Can automatic algorithms completely eliminate the impact of Starlink and other satellites?

No, automated algorithms won't ever be able to completely eliminate the effect of passing satellites.

Some of the light reflected from the satellites into the atmosphere will then scatter and cause some amount of light pollution.  The pollution isn't limited to just in the direction of observing.  See: https://photo.stackexchange.com/questions/41722/can-you-photograph-the-milky-way-with-a-full-moon-out.

Whenever automated algorithms remove the presence of satellites from images, some of the underlying data will be removed as well.

As the number of satellites increases, eventually it will render some data for stellar occultation light curves unusable as the occultations from satellites coincide with the occultations from asteroids.  See: Is 486958 Arrokoth (2014 MU69 aka Ultima Thule) the only solar-system object determined to be binary by occultation?.  This particular effect will occur even if the satellites have zero reflectivity.

Elon Musk says:

I am confident that we will not cause any impact whatsoever in
astronomical discoveries, zero [...] That's my prediction, we will
take corrective action if it's above zero.

But he is confident in an impossibility.  We can minimize the effect by reducing satellite reflectivity, but we can't make the effect zero, even with advanced algorithms.
Note: The above list is not meant to be comprehensive.


total energy and apparent bolometric magnitude

You have been given a terrible question and your confusion is totally justified.

The magnitude system is a measure of the brightness of objects which is a measure of the energy received per unit time interval.

Without knowing a characteristic timescale over which the energy is released in each case, then it is not answerable. Note that the timescale for energy release in the case of supernovae and bombs are different by many orders of magnitude - a type Ia supernova has a bolometric light curve (e.g. see Fig.6 of Scalzo et al. 2014) that lasts 2-3 weeks, peaking at around $10^{36}$ Watts.

Given this, we can compare with the Sun's bolometric luminosity of $3.8\times 10^{26}$ Watts and apparent bolometric magnitude of -26.8, to estimate that the peak apparent bolometric magnitude of a type Ia supernova at the distance of the Sun would be -50.3.

You cannot do the same thing with nuclear explosions. They last a fraction of a second (in terms of their light output). Or are you meant to assume that you take the average flux over the 70 years that mankind has had nuclear weapons?? The question is a nonsense unless you are given some sort of timescale over which this explosive energy is released. About the only sane things you can do are (a) use ~70 years (actually, most of the energy was released during the 1950s and early 1960s) (b) use the duration of a single nuclear explosion or (c) (and I imagine what was intended) use a similar duration as for a typical type Ia supernovae so that the question should have begun - "Imagine that X and Y amounts of total energy are emitted over a similar timescale..."


No. Basically, there is no such thing as a stable orbit in a realistic solar system.  All planetary orbits are unstable due to gravitational interactions between planets, but how long it takes the instability to make a big change in a planet's orbit varies a lot.  It can be a few of the planet's years, thousands, millions or billions. (Earth's orbit has been stable, so far, for 4+ billion years.)  But stability going forward can't be guaranteed beyond at most a hundred million or so years.

The main source of instability is resonances. A resonance is where two planets are in orbits where the ratio of their periods is a simple fraction like 1/2 or 2/3 or 5/3 or whatever.  Because of the simple relationship between periods, planets repeat the same configuration in space repeatedly, and gravitational interactions can build up. (Without the resonance, the interactions are just as strong, but in random directions and thus mostly cancel.)

Resonances can disrupt an orbit or can stabilize it. The Hilda group of asteroids are in a stable 3:2 resonance with Jupiter, for example, and Pluto is in a 2:3 resonance with Neptune.  Both of these resonances act to stabilize the orbits.  But the Kirkwood Gap in the asteroid belt is a 2:1 resonance with Jupiter which tends to keep the gap empty by disrupting the orbits of any asteroids that stray into the gap.

After the interaction, Earth will be in a new orbit and that orbit may also be stable over an extended period or it may be one where interactions with other planets quickly render the new orbit unstable.  There's no way to tell without detailed orbital information on the encounter.

In any event, there is no restoring force to push Earth back to its old orbit.


How difficult is adaptive optics computationally?

If you're only trying to correct a small field, using a single guide star (or a single laser guide star), then the requirements aren't very arduous. The MACAO adaptive-optics module used on the Very Large Telescope at the European Southern Observatory dates from the early 2000s and is based around two 400 MHz PowerPC 604 chips (one for the "supervisory computer", the other for the "real time computer"). You can read more about it here (PDF file).

For next-generation multi-conjugate adaptive optics, which tries to correct a larger field using multiple guide stars, the requirements are bit more stringent: my impression is that you need several GFLOPs in the real time computer (some of this may be handled by using specialized DSP chips instead of general CPUs), and additional processing power for external monitoring and analysis.

So, crudely speaking, your smartphone could probably handle single-guide-star AO (except it would need to be re-engineered a bit and running a real-time operating system, and you'd need an extra computer for external monitoring); for multi-guide-star AO, you'd end up needing the equivalent of a powerful (multi-CPU) server or (small) cluster.


Could S5-HVS1 conceivably still hold on to any planets?

According to Ginsburg, Loeb & Wegner (2012) "Hypervelocity planets and transits around hypervelocity stars", planet-hosting hypervelocity stars could form from the disruption of a stellar binary by a massive black hole, and the probability that the planet remains bound to the hypervelocity star is fairly high for suitable initial conditions.

In their simulations, this required initial planetary separations of 0.02–0.05 AU. The star S5-HVS1 is described in Koposov et al. (2019) "The Great Escape: Discovery of a nearby 1700 km/s star ejected from the Milky Way by Sgr A*" as a 2.35 solar mass A-type star. Planets around similar stars include ultra-hot Jupiters such as KELT-9b, which may at first appear promising.

The initial binary separation is another matter. The systems Ginsburg et al. investigated had binary separations in the range 0.05–0.5 AU. For the S-type planetary systems considered (the planet orbiting one of the stars as a satellite), this is a much smaller separation than any known planet-hosting binary: the planets known in such systems are in circumbinary systems which the paper does not explore.

It's therefore unclear whether the kind of progenitor system required would be capable of forming planets in the first place.


There is definitely not a catalogue of all meteorites hitting Earth. For instance, the ones falling in desert areas and in the ocean aren't found, and even the ones falling in more populous regions are easily mistaken for normal rocks.

Meteorite rate

Meteorites come in all sizes, from sand grains to dinosaur-annihilating rocks.
The size distribution of these rocks (as well as most other things in space) follow an approximate power law, meaning that the smaller the meteorites you consider, the more often they hit Earth.

Estimating the total number of meteorites hitting Earth is difficult, but one study (Bland et al. 1996), looking at the composite of the earth on three arid site and considering the weathering over time, finds that between 36 and 116 meteorites over 10 g hit Earth every year per $10^6\,\mathrm{km}^2$. Multiplying by the surface area of Earth $A_\oplus$, that's roughly 20,000 to 60,000 meteorites hitting Earth per year.

Meteor rate

UPDATE: You want to know the rate of incoming meteors, i.e. all rocks that hit Earth's atmosphere. By definition, a meteorite is a meteor that lands on the surface of Earth. Before it enters the atmosphere and becomes a meteor, you can call it asteroid, meteoroid, rock, pebble, or dust grain, depending on its size and your mood.

Taylor et al. (1996) have monitored incoming meteors with the AMOR radar in New Zealand and detected in one year 350,000 trails from incoming meteors of size 10–100 $\mu$m vaporizing in ~100 km's height. I don't know exactly how large an area AMOR monitors, but from simple trigonometry, and as an order-of-magnitude estimate, if the events happen in $h \sim 100\,\mathrm{km}$'s height, the radar can "see" a distance $d=\sqrt{h^2+2R_\oplus h} \sim 1100\,\mathrm{km}$, and thus covers an area $A\sim4\times10^6\,\mathrm{km}^2$. Thus, again multiplying by $A_\oplus$, this amounts to the order of 40 to 50 million meteors per year.

Killer meteor rate and the Sudan asteroid

The meteors discussed above is probably not what you had in mind. As mentioned, the bulk of these "rocks" are actually more like sand grains. So what is the rate of devastating killer-rocks from space? Brown et al. (2002) fit a power law to several different data sets (LINEAR, Spacewatch, NEAT and others), and find that the number $N$ of meteors above a diameter $D$ in meters is roughly
$$
N(>D) \simeq 37 D^{-2.7}.
$$
The following plot is from their paper, with the long black line showing their fit, and the short black lines indicating the uncertainty:

You mention the asteroid 2008 TC3, which had a diameter of 4.1 m.
According to the fit, such an event happens roughly $1/(37[4.1\,\mathrm{m}]^{-2.7})\sim$ once per year.

You can also see that the largest meteor hitting Earth in a century would be $(37\times100\,\mathrm{years})^{1/2.7}\sim20\,\mathrm{m}$, equivalent to $\sim500\,\mathrm{kiloton\,TNT}$.
Similarly, hundred-meter-sized objects ($\sim1\,\mathrm{megaton\,TNT}$) should hit us roughly once per 10,000 years. By extrapolation of their fit beyond observational data, objects of $D>1\,\mathrm{km}$ — or 100 gigaton TNT — should hit Earth on average once every ~3.4 million years.


