reason we are unsure about when Betelgeuse will explode is because we can't actually know its true composition, since we can't see directly inside the star. If we could, we would be able to predict when it would finish fusing remaining heavier elements and then extrapolate when it might go supernova. All we can really see is what is coming off of the surface of the star, infer its age from other data, and then make an educated guess about when it will explode.


Why is sun's photosphere a million times less dense than air at the surface of the Earth?

I think there is a misconception underlying your question. You write

But Solar surface gravity is about 28 times terrestrial surface gravity.
So, shouldn't the density of gases at the photosphere of the sun be comparable to that on the surface of the Earth?

It does sound natural that a stronger gravity would imply a thicker atmosphere, but no matter how strong the gravity is, there is a point where the atmosphere fades away almost to nothing. At the height of the ISS, earth's gravity is around 80% of the gravity at sea level. But yet the ISS is orbiting in the vacuum of space. Similarly, above Jupiter's atmosphere, the gravity is still more than twice that of earth, but there is still vacuum. With the sun, it just so happens that the part of the atmosphere that produces the light we see is close enough to the (imprecisely defined) edge of space that its density has gone down almost to nothing.
Certainly, if it were cooler, it would be denser. But some day the sun will be a white dwarf, about the size of the earth but with not that much less mass than it has now. Its gravity will be hugely strong, but there will still be a top of the atmosphere were it is less than a millionth of earth's atmosphere. As it cools, that atmosphere will cool so that it is no warmer than earth's atmosphere, but there will still be a point where the atmosphere is that thin. And even above that height, gravity will be much higher than even at the current surface of the sun.


The most commonly used packages for generating graphics in astronomy are probably

IDL -- this has long been the most popular, in part because there is a lot of astronomical data reduction and analysis code written in IDL.
SuperMongo
The Python package matplotlib

The apparently uniform style you refer to is probably a historical combination of IDL and SuperMongo.

Matplotlib (and other Python packages that act as wrappers around it) is on a trajectory to displace the others; I strongly suspect it is already much more common than SuperMongo, and may be roughly as common as IDL by now.

You will sometimes see figures produced by the PLplot library as output from programs written
in Fortran, C/C++, etc. Plots produced using R are also starting to appear.


What caused instant jumps and exactly flat periods in Kepler's light curves?

There are two separate points of interest you're looking at so I'll separate this into sections.

Sudden Drop at Day 1559

As near as I can tell, this is the result of a quarterly roll of the satellite, specifically the end of quarter 17. Every 90 days, which NASA calls a quarter, the space craft does a 90 degree roll to optimize the solar panel efficiency. What this means for the data is that when this roll occurs at the end of the quarter, the photons from the star now fall on a different CCD channel which is calibrated slightly differently, causing a slight discontinuity in the data. You can read about "Quarter Stitching" here. I've extracted some relevant text and an image for posterity.

  After each roll maneuver, most Kepler targets fall on a different CCD channel and the target’s point-spread function will be distributed differently across neighboring pixels. Naturally this redistribution requires a new computation of the target mask and optimal aperture size within the Kepler pipeline, taking into account the point-spread function, new CCD characteristics, and new estimates of nearby source crowding. The operational outcome is often a different mask shape, with differing amounts of flux within the optimal aperture from both the target and contaminating sources.

Light curves extracted from all single pixels within the quarter 4, 5 and 6 Target Pixel Files for KIC 9603833 (the symbiotic star StHA 169). Gray pixels comprise the optimal apertures that yield the archived light curves. The target point-spread function is distributed across neighboring pixels differently from quarter to quarter and hence the optimal aperture varies in size and shape from quarter to quarter.

Flat line between days 1581 to 1587

I'm less certain about this one. I believe the flat line is just missing data. To keep the line continuous, they just connected the data points they had with a straight line.

Why the data is missing, I can't be sure. I know that Quarter 17 ended with the failure of Kepler's second reaction wheel so my guess is that this missing data is in some way related to these reaction wheel troubles. I do know they tried,in the past, to put the spacecraft in 10-day "wheel rest" mode in hopes that it would give the reaction wheels a break, let them get lubed up again, and be better when they restarted. Perhaps, by this stage, they had already suspected a second reaction wheel problem and were trying out a wheel rest mode. I can't find any evidence a wheel rest occurred during Quarter 17 though, so this is just my guess.


Why is dark energy preferred to the cosmological constant?

A cosmological constant should be considered a special case of dark energy. The effective stress-energy tensor for a cosmological constant is proportional to the metric $g_{\mu\nu}$, so in a local inertial frame will be proportional $\mathrm{diag}(-1,+1,+1,+1)$. This is equivalent to perfect fluid with energy density and pressure directly opposite one another, but more importantly, it is the only possible form for the stress-energy that would give the exact same energy density and pressure in all local inertial frames.

If by 'dark energy', we understand it to mean all the contributions to stress-energy in the above form, then there is no reason for this to be constant, and plenty of reasons why it might not be, as this situation is not exceptional in fundamental physics. For example, there could be false vacua with various different energy densities, and they must be invariant across inertial frames.

In particular, the basic idea of inflation considers a flat FRW universe with expansion driven by a scalar field $\phi$ at a local extremum of its potential, $V'(\phi_0) = 0$, which yields an exponential expansion with constant energy density $T^0{}_0 = V(\phi_0)$. More refined models, such as slow-roll inflation, could therefore be directly interpreted as a time-varying dark energy density, while eternal inflation would also include spatial variability. There's plenty of other inflationary models besides.

One the interpretational flip-side, one could always have a $\Lambda$ that corresponds to the energy density of the true vacuum, and the rest as separate contributions on top of that. It's just not as useful in a cosmological context compared to grouping all 'dark energies' together, as all stress-energy gravitates equally.


What is the degree of reflection of light from sun to the moon?

Basically, the sunlight is reflected from the Moon at different angles. It works the same as on Earth. For instance, during the sunrise & sunset, the sunlight is reflected at a bit above 0 degrees angle, whereas at the noon in the tropical areas (in general) where Sun passes the zenith, the sunlight is reflected at 90 degrees. On Moon and other planets, it works exactly the same! This is how the angle of reflection works. Obviously, because we have a rough surface built with different rocks, we can't treat the celestial body as a mirror. However, the phenomenon is analog, as each of them has its own albedo - ability to reflect the light. Lunar albedo is 10%.
https://en.wikipedia.org/wiki/Albedo#:~:text=Albedo%20(%2F%C3%A6l%CB%88b,that%20reflects%20all%20incident%20radiation.
https://scienceworld.wolfram.com/physics/AngleofReflection.html#:~:text=The%20angle%20of%20reflection%20of,same%20point%20as%20the%20ray.

Obviously, it's worth mentioning that as the angle of light reflection is smaller, then the light beam is weaker (less intensive). It arises out of the inverse square law:
https://en.wikipedia.org/wiki/Inverse-square_law

which can be explained by the image above.
The angle of incidence is always equal to the angle of reflection, but it's perpendicular to the surface exactly, where the beam hits this surface.

http://www.mkrgeo-blog.com/the-aspect-of-light-reflection-in-the-long-distance-observations/
https://en.wikipedia.org/wiki/Angle_of_incidence_(optics)
Otherworldly, the angle of incidence is related to the plane of incidence. The plane of incidence is the "place" where an observer can see the reflected solar beam on the surface.
Along with the angle of incidence, we can mention the light glitters produced when our angle of reflection is low enough.


What are Trojan Asteroids and where are they located?

A Trojan Asteroid is defined by the Swinburne University Astro page "Trojan Asteroids" as:

  Asteroids sharing an orbit with a planet, but which are located at the leading (L4) and trailing (L5) Lagrangian points.

These are often divided into 'leading' and 'trailing', as per the diagram below (from the source linked above):

Even though the term Trojan asteroid is often associated with the 4800 or so that lead and follow Jupiter, it actually applies to any asteroid that displays this behaviour, associated with any planet.

There are numerous Trojans co-orbiting with Mars and Neptune.  Recently, single Trojans have been found co-orbiting Uranus and even Earth.