A lot of different "alternative biologies" have been considered, but if you analyze their required chemistry, this always falls apart at some point, some mechanisms necessary for life are simply impossible. Replace carbon with silicon, increase ambient temperature, and you're getting a lot of interesting reactions that could be conductive to life. But instead of CO2, which is a gas, you have SIO2 which is a crystal. You can't breathe crystal. It's awfully insoluble, so no simple hemoglobin counterpart. And so on. Won't work.

Water and carbon based biology is one most plausible. There are odd variants, like bacteria thriving on sulfite, manganese and copper oxide, but they still require water to live, and still use carbon as the fundamental building block.

While others might exist, chance for them to form is much lower than for one similar to ours.  And since we haven't designed even one fully plausible alternate biology that doesn't involve water we don't even know what sort of detectable signatures it would produce. So we search for something we know might yield results; hint at possibility of life somewhat similar to ours - instead of a wild goose chase after the unknown.

Actually, there exist one "alternate biology" which is fully plausible even if its early origins would be Earth-like: Silicon based, artificial intelligence, robotic physiology, electrical power, proliferation through fully automated manufacturing processes. It's unlikely to produce any detectable chemical signature (at least distinguishable from half of the natural celestial bodies out there), but we are fairly sure its electromagnetic spectrum - radio waves signature - would be quite specific. Well, The SETI project is on it.


Naively, I would say that you're looking for a periodicity in the luminosity.  What about multiplying the probability with the photon energy, and binning those data?  That would allow you to use an FFT.


Why don't multiple eclipses happen over the course of a few days?

The reason is because the moon moves much faster (in terms of angle per day) than the sun. In fact it moves on average 12.5 degrees per day, and all in the same direction.  Given that the sun is only about 0.5 degrees in diameter, it doesn't take long for the moon to overtake it, and once it's past  it doesn't go backwards.


How big and how far away would the moon be if earth was the size of Jupiter?

The basic numbers are:

Radius of Earth 6370km
Radius of Moon 1740km
Earth-moon distance 384000km and
Radius of Jupiter 71500km

So the scale factor is 71500/6370 = 11.22
The radius of the moon scaled is 1740×11.22 = 19500 km, and its distance becomes 384000×11.22 = 4310000 km
For comparison, Neptune has a radius of 24600 km and a distance from Jupiter of that is always more than 3700000000 km
The size of the "moon" is comparable to Neptune, but the distance from Jupiter to Neptune is nearly 1000 times greater than the distance to the "moon".


My question: is there a way to determine the "stable" region in each Lagrange location? Particularly in the L4 and L5 regions.

tl;dr: Yes but it's usually a fancy version of "trial and error".

We are sometimes wrongly told that this abominable zero-velocity potential diagram show areas that are "bound" or "unbound" to Lagrange points, but it does not and that's obvious, because while the surface dips near L1, L2 and L3 it is a maximum near L4 and L5 and yet we know that for some mass ratios these regions can collect and accumulate trojan asteroids.

  Source

The problem is that these are 2D plots of the pseudo-potential at one specific velocity. The pseudo-potential is given by

$$C = \omega^2 (x^2+y^2) + 2\left(\frac{\mu_1}{r_1} + \frac{\mu_2}{r_2} \right) - (v_x^2 + v_y^2 + v_z^2)$$ and these zero velocity surfaces which are contours when drawn in a 2D slice don't depict orbits at all. If you put a particle on any contour other than $C=0$ it will accelerate rather than follow one of these lines.

So instead what people do is to run numerical simulations with a whole range of starting position and velocity vectors and find out by a fancy version of "trial and error" which particles will end up back at the same position where they started with the same velocity vector they started with. Every time they find a closed, periodic orbit it helps map out the "stable region".

One example of a method to find regions of 4D or 6D phase space ($x, y, v_x, v_y $ or $x, y, z, v_x, v_y, v_z$) that have closed periodic orbits is the shooting method which can be made faster by use of the transition matrix.

In Figure 1 of F. Marzari et al. Origin and Evolution of Trojan Asteroids you can see how tempting it is to use a zero velocity contour to indicate where the region of stable orbits lay with out actually saying it. Instead the the words "are closely related to" are used.

  Fig. 1. The location of the five Lagrange equilibrium points in the circular-restricted three-body problem. The primary and secondary masses (the Sun and planet in our examples) are denoted by the large and small filled circles respectively. The selected zero velocity curves (see text) are closely related to the types of orbits that can occur in the system. The letters P, H, and T denote passing, horseshoe, and tadpole orbits respectively. Note that the two masses and each of the L4 and L5 points form an equilateral triangle.

If you want to go beyond just a circular restricted three-body model and to the real solar system, numerical simulations become much more complex because there is long term evolution of these orbits due to all of the devitaions of reality from the simple CR3BP model. See just for an example The resonant structure of Jupiter's Trojan asteroids – I. Long-term stability and diffusion


Have any planetary systems been found around Population II stars?

Kapteyn's star has 14% the metallicity of the Sun and is about 11 billion years old and is in the galactic halo. It has two reported planets.
I don't know if 14% is low enough to count as  Population II, but according to the NASA exoplanet archive it is the lowest metallicity star known to have planets.


At this stage of our solar system it is impossible for something with size of moon to fall in inner orbit. But smaller things like asteroids do fall in, and become captured by earths gravity. but they aren't big enough to see with naked eye. For example 2006 RH120 is a near earth asteroid which orbited Earth from September 2006 to June 2007.


I'd like to photograph NEOWISE from Los Angeles, what is the optimal time?

I don't know if this is helpful but using the Python package Skyfield one can calculate the positions of things including comet C/2020 F3 (NEOWISE) and the times that they rise and set for the month of July:

Here are the rise and set times of the Sun and the comet for reference for the next week
sunset:  2020-07-12T03:06:17Z
sunrise:  2020-07-12T12:51:20Z
sunset:  2020-07-13T03:05:55Z
sunrise:  2020-07-13T12:51:56Z
sunset:  2020-07-14T03:05:32Z
sunrise:  2020-07-14T12:52:33Z
sunset:  2020-07-15T03:05:07Z
sunrise:  2020-07-15T12:53:10Z
sunset:  2020-07-16T03:04:41Z
sunrise:  2020-07-16T12:53:48Z
sunset:  2020-07-17T03:04:13Z
sunrise:  2020-07-17T12:54:27Z
sunset:  2020-07-18T03:03:43Z
sunrise:  2020-07-18T12:55:06Z

cometset:  2020-07-12T04:15:12Z
cometrise:  2020-07-12T10:46:43Z
cometset:  2020-07-13T04:34:10Z
cometrise:  2020-07-13T10:48:01Z
cometset:  2020-07-14T04:54:27Z
cometrise:  2020-07-14T10:51:29Z
cometset:  2020-07-15T05:15:33Z
cometrise:  2020-07-15T10:57:33Z
cometset:  2020-07-16T05:36:48Z
cometrise:  2020-07-16T11:06:45Z
cometset:  2020-07-17T05:57:17Z
cometrise:  2020-07-17T11:19:29Z
cometset:  2020-07-18T06:16:07Z
cometrise:  2020-07-18T11:36:02Z

Here's the python script:
from skyfield.api import Topos
from skyfield.api import Loader
from skyfield.data import mpc
from skyfield import almanac
import numpy as np
import matplotlib.pyplot as plt
from skyfield.constants import GM_SUN_Pitjeva_2005_km3_s2 as GM_SUN

from skyfield.api import load
loaddata = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files

ts = loaddata.timescale() # include builtin=True if you want to use older files (you may miss some leap-seconds)

with load.open(mpc.COMET_URL) as f:
    comets = mpc.load_comets_dataframe(f)

comets = comets.set_index('designation', drop=False)

print(len(comets), 'comets loaded')

row = comets.loc['C/2020 F3 (NEOWISE)']
print(row)

eph = loaddata('de421.bsp')

earth, sun, moon = [eph[x] for x in ('earth', 'sun', 'moon')]

comet = sun + mpc.comet_orbit(row, ts, GM_SUN)

times = ts.utc(2020, 7, range(1, 32))
ra, dec, distance = earth.at(times).observe(comet).radec()

RA, Dec = [thing._degrees for thing in (ra, dec)]

plt.plot(RA, Dec, 'ok')
plt.gca().set_aspect('equal')
plt.show()

t0 = ts.utc(2020, 7, 12)
t1 = ts.utc(2020, 7, 19)
LA = Topos('34.05 N', '118.25 W') # 34.05, -118.25

times, is_sunrises = almanac.find_discrete(t0, t1, almanac.sunrise_sunset(eph, LA))
for time, is_sunrise in zip(times, is_sunrises):
    if is_sunrise:
        print('sunrise: ', time.utc_iso())
    else:
        print('sunset: ', time.utc_iso())

times, is_rises = almanac.find_discrete(t0, t1, almanac.risings_and_settings(eph, comet, LA))
for time, is_sunrise in zip(times, is_rises):
    if is_sunrise:
        print('cometrise: ', time.utc_iso())
    else:
        print('cometset: ', time.utc_iso())


Are there any double stars that I can actually see orbit each other?

Vir (12h 42m, –01° 27′)

Probably Porrima, $\gamma$ Vir, is the best candidate for most observers in the Northern Hemisphere to see changes in a binary orbit, particularly using a small telescope. It is a pair of stars with similar size and visual magnitude, of about 3.6. Their orbital period is about 169 years, but the orbit is eccentric, e = 0.88. They are also relatively close at about 40 ly. Periapsis was in 2005, so the stars are now moving away from each other, but their rate of separation is decreasing. Separation at periapsis was about 0.4 arcsec, so would not have been resolved using a small telescope in 2005. By 2015 their separation was ~2.5 arcsec, and will increase to ~3 arcsec by 2020. I estimated a position angle change between 2015-2020 of ~7 degrees. These changes should be detectable with a 100-200 mm (4-8 inch) telescope.

Since most short period binaries are close together, with nearly circular orbits, and often more distant, they are very difficult or impossible to resolve with a small telescope.