Most of the absorption line comes from the photosphere in the region where the temperature is still dropping.  Nevertheless, you can also get an absorption line where the temperature is rising, due to scattering effects.  The Kirchoff-Bunsen law you describe is a simplified rule that assumes the lines form in thermal equilibrium with the local temperature, but sometimes scattering produces non-thermal-equilibrium ("non-LTE") effects.  With scattering, line photons that get absorbed get re-radiated right away without coupling to the local temperature, and their re-emission has a 50% chance of being radiated back downward, where they will be reabsorbed deeper in the atmosphere.  This contributes to absorption, especially near line center (and can cause a feature called "central reversal" there).  But it is true that you can also get emission lines from very strong chromospheres where the Kirchoff-type effect does occur, see the profiles from chromospherically active dwarf stars (called "dMe stars").


Could Dark Energy be a "Cosmic Gravity Background"

There is in fact a cosmic gravitational wave background. These waves are expected to be stochastic, having originated in the early universe (much earlier than the cosmic microwave background). Random fluctuations were subsequently stretched during inflation, making them observable over many wavelengths. A good and reasonably up-to-date introduction I read is Lasky et al. (2016), which gives some of the most recent constraints on the parameters of the frequency spectrum $\Omega_{\text{GW}}(f)$. The great thing about the wide spectrum is that a number of different tools (pulsar timing arrays, interferometers like LIGO, etc.) can be used to constrain different parts of it, both directly and indirectly.

That said, dark energy is almost certainly not a result of the gravitational wave background. First, there's no theoretical mechanism to explain how such an effect would arise. Second, the background is probably not strong enough to have an influence on the current expansion of the universe (see the graph here to see how weak it is). Finally, dark energy is a relatively recent phenomenon, arising several billion years ago. The background was around long before then, and yet had no effects.


Why do satellites arcs end abruptly when observed from Earth

Because satellites are only visible when they are in sunlight, they are not visible when they go into the Earth's shadow. The app most likely predicts where this occurs and ends the arc.

In other words, it does not make sense for an observer to look for a satellite when it is not visible, so there is no need to draw the path when it is in the shadow.


As Andy pointed out, the spiral galaxy is IC 1296. You're correct that it's not in the Messier Catalog, but it's obviously in the revised Index Catalogue. NED and SIMBAD have some more information on it:

Morphological type: SBbc, as classified according to the Hubble sequence. It's a spiral galaxy with two clearly visible arms.
Redshift: 0.017075 $\pm$ 0.000070, according to Marzke et al. (1996).
Distance: 66.6 Mpc to 68.9 Mpc after corrections, depending on your distance definition of choice (angular diameter, luminosity, etc.). Large uncertainties arise when calculating the difference from redshift. For comparison, Andromeda is approximately 0.78 Mpc away, and the Ring Nebula is $10^{-3}$ times that distance.
Apparent magnitude: 14.80 $\pm$ 0.30, at visible wavelengths. For comparison, the Ring Nebula's apparent magnitude is 8.8.


Well, there are three parameters of interest to be evaluated to get intensity of sunlight.

Sinusoidal variation of sun vector on earth's surface througout the year
The latitude variation. Which is also sinusoidal
The time variation. Which is also sinusoidal

$$ Intensity = A_0 \sin(\theta - \theta_0) \sin(\pi/2(1 - |t-12|/6))$$

Where,

$$ \theta_0 = A_1\sin(2\pi|d/365.25|)$$

Here $d$ is the day and starts from vernal equinox. $t$ is the time from 00:00 - 23:59. $A_0$ and $ A_1$ are constants. Also when Intensity is negative, obviously there it means darkness.

This is very crude approximation. It assumes circular orbit.


Why is the flux density and amplitude different for galaxies than stars?

A galaxy is obviously going to be much more distant than the stars. On the other hand the galaxy contains a lot of stars which will contribute to the overall spectrum.
Whether the galaxy or a star will be brighter will depend roughly on the ratio of $N/d^2$, where $N$ would be the number of stars and $d$ would be the distance to the galaxy.
What your plot has done is empirically remove this factor by linearly scaling the galaxy spectrum so it matches with the stellar spectrum over most of the wavelength band shown.
On the other hand, the light from a galaxy is not only made up from the stars that are within it. There are contributions to the light from (for example) hot gas near any active galactic nucleus and also relatively hot gas in the interstellar medium, possibly associated with bursts of star formation, supernova remnants and nebulae ionised by hot stars. These latter contributors are responsible for the emission lines seen in the galaxy spectrum.


What's the name of outside the cosmic horizon?

The outside is also the universe, the inside is just the observable universe.


How can a rotation period of a planet's satellite/moon be "chaotic" (like Hyperion around Saturn)?

According to wikipedia (see info below image) Hyperion's rotation period around Saturn is "chaotic". What does that mean?

The wikipedia article refers to Hyperion's rotation being chaotic rather than its orbit about Saturn. I'll assume you are asking how Hyperion's axial rotation can be chaotic. (Its orbit is not.)

An object that is not subject to any external torques will have a constant angular momentum. This does not mean that such an object will have a constant angular velocity. Place a rubber band about a book and toss it in the air, giving it a nice spin rate before letting it go.

See the question Stability of rotation of a rectangular prism for details.

Hyperion is small enough and rigid enough that it's not quite capable of pulling itself into a roundish shape by its self gravitation. It instead has a markedly triaxial shape. Designating the principal moments of inertia as $A < B < C$ (corresponding respectively to the rotations in subfigures b, c, and a in the above image), the longitudinal libration parameter $(B-A)/C$ is about 0.21 for Hyperion. That's huge, and is a key factor that makes Hyperion's rotation chaotic. If Hyperion was a perfectly rigid body that wasn't subject to external torques and was instantaneous rotating about an axis close to that depicted as c in the above figure, it's rotation would essentially be chaotic.

None of the above conditions is true. Hyperion is not perfectly rigid and it is subject to external torques. Almost every regular moon in the solar system exhibits synchronous rotation and revolution. This is a result of tidal torques on the moon by the parent planet and the moon exhibiting a bit of plasticity (non rigid behavior) in response to those torques. The torques coupled with the plasticity are the mechanism by which those moons eventually became tidally locked.

There are other possible spin:orbit couplings than the 1:1 coupling exhibited by those tidally-locked moons. Mercury, for example, is gravitationally locked with the Sun in a 3:2 spin-orbit resonance. Mercury has a rather low longitudinal libration parameter of about $10^{-8}$. The high eccentricity of Mercury's orbit together with its nearly spheroidal nature makes it more likely to have been captured in the observed 3:2 spin-orbit resonance than in a 1:1 resonance.

With most moons and planets/exoplanets that orbit sufficiently close to the parent planet or star, the torque, plasticity, and low longitudinal libration parameter will eventually result in the orbiting body slipping into one of those small integer resonances. This is very unlikely to happen with Hyperion thanks to that very large longitudinal libration parameter. When Hyperion's spin and orbit happen to get close to a small integer ratio, the torque from Saturn is more likely to kick the moon into a rotation that is further away from the two stable axes than toward either one. It tumbles away from the resonant spin.


What is the right ascension overhead at Greenwich at noon on 21 march

At the vernal equinox, the sun crosses the celestial equator, and by definition, it has an RA of 0h (at the current epoch).  At mid-day on that day, the sun will be more or less due south, and as the line of zero RA forms a great circle, the RA "overhead" will also be zero.

The sun won't be exactly zero RA except at the moment of the equinox, and also the length of the solar day varies throughout the year, which results in the time at which the sun is due south is not quite 12 noon by clock time.

All the above is independent of whether or not you are observing from Greenwich.


Has a near earth object in heliocentric orbit ever been bright enough to be visible to the unaided eye?

The Near-Earth close approches website shows close approaches to the Earth by near-Earth objects (NEOs). The table showing all close encounters indicates the absolute magnitude.

The data can be exported to a CSV file to estimate the apparent magnitude for each object, using the following equation.

$$
m = H + 5 \log_{10} \bigg( \frac{d_{BS}d_{BO}}{d_0^2} \bigg) - q(\alpha)
$$
where $H$ is the absolute magnitude, $m$ is the apparent magnitude, d_{*} are the distances between the objects and $a(\alpha)$ is the reflected light. $q(\alpha)$ is a number between 0 and 1.

I only want to know what happens when the object is closest to the Earth, so I use the approximation that the distance from the Sun to the NEO is 1AU.

$q(\alpha)$ is complicated to compute, so I just compute $m$ using $q=0$ and $q=1$. This leads to

min value $ = H + 5 \log (d_{BO}) - 1 < m < H + 5 \log (d_{BO}) = $ max value

with $d_{BO}$ the distance between the Earth and the NEO expressed in astronomical units (AU).

The server is unhappy when I try to get the entire database, so I limited my export to the objects that come reasonably close to Earth (d<0.05 AU), with no time limit.

Among these 24588 objects, 4 have a maximal magnitude less than 6, and 16 have a minimal magnitude less than 6. So between 1900 and 2200, no more than 16 NEOs are visible by the naked eye.

In particular, 99942 Apophis (2004 MN4) has an apparent magnitude between 1.7 and 2.7 based on these estimates. Its close approach date is April 13 2029.

But this doesn't say anything on NEOs from before 1900 or after 2200.


Could two stars collide without a resultant explosion?

Are there any scenarios in which two stars could collide, and simply fuse without triggering a supernova

As said in the comments, binary evolution of stars is very uncertain, but there exist numerous codes (i.e. MESA, STARTRACK, BSE, etc...) which incorporate LOTS of physical processes over the course of millions of years (and if they evolve to black holes billions of years) of binary evolution. Ono Pols has a nice textbook free on his website here.

Stellar binaries usually experience at least a couple mass-transfer (MT) phases after the binary components fill their Roche lobes. To put it simply, MT is either stable or it is unstable, but either way it results in the donor star losing it's envelope so that it's core emerges (which will later turn into a compact object). In stable MT, the donor star transfers some mass to its companion and thusly loses rotational angular momentum (it's spin decreases) while the accretor's spin increases. In unstable mass transfer, the mass transfer rate is too much for the accretor to handle so that the binary is engulfed by the envelope, this is also called a Common Envelope (CE). The CE results in either 1) the envelope is ejected from the binary resulting in a drastic decrease in the binary separation, or 2) the binary merges leaving a single star which accretes the leftover envelope.

So, I think the case 2) is an answer to your question. Keep in mind that CE is very very uncertain, as Postnov explains, and it is a relatively quick process (less than 1 million years). It's worth noting that case 1) of CE is crucial for forming compact binaries that have small separations so that they merge within the lifetime of the universe (so we can measure they're gravitational waves).

For your nutrition, here in Postnov's review on compact binary evolution the section on CE is lovely (section 3.6 on page 35).

Of course, if you're willing to include black hole binaries in your question, then their merging due to the emission of gravitational waves is certainly a calmer scenario.


Do Enceladus' geysers fall back to its surface or do they achieve escape velocity?

We start by calculating the moon Enceladus surface escape velocity $v_e$ as
$$v_e=\sqrt{\frac{2GM}{r}}$$
where $G = 6.674×10^{-11}m^3kg^{-1}s^{-2}$ is the universal gravitational constant, $M=1.08×10^{20}kg$ is the moon mass, and $r=2.52×10^5m$ is the radius of the moon. Evaluation of the above equation gives us an escape velocity of about $239m/s$.
Perry et al. provide high fidelity observations of Enceladus vapor plumes from the Cassini spacecraft.

The above graph is just for the high velocity plumes as described in the paper:

The H2O neutrals measured during E8 had a Mach-4 distribution centered
on 1.2 km/s with a width of  ±300  m/s,  which  corresponds  to  a
temperature  of 65K

Since the average high speed plume velocity of 1.2km/s is about five times greater than our calculated escape velocity of 239m/s, we can conclude that some of Enceladus's plumes certainly achieve escape velocity.
Teolis et al. agree:

The expression assumes radial expansion of the gas from the surface
sources at constant speed, neglecting gravity since the mean molecular
speed in the jets significantly exceeds (by at least a factor two) the
240 m/s Enceladus escape speed.

Enceladus is losing mass through it's (mostly water) jets. I couldn't find a conclusive analysis of mass loss rate. Eventually I would expect that enough water will be lost that the current mechanism for creating the jets under the 10 km ice crust will no longer be in effect, but this is just my intuition.  I don't have a credible source or calculation to back this up.


Why was the size of the solar system not defined by Mercury Transits?

It is indeed possible to measure the AU using transits of Mercury, and Edmund Halley tried to do just that in 1677.  However, there are two advantages to a transit of Venus.  The first is that during the transit, Venus is only 0.28 AU from Earth, whereas Mercury is about 0.7 AU away.  This makes the parallactic effect twice as big.  The second difficulty is that you need to measure the time between second contact and third contact (i.e., the first and last time that the planet is entirely in front of the Sun).  Mercury is so small that it is hard to determine when second and third contact are.  (In fact the limiting effect for the measurement of the AU during the transit of Venus was the so-called "black drop effect", which made it impossible to measure the times of second and third contact to a precision less than about ten seconds.)

A former professor of mine from when I was in graduate school has a really excellent writeup of the history of these measurements:

http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/venussun.html


Is there anything like a timelapse showing the Alpha Centauri stars orbiting around each other?

Unfortunately, I couldn't find any infos on this specific case.
But as I remember, there was a similar study from NASA a while ago, observed with the Hubble telescope:
https://hubblesite.org/contents/news-releases/2013/news-2013-22.html#section-id-2
There is a nice animation that show the proper motion of Proxima Centauri relative to background stars (not specified) between 2000 and 2010. The attempt was to detect the presence of planets around Proxima Centauri.


When will the last total solar eclipse occur?

I will take a stab at it.  Assuming

the last eclipse happens when the moon is at perigee and the earth is at aphelion.
the moon's perigee is increasing by 4 cm per year,
the earth's aphelion does not change with time,
the sun and the moon are perfect spheres,
the radius of the sun and moon don't change, and
a total eclipse occurs when $r_{\mathrm{sun}}/d_{\mathrm{sun}} < r_{\mathrm{moon}}/d_{\mathrm{moon}}$ where $r$ is radius and $d$ is distance,

I get the following results:

The last total eclipse at sea level will occur in 721,587,917 years at sea
level  (obviously too many significant digits).
The last total eclipse at mount Everest will occur in 721,807,917 years.
The difference is 220,000 years = (height_of_Everest) * 25000 years/km.

Mathematica Source Code:

dSun = 152097000; rSun = 695508; dMoonNow = 357347; rMoon = 1737;
rEarth = 6371; hEverest = 8 + 8/10;
sol2 = Solve[ rMoon/(d + t 4/100/1000) == rSun/(dSun - rEarth), t][[1]];
Print[ t /. sol2 /. d -> dMoonNow - rEarth  // Round];
Print[ t /. sol2 /. d -> dMoonNow - rEarth - hEverest // Round];


The fact that the perihelion is close to the the winter solstice is purely coincidental. The perihelion might occur in any season. It wanders around in a somewhat chaotic fashion because of the gravitational influence of the other planets. At the moment the perihelion moves a day every 58 years.

Let A = Earths axis, and E = the line perpendicular to the orbital plane. The angle between A and E is called the axial tilt. Let S = the line between the centers of the Earth and the Sun. Let P = the plane containing S and E.

Q1 is (maybe) asking what is the angle between A and P on Jan 2. It will be about 23.4 times sine(2*pi*12/365).  Because there are 12 days from Dec 21 to Jan 2.

Q2. yes.


Assumption of our universe being the surface of a 4-D sphere in order to describe relative uniformity in the cosmic microwave background?

To a high degree of accuracy, the cosmic background radiation we observe is homogeneous and isotropic... except for the CMBR dipole anisotropy, which is there because we are moving at $369\pm 0.9\;\mathrm{km/s}$ relative to the CMB rest frame, in which this anisotropy would vanish.

Relevantly, the Friedmann-Robertson-Walker family of solutions of general relativity used to model the Big Bang cosmology assume that the universe is, on the large scale, homogeneous and isotropic.

  It seems as if this is necessary to assume in order for the CMB to make an sense. ... Our position in space would dictate what we would observe in terms of background radiation.

That's not the case. There are four distinct homogeneous and isotropic spatial geometries: the Euclidean space $\mathbf{E}^3$, the hyperbolic space $\mathbf{H}^3$, the real projective space $\mathbf{RP}^3$, and the sphere $S^3$. In any of those cases, the universe would look the same all around us and be independent of our position. Thus, the sphere is possible but not necessary. The first two of those cases correspond to the "flat" and "open" Big Bang cosmologies, respectively, and the last two are variations of the "closed" cosmology.

And if we don't require the universe to be completely homogeneous and isotropic, but merely look like it in the part of it we observe, then much more exotic geometries are also possible. With cosmological inflation, virtually any large-scale geometry can be "blown up" to look flat, homogenous, and isotropic within our observable portion--even though it might not be s beyond the our cosmological horizon.

  Otherwise, an instantaneous infinite Euclidean space model would work.

It does work, in the sense of being consistent with observations. In fact, in the standard ΛCDM cosmological model, space is infinite and Euclidean.


Are red moons possible without eclipses or smoke from fires?

What causes the moon to be red is our own atmosphere.  It's the same effect that causes a sunset to be red.  As the sun nears the horizon, the amount of air that we're looking through increases - and with it everything in the air such as dust, pollution, humidity, etc....  If you look at the moon near the horizon, you'll notice it is significantly yellower/oranger than when it's high in the sky, where it is normally whiter and bluer.  Again, this is all caused by the air and the stuff in it.

When the moon turns read during an eclipse, it is because it is passing through the shadow of the atmosphere, which filters the light passing through and on to the moon, leaving more red light to make its way to the moon and, then, to reflect off of it.

A similar action happens when smoke from fires gets into the air - it filters out the shorter wavelength, bluer light and lets the red pass.  Other pollutants besides smoke particles can have the same effect.

So, the only way for this to happen is for something to pass between the sun and the moon, or the moon and the earth - between the source of light and where it is received.

As for the moon appearing larger: this is an optical illusion.  No one is entirely sure why it happens, but the moon does APPEAR larger on the horizon than when it is high in the sky, but this is merely an illusion.  To prove it, next full moon, go outside with a ruler and hold it up at arm's length and try to measure the apparent size of the moon (this will vary from person to person due to the length of your arm).  Then, wait until it's high in the sky and do the same thing.  It will measure the same size (give or take a small amount due to the change in position affecting the distance from your eye to the ruler).

If you have the right kind of telescope and mount, you could connect a camera, take a picture on the horizon, wait for it to pass close to its highest point and take another picture, then compare.  Assuming no changes were made to the camera or telescope, the images should show the same apparent size.


Why is the moon's motion more complicated than that of those planets?

Keplerian motion would be a correct way to explain the Moon's orbit about the Earth if the Sun, Venus, Jupiter, and all the other planets weren't present, if the Moon wasn't slowly receding from the Earth due to tidal acceleration, and if Newtonian gravity properly described the underlying physics. None of these is the case.

In the case of modeling the behaviors of the major objects that comprise the solar system (i.e., the Sun and the eight planets), perturbations involving planets interacting with one another, particularly with Jupiter, make the orbits of the planets slightly non-Keplerian. That the universe is relativistic rather than Newtonian famously makes Mercury's orbit precess in a way that even accounting for planetary perturbations cannot explain. That said, Keplerian motion is a fairly good approximation with regard to the Sun and the eight planets.

The same is not true for the Moon. It's orbit is notably non-Keplerian, even over the course of just a month. Ptolemy (after the fact) and Brahe (also after the fact) both noticed irregularities in the Moon's motion that cannot be explained in terms of Keplerian elliptical orbits in which area swept out per unit time is constant.

Newton explained some of these effects by noticing that the Sun is a huge perturber of the Moon's orbit about the Earth. Newton was not quite successful in this endeavor; the mathematical tools needed to model the Moon's motion hadn't been developed. The full development of classical lunar theory took two hundred years after Newton's time.

Another complicating factor is that the Moon is gradually receding from the Earth due to tidal acceleration. That the Moon is slowly receding from the Earth was first hypothesized by Halley, one of Newton's contemporaries. A century after the first publication of Newton's Principia, Laplace erroneously attributed this lunar recession solely to perturbations from the Sun. In the next hundred years, others corrected Laplace's calculations and found that solar perturbations are unable to explain this recession. George Darwin (Charles' son) explained the recession in terms of the tides. The full explanation? People are still writing papers on this subject.

Relativity makes for yet another complication. Relativistic effects on the Moon are very hard to see given the many orders of magnitude larger complications due to perturbations by the Sun, Venus, Jupiter, ..., and tidal acceleration.


You are missing one last ingredient: the correlation between the virial radius $r_{\mathrm{vir}}$ and the virial mass $M_{\mathrm{vir}}$. The idea is that the orbital time scale of a particle orbiting in the halo becomes roughly equal to the age of the universe at $r_{\mathrm{vir}}$. This implies a relation between the (mean) density of the halo and of the universe:
$$
    \bar{\rho}_{\mathrm{vir}} \equiv \frac{3M_{\mathrm{vir}}}{4\pi r_{\mathrm{vir}}^3} = \Delta \rho_{\mathrm{crit}}(z)
$$
where $\bar{\rho}_{\mathrm{vir}}$ is the halo's mean density interior of $r_{\mathrm{vir}}$, $\rho_{\mathrm{crit}}$ is the critical density of the universe (as function of redshift $z$), and $\Delta$ a parameter that depends on the cosmology. Often $\Delta=200$ is used and the corresponding quantities labeled $r_{200}$ and $M_{200}$ instead. See also this Wikipedia page.

So the way to construct a halo is as follows:

Pick $M_{\mathrm{vir}}$ (or $M_{200}$) and redshift $z$ (and your favorite cosmology)
Use above relation to find $r_{\mathrm{vir}}$ (or $r_{200}$)
Use the relation from Napolitano et al. (2005), or equivalent, to find $c$ and hence $r_s$
Obtain $\rho_{\mathrm{DM}}$ (if you still need it)


Day,Night, and Year in the Ptolemaic Model

As I remember, the Ptolemaic model involved a series of concentric transparent spherical shells around the Earth.
The stars were attached to the outermost shell or sphere which revolved about once a day, but not exactly once a day since there was also a yearly cycle to which stars were visible at night.
The innermost sphere was the sphere of the Moon, which sometimes passed in front of the Sun and thus was closer, and the next innermost sphere was the sphere of the Sun.
The Sphere of hte Moon revolved about once a day, but not exactly, since there was a difference which added up to a full circle every 28.5 days.  So, if i remember correctly, the sphere of the Moon would revolve about 360 degrees minus 12.63 degrees every day, or about 347.37 degrees per day.
The sphere of the Sun made a full and complete revolution of 360 degrees in exactly one day.
And the spheres of Mercury, Venus, Mars, Jupiter, and Saturn also revolved about 360 degrees per day, but not exactly, the differences accounting for their various synodic periods.
Since the geocentric model was totally different from the actual heliocentric model, and the giant transparent spherical shells didn't really exist, there were a lot of problems fitting the directions to various planets at various times predicted by the system with the actual observations of the directions to those planets at those times.  Even though the ancient Greeks had to rely on naked eye observations, they were able to notice deviations of the planets from their predicted paths across the background of the stars.  Which meant that the theories had to be improved to make the predictions better.
So Ptolemy had to do a lot of thinking and explaining to account for those deviations and to make correct future predictions of the movements.
So Ptolemy took the basic geocentric model I described above and elaborated it with various modifications which are called eccentrics, epicycles, deferents, and equants.  And if that sounds complicated, that is correct.
There is a famous story that King Alfonso X The Wise (1221-1284), King of Castile, etc., and rival King of the Romans, once said of the Ptolemaic model of the universe that if the creator had asked Alfonso's advice during the creation the universe would be simpler.
And the Ptolemaic model was able to predict the apparent motions and apparent positions of the planets well enough to satisfy almost all European and Muslim astronomers for over a thousand years.


Why aren't globular clusters disk shaped

In order to result in a disk-like stellar system there are two conditions that need to be satisfied. (a) The initial gas from which the stars form must have a significant ratio of rotational to gravitational energy. (b) The star formation would have to occur slowly enough, that the gas collapses to a disk before star formation is complete.

The formation of a disk from a spherically symmetric state requires a reduction (more negative) gravitational potential energy, whilst conserving angular momentum. This can only happen if there is some means to lose energy from the system in the form of dissipative interactions.

A pure stellar system is almost collisionless and there is no means to dissipate energy. That means once a spherical system of stars has formed, there is no way for it to become more disk-like. Thus if star formation takes place in a spherical way, then those stars will end up in a spherical system. Only in cases where the gas collapses to a disk before star formation occurs do we end up with disk-like systems. This is true of the disk of our Galaxy (and of the solar system).

In the case of globular clusters (and other types of star cluster in the Galaxy) it appears the star formation is rapid enough that the gas cannot have collapsed to a disk before most star formation is complete. NB. Many globular clusters do have measurable rotation.


Do origin theories imagine each planet to first orbit the Sun very irregularly before stabilizing?

Actually, the sun, planets and other small bodies are all created from the cloud of dust and gas that existed before the solar system. As the gas and dust collect towards the center, the mass rotates and soon an accretion disk forms (which is why the planets are nearly on the same plane).

So your notion of the beginning of the universe is incorrect. It did not start with a bunch of planets (or stars) randomly thrown into space. Also all the stars in a globular cluster are rotating in the same direction; the stars are in orbit in the cluster.

Added:
Yes, they did have irregular movements. It started with a large cloud of gas and dust, some from the big bang, mostly from exploded stars systems. The particles all had different directions and speed.

When gravity influenced enough particles to group and start accumulating is when the mass began to spin (conservation of momentum). The spinning mass flattens out to an accretion disk and planets can start to form (along with the sun).


Why does the Chandrasekhar limit affect white dwarfs differently?

There are a variety of white dwarfs with various compositions, and analysing how they detonate in a supernova (or not) is an topic under investigation. A simple model, described in "How is the first detonation in Supernove type Ia triggered?" is of a helium shell initially igniting and that setting off the carbon in the core.

In this type of Type 1a supernova there is no neutron star formed, as the star is completely destroyed. The mass at which a white dwarf will undergo a type 1a supernova is just below the chandrasekhar limit, so neutronisation won't occur.

However in some white dwarfs with an atypical composition (as you note a Mg-Ne-O white dwarf) it is possible for the star to avoid detonation, and reach the chandrasekhar limit, and so for electron capture to occur, and a neutron star to form. It is worth noting that there is not definite observation of a white dwarf collapsing to a neutron star (where as there a lots of observations of type 1a supernovae) however these "accretion induced collapse" scenarios may explain some magnetars and short gamma ray burst

Progenitors of the Accretion-Induced
Collapse

So the two scenarios are,

Accretion occurs onto a Carbon-oxygen WD. The pressure and temperature in the core of a white dwarf increases until thermonuclear reactions begin (at about 1.38 solar masses). Since the white dwarf is degenerate it can't expand to reduce the rate of thermonuclear reactions, and the entire star detonates and is destroyed.
Accretion occurs onto a ONeMg WD. The star reaches 1.44 solar masses, electron degeneracy is no longer sufficient to prevent collapse. Electron capture occurs and the star collapses to neutron star.


How would a retrograde satellite experiencing tidal deceleration affect the rotation of the primary?

For a retrograde satellite, you are right that the satellite will migrate inwards towards the planet. Contrary to a prograde orbit, the rotation of the primary will slow down.
Think about it in terms of angular momentum. Let the primary have a positive angular momentum and the satellite a negative one (since they rotate/orbit in opposite directions). Since the satellite is being pulled inwards, its angular momentum is being lowered in magnitude (becoming less negative). In that case the primary's spin angular momentum needs to become less positive to conserve the total angular momentum. This means the primary's rotation will slow down.
An example with numbers (unitless):
Primary(initial) = 10 // Satellite(initial) = -5 ///
Primary(final) = 7 //   Satellite(final) = -2 ///
Thus a transfer of 3 angular momentum units has occurred.
Eventually the satellite will slow down the primary until the primary is no longer rotating (assuming the satellite is not lost by this point). Then it will cause the primary to start rotating in the same direction as the satellite is orbiting.
Hope this helps!
EDIT:
In this paper, see equation (7), the torque on the primary due to the satellite. If we only care about the sign, we can note that $N_m \textit{~}\, (n_m -\Omega_p)$ where $n_m$ is the orbital frequency of the satellite and $\Omega_p$ is the rotational frequency of the primary. Let's take our previous example of the $\Omega_p$ positive and $n_m$ negative. This would make the torque $N_m$ negative, no matter what the magnitudes of the frequencies are. Consequently, the positive $\Omega_p$ would be reduced, signifying the slowing of the primary's rotation.

$$N_m = 3 k_2 \tau (n_m - \Omega_p) \frac{GM_m^2 R_p^2}{a_m^6} \tag{7}$$

where the subscripts $m$ and $p$ refer to the moon and planet respectively, $k2$ is the Love number of the planet that depends on its rigidity and $\tau$ is the tidal time lag of the planet.


How to convert ra&dec between different epoches?

The difference between B1950.0 AND J2000.0 is not really concerned with proper motion., they refer to the definition of the coordinate system. Thus a star with zero proper motion would still have different B1950 and J2000 coordinates.

Usually (but not always), if you quote a position as equinox 1950 you are implicitly assuming that this is epoch 1950 (unless otherwise stated). And similarly for J2000, the implicit assumption is that the coordinates refer to epoch 2000 unless you specify otherwise.

Thus there are two things you needs to do. One is to apply a correction for (roughly$^{*}$) 50 years of proper motion (if indeed you only have the epoch 1950 position; or maybe 66 years if you want the coordinates of the star now). Second is to apply (roughly) 50 years of precession to the coordinates to put them in the J2000 system.

This description seems reasonable.

$^{*}$ There is some detail about B= Besselian years and J = Julian years which you can read about here.


What is the maximum possible separation between sun and moon in the earth sky for an eclipse to occur?

We'll need the angular diameter of the Sun: 31.6-32.7 arcminutes; the Moon's is between 29.3 and 34.1 arcminutes (found on Wikipedia).

The solar eclipse case is 0 degrees; it can only be an eclipse if the Moon is touching the Sun. In that case, their centers are at most (32.7 + 34.1) / 2 = 33.4 arcminutes apart.

A lunar eclipse would occur if the Sun and Moon are diametrically opposite in the sky; the separation of their centers is then 180 degrees. The actual separation is at least (31.6 + 29.3) / 2 = 30.45 arcminutes less than 180 degrees.


how to get image coordinates of source points in saoimage ds9?

OK, it seems to me that what you have is a fits image (from where?) that has some world coordinate system information attached. When you load the 2MASS catalogue, SAOimage is able to use the RA and Dec in the catalogue to calculate the x,y positions of the catalog sources in your image (and marks them as green circles).

Is what you are asking - how do I get the x,y coordinates of these 2MASS sources in my image? Or do you wish to know what the x,y coordinates of the actual sources in your image? If the latter then you will need some other software like Sextractor or DAOphot to do source-searching and parameterisation in your image.

If the former, here's a way of doing it. When you load the catalog you should see a window open entitled "2MASS Point sources" which lists all the sources in your image. From the "File" dialogue, select "Copy to regions". Then, in the main image window open the "Regions" dialogue and select "List regions". Another dialogue box will open asking you to select the format and coordinate system. From this, select "x,y" for the format and "image" for the coordinate system. This will give you a list of x,y coordinates.

If you want to match this up with the 2MASS sources then you have to "list regions" again, but this time select "fk5" as the coordinate system. This will then give you a list of the RA, Dec of the same sources. Use you favourite editor to attach these lists to each other.

Quite probably there is a more elegant solution, but this works.


How do you measure something that has already left the observable universe?

First off, you need to understand what "Redshift" is.  Light that is redshifted means that its wavelength has increased, making it appear more red.  This can come from a variety of sources, including Doppler shifting.  What we're more concerned with here is "cosmological redshift".  As light travels through space, its wavelength increases due to the expansion of space.  We can use a measurement of how much the light has redshifted to calculate the distance to its source.  There is nothing "beyond" redshift, because redshift is just a lengthening of the wavelength.  And nothing is happening to the objects themselves, they just appear to get redder to us.

Next, the definition of the "observable universe".  Because light travels at a finite speed, it takes a certain amount of time to reach us.  The universe had a beginning in the big bang, meaning that light has only had a certain amount of time to travel.  Therefore we can only see things out to a certain distance.  Light coming from objects further than that has not had enough time to reach us yet, so we can't see them.  That's what we mean by the "observable universe".

Galaxies do not "phase out of existence" because of redshift.  As we look outwards, we're looking at older and older objects.  Because the light takes a certain amount of time to reach us, we're essentially looking back in time.  As time passes, however, we can see more and more because light from further away objects has more time to reach us.

Dark matter and dark energy cannot be directly observed because they do not interact with light.  They do not absorb it, emit it, reflect it, or anything else.  However, they are still within the observable universe, because the observable universe is basically just the size of the universe.  There are many different theories as to what dark matter and dark energy actually are, but they are not ordinary matter that has been "blackshifted".


Is the size and heat of a star important in the development of a solar system?

I think the quick answer to this is no, at least, any solar-system around a Wolf-Rayet or similarly large star wouldn't have time to develop.   I would think that large stars have solar-systems cause I see no reason why they wouldn't, but the stars don't last long enough for their solar-systems to develop much.   The formation period, including bombardment period(s) and time for the planets to cool down from the heat of formation takes a significant amount of time, perhaps a couple hundred million years minimum.

Our solar system's late heavy bombardment occurred between about 4.1 and 3.8 billion years ago, ending some 700 million years after our solar-system formed and Earth didn't cool down enough to have a magnetic field until about 3.5 billion years ago, roughly 1 billion years after formation.   Granted, that's a sample size of one, so numbers could vary quite a bit, but I have a hard time imagining that planet's could cool down and complete most of their early bombardment in the lifespan of a very large star.

For developed solar-systems where the planets had time to cool and the solar-systems had time to mostly coalesce and get past early-bombardment, I don't think there's much of a window beyond maybe 2 solar masses  (a 2 solar mass star would have roughly 1/16th the lifespan of our sun, or a bit less than a billion years in main sequence), based on an exponential ratio to the 4th power of the star's mass.  Source.

Now, as to whether a solar-system could survive a Wolf-Rayet star's enormous ejections, that's another question.   I'm not 100% sure but large planets are pretty resilient, so my guess is yes, but it's just a guess.

Another interesting question is gravity vs luminosity.   Our solar-system the furthest planet Neptune, about 30 astronomical units from the Sun, and, presumably a planet could orbit a fair bit further out than that and be in a stable orbit, but lets go with 30 AU.

A star with 10 times the mass of our sun during it's main sequence has roughly 3,000 times the luminosity (using the exponential relationship of 3.5 Mass to Luminosity - see here)

The gravity for 10 solar masses, gravity reduces by the square of the distance, so square root of 10, a Neptune equivalent orbital velocity would be about 95 Astronomical Units, so, at 95 astronomical units from a star with 3,000 times the luminosity of our sun, the energy received from that star at that distance would be roughly 1/3rd the energy the Earth gets from the sun.

Too much larger than that, somewhere around 20 solar masses or, not much more than that, the star gets too hot to have much of a solar-system within it's theoretical Goldilocks zone.   Not that that matters much, cause the short lifespan of the star kind of answers the question anyway, but my guess would be, if planets survive around a Wolf-Rayet star, even the outer parts of such a solar-system should be hot unless they were enormously distant, like a couple hundred AUs.   There might be no ice or water planets around a star of that size.


Is it safer to be near a star or a black hole?

the candidate star is an "average" 5 solar mass star, and the black
  hole is a 5 solar mass black hole

Then their gravity is identical. Black holes don't have magic powers. A 5 M☉ star and a 5 M☉ black hole exert the same attraction from the same distance. The only difference is that the black hole would be much, MUCH smaller (about 30 km diameter in this case), so you could get a lot closer to the center, which is where the extreme gravity happens. But at cosmic distances they are the same.

Hawking radiation from a 5 M☉ black hole is negligible, likely too small to measure - its temperature is about 10^-8 Kelvin.

If it has an accretion disk, that might be a problem, but usually it only generates two relativistic jets at the poles - if you're not hit by the jets, you're fine.

The overall thermal glow from the accretion disk of a 5 M☉ black hole might be intense, but it can't last very long. It's not a continuous burn like a star, unless it's a much bigger black hole with a huge accretion disk and plenty of source material nearby to feed it - which is not the case here.

Your conclusion is technically correct. Gravitationally they are the same, but the star has the added problem of generating extra heat. Not that it would matter anyway, because both would completely disrupt orbits in the solar system, and the Earth is going to be flung out into space and would freeze anyway. It's just that the star would bake us first, and THEN we would freeze solid.

Or we would settle into a highly elliptical orbit around either the Sun or the invader body, which will have us alternately baking to death and freezing to death.

There is also the very tiny chance that, due to orbital disruptions, we would collide with something else, either with the invader, or with the Sun, or with another planet. This would mean a quicker death and might be overall "preferable".

Regardless, having a 5 M☉ body entering the system is not a good scenario for life on Earth.

http://xaonon.dyndns.org/hawking/


Is there anywhere I can find some kind of database on all known stars and their properties (mass, surface temperature, radius, and luminosity)?

There really isn't anything as comprehensive as you want. And, we don't have accurate information for all of these variables for a large number of stars.

The two main problems are masses and radii.

Really the only set of stars that have both measured masses and radii are those in eclipsing binary systems. If these binary stars also have distances, measured by parallax, then the components could also have luminosity estimates and estimates of effective temperature.
The best catalogue I can recommend is that compiled by John Southworth, containing perhaps a 100 systems.

http://www.astro.keele.ac.uk/jkt/debcat/

If you just want a mass, but can cope without a radius, then you can also include components of astrometric binary systems. There is a well-known paper by Delfosse et al. (2000) containing stellar masses and luminosities.


Retrofitting a DSLR camera to a super-cheap Tasco reflector telescope

The problem you have with many newtonian reflectors - especially smaller ones - is that to use them for prime focus photography (where you're effectively using the telescope like a big telephoto lens) , you need to set things up so that the camera sensor ends up at the focal plane of the telescope. And in a DSLR, the sensor is some way inside the camera body (behind the shutter, which is behind the mirror.

With a newtonian scope, the focal plane is usually somewhere inside the focuser tube, and many newts don't have enough focuser inwards travel to let you get the focal plane the 40mm or so beyond the end of the focuser that you need to get it to reach a DSLR sensor.

(With other types of scope that use a separate star diagonal, things are usually easier - you can leave the star diagonal off, which has the same effect as a big chunk of inward focus).

With a newtonian, if you can't reach prime focus, there are three basic options - either take photos through the eyepiece (afocal photography) as you already tried,

or move the main mirror further up the tube (which moves the focal plane further out, giving you more in focus). It also means the light cone is fatter when it reaches the secondary mirror, so you may lose some light as the outer part of the light cone misses the secondary.

or you can try using a barlow lens - that replaces the last bit of the light cone with a longer, narrower angle version (which increases the magnification and moves the focal plane further out). That MAY or MAY NOT let you reach infinity focus; it's not guarenteed, but may help. You can get barlows that have a T-Thread on the output end, so you can attach a T ring directly to the barlow (yours doesn't look like one of this type) and then hook the camera up that way. (as an added benefit, since you end up with a closed tube setup (the barlow lens closes off the access to the camera, it also keeps spiders and dust out of the camera :) ).

If you have a look at the bottom bit of the barlow (with the lens) you may find that you can unscrew it. You might be able to make a short top section with a T thread for the T ring at the back, and a suitable thread for the lens at the front; you'd probably need to make it the same diameter as the eyepiece barrel (to fit in the focuser) with a fatter bit at the back for the T thread and to stop it sliding all the way into the focuser.
I've used a commercial 1.25" barrel T threaded barlow (somewhat shorter than yours) for lunar and solar (WITH A SUITABLE FRONT OF SCOPE FILTER - DON'T TRY LOOKING AT THE SUN WITHOUT ONE) imaging with an APS sensor camera.

(You may need to check your scope - older ones often used a .965" barrel rather than the current 1.25" (or 2") barrel size used by most modern astro scopes.)


It wouldn't look much different. The atmosphere absorbs some light and "seeing" reduces the effective resolution, but these have little effect on the visual impact of a lunar eclipse.

From a sufficiently high orbit, you might be able to see the terminator between the partially lit "red" side of the moon and the (nearly) completely dark side on the far side from the Earth (and the sun).

A lunar eclipse looks pretty cool from the Earth, and it would look equally cool from space.


Why do satellites with greater altitude travel at a slower speed?

Kepler's third law in this situation tells us that
$$ \frac{R^3}{P^2} = \frac{GM}{4\pi^2}, $$
where $R$ is the orbital radius (or semi-major axis for an elliptical orbit), $P$ is the orbital period and $M$ is the mass of the Earth (actually, it is the mas of the Earth plus the mass of the satellite, but we can neglect the latter). Thus the orbital period $P \propto R^{3/2}$.

To calculate the orbital speed, we divide the orbital circumference by the orbital period
$$ v = \frac{2\pi R}{P}. $$

Then substituting for $P$ from Kepler's third law, we get
$$ v = \sqrt{\frac{GM}{R}}$$

Thus, the larger the orbital radius, the slower the orbital speed.

Or you can just use Newton's second law and equate the centripetal acceleration to the gravitational force:
$$ m \frac{v^2}{R} = G\frac{Mm}{R^2},$$
where $m$ is the mass of the satellite, which leads directly to the same result.


Recently NASA has revealed that they have recorded the sound of Sun. They say that it produces a sound like "Om". I can't understand how they can hear it.

It's not that recent (it was 2010), it wasn't NASA (it was researchers at the University of Sheffield who used data from a NASA satellite), it wasn't sound per se (it was instead sonified data), and neither NASA nor those researchers said that the sound produced was "Om".

Data sonification is a scientific tool of growing importance. It is the sound equivalent of visualization techniques that have been used for centuries. (Graphs are a rather old visualization technique.)

In this case, researchers at the University of Sheffield used sonification techniques to translate imagery of solar coronal loops into sound. A faithful representation of that generated sound would not have been audible; the frequency would have been too low for a human to hear it. Those researchers had to speed up the playback to make the generated sound audible.

That's not to say that those sounds are a complete fiction. The surface of the Sun is not a vacuum, nor is its interior. There truly are low frequency sounds associated with events on the surface of the Sun. The Sun rings with sound (low frequency sound). Studying the sounds produced by the Sun gives scientists insights into what is happening inside the Sun.

There's a small downside in making scientific data public: Non-scientists who have a penchant to read/see/hear things into that scientific data will do just that. To me, the people who say that this is proof that the universe does indeed chant the sacred "Om" are no different that those who see cities on the Moon or Mars in NASA photography.


Why do we see galaxies with their actual spiral shapes?

Basically, because the relative speeds of stars within galaxies are much less than the speed of light.

Imagine a structure with dimension 100,000 light years (about the size of a galaxy). Now suppose the components of that structure move with speeds relative to each other of around 100 km/s (or 0.03% of the speed of light). In the time it takes light to cross the structure (100,000 years of course), the components within it will have moved by just 30 light years (0.03% of the size of the structure).

There would therefore be no significant "blurring" of our snapshot of a galaxy caused by the differing light travel times from its structural components.

The other thing you may have not understood, is that the lives of stars are generally much longer than 100,000 years. So the stars that emit the light on the far side of a galaxy are still in  existence 100,000 years later, when their light passes stars that are present at the near side of a galaxy.


What's the critical mass of a rocky planet before its gravity traps enough gas in its atmosphere to start becoming a gassy planet

In most planet formation theories, the boundary is around 10 Earth masses - the build up of the core mass before that is relatively slow, but once it crosses that threshold, the planet gains mass quickly via attracting gas from the surrounding nebula via the core’s gravitational pull, a process called “runaway accretion.”  As this summary shows, you can still end up with some planets in between.
One interesting result that has emerged from recent studies of exoplanets is that small planets seem to be divided into two groups, those with radii around 1-1.5 Earth radii, and those with radii of ~ 2-3 Earth radii, but with a gap in between where there are few planets:

The current theory explaining this is that the larger group of planets managed to accrete some atmosphere, while the smaller group are bare, rocky worlds with little or no atmosphere.


Any information on the salt content of Mars' atmospheric dust?

[WARNING: I may be wrong. Astronomy is my hobby and I will answer to the extent that I know of. I may be a bit off.]

Mars' atmosphere contains I guess what you can say 2 major parts. Gases, and debris picked up by winds on the lower levels of the atmosphere (like Earth).

Gases

Mars' atmospheric gas composition mainly consists of CO$_2$.

More specifically, in an article page by NASA, Mars' atmosphere contains

  Carbon dioxide: 95.32 percent

  Nitrogen: 2.7 percent

  Argon: 1.6 percent

  Oxygen: 0.13 percent

  Carbon monoxide: 0.08 percent

  Also, minor amounts of: water vapor, nitrogen oxide, neon, hydrogen-deuterium-oxygen, krypton and xenon.

    http://www.answers.com/Q/What_are_the_contents_of_Mars%27_atmosphere

Dust/other stuff

Mars' atmosphere, although 99~100 times thinner that that of Earth's, can support winds, weather/storms, and clouds. The winds are a major criterion to Mars' permanent atmospheric dust composition. Dust particles, airborne or not absorbing sunlight to some extent, which in turn warms up the atmosphere in and near their vicinity. Warm pockets of air flow toward colder regions, generating winds; the same effect here on Earth. Strong winds lift more dust off the ground, which in turn heats the atmosphere, raising more wind and kicking up more dust. This is amplified by Mars' axis tilt and momentum, which can contribute to Mars' massive dust storms encircling the planet.

In addition to the winds, the Martian atmosphere is dense enough to support vortexes. Dust devils routinely kick up the rusty dust that covers Mars' surface.

So we can see now the causes for such permanent content of dust in the atmosphere of Mars, with...

  ...higher amounts of dust in the northern fall and winter, and lower
  amounts in the northern spring and summer.

    https://www.space.com/16903-mars-atmosphere-climate-weather.html

Let's now take a look at Mars' surface dust composition.

Its dust mainly is composed of Perchlorate salts; Chlorine; Iron(III) Sulfide; NiFe; Sulfur; Iron Sulfide; Nickel; Silicon; Oxygen; Iron; Magnesium; Aluminum; Calcium; Potassium; Pyroxene; Plagioclase; Glass; Olivine; and Feldspar. I would imagine that the dust in the air would contain many of these same components as the surface dust, although some heavier salts and chemicals would only last a brief time in the air.

According to Wikipedia,

  The NASA Phoenix lander first detected chlorine-based compounds such as calcium perchlorate. The levels detected contain 0.6% of the salt perchlorate.

and

  The soil has a basic pH of 7.7

    https://en.wikipedia.org/wiki/Mars

At such high levels, there should be definitely perchlorate salt in the air, but not necessarily permanent though, as the atmosphere may not be able to support the density of the salt.

However, a recent study of terrestrial dust devils has shown that around two thirds of the fine particles lifted by these vortices can remain suspended in the atmosphere and be transported around the globe. I would theorize that lighter salts would be able to remain suspended in the air, while heavier ones would slowly settle to the surface, given enough time. But hey, this is MARS! With those huge dust storms and continuous vortexes, I doubt there is enough time.

All of this suggests that during the times when there are rather windy conditions, such as in a close proximity to a dust devil, or during one of the dust storms, the atmosphere should contain almost the same concentrations of salts as in the soil itself.

During calm periods, the air dust content should still remain the same.


Finding Latitude and Longitude of a person or of where a picture was taken

EDIT: It turns out I answered this question earlier (in a slightly different format):

Cancelling out earth rotation speed, Altazimuth mount

Imagine you are at 35N,106W at 5h local sidereal time.

The celestial north pole is at (0,90) in the celestial frame and
(0,35) in your frame.

The point (5h,0) is (75,0) in the celestial sphere (1h = 15 degrees),
and (180,55) in your frame

The point (11h,0) is (165,0) in the celestial sphere and (90,0)
(setting in the east) in your frame.

Converting to rectangular coordinates, we need the matrix that does this:

{0,0,1} -> {Cos[35 Degree], 0, Sin[35 Degree]}

{Cos[75*Degree],Sin[75*Degree],0} -> {-Sin[35 Degree], 0, Cos[35 Degree]}

{Sin[-75*Degree],Cos[75*Degree],0} -> {0,1,0}

or, generalizing a bit (lst = local sidereal time as an angle, lat =
latitude, notice that lon is irrelevant since we're using local
sidereal time)

{0,0,1} -> {Cos[lat], 0, Sin[lat]}

{Cos[lst],Sin[lst],0} -> {-Sin[lat], 0, Cos[lat]}

{-Sin[lst],Cos[lst],0} -> {0,1,0}

Solving (Mathematica):

m0 = Table[a[i,j],{i,1,3},{j,1,3}]

(* Some simplifying assumptions, intentionally avoiding corner cases *)

conds = {-Pi/2 < dec < Pi/2, -Pi/2 < lat < Pi/2, 0 < ra < 2*Pi, -Pi < lst < Pi}

m = Simplify[m0 /. Solve[{
 m0.{0,0,1} == {Cos[lat], 0, Sin[lat]},
 m0.{Cos[lst],Sin[lst],0} == {-Sin[lat], 0, Cos[lat]},
 m0.{-Sin[lst],Cos[lst],0} == {0,1,0}
},Flatten[m0]],conds]

m = m[[1]]

This gives us

$
   \left(
   \begin{array}{ccc}
-\cos (\text{lst}) \sin (\text{lat}) & -\sin (\text{lat}) \sin (\text{lst})
  & \cos (\text{lat}) \\
-\sin (\text{lst}) & \cos (\text{lst}) & 0 \\
\cos (\text{lat}) \cos (\text{lst}) & \cos (\text{lat}) \sin (\text{lst}) &
  \sin (\text{lat}) \\
   \end{array}
   \right)
$

The inverse of this matrix:

$
   \left(
   \begin{array}{ccc}
-\cos (\text{lst}) \sin (\text{lat}) & -\sin (\text{lst}) & \cos
  (\text{lat}) \cos (\text{lst}) \\
-\sin (\text{lat}) \sin (\text{lst}) & \cos (\text{lst}) & \cos (\text{lat})
  \sin (\text{lst}) \\
\cos (\text{lat}) & 0 & \sin (\text{lat}) \\
   \end{array}
   \right)
$

will transform rectangular azimuth and elevation back to right
ascension and declination and right ascension. This would be an
interesting alternative approach to this problem, but I won't be using
it.

OK, if we convert right ascension and declination to rectangular
coordinates, multiply by the first matrix above and convert the result
back to spherical coordinates, we will have azimuth and
elevation. Let's do that...

The spherical coordinates for (ra,dec) are:

{Cos[dec] Cos[ra], Cos[dec] Sin[ra], Sin[dec]}

Multiplying by the matrix, we get

$
   \{\sin (\text{dec}) \cos (\text{lat})-\cos (\text{dec}) \sin (\text{lat})
\cos (\text{lst}-\text{ra}),-\cos (\text{dec}) \sin
(\text{lst}-\text{ra}),\cos (\text{dec}) \cos (\text{lat}) \cos
(\text{lst}-\text{ra})+\sin (\text{dec}) \sin (\text{lat})\}
$

(note that lst-ra is sometimes called the "hour angle", which would
simplify the above)

Converting back to spherical coordinates and simplifying:

{x,y,z} = Simplify[m.{Cos[dec] Cos[ra], Cos[dec] Sin[ra], Sin[dec]},conds]

(* NOTE: the use of single argument arctangent here introduces
ambiguity, and the final answer must compensate for this *)

az = Simplify[ArcTan[y/x],conds]

el = Simplify[ArcSin[z],conds]

yielding:

$
   \text{azimuth}=-\tan ^{-1}\left(\frac{\cos (\text{dec}) \sin
(\text{lst}-\text{ra})}{\sin (\text{dec}) \cos (\text{lat})-\cos
(\text{dec}) \sin (\text{lat}) \cos (\text{lst}-\text{ra})}\right)
$

$
   \text{elevation}=\sin ^{-1}(\cos (\text{dec}) \cos (\text{lat}) \cos
(\text{lst}-\text{ra})+\sin (\text{dec}) \sin (\text{lat}))
$

Assuming I haven't made any mistakes, the next steps would be:

Converting local time to local sidereal time
Inverting these equations to get latitude and longitude from
azimuth, elevation, and time.

I may or may not edit this answer to add those steps, but this
hopefully provides a start.


Maxwell's Smith Prize Essay on the formation of the Rings of Saturn is relevant. Using structural stability as his criterion, he deduced that the 'rings' could only consist of a single almost infinite mass of very small particulates each orbiting in accord with Newtonian dynamics--the correct result as verified by NASA/ Cassini.
 Applying the same approach to a forming  solar system this analysis shows that rings of particulates are a structurally stable attractor for the dynamics, provided the central star is much larger than the mean particulate size.
I am now  well outside my comfort zone so will leave further inferences to others :).
Maxwell's essay on rings of saturn


Can we know whether any distant galaxy is moving away in same direction or in opposite direction w.r.t. our galaxy?

Yes, it can. For instance the Andromeda Galaxy is moving towards us, due to local gravitational attraction but that does not affect the validity of the theory of the expansion of spacetime because one result based on local factors is perfectly compatible with a cosmological theory which, by definition, includes every galaxy.


Are galaxies moving in each direction with same speed?

Are the galaxies moving towards each direction with same speed ?

Short version : Locally, No.  On the cosmic scale, yes-ish.

The motions of galaxies are more complex than this questions suggests.

Let's start with the Milky Way.  From our point of view stationary.

But the Milky Way is part of a group of galaxies (the local group).  They essentially move in their mutual gravitational fields.  Even that field is complicated by the gravitational field of dark matter halos, which are invisible (unobservable directly).

But the local group is not the only group out there.  The local group is part of an even larger grouping known as the Virgo Supercluster.  Again motions of these groups are complex and again dark matter plays a part.

The local group contains 54 galaxies, and the super cluster contains something on the order of 100 such groups.  That's about 5400 galaxies in a fairly complex motion.  There's no one easy direction to spot here.

But the larger scale universe is thought governed by what is known as the FLRW metric, which (with dark matter, dark energy and some adjustments) fits the overall expansion of the universe we see.

It's only on that scale (essentially the entire universe excluding the nearest bits) that you see the kind of pattern you mean.  So the local bits are maybe 5400 galaxies, but the rest of the universe is very roughly two trillion galaxies.  So our local part is on our scale vast and unimaginable, but on the scale that FLRW applies it's a microscopic dot lost in the averages.

FLRW is a result of using General Relativity to model the expansion of the universe.  It does predict (when you match it to our universe's conditions) that from any point everything else will be moving away from you (and we now think accelerating away).

There is an apparent contradiction between these two extreme (local group and universe away from that).  The resolution of this is that FLRW ignores local grouping and irregularities and assumes a nice continuous distribution (galaxies don't even feature on this grand scale).  On that local scale gravitational fields (currently) are large enough to overcome the expansion, hence our closest galaxy (Andromeda) is approaching us, not moving away.  But the further away you get, the larger the overall effects predicted by FLRW become and dominate "normal" gravitational effects.

FLRW does not say that they all move away at the same speed.  It's distance dependent - further away = faster moving.  So not the same speed, but from every viewpoint in the universe the same pattern of faster when further away.


Question about $3p/c^2$ in Friedmann equation

There are two Friedman equations.

One related to the energy-density component in the Einstein field equations.
One related to the pressure component in the Einstein field equations.

The Einstein field equations are the following ten equations (16 equations for 4x4 tensor components but only ten of them are independent due to symmetry)

$$G_{\mu\nu} + \Lambda g_{\mu\nu} = \kappa T_{\mu\nu} = \frac{8 \pi G}{c^4} T_{\mu\nu}$$

where $G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} g_{\mu\nu}R$ is the Einstein tensor, $g_{\mu\nu}$ the metric tensor, $\Lambda$ is the cosmological constant, $\kappa = \frac{8 \pi G}{c^4}$ is the Einstein constant, $T_{\mu\nu}$ is the energy-momentum tensor

Friedman was actually assuming that only the matter/energy density component is non-zero $T_{44}= \epsilon = \rho c^2$. He assumed "Die Materie ist inkohärent" (ie non interacting dust with $p=0$) by which the components $T_{ii}$ are non-zero for $i \neq 4$. The other components (energy flux and momentum) are set equal to zero based on the assumption that the motion of energy and matter is considered to be negligibly small.

LeMaître kept the pressure in place arguing that the pressure associated with the kinetic energy of matter might be negligible but this may not be the case for radiation pressure. So he used also $T_{11} = T_{22} = T_{33} = -p$

Using the Robertson Walker metric metric $d s^2 =  dt^2 - a(t)^2 \left[ \frac{dr^2}{1-k r^2}+r^2d\Omega^2\right]$ the equations become expressions in terms or the scale factor $a(t)$:

$$\begin{array}{rcrcl} G_{00} &=& c^{-2}\frac{3k}{a^2} + 3c^{-2} \frac{\dot a^2}{a^2} &=& \frac{8 \pi G}{c^4} \rho c^2 - \Lambda  \\ -G_{11} = -G_{22} = -G_{33} &=& c^{-2}\frac{k}{a^2} + c^{-2}\frac{\dot a^2}{a^2} + 2 c^{-2} \frac{\ddot a}{a} &=& -\frac{8 \pi G}{c^4} p + \Lambda \end{array} $$

If you subtract the second equations from the first equations

$$G_{00}+G_{11}+G_{22}+G_{33} = -6 c^{-2} \frac{\ddot a}{a} =
  \frac{8 \pi G}{c^4} (3p+ \rho c^2) + 2\Lambda   $$

and rearrange some terms, then you could also write

$$
 \frac{\ddot a}{a} = -\frac{4 \pi G}{3}  \left( \rho + \frac{3p}{c^2} \right) + \frac{ \Lambda c^2}{3}   $$


Why is the equinoxes only 2 days and not more?

First, don't think that equinox is when the day and night are the same. It is the moment, when the sun's declination is zero, or when the sun is directly over the equator. Let me explain ...
Earth is always rotating around the Sun, so declination of the Sun is always changing (except at the solstices when it stops for a minute and goes in the other way). As @JohnHolst has pointed out, Sun's declination is changing for 1 angular minute per hour (mark 1 mm on the 3 m of distance – this is nearly 1 angular minute). An equinox doesn't last for a day. It lasts for a moment when the Earth is totally aligned and straight with the sun:
(this figure shows the geocentric model, which is easier to represent and imagine)

Also, Wikipedia says:

An equinox is commonly regarded as the instant of time when the plane (extended indefinitely in all directions) of Earth's equator passes through the center of the Sun. This occurs twice each year, around 20 March and 23 September. In other words, it is the moment at which the center of the visible Sun is directly above the equator.

P. S.
You can look this simulation.


When does a solar eclipse become noticeable?

It's hard to say because the situation is very similar to this question:
How much of the Sun's disk must be covered for a visible shadow to be cast?
The major problem lies in the sensitivity of the human eye, which is able to detect illumination changes sooner or later. Basically, there are people, who might not know about the ongoing solar eclipse until 98% obscuration! This is driven by the aforementioned human eye sensitivity and the surrounding scene, which keep seemingly the same level of illuminance for an observer.
I did a few observations both in solar and lunar eclipses and I have basic advice for you: try to remember the brightness of your scene at the very beginning and watch carefully how it changes against objects illuminated by other light than sunlight.
If, for instance, you are watching the eclipse outside, but close to the building, which rooms have a light turned on, you can try to detect this artificial light on the surface underneath this window (especially when it's placed in the shadow).
However, I assume, that for this purpose you will use a camera. In your device, you can set fixed image parameters and take the snaps at the same time intervals. Afterward, you can see how they vary from each other as the eclipse progresses.
See what I did for the lunar eclipse

The solar eclipse is analogous.
In the detection of the "visual start" of the solar eclipse, some amateur measurements can be helpful. You can install some light measurement applications on your phone. Some examples of use and pros & cons you can find here:
http://www.mkrgeo-blog.com/light-around-us-and-how-to-measure-it/
The smartphone headed towards any part of the sky should show how the level of light in your area changes as the eclipse progresses. With help of the images mentioned above, you will be able to detect the "beginning of visual eclipse visibility". Another example (also for lunar eclipse) you can find here:
http://www.mkrgeo-blog.com/remote-observation-of-the-total-lunar-eclipse/
and for a solar eclipse is here:
http://www.mkrgeo-blog.com/deep-partial-solar-eclipse-in-scotland/
http://www.mkrgeo-blog.com/light-level-measurements-during-total-solar-eclipse/
with visual examples I did myself

while also with artificial lights, which are highly recommended during observation such this. I used car dimmed headlights, but streetlamps are undeniably the best, because you can see also their light reflected from the ground, when gets darker.

http://www.mkrgeo-blog.com/light-level-measurements-during-total-solar-eclipse/
Bear also in mind the cloudiness. The observation above was carried out under the mostly overcast sky. In this case, clouds reflect the sunlight, hindering it from further way down to the ground. In turn, it's darker than it could be under clear skies. When the eclipse occurs along with some thunderstorms passing above your observation place, then it's extremely hard to define, when the eclipse-driven conditions started to be noticeable, as very thick clouds can reduce the sunlight up to 1000 times. Read more details here:
http://www.mkrgeo-blog.com/light-scattering-in-the-earths-atmosphere-part-3-clouds-haze-and-surface/


What percentage of gas giant exoplanets are in the habitable zone?

Of the