Sirius B (06h 45m, −16° 43′)

As mentioned in @MichaelWalsby's answer, it is also possible to observe the orbit of the white dwarf binary companion to Sirius, the brightest star visible in the night sky. Sirius is only 8.6 ly away, and their orbit has a semi-major axis of about 7.5 arcsec, an eccectricity of e = 0.59, and a period of about 50 years. If this pair were similar in brightness, they would be an easy answer to this question. Unfortunately, Sirius B, or the Pup (as the companion is known to amateur astronomers), is ~10 magnitudes dimmer than Sirius, and usually lost in its glare. It takes a night with excellent seeing, i.e. a stable, non-turbulent atmosphere, especially since Sirius never gets much above 30 degrees elevation at the mid-northern latitude where I live. I have only seen the Pup 4 or 5 times (one view was probable but not certain) over almost 6 decades of observing, and I have never seen it in a telescope with aperture under 300 mm. I know other amateurs that have seen Sirius B in 150-200 mm telescopes, but mostly at lower latitudes. However, by seeing Sirius B at intervals separated by decades, I have observed its polar angle change.

I believe the separation of Sirius B is now over 10 arcsec, and still increasing slightly. So for the next couple of decades, observing it might be a bit easier. In recent winters I have  tried with telescopes from a 120 mm refractor to a 250 mm Dobsonian, and occasionally larger, but still have not seen it for several years. This Hubble photo of Sirius gives some idea why Sirius B is hard to observe in small telescopes.

Indirect methods

Also, many eclipsing binaries are often observed, and their light curves measured. By analyzing light curves, orbital elements can be estimated. However, these indirect orbital observations are  probably stretching the intent of the original question.


Can I get help getting image of Celestial Body at RA 2.220 and Declination +15.59?

I think your star is an 8th magnitude star in Aries: properly catalogued as HD 14610. It is an F5 white dwarf, a little larger and brighter than the sun, but about 300ly away, and too dim to see without a telescope.

It is located at RA 02h 22m 01s, dec +15* 59' 30''

An image from the sky surveys: http://simbad.u-strasbg.fr/simbad/sim-id?Ident=HD++14610 (click on "Interactive Aladin-lite to get a wider view)

Others have touched on the unofficial nature of these "star naming" services, which some feel are fraudulent.  However if it makes your mum happy.


Well, currently, we don't have a shred of evidence that they exist.

White holes are hypothetical objects predicted by certain solutions to Einstein's theory of general relativity. To be specific, they  come from the maximally extended Schwarzschild solution. We can deal with them a little mathematically, but nobody will really believe in objects like these unless we have experimental evidence for them.

Do we? I quote NASA here

  White holes are VERY hypothetical.

and here

  There is no observational evidence for white holes.

This supports this position:

  Unlike black holes, there is no real astronomical evidence to suggest that white holes actually exist.

Also, the idea is put nicely here:

  A white hole is only a concept for higher levels of thinking. No one has every observed one and no one probably ever will.

Why? This gives an interesting answer:

  Once even the tiniest speck of dust enters the part of space-time which includes the black hole, the part which includes the white hole disappears. The universe has been around for a long time and so even if it did start with white holes, they would have all disappeared by now. (Emphasis mine)

Essentially, white holes are unstable. Even if a particle did try to enter one from a black hole via an Einstein-Rosen bridge, the bridge would collapse before the particle could cross, and the particle would hit the singularity.


Why are the moons of Neptune sorted by diameter?

Likely there's two different explanations, which can even work in conjunction:
a) An explanation along the same as goes for the size sorting of the planets in the solar system. For those regular moons which formed along with Neptune or in orbit of the protoplanet we see that the further outward the moon forms in the accretion disk around proto-Neptune, the more material is present due to the greater circumference of the orbit. Thus it can grow larger in size.
b) Neptune exerts tides onto the moons. Typically no moons can exist within the Roche limit. How close a moon can exist depends on the rigidity of the body, though. Big moons (mass $M_m$) cannot exist as far inward (distance $d$ to Neptune) as a small moon (Neptune's radius $R_N$ and mass $M_N$):
$$ M_m = \left(\frac{R_N}{d}\right)^3\cdot2M_N$$ or easier expressed as ratio of the moon's radius $r_{moon}$ and the moon's distance to Neptune $d$ as well as its density $ \varrho_{moon}$: $$ \left(\frac{r_{moon}}{d}\right)^3\cdot\varrho_{moon} = constant $$
Thus the further out, the larger the moon can be without falling apart due to tidal forces. Wikipedia has a page on Neptune's moons and references which indicates that the inner moons are somewhere at the tidal limit for bodies which are of somewhat granular nature as is expected for at least the smaller moons or maybe even all with the exception of Triton.


As @barrycarter mentioned you can use the Horizons web interface from JPL: http://ssd.jpl.nasa.gov/horizons.cgi

You can change the setting in the web form by clicking on the [change] links for the different settings (for instance the time span or the planet). There are several different settings that can be changed to get different outputs (table columns) and such.

If you really want the orbital elements (and not just the Right Ascension and Declination) you will have to select ELEMENTS in the Ephemeris type settings.

You then get a table with the following elements:

JDCT     Epoch Julian Date, Coordinate Time
EC     Eccentricity, e
QR     Periapsis distance, q (AU)
IN     Inclination w.r.t xy-plane, i (degrees)
OM     Longitude of Ascending Node, OMEGA, (degrees)
W      Argument of Perifocus, w (degrees)
Tp     Time of periapsis (Julian day number)
N      Mean motion, n (degrees/day)
MA     Mean anomaly, M (degrees)
TA     True anomaly, nu (degrees)
A      Semi-major axis, a (AU)
AD     Apoapsis distance (AU)
PR     Sidereal orbit period (day)


What more could be learned from a rare astronomical event if we knew precisely when it would occur?

I would argue that simply knowing that a star was about to undergo a supernova could be information enough.

It's not always easy to figure out what type of star a given supernova progenitor was. Sometimes, there were no observations of the relevant region of sky prior to the supernova, and so we can't simply look back at prior observations and extrapolate from that. We can definitely look at certain spectral lines and other information from the remnant, but it's rarely as sure as actually observing the progenitor.

There are some classes of stars which have yet to be conclusively identified as supernova progenitors, though there may be evidence suggesting that they should evolve as such. Red supergiants are one group, leading to the so-called "red supergiant problem" (see Smartt et al. (2009) and this question). If we knew that a red supergiant was a supernova progenitor, that would possibly solve the problem.

Wolf-Rayet stars, thought to produce Type Ib/c supernovae, are another group (see Yoon et al. (2012). Evolutionary models and indirect observational evidence predict that they should lead to these supernovae, but no observations of Wolf-Rayet supernova progenitors have been observed.

Eta Carinae A - which I assume is the star you're referring to - appears to be a luminous blue variable (LBV). We do know that LBVs lead to supernovae (see for instance the case of SN 2006jc, which underwent a major outburst two days prior). However, it can be harder to distinguish between supernovae and certain supernova imposter events, caused by LBVs. I'd speculate that we could try to observe Eta Carinae A in the period leading up to the supernovae, to try to observe the differences between outbursts and the actual supernova, which could give us better information on the nature of the evolution of these stars. Additionally, there is neutrino emission in the periods leading up to supernovae (see this question); perhaps detectors on Earth could monitor the progenitor and see if they can find anything.

We could also, given enough warning time, attempt to study the progenitor in the final phases of stellar evolution. This page gives a table for the time periods of various fusion phases in the life of a $25M_\odot$ star:
$$\begin{array}{|c|c|}
\hline \text{Fusion phase}&\text{Length of phase}\\
\hline \text{Hydrogen} & \text{7 million years}\\
\hline \text{Helium} & \text{500,000 years}\\
\hline \text{Carbon} & \text{600 years}\\
\hline \text{Neon} & \text{1 year}\\
\hline \text{Oxygen} & \text{6 months}\\
\hline \text{Silicon} & \text{1 day}\\
\hline
\end{array}$$
This would let us figure out where the star is in its life and figure out what effects different fusion pathways could be having, if any.

Finally, I'm assuming that your question means that we know when the light from the supernova reaches Earth, and that this would happen in the near future.


Do we make predictions in our time, or local time?

We express when events are going to happen (or have happened) in our observed time frame.

For example, we say SN 185 went supernova in the year AD 185; but it didn't. The star actually physically exploded about 9,100 years earlier.

When we say the sun will set tonight at 8:07 PM ET, that is when we observe the event. But our actual position relative to the tangent from the sun will have already passed about 8 minutes 20 seconds earlier (plus several minutes due to atmospheric refraction).

So when you say Betelgeuse is predicted to explode sometime in the next million years, that's how many more Earth years are expected to pass until we see it happen… from here. From Betelgeuse's point of view, it would have happened about 642 years earlier… or it may have happened already.


Luminosity value to use for the calculation of the habitable zone?

The number you need is the bolometric luminosity. This tells you how many Watts per square metre are incident upon the planet. That is true for the Sun and it is true for GL 570.

Beyond that, you will need to make assumptions about an atmosphere and an albedo if you are trying to calculate a realistic habitable zone. If you just assume the planet is like the Earth then using the same albedo value will probably do. But, because Gl 570 has a spectrum that peaks at 50% longer wavelengths than the Sun, then the albedo values at longer wavelengths are more important for a planet orbiting Gl 570. Whether that is bigger or smaller than the Earth would depend on exactly what sort of atmosphere/clouds/terrain were on the planet.


This is the frequency that the observations were made at. These are sub-mm (the wavelength is given as 1.3mm) observations (between the far infrared and the radio parts of the spectrum) and so it is more usual to talk in terms of frequency rather than wavelength (although there are exception e.g. the 21cm line of neutral hydrogen which is in the radio at 1427 MHz).

In the optical and infrared , we would normally say "observations were made at 550nm" but through the relationship $c=\frac{f}{\lambda}$ (where $f$ is the frequency, $\lambda$ the wavelength and $c$ is the speed of light), we could also say "these observations were made at 545077 GHz (545 THz)" although almost nobody ever would.

The "bandwidth of 320 MHz." part refers to how much of the spectrum is being sampled and digitized around the parts of interest, the CO J=2-1 line at 230.538 GHz (molecular spectra in the radio reference) and the continuum on side. The continuum is measured to get the strength of the background "underneath" the line and is then subtracted to get the true strength of the CO line.


How many planets are there in this solar system?

In addition to Undo's fine answer, I would like to explain a bit about the motivation behind the definition.

When Eris was discovered, it turned out to be really, really similar to Pluto. This posed a bit of a quandary: should Eris be accepted as a new planet? Should it not? If not, then why keep Pluto? Most importantly, this pushed to the foreground the question

  what, exactly, is a planet, anyway?

This had been ignored until then because everyone "knew" which bodies were planets and which ones were not. However, with the discovery of Eris, and the newly-realized potential of more such bodies turning up, this was no longer really an option, and some sort of hard definition had to be agreed upon.

The problem with coming up with a hard definition that decides what does make it to planethood and what doesn't is that nature very rarely presents us with clear, definite lines. Size, for example, is not a good discriminant, because solar system bodies come in a continuum of sizes from Jupiter down to meter-long asteroids. Where does one draw the line there? Any such size would be completely arbitrary.

There is, however, one characteristic that has a sharp distinction between some "planets" and some "non-planets", and it is the amount of other stuff in roughly the same orbit. This is still slightly arbitrary, because it's hard to put in numbers exactly what "roughly" means in this context, but it's more or less unambiguous.

Consider, then a quantity called the "planetary discriminant" µ, equal to the ratio of the planet's mass to the total mass of other bodies that cross its orbital radius and have non-resonant periods (so e.g. Neptune doesn't count as sharing Pluto's orbit) up to a factor of 10 longer or shorter (to rule out comets, which has little effect in practice). This is still a bit arbitrary (why 10?) but it's otherwise quite an objective quantity.

Now take this quantity and calculate it for the different bodies you might call planets, comparing it to both the objects' mass,

and their diameter,

or with an arbitrary horizontal axis, in order of decreasing discriminant,

Suddenly, a natural hard line emerges. If you look only at the mass and the diameter of the objects (shown in the insets above the plots), then there is a pretty continuous spread of values, with bigger gaps between the gas giants and the terrestrial planets than between Mercury and Eris/Pluto. However, if you look at the planetary discriminant, on the vertical axis, you get a very clear grouping into two distinct populations, separated by over four orders of magnitude. There's a finite set of bodies that have "cleared their orbits", and some other bodies which are well, well behind in that respect.

This is the main reason that "clearing its orbital zone" was chosen as a criterion for planethood. It relies on a distinction that is actually there in the solar system, and very little on arbitrary human decisions. It's important to note that this criterion need not have worked: this parameter might also have come out as a continuum, with some bodies having emptier orbits and some others having slightly fuller ones, and no natural place to draw the line, in which case the definition would have been different. As it happens, this is indeed a good discriminant.

For further reading, I recommend the Wikipedia article on 'Clearing the neighbourhood', as well as the original paper where this criterion was proposed,

  What is a planet? S Soter,  The Astronomical Journal 132 no.6 (2006), p. 2513. arXiv:astro-ph/0608359.

which is in general very readable (though there are some technical bits in the middle which are easy to spot and harmless to skip), and from which I took the discriminant data for the plots above.

Edit: I must apologize for having included, in previous versions of this post, an incorrect plot, caused by taking data from Wikipedia without verifying it. In particular, the planetary discriminant for Mars was wrong (1.8×105 instead of 5.1×103), which now puts it below Neptune's instead of just below Saturn's, but the overall conclusions are not affected. The Mathematica code for the graphics is available at Import["http://goo.gl/NaH6rM"]["http://i.stack.imgur.com/CQA4T.png"].

... and, as a final aside: Pluto is awesome. It was visited in July 2015 by the New Horizons probe, which found a world that was much more rich, dynamic, and active than anyone expected, including what appear to be churning lakes of solid nitrogen ringed by mountains of water ice, among other marvels.

(Note the image has been colour-enhanced to bring out the variety of surface materials; the true-colour version of this image is here.) I, personally, don't feel it's at all necessary to 'grandfather' Pluto into the list of planets to really feel the awe at the amazing place it is - it's perfectly OK for it to be a cool place with cool science, that is also not a planet.


What is the name of the cycle whereby the lunar azimuth and altitude recur on a given day?

You are looking for the Metonic cycle.  This is almost exactly 19 years, and represents 235 synodic months.

Now you have edited the detailed question so that it is out of sync with the original Title.  The other period you need is the Draconic month, which is the period between the moon crossing the ecliptic (in the same direction).


What is the temperature on surface of Pluto facing Sun?

According to this site, Pluto may experience temperatures of up to 55K, which I assume happens during the daytime. This low temperature is due to Pluto's distance from the sun, its high albedo (about 0.5), and its extremely thin atmosphere.


Transit Depth Color Versus Transmission Spectroscopy

But this is already possible. Spitzer has performed "low-resolution" transit spectrophotometry of several exoplanets in its near infrared bands - see for example Tinetti et al. (2007);  Knutson et al. (2011); Desert et al. (2011).

I dare say you could get better data on more planets with a dedicated telescope with a larger mirror and the ability to image in several bands at once. All that is required is money.


How can gravitational waves escape from the gravity of black hole?

I see this phrase all the time, and I have to say I've come to greatly dislike it because it's a very bad misnomer. Nine times out of ten, when someone is talking about a black hole, they describe it as an object with such strong gravity that "not even light can escape".

However, this unqualified statement presents a strong misconception as to what black holes actually are and how they work and accomplishes nothing but confusing innocent bystanders such as yourself. The gravity of a black hole is no more or less strong than any other object in the universe. Black holes are not cosmic vacuums which use their powerful gravitational forces to suck up all nearby matter, light, etc. In fact, if you replaced our Sun with a black hole of exactly the same mass, all the planets in our system would go about orbiting exactly the same way and wouldn't notice a difference at all (aside from the mass extinction on Earth due to no longer receiving any energy from the Sun).

That being said, let's paint a better picture of what a black hole is and how it works. A black hole is a clump of mass which has become so massive that the gravitational force of that mass on itself, trying to pull it together, actually collapses the mass into a singularity. The singularity is a point-like region of space where all the mass becomes contained. Slightly outside this singularity, the physics gets weird. For example, if you're right next to this singularity, and you calculate the speed needed to get away from that singularity (e.g., you need to travel ~11 km/s to get away from Earth) you find a speed which is much greater than the speed of light. That's the origin of the phrase "not even light can escape". But, if you start farther away from the singularity, you need less speed to escape it because you feel less gravitational pull from it (gravity decreases with distance). This means, at some distance from the singularity, the speed of light actually is fast enough to escape the black hole. This distance is so important that scientists have given it a special name, the event horizon. It can get a lot more complicated than the simple picture I've painted above, but that's the general idea.

If you put all that together, then that tells you that any light which is outside the event horizon has no trouble escape the black hole. It is only the light inside this event horizon which cannot escape. Likewise, any gravitational waves outside the event horizon can escape just as easily. This is what the answer by StephenG meant by saying they were "outside" the black hole. By outside, he meant outside the event horizon. And it is true that as long as the creation of the gravitational wave happens outside the event horizon it will escape the black hole.

And just for reference of size, the supermassive black hole in the center of our galaxy, which is 4,000,000 times more massive than our Sun, has an event horizon which only extends ~10,000,000 km. That's barely out to the orbit of Mercury if it were in the position of our Sun. So you can see, it's not very hard to be outside the event horizon as the event horizon isn't that big in astronomical terms.


stellarium software: selective artificial satellite display

Satellites are provided by a plug-in (that is enabled by default)
Configuration (the spanner-and-star button) -> plug-ins -> Satellites -> configure.  You can configure which satellites to display from here.
It might be worth noting that meteosat-8, a geostationary satellite has a magnitude of about 19, and so is not a naked eye object, and is in fact beyond the capabilities of most amateur equipment.


Do the earth and the moon reflect the different fractions of total solar radiation proportionally?

The obvious answer is: "of course". Different colour simply means different wavelength-dependent reflection; different brightness means different wavelength-independent reflection. Earth and moon have different colours when viewed from space.
The real question is by how much does it differ as a function of wavelength. There are already here nice comparisons of the solar spectrum and the Moon's spectrum.
There's a relative measurement of Earth's reflectance by Björn et al (2009). Similarily Turnbull et al (2006) discuss the reflectance of Earth compared to the Moon. Reality is complicated as exact reflectance depends on the observation angle and angle of incidence - and the exact materials looked at which varies slightly for the Moon and can vary a lot for Earth, especially when considering cloud and ocean coverage. Earth might be slightly redder due to increased thermal radiation, especially in the absorption bands of water in the atmosphere.


Let's assume a uniform laminar emitter, oriented so that it is at right angles to the line to the detector.

The specific intensity emitted by a blackbody, at the surface of the blackbody is
$$B_{\nu} = \frac{2h\nu^3}{c^2} \frac{1}{\exp[h\nu/kT] -1}\ {\rm Wm}^{-2}{\rm Hz}^{-1},$$
where $B_{\nu}$ is the "Planck function".

Using the definition of specific intensity we can calculate the flux (energy per unit time, per unit frequency) received at the detector as
$$ f_{\nu} \simeq B_{\nu}A \Delta \Omega \cos \theta\,$$
where $\Delta \Omega$ is the solid angle of the detector subtended at the laminar surface, $A$ is the area of the laminar surface, and $\theta$ is the angle with respect to a normal from the surface and here, $\cos \theta \simeq 1$.

The solid angle subtended by the detector at the surface is $\Delta \Omega = a/D^2$, where $a$ is the area of the detector, but we can also define a solid angle $\Delta \omega = A/D^2$, which is the solid angle of the surface at the detector. Thus
$$\frac{f_{\nu}}{\Delta \omega} =  B_{\nu}a  $$

If the detector has a fixed sensitivity per unit solid angle then unless the object is unresolved, then it does not matter how far away it is; the value of $D$ does not come into it. It is a well-known fact in astronomy that resolved, extended objects have an observed surface brightness (flux density per unit solid angle) that is independent of distance.


Why does the Parker Solar Probe slow down as the distance from the Sun increases?

Why does the “parker solar probe” lose speed as the distance from the sun increases?

Because energy and angular momentum are individually conserved quantities in the two body problem. Except for where the Parker Space Probe has close fly-bys with Venus, the gravitationally interactions between the Parker Space Probe and the solar system are very closely modeled as a two body problem (the Sun and the probe), plus very small perturbations from the planets.

One way to express conservation of energy in the two body problem is the vis-viva equation, $$v^2 = \mu\left(\frac2r - \frac1a\right)$$

where

$\mu = G(M+m)$ is the sum of the central body's standard gravitational parameter and that of the orbiting body,
$r$ is the distance between the two bodies,
$a$ is the semi-major axis length (a constant), and
$v$ is the magnitude of the velocity vector.

Note that the mass of the Parker Space Probe is so much less than that of the Sun that one can drop the Parker Space Probe's mass from the expression $\mu = G(M+m)$, resulting in $\mu = GM_{\text{sun}}$.

Note that the only variable on the right hand side of the vis-viva equation is the radial distance. As radial distance increases the square magnitude of the velocity vector (and thus the magnitude of the velocity vector) decreases.

Without mathematics, conservation of energy dictates that the sum of an orbiting body's kinetic energy and gravitational potential energy must remain constant. As the orbiting body moves further from the central body, the orbiting body's potential energy increases, which means its kinetic energy must correspondingly decrease. This in turn means the orbiting body's velocity must decrease.


Why is the moon phase not in sync with the moon's illuminated fraction?

This is a partial answer since your question is still unclear.  I would recommmend finding an astronomy package for tools like this rather than using what's basically a "bonus feature" from an oceanographic package.

From the help pages for package oce,

  illuminatedFraction
   fraction of moon’s visible disk that is
   illuminated
  phase
   phase of the moon, defined in equation 32.3 of Meeus
   [1982]. The fractional part of which is 0 for new moon, 1/4 for first
   quarter, 1/2 for full moon, and 3/4 for last quarter.

Perhaps you are missing the fact that marginally more than 50% of the moon is illuminated, and thus at first/last quarter marginally more than 50% of the visible disk is illuminated?  I would also recommend looking up the referenced "equation 32.3" to see what the definition is and how it applies to a zero-altitude viewer.


Python 2D Jean Instability in Spiral Galaxy Simulation

First a warning, then a hack:

1) If you didn't run this simulation, if you are not a trained theoretical astronomer of undergraduate level or above, and if this simulation doesn't have the kind of data you need to answer the question you are asking you should not trust the results at all.

It takes years of study to understand the weird numerical effects (things that happen in the simulation because of the way it is set up that would never happen in reality), background physics and loads of assumptions and approximations that go into this kind of stuff.

We often say 'garbage in, garbage out'. The answers you get from this will probably not be very illuminating because they say more about the simulation than real space.

2) That said, if you are just doing this for fun or to learn the skills, GREAT! Keep playing! And a very basic assumption you can make in order to get some numbers out (that will be quite wrong) is that all gas in space is ~10K.


Is there a difference between the solar elevation angle and sun declination?

Elevation and Declination are from different co-ordinate systems, so solar elevation is given in Alt/Az co-ordinates and refers to the elevation above the local horizon. Sun declination is measured in RA/Dec co-ordinates (which is an equatorial co-ordinate system) and measures the Sun's inclination above or below the equator on the celestial sphere.


You are correct that the main way to determine planetary surface ages is through crater dating. This comes in two versions, relative and absolute dating. Both require counting the number of craters per area and so you need good resolution imaging since the number of impactors and craters follow a power-law (more small impactors/craters than large). It then basically boils down to older surfaces have had a longer time to be hit by more impactors and so there will be more craters per unit area.

With relative dating, all you can do is put different surfaces in order of amount of craters and say "surface A is older than surface B as it has more craters". If you have another method of dating, then you may be able to say where on the timeline a surface may be but you would still only be able to "surface A is older than surface B (dated to be 250 Million years)" but you don't know much older.

If you have an idea of the rate of cratering bodies and the rate of crater formation in the system you are trying to study, then you can perform absolute dating. This allows the cratering rates to be compared across different bodies. This has been done for the inner Solar System (Mercury, Moon, Mars) but is tricky where the surface can be altered (weather, lava flows etc). There are also problems with crater saturation with very old surfaces where new craters overlap and overwrite older ones and issues with understanding the lifetimes of the craters and what processes can fill-in and erase craters and how long this process takes.

In the case of the Saturn system, determining the number of potential impactors and performing absolute dating is likely to be pretty tricky as the environment is dynamically complex. There are a lot of large Moons and you also have Jupiter and the other gas giants planets affecting things. In addition, the positions of the giant planets are thought to have changed dramatically around 500 Mya after formation (the Grand Tack model and hypothesis).

The images of the Pluto and Charon system returned by New Horizons have also shown some of the difficulties with crater counting (blog article on this by a member of the New Horizons team and crater dating expert). Some parts of the system such as Charon appear to be very old and have large numbers of craters; other areas such as Sputnik Planitia appear to be very young and no-one can find any craters of any size anywhere on that region.


Why don't storms on gas giants move to the poles, like hurricanes on Earth do?

The answer is the Coriolis effect, on Earth this produces cells within which storms move, converging towards the cell boundaries as you can see below.

Jupiter however spins much faster that the Earth which produces a stronger Coriolis effect and thus more cells. This is the reason why there are so many different coloured bands on Jupiter (see image below). So storms like the big red spot don't have as far to move laterally within these cells compared to hurricanes on Earth (in comparison with the planets size).


The non-thermal S-Z effect is caused by inverse Compton scattering of the CMB photons from a non-thermal population of electrons - i.e. electrons that have high energies not because they are hot, but because they have been accelerated non-thermally. The usual mechanisms are accelerating by electromagnetic fields and the Lorentz force.

The rest-mass energy of an electron is 0.511 keV. In order to attain "relativistic energies" then the electrons must have kinetic energies similar to this or higher. The average kinetic energy of a particle due to its temperature is just $3k_{B}T/2$. If we equate this to 0.511 keV then the temperature required to have thermal relativistic electrons is $T > 4 \times 10^{6}$ K.

Gas at this temperature does exist in the intracluster medium of massive galaxy clusters. But in addition there are populations of non-thermal electrons that are accelerated to relativistic speeds for instance in the jets and radio-emitting lobes of active galaxies.

The distinction is important when studying the S-Z effect because the energy distributions of thermal and non-thermal populations can be quite different. Non-thermal populations are usually characterised by a power-law, whilst thermal electrons have a Maxwell-Boltzmann distribution. This results in a different inverse Comptonisation spectrum from electrons of the two populations that are often (usually) spatially unresolved. The non-thermal S-Z effect is essentially a contaminant that needs to be accounted for when using the thermal S-Z effect to investigate the structure and parameters of the intracluster gas and using the S-Z effect as a cosmological probe.


How to find which part of the sky a photo contains?

Have you tried nova.astrometry.net?

They set up a web service for doing more or less what you're talking about.


What kills us in the event of two galaxies combining?

There are two reasons why nobody will die when the Milky Way merges with Andromeda:

The merger won't take place for another 4.5 billion years, by which time solar evolution will have long since made Earth uninhabitable (and even if humans did colonize the outer planets, we'd have long since gone extinct or evolved beyond recognition).
The merger itself will be a slow and not very dramatic process - stars aren't going to collide, planets aren't going to be ejected from their solar systems, and while it will look dramatic from a distant galaxy, other than Andromeda looking larger in the night sky not much will change for any intelligent life that's around at the time.

However, there's one potential wrinkle. Both the Milky Way and Andromeda host supermassive black holes at their cores. We've never observed a merger between two supermassive black holes, and modeling the process is theoretically challenging (look up the "final parsec problem" if you're interested). When this does happen, though - long after the merger of the stellar portions of the galaxies - there would be a massive release of energy. Most of this will be in the form of gravitational waves, but if enough of it comes as high-energy radiation it could be bad news for any remaining life.


Rings around a smaller, close-in planet?

I think the related answer that James Kilginger provided is very good and worth a read.  I'll touch on a few points related to your question.

There's 2 parts to this, how do ring systems form and where are they more likely to be stable and long lasting.   I doubt that any ring system would be permanent, but some should be much longer lasting than others.

So, lets talk stability first.

As you pointed out, as ices melt inside the frost line (somewhere around Ceres, though different ices have different melting points, but in general, around Jupiter distance or greater ice particles are largely stable and can be part of a ring system (I'm not sure about Jupiter's gonzo magnetosphere though).   Closer to the sun, certainly by the time you get to Mars, probably a bit further out than that, solar radiation melts ices and it turns into gas which the solar wind blows away from ring systems pretty quickly.   What's more, any smaller rocky debris mixed with ice could experience spin or acceleration as ice is melted and escapes from one side as gas.   Those local accelerations could blow some material from a ring system too.

Smaller dust particles in a ring system near a star would get blown away too by the Poynting-Robertson effect which is much more significant closer to a star.   Some of these problems would be avoided if the star is in it's white dwarf stage and quite cold, but a main sequence star is much more friendly to distant ring systems than closer ones.

Ring systems are only stable inside the Roche Limit.  That's covered in the other question and outside the Roche Limit, the material would tend to reform into a moon.   Solid objects however can orbit inside a liquid or fine particle Roche limit and maintain their shape.

Planets close to the sun still have Roche limits usually well inside their Hill Spheres, but long term stability is still a problem for planets close to their stars.   Here's an article that touches on this regarding Mercury and one on Venus.  The same principal would apply to ring systems where the significantly larger solar tide would have a destabilizing effect on all objects in orbit around a planet close to it's star.   It wouldn't destabilize immediately but over many orbits, objects could either spiral into the planet or outside the Roche limit, so Ring systems would be significantly shorter lived as planets get closer to their star.  The tidal force grows with the square of the distance.  Earth, for example experiences 90 times the tidal force from the Sun that Saturn experiences.

Saturn is also about 95 times the mass of the Earth, so it's not hard to see why Saturn is a much better candidate for a long term ring system than one of the inner planets.   Much more mass and much further form the sun.

Formation of ring systems can happen by 3 methods, (that I can think of).  Large collision, break-up of a moon that travels inside the Roche limit or accretion.   The collision can be on a planet's surface where debris is blown into orbit or similar high speed impact on a moon in orbit around the planet.    Collisions of this size are rare, but likely more common in young solar systems.

A moon that moves too close to a planet can get broken up and form into a ring system if it travels inside the Roche limit see pictures top right.   There's no reason why that couldn't happen from time to time on a number of planets.   It might one day happen to Phobos as it spirals in closer to Mars.   Moons tend to move towards or away from planets quite slowly so this doesn't happen often, but it can happen from time to time, even when solar-systems are billions of years old and mostly settled into stable orbits.

Finally, accretion, Enceladus see here is feeding Saturn's e-ring and may have fed other rings in the past.   Enceladus is unique among all the moons in our solar-system in that it's able to steadily feed a ring system.   This is possible because it's quite small, so volcanoes on it's surface are able to eject material at greater than it's escape velocity and because it probably has a liquid water-like layer under it's surface, maintained by tidal heating because it's close enough to Saturn to experience enormous tidal forces.   Enceladus is likely responsible for only a tiny part of Saturn's rings, but this is still a viable method for ring formation.

Europa on Jupiter experiences similar consistent volcanism due to large tidal forces that also keep some of it's inner layers liquid, but Europa's a much larger moon, so far less material gets ejected at or above escape velocity than with Enceladus.   Only small moons close to large planets can supply material to a ring system in this way.

But, because moons may play a role in ring formation and Large planets are more likely to have more moons due to greater gravity and Planets far away from their star are also more likely to capture more moons because they have larger Hill Spheres and objects don't orbit the sun quite as fast, that's probably the gist of it.

Large planets are more likely to have moons and rings than small planets and further away from the sun, a planet is more likely to maintain a ring system then close to the sun.    It's not impossible that planets close to a star or smaller planets could have a ring system, but quite a bit less likely.

  A smaller planet means the rings would have to be closer to the
  planet, which means the radial shear (difference in orbital speed vs.
  radius) in the rings would be higher. A higher radial sheer would be
  more likely to cause turbulence...or so I would think.

High orbital speed, I suspect, makes little difference, in fact, a larger planet would have higher orbital speed in it's rings than a smaller one.  A heavy Jupiter, 8 or 10 Jupiter masses and 2 or 3 times denser than the Earth could have a large orbital region inside it's Roche limit and potentially, very large, very fast orbiting rings.

Hope that wasn't too long.


The deduction of the Hill Sphere formula

Hill sphere is named after John William Hill (1812–1879) and its simple logic follows from the presence of three bodies (let's assume Sun is the largest mass with Earth as the secondary mass and a satellite of negligible mass orbiting the Earth as the third mass), where the radius of the Hill sphere will be the largest radius at which a satellite could orbit the secondary mass (Earth in this case). If its orbit exceeds the Hills radius, then it will fall to the gravitational influence of the first body (sun) and hence will no longer be a satellite of the secondary body.

One could write Newton's equations using the idea that the satellite has the same angular velocity as the secondary object. This is that, the angular velocity of the Earth around the sun equals to the angular velocity of the satellite around the sun. A demonstration about the derivation is given in the following link as well as that of the Roche limit:

http://www.jgiesen.de/astro/stars/roche.htm


To really get exact times, you would have to take into account the opacity of the atmosphere, but I will be ignoring this here, and consider the light flux outside of the atmosphere of any planetary body.

There is no [planet] time for planets closer to the Sun than Earth, or for exoplanets orbiting brighter stars than Earth at the same distance. This  because they receive more light flux than we do. On a side note, there is an Earth time on Mercury.

For planets in further orbits than Earth around the Sun, [planet] time can be determined by comparing the solar flux at any place on Earth to the flux received by the planet at midday.

Flux received by the planet at midday

The average light flux on a planet can be determined based on the fact that the  flux varies inversely proportional to the square of the distance.
Outside Earth's atmosphere, the Solar flux is $S_0 = 1370$ Watts per square meter, for a surface placed perpendicular to the Sun rays.

Since Earth is 1 AU from the Sun, the received flux for a planet at a distance of $x$ AU from the Sun is then given by

$
F_{planet} = S_0 * \Big(\frac{1}{x}\Big)^2
$

Solar intensity on Earth

Based on this article, the flux received from the Sun at a given location at latitude $\phi$ on Earth, and when the Sun is at declination $\delta$ in the sky can be determined to be

$
F_{location} = \frac{S_0}{\pi} \Big(H \sin \phi \sin \delta  + sign(H) \cos \phi \cos \delta \Big)
$

where $H = \arccos \big( -\tan (\phi ) * \tan (\delta ) \big)$, so $H/\pi$ is the length of a day.

Putting it all together

So given your latitude $\phi $ on Earth, you want to find the declination $\delta$ of the Sun by saying that $F_{location} = F_{planet}$.

This gives you a trigonometric equation that you can solve numerically with a tool such as this, or any other method you fancy.


The escape velocity is c at the event horizon of a black hole. Above the event horizon the escape velocity is below c. Escaping particles of the Hawking radiation form above the event horizon; that's why they can escape, if they are pointing towards a sufficiently narrow angle to vertical upward, and if they are sufficiently energetic.

Escaping particles form as virtual particle-antiparticle pairs in the "infalling" coordinate system: One of the two particles forms outside the event horizon; the counterpart forms below the event horizon. Thus the originally virtual particles cannot annihilate, and therefore become real particles; one particle can escape; the counterpart falls toward the singularity.

The energy needed to form the escaping particle, and its remaining kinetic energy after escape, is subtracted from the mass of the black hole.

The described mechanism probably works also for both particles forming very close above the event horizon if the tidal forces are high enough to separate the virtual particle pair, before it can annihilate.

Formation of virtual particles is due to the Heisenberg uncertainty applied to time and energy: Very short time intervals require energy uncertainty, leading to short-lived particle-antiparticle pairs.


As mentioned above the density of vacuum energy in empty space is absolute constant, even as the volume of a region of space grows as the universe expands. So the total energy, density times volume, goes up.

More information here.


How much of carbon, sodium, silicon, and magnesium does the Sun have?

Stars are mostly hydrogen and helium. The presence of small amounts of other elements does change their luminosity, and their spectra. The heavy elements were formed in other stars.

After the big bang there was a short period in which the conditions were right for hydrogen to fuse into helium. The conditions were not suitable for any heavier elements to form, the state of the universe right after this time was about 3/4 hydrogen, 1/4 helium.

After some time clouds of hydrogen and helium began to form (around clumps of dark matter) and inside some of these clumps, smaller clouds collapsed to form the first generation of stars. These were large and they had no elements other than hydrogen and helium. These stars produced in their cores heavier elements like Carbon and Oxygen, and when they had reached the end of their lives, being big stars, they exploded in supernovae, and these elements were returned to the interstellar space, along with other heavier elements that were formed by the supernova.

The next generation of stars formed from a mixture of hydrogen, helium and the heavier elements formed in the first generation. These stars still had less heavy elements than modern stars, but were more like normal stars. Some of these stars are still around, they are known as Population 2 stars. They are found mostly in globular clusters http://hyperphysics.phy-astr.gsu.edu/hbase/Starlog/pop12.html.

After the population 2 stars had lived their lives and died, either in a supernova or in a planetary nebula, elements that had been made in their cores was returned to the intersellar space, futher enriching it with heavy elements. It was from one of these clouds, seeded with elements from stars that had gone before it, that the sun and the solar system formed.

This is why Carl Sagan says We are star stuff.


Why does Kepler's "Big Picture" of comet 67P look so strange? And what is the significance?

One thing to keep in mind is that the Kepler instrument is not a telescope like Hubble.  It is a photometer and though it uses CCDs to look at the sky, it doesn't return a picture in the usual sense.  The way it operates is that you only look at the pixels around the object you're interested in because otherwise you'd never get all those pixels transmitted to the ground because there are so many of them.
To observe the comet, they looked at the track it would make across the Kepler field of view.  They chose only the pixels you'd need to look at the comet and its tail, and this turned out to be a 15-pixel-wide swath across the detector.  That band across the image you see is that swath.  You'll notice that most of the picture is black, which are all the pixels that they didn't look at.
The stars in the scene where the image isn't back are the stars that Kepler was also looking at.  However, I'm not intimately familiar with raw Kepler data so I can't tell you why the stars are in clusters of four.  Because all the stars are lined up in the same direction, I would surmise that this is because Kepler was tracking the comet and the stars were moving in the background.  From the way they were going to take the measurements, each 36 hour observation was made up of three observations, so I am at a loss as to why there are four dots per star when this would suggest there would be three.
The answer to the science question can be found from the observation proposal.  To summarize:

Kepler photometry can provide the following key measurements at a time
when no other platform can observe the comet:

Total brightness of the comet. This relates directly to the amount of dust being released and therefore gives a critical ‘total’ activity
measurement, which can be compared with the instantaneous (but local)
measurements from various instruments on Rosetta.
Variability of the comet. There are two sources of variability; changing amounts of dust in the coma, and the underlying light-curve
of the nucleus (its non-spherical shape reflects more or less sunlight
as it rotates).

The measurements from Kepler will extend the total activity curve to
almost the end of the Rosetta mission, meaning that data in the last
months of Rosetta will not rely on extrapolation of the earlier
Earth-based observations. As the final months will be some of the most
exciting – as Rosetta spirals down and makes local measurements from
incredibly close to the nucleus – this context information is very
important in this phase.


Can we discern black holes with visible light?

There are two "usual" ways to detect a black hole:

If it is part of a binary system (two stars closely rotating around each other) and dragging matter from the other star into an accretion disk, that disk will heat up. Energy distribution will tell us whether the accreting object is indeed a black hole or a neutron star instead. This is primarily visible in X-ray, so not an answer to your question.

If the black hole has a mass of 100.000 to millions of times that of the sun (supermassive black hole), it will sit in the centre of a galaxy, and sometimes accrete matter. This is mostly visible in Radio, and again in X-ray. Still not an answer.

So, how can we detect a black hole in the optical? Best chance is an indirect detection, like in the images you linked above. The one with the distorted stars in the background is not "just fantasy", but a simulation based on our understanding of gravity. The problem when we try to see this is scale, and hence resolution. The event horizon of a stellar mass black hole is on the order of a few kilometers.

Let's replace our sun by a black hole. At a distance of $1.5\cdot10^8$km, with a black hole diameter of 6km, we'd see it occupying an angle of
\begin{equation}
 \alpha = \tan{\left(\frac{6}{1.5\cdot10^8}\right)} = \frac{6}{1.5\cdot10^8} = 2.3^{\circ}\cdot10^{-6} = 8''.3 \cdot 10^{-3}\,.
\end{equation}
So a solar mass black hole at the position of the sun would be 8 milli-arcseconds (mas) in size. A single 8m Telescope at the  VLT has a resolution of 50mas. If we consider that the gravitational lensing effect takes place on scales larger than the Schwarzschild radius, it could just manage to see this effect. (The theoretical resolution limit for an 8m telescope at 500nm wavelength is $1.22 \lambda / d = 15$mas.)

For real black holes at distances of 10s or 100s of lightyears, you would need a truly gigantic telescope.

Next option, again with gravitational lensing: The microlensing effect causes a background star to be temporarily magnified - it shines brighter for short period of time. A black hole passing in the foreground can cause this, and it is an optical observation, but it is very difficult to know that the lens was indeed a black hole and not some star or other compact object.

Last option, Hawking radiation: Hawking postulated that black holes might emit radiation inversely proportional to their size. For small enough black holes, this is a runaway process - they emit more as they get smaller, and because the emission causes them to shrink, they emit yet more again. Since Hawking radiation has a black body spectrum, the maximum of this emission will lie in the visible range for a brief period of time before going into UV, X-ray, and gamma radiation. The problem here is that no one has observed it yet, and you'd need a fairly small (but not too small) black hole close by. Also, you might get killed as the radiation gets harder.

To your questions:

Possible, but probably not. You would need to get very close, but not close enough to get killed by tidal forces. Or you'd need to be able to identify a microlens as a black hole by some means other than lensing.
Either lensing effects (see above), or a brief flash going from red to blue (probably too short to observe).
These black holes would be mostly visible by the absence of other particles. You'd get less elementary particles than you would expect for the reaction,some energy and some momentum would be missing, and there would be characteristic gamma radiation. However, for CERN to announce this, this would have to have happened often enough to not be a statistical fluke (this turned out to not be a detection after all.)


How would we detect a planet behind the Sun?

According to the Wikipedia article Counter-Earth, this idea has been floated around for millennia.  However, modern day earth-bound techniques would easily detect a 'Counter-Earth' via the following (but not limited to) these methods:

Gravitational perturbations would have been observed to have occurred on the other planets (terrestrial)
As Earth's orbit is not circular, it is elliptical, it would not be consistently 180 degrees around the orbit, as below

(Image source)

Most of all, it would have been detected/observed by the satellites we have around the Earth, Sun and other planets (many examples here).


What distance has the moon been from earth at the closest?

If we accept the Giant impact hypothesis, which says that the Moon was once a part of the Earth pushed away by a big impact, the answer is obviously 0.


The Astronomy Picture of the day for the 23rd November 2015 has all the neblosity around Orion, in fantatic detail. The result of 212 hour exposure.


First, some clarifications:

  I know that light is considered as energy

No, light is a group of photons and those have energy

  So it is certainly taken into account in universes models and calculations.

The energy of radiation is minimal, reaching about $0.01\%$ (yes, what you said is correct, kingW3) of the energy of the Universe, and that's because of two phenomena causing it to:

Decrease in absolute amounts: because of redshift.
Decrease in relative amounts: because of more dark energy, because it doesn't "dilute" with the expansion:

Think of it in this way: even if there is more space (because of expansion) the ratio between the dark energy and volume of the observable universe is constant (almost, this is more like a metaphor).
Another way to think about (and was one of the possibilities of what is dark energy) is to think of it as a property of space. So you would see that the ratio remains constant.

  Is it positive or negative?

If it were negative, we would be testing some kind of light propulsion engine or using it for building an Alcubierre warp drive. You don't see that happening. Also, light would be a tachyon (which are bad: if you ever see a theory produces a tachyon, throw the theory out of the window, as it's equivalent to saying it doesn't work)
If it were positive it would have mass, we wouldn't have discovered a lot of things, such as Special Relativity because of it traveling at subluminal speeds, and possibly it would hurt we could "feel it" (because of touch) it depending on the exact value of its mass.

  it should decrease expansion somehow

It does, but $0.01\%$ of the energy of the Universe clearly is negligible, and as I said, that number decreases over time.

The mass of light is equal to...?

Light is massless, meaning that its rest mass ($m_0$) is zero. But that doesn't mean it can't interact gravitationally: the reason objects have gravity is that they have mass (and momentum, if not you would have what I think is about $92\%$ less mass), and in this case, mass is equivalent to energy in the sense that it curves Spacetime similarly.

Other way to phrase that last sentence is as StephenG put it:

  [...] Energy (i.e. non-mass energy) can contribute to the distortion of space-time, but that's not the same as it having mass.

Summary of what I said, answering other questions:

How much of the total mass of the universe does light and electromagnetic waves accounts for?

About $0.01\%$.

And how much does it [energy of radiation] contributes to its [the Universe's] expansion?

Almost nothing.


What would happen if an ice cube is left in space?

It depends on where in outer space you are.

If you simply stick it in orbit around the Earth, it'll sublimate: the mean surface temperature of something at Earth's distance from the Sun is about 220K, which is solidly in the vapor phase for water in a vacuum, and the solid-vapor transition at that temperature doesn't pass through the liquid phase.  On the other hand, if you stick your ice cube out in the Oort Cloud, it'll grow: the mean surface temperature is 40K or below, well into the solid phase, so it'll pick up (or be picked up by) gas and other objects in space.

A comet is a rough approximation to an ice cube.  If you think of what happens to a comet at various places, that's about what would happen to your ice cube.


If Jupiter would stop rotating, would its atmosphere become homogeneous and single coloured?

They're often called jet streams, but the visible bands you see in Jupiter are probably more like Hadley cells.  Convection currents that move heat from lower in the planet's surface to higher before cooling and circling back downward.   The darker bands are rising warmer atmosphere, the lighter bands, made lighter in color by more ice, are falling.

https://en.wikipedia.org/wiki/Hadley_cell#/media/File:Earth_Global_Circulation_-_en.svg

https://en.wikipedia.org/wiki/Atmosphere_of_Jupiter#Zones,_belts_and_jets

Jupiter's fast rotation gives it several Hadley cells per hemisphere.   By comparison, Earth has 3 and Venus has 1.   Faster rotation and perhaps Jupiter's larger size as well creates more Hadley cells.

That said, if Jupiter didn't rotate, it would still have heat from below, rising and escaping.   It would still shed temperature from below, because of internal heat.

What it might look like is a complicated question but I think a uniform brown is unlikely.   It would still have regions of darker, warmer rising air and lighter more ice filled, colder falling air.   Jupiter's bands aren't because it's atmosphere doesn't mix, it's because warm Jupiter air is a different color than cold more icy Jupiter air.

Also, no rotation effectively gives the planet a sun side and a night side.    That would vary some depending on no rotation at all, vs a slow tidally locked rotation, but even so, the Sun side would be warmer.    Jupiter's upper atmosphere is more driven by heat from below than heat from the Sun, but the Sun would still have an effect, especially if it only shone on one side of the planet.

I don't think we have a good sense of what non-rotating Jupiter type planet might look like.  The upper atmosphere temperature would be a primary factor, but with no rotation, I think it would probably require some high level mathematical modeling to work out an answer.

Some kind of grand current might be possible, perhaps with patterns perhaps similar to the "Y" shape observable in UV in Venus' atmosphere, or perhaps an ocean current like pattern, but that's just speculation.

https://www.space.com/28835-venus-y-shape-mystery-solved.html


Are red dwarfs really 30-100 times our Sun's density?

Red dwarfs, depending on your definition, can range from 2.5 to 150 times more dense than the Sun.

What is the cause of this discrepancy?

They give no calculations, so I can only guess.

The article is from 1946 and we've gotten a lot better at science.
It's 1946 and information exchange is limited. No internet, no TV, and long distance calls are expensive.
Red dwarfs are tricky to observe and we've gotten a lot better at that.
Their hypothesis is that red dwarfs do not fuse hydrogen which is incorrect.

Edgeworth's other contemporary works might provide insight.
"Red dwarf"
There is no strict definition of a red dwarf. Some use class M stars below a certain temperature and mass. Others include some class K stars. Fortunately they give a definition.

low luminosity (say, not more than one tenth that of the sun), small mass (say, not more than three quarters that of the sun) and high density (perhaps 30–100 times the density of the sun).

Luminosity and mass are basically correct for the modern definition of a red dwarf, M-class, but red dwarf density runs a considerably larger range from 2.5 to 150 times more dense. Still, they're inside the range.
They give no rationale for their density calculations. Being 1946 its probable their mass and radius calculations were considerably off. Red dwarfs, being so faint, are very difficult to observe.
Red dwarfs fuse hydrogen

In the case of stars belonging to the main sequence, it seems to be necessary to assume that the general conditions in the interior of the star are such that convection currents occur which are of sufficient importance to convey hydrogen in adequate amounts from the outer layers to the central core...

Main sequence stars are convective, but it's complicated.

...although this does not imply that the convection currents are of
sufficient magnitude to be the dominating factor in determining the
distribution of temperature.

Incorrect. The lower the mass the more convection is the "dominating factor" over radiation.

...red dwarfs probably contain a normal proportion of hydrogen...

Correct.

...but that convection currents in these stars have never been of
sufficient magnitude to convey the hydrogen in adequate amounts from
the outer layers to the central core...

Incorrect. The lower the mass the more convection. Low-mass red dwarfs are fully convective with no helium accumulating in the core.

When the mass of the star is less than [0.75 M☉], the hydrogen which
the star contains never reaches the core in sufficient quantity to
initiate and sustain the process of transforming hydrogen into helium,
and dynamical equilibrium is secured only by contraction involving
comparatively high densities.

Incorrect. Red dwarfs fuse hydrogen. What they're describing is more like a brown dwarf.

On this theory the red dwarfs are not stars of great age which have consumed their substance in riotous living; they are simply feeble things which have never been able to make use of the supplies of fuel which they actually possess.

Red dwarfs are of great age, but they aren't bigger stars which have run through their fuel. They also are not "feeble things" which can't fuse hydrogen. They are low-mass stars which use their fuel very efficiently and so live to a great age.
With an incorrect model they will get incorrect answers.


Why do stars rotate slower than they're expected to?

There are two phases to this problem.
In order to accrete into stars, a huge amount of angular momentum must be lost to allow so much mass to gather into a small volume.
A second problem is how stars like the Sun end up rotating so slowly, when younger versions of stars similar to the Sun rotate much faster.
The solution to the first problem may be solved by the outward transport of angular momentum in accretion disks. The angular momentum may ultimately be shed by winds originating from the disk surfaces. Another possibility is the loss of angular momentum in jets and bipolar outflows, which are accelerated from the central star-forming core by poorly understood processes. Some of the angular momentum can of course end up in planets.
Stars themselves can then lose angular momentum throughout their lives. Initially this can be through magnetic coupling to the protostellar accretion disk. Later, the principal angular momentum loss process is through stellar winds that become controlled by coronal magnetic fields. The plasma decouples from the co-rotating field at many stellar radii, taking away angular momentum.


Could we detect an earth-like planet around another star?

The short answer is yes, the Kepler mission has used transits to detect nearly "Earthlike" planets, one of the classic examples is Kepler-22b is only about 2x the radius of Earth and has a year that is 290 (Earth) days. Kepler has detected many planets smaller than Earth and closer to the star. With more observing time Kepler could discover similarly sized or smaller planets at comparable orbits, so technologically transits can already detect  "Earthlike" planets. Practically, with Kepler's lost reaction wheel this has been difficult, but more Earth sized planet detections are likely from the K2 Kepler reboot.

Microlensing is more sensitive to planets further from stars, a next-generation space telescope like the WFIRST space telescope, which might launch in the 2020s, expects to be able to detect planets smaller than Earth at slightly larger distances from stars.

Taking a photograph or spectra, or directly imaging, a small planet far from its parent star is much more difficult, light from the Sun is around 10 billion times brighter than the sunlight the Earth reflects (a $10^{10}$ contrast), the sensitivity of current ground based instruments (like GPI) and the Hubble space telescope are closer to $10^7$.  WFIRST project will be be able to detect giant planets and may be able to detect Earthlike planets but the technology is not quite there yet.

Because of instrument stability and stellar variability, radial velocity instruments are not yet sensitive enough to pick out the incredibly small wobbles caused by a small planet in an Earthlike orbit. Future instruments, like the 100 Earths Project may get to this sensitivity in the moderately near future.


What is the acceleration for the expansion of the Universe?

The figure below shows the evolution of the Hubble parameter $H$ from 10 billion years (Gyr) ago, to 10 Gyr into the future:

As you can see, the change in $H$ is modest nowadays, compared to the past. The "acceleration" of the expansion at any point in time is given by the tangent to the curve at that time. Today, the change is $dH/dt \simeq -1.2\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}\,\mathrm{Gyr}^{-1}$, or roughly $-10^{-17}\,\mathrm{km}\,\mathrm{s}^{-2}\,\mathrm{Mpc}^{-1}$.

In other words, it will take roughly a billion years before the expansion rate has decreased by $1\,\mathrm{km}\,\mathrm{s}^{-1}$, and it asymptotically approaches some $57\,\mathrm{km}\,\mathrm{s}^{-1}$.

Since $H$ is defined as $(da/dt)\,/\,a$, where $a$ is the scale factor (the "size" of the Universe), a constant $H$ implies that $a\propto e^{Ht}$; that is, the size of the Universe increases exponentially.


What is the condition for the number density of a gas to be ultra-relativistic or non-relativistic and degenerate or ideal

This is fairly standard bookwork that I am not going to reproduce and you should understand that the lines on your diagram are "fuzzy" in the sense that they mark the loci where you can neither use one approximation or another.

So first, the division between a classical gas and a quantum gas (note that all the gases in your plot are ideal - the term ideal refers to the particles being point-like and non-interacting, which is as true for the electrons in a white dwarf as those in the atmosphere of the Sun). The condition here is that for a classical gas, the density/temperature combination should be such that the phase-space (3d momentum $\times$ volume) occupied by each particle is much greater than $\hbar^3$. This ensures that there are more available quantum states than indistinguishable electrons to fill them.

The second criterion is to compare the momentum of particles with $mc$. Clearly if these are comparable then the particles are relativistic. In a classical gas, the momentum depends on the square root of temperature, as you quote. In a Fermi-Dirac quantum gas, the maximum momentum only depends on the density of particles (as $n^{1/3}$). Hence the differing criteria.


Where is the materials of comet after it crashed into the sun?

It depends on the size of the comet. If it's only $0.1$–$1\,\mathrm{km}$, it will evaporate before even reaching the Sun (Li. et al. 2013; Knight & Walsh 2013), and the released gasses will be ejected by the Sun's radiation pressure. Larger comets, so-called Sungrazing comets, can survive close passages, although they probably will be torn apart by tidal forces.

If the comet comes so close that it crashes into the Sun (I'm not sure this has ever been observed, despite various blogs discussing this, but there's no reason it couldn't happen), it would reach the escape velocity of the Sun, i.e. $\sqrt{2 G M_\odot / R_\odot}\sim600\,\mathrm{km}\,\mathrm{s}^{-1}$. This New Scientist article claims that the impact would cause an explosion and possibly a coronal mass ejection, but the underlying scientific article by Brown et al. (2015) seems to indicate that the large ablation of the comet will result in less dramatic velocities.

Anyway, some of the released gas would stay at the surface of the Sun, just like some of the Sun's "own" gas stays. And since the convection cells, or granules, are quite shortlived (of order 10 minutes; Bahng & Schwarzschild 1961), the gas will soon be engulfed by the Sun, being carried deeply into the body of the Sun.

The convective zone of the Sun is in the outer ~30% of the radius, and there isn't much mixing of material between that and the inner, radiative zone. In the last phases of the Sun's life, however, when core hydrogen burning stops and the Sun becomes a red giant, convection reaches so deeply into the Sun that some mixing takes place. This is called dredge-up. Eventually, roughly half of the Sun will be ejected due to the large mass-loss during the red giant phase, while the other half becomes a white dwarf. If the dredge-up didn't happen, the comet material would all be a part of the ejected material, but due to the dredge-up, some of it will end up in the white dwarf.


Does "spectral type" and "stellar classification" refer to the same thing?

Spectral type is only based on the star's color. You need additional the mass or luminosity to make a stellar classification form it.
Example: Giants may belong to the same spectral class as main sequence stars.
Of course, if you don't simply look at the color but measure the spectra you see the differences. Spectroscopy is one of the most important tools for astronomers.


TL;DR: No. Well, yes, but they are negligible.

Full answer: a conjunction like this 'nearly' happened on May 5th, 2000. NASA published an article about this event, stating:

For example, the combined gravitational effect of all the planets together is much less than the effect of the Sun or the Moon on the Earth.

and

While unusual, such alignments have happened in the past without any consequences. The planets are simply too far away to have an effect on anything here on Earth - except our imaginations.

For hard numbers, see this article by Truman Collins which is linked there:

Conclusion
What I discovered was that the tidal forces exerted on Earth from the five planets on the other side of the Sun are vastly smaller than those exerted by the Moon or the Sun. In fact, with the planets in the position they will be in on the big day, the largest tidal force from one of them will be from Jupiter, and it is about one five hundred thousandth the size of the tidal force exerted by the Moon on an average day!


Why is dark energy density given in grams per cm$^3?$

This is a case of using a sort of natural units (similar to geometerized units), which are often convenient in relativity (special and general). Take, for instance, the Minkowski metric of special relativity. In dimensional units, it is
$$\mathrm{d}s^2=-c^2\mathrm{d}t^2+\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2$$
Setting $c=1$ makes it a bit more convenient, and gives time and space the same units - which is more intuitive, if you think of time as another dimension.

You also brought up $E=mc^2$. This is a special case of the energy-momentum relation:
$$E^2=p^2c^2+m_0^2c^4$$
Setting $c=1$, we have the much more convenient form
$$E=p^2+m_0^2$$
This comes up again and again. Add to this the mass-energy equivalence that is also important in relativity (especially the right-hand side of the Einstein field equations), and it simply makes sense to treat energy and mass as the same thing. In fact, it becomes perhaps even more useful than in special relativity, because we also often set $G=1$ - and that comes up a lot in general relativity.

Now, the units of density under this system are $\text{Length}^{-2}$, and the density above is clearly not in those units. I'm guessing that this just isn't a case of setting $G=1$ - and that makes sense, because in cosmology and astronomy, many types of densities are measured in g/cm3: dark matter, interstellar gas and dust, and sometimes dark energy. Keeping it in $\text{Mass}\cdot\text{Length}^{-3}$ makes it easy to compare these, working in a system that is more commonly used.

In a nutshell, then: Cosmologists sometimes like $c=1$ (and sometimes $G=1$) because it makes a lot of equations a lot easier to write out. Here, it becomes really easy to compare mass densities and energy densities.


What resources do C-Type(Carbonaceous) Asteroids contain?

Carbonaceous asteroids contain carbon in the form of chondrites (small marble-sized nodules),tiny crystals and other sorts of carbon dust. This is mixed in a sort of conglomerate with dust and nodules if silicate material, sometimes even nickel-iron nodules. What resources do they contain? They don't contain diamonds, because diamonds require enormous heat and pressure to form, so sometimes micro-diamonds are found in the craters of large meteorite impacts on Earth. Some of them probably contain valuable materials, but they would have to be very valuable indeed to be worth the trouble and expense of retrieving them from the asteroid belt. Prospectors there wouldn't restrict their attentions solely to carbonaceous types, and would probably explore large asteroids like Ceres. Asteroids and meteorites couldn't possibly contain coal, which is a biological product formed from thick layers of plant debris transformed by heat and pressure over millions of years in the bowels of the Earth.


How space scientist keeps the track of the stars?

Scientists make lists of stars in a catalogue. In the past this would be in a book nowadays it is more likely to be a computer database.

The stars don't move (much) so it is possible to make a list of them by position Modern catalogues are online (see SINBAD older ones were on paper (see the 1918 Henry Draper catalogue)

The Henry Draper catalogue is typical and easier to follow. It is just a table, the first column is the catalogue number (used instead of a "name")  the second column is a cross reference to another database. Then there is information on the location of the star in the sky using the RA and Dec system. Next comes information on the brightness of the star (estimated in two ways). Some letters come next to describe the spectrum (or "colour") of the star and finally some short "remarks", and a "plate number (the number of the photograph that contains an image of the star"

Scientists need to be systematic and careful to avoid missing anything.


Items at the Event Horizon vs. colliding Black Holes

I don't know if this question is a good fit for the Stackexchange format, because it's a complex topic that would require a lot of discussion. But I'll try to provide an answer.

I think you're getting the wrong ideas from the image of the event horizon being frozen in time for the remote observer. While this is true, it also means the remote observer never actually gets any information from the event horizon proper.

If you're a remote observer, and you see any change that happens related to a black hole, by definition that change is not happening at the event horizon or inside it. It happens outside that region. The event horizon might as well not even exist from this p.o.v.

The merger of the two black holes is what the remote observer sees. But a hypothetical local observer, inside the event horizon, would have a different perception.

For the local observer, the whole history of anything outside the black hole takes place in an instant. The whole evolution of the universe, including the black hole colliding with another black hole, and the resulting object evaporating via Hawking radiation - all this stuff happens in an instant.

You really need to think of it this way: the remote and the local observer experience two completely separate realities, where the flow of time in one is completely disconnected from the flow of time in the other.


Using basic circular maths: where $d$ is the distance of the star from the observer in AU, and $r$ is the star's radius in AU, and $a$ is the angle encompassed by the radius of the star in degrees:

$
r = \frac{a}{360}2\pi{d}
$

Now rearrange it to make $a$ the subject:

$
a = \frac{180r}{\pi{d}}
$

To get $a$ in arcseconds, you need to multiply the result by 3600 (because there are 3600 arcseconds in a degree):

$
a = \frac{648000r}{\pi{d}}
$

Now, convert your numbers from cm and parsecs to astronomical units:

$r$ = 3.18e13cm = 2.126 AU

$d$ = 220pc = 4.538e7 AU

Put them into the equation:

$
a = \frac{648000\times2.126}{{4.538\times10^7}\pi}
$

To get:

$
a = 0.00966327′′
$

Hope that helps!

Edit:

As Mike has pointed out, the final equation I've ended with can be further simplified if you use different units, to:

$
a = \frac{r}{d}
$

where $r$ is in AU and $d$ is in parsecs, giving an answer $a$ in arcseconds.

This is because the equation $a = \frac{648000r}{\pi{d}}$ has $\frac{648000}{\pi}$ in it - which is the definition of a parsec in AU.  Therefore, multiplying $d$ by this number to convert it from AU to parsecs, we get $a = \frac{648000r}{648000/\pi\times\pi{d}}$. Cancel out the pi on the bottom, and cancel the ${648000}$ on the top and bottom, and you're left with $a = \frac{r}{d}$ where $r$ is in au and $d$ is in pc.


Mass.

The more massive a body, the larger the gap between its lowest and highest orbit; the range of speeds at which a random body entering its gravity is likely to remain as its satellite. Sun has millions of satellites if you count all the asteroids; smaller planets tend to have one or two moons at most (Pluto with five being a notable and not fully explained exception)

To a lesser degree there's a matter of shape too. A regularly round body will have more regular and stable orbit than a potato-shaped one. Jupiter, being a gas giant is perfectly round. This doesn't play that much of a role though, especially with higher orbits.

And last but not least, no destabilizing influence of other bodies. It's very hard to maintain a lunar orbit - artificial satellites around our Moon last only a couple years each, because relatively close neighborhood of Earth tends to destabilize orbit of anything orbiting the Moon. Jupiter being a single massive planet with relatively tiny (relatively to its mass) moons doesn't have them influence each other all that much.


When a fast-moving star goes supernova, how will the resulting remnant evolve?

A spherical supernova in free space will be a sphere no matter the velocity of the star (ignoring relativistic flattening). Just consider the explosion in the rest frame of the star.
If there is a medium a lot depends on the actual fluid dynamics, which can get rather involved (especially since there is plasma and magnetic effects). The initial shock front moves faster (>10,000 km/s) than the local speed of sound and would be spherical, but soon Rayleigh-Taylor instability sets in and things curl up. Still, momentum conservation means that a particle ejected "upwind" will be slowed a bit more than a particle going "downwind".
However, the relative velocity difference is small (400 km/s) compared to the ejecta velocity so the effect is going to be small in the early stages. It will begin to matter once the ejecta has slowed to such velocities, about the time it has swept up its own mass ahead of it. However, at this time the hydrodynamic instabilities will also have turned the sphere into a fractal. It might well be roughly elliptic with the star at one foci, but the overall shape is still likely close to spherical.


Are any galaxies partially in the observable universe?

Perhaps, on rare occasions this will occur.  Quoting from Phil Plait in an  astronomycast ,

  Dr. Plait:    Yeah, because the universe is expanding ever faster every
  day, basically the distance we will be able to see, away from the
  earth, is going to get closer and closer to us. Anything past the
  distance will be moving away from us too rapidly to see and so
  eventually, all that is going to be left is some sort of super galaxy
  that is us and the Andromeda galaxy and all of the little dwarf
  galaxy’s, will have all combined into, sort of, one thing. Of course,
  all of the stars will be dead and then that’s it.

  So, you know, if you are a civilization, somehow, that evolves a
  hundred trillion years from now, the universe is going to look way
  different than it does now. You are not going to see any other
  galaxies; it’s just going to be you.

  Frazier:  You are not going to see any stars, because it will all be
  dark.

  Dr. Plait:    That’s right and for a long period of time that is all
  there is going to be, it’s just going to be that. Even on the scale of
  what we are talking about, we are talking about a long period of time,
  now instead of an octillion years, which is what, ten to the twenty
  seven, now we are starting to talk about ten to the fifty and sixty
  years, so far into the future, that the present age of the universe,
  13 billion years, is so small, that I don’t even have, ten to the
  sixty divided by ten to the thirteen is ten to the forty seven, that
  is such a huge number, that there is no ratio that exists today.

On those expansion scales, a galaxy is small (!!) enough that it'll pretty much disappear all at once.


What effect does a natural satellite have on a planet's rotation and revolution?

A body and its satellite in truth form a system and both rotate around their common centre of gravity. Hence both the Earth and Moon rotate around this centre - however because the Earth is significantly more massive than the Moon this can be approximated as the Moon orbiting the Earth.

The Moon does have a measurable impact on the Earth - tides are the obvious example, and tidal friction is also slowing down the Earth's rotation by a very small amount (around 2.3ms per century).

In the Moon's case, though, the Earth's gravitational pull has already effectively locked its rotation period with its orbital period around the Earth - so we essentially only ever see the same face of the Moon on Earth.

The Earth and Moon are much closer in mass than most other planet/satellite systems in the Solar System - think of the huge mass of Jupiter - and so it is difficult to show how these planets' satellites have a similar impact on the planet - but the same effects are there, albeit in much smaller scale.


How much helium has been created since the Big Bang?

We can estimate the answer by considering the "mass to light ratio" for baryons, and assuming that ratio hasn't changed too much over the time that there has been galaxies, say for ten billion years.  The Wiki (https://en.wikipedia.org/wiki/Mass-to-light_ratio) on the mass to light ratio says this ratio for the universe as a whole is about 100 times what it is for the Sun right now, but that includes dark matter.  If we restrict to baryon mass-to-light, that would be more like 16 times the Sun right now.  The Sun's main-sequence lifetime is about ten billion years, conveniently, and in that time it will convert perhaps 20% of its hydrogen, so a mass-to-light ratio 16 times larger would convert only about 1% of its mass from hydrogen to helium.  So that says roughly 1% of all the hydrogen in the universe has already been converted to helium.

Since it started out 75% hydrogen and 25% helium by mass, now it should be more like 74% hydrogen and 26% helium by mass.  So although the estimate is crude, I think it's fair to say we are just now beginning to see a trace of an effect, which would be hard to see after only 1 billion years, and will be quite obvious after 100 billion years.


Is star a necessary phase in black hole lifecycle?

There are speculated objects known as "primordial black holes" which could have been present at the origin of the universe, or formed in its early stages via presently unknown processes, but there is no observational evidence that such objects exist. Also, some models for forming supermassive black holes involve the coalescence of such a huge amount of matter that you would not call it a star, unless some kind of supermassive superstar.

Alternatively, the mass could form stars first, and then coalesce into the supermassive black hole, but those stars themselves would not necessarily need to collapse first. Indeed, since much of the mass that goes into stars goes into lower mass stars that don't make black holes, it seems likely that a significant fraction, or even a majority, of the mass in supermassive black holes was not in a black hole prior to falling into the supermassive black hole.

Finally, since we don't know much about dark matter, one presumes it might be possible for dark matter to form black holes via some presently unknown mechanism, but there is no evidence dark matter would do that, and the idea that dark matter might itself be already in the form of black holes has not been supported by observational searches.

So I'd say our current expectation is that supermassive black holes are largely made of matter that did not collapse in a massive star, and stellar-mass black holes could be mostly or entirely matter that did.


What are the large round dark "holes" in this NASA Hubble image of the Crab Nebula?

I think my deleted answer to your previous question covers this well, so I'll add it here.
These two spots are known as the east and west bays of the Crab Nebula. They appear to be the result of a torus partially encircling a section of the nebula. The pulsar's magnetic field interacts with the gas and dust in the torus, which blocks synchrotron radiation from gases in the nebula. It has been conjectured that the torus is the result of a $\sim0.1M_{\odot}$ ejection of material from a circumstellar disk prior to the supernova, based on studies of the proper motion of gas on the edges of both bays.
References

Fesen et al. (1992)
Hester et al. (1995)

We can also rule out that this is the result of a defect in Hubble, because these features have been observed in different wavelengths, including infrared, visible, ultraviolet, and x-ray (though not in radio or gamma-ray wavelengths):

Image courtesy of Wikipedia user Hunster under the Creative Commons Attribution-Share Alike 3.0 Unported license.
The infrared, ultraviolet, and x-ray images come from the Spitzer Space Telescope, the SWIFT observatory, and the Chandra observatory, respectively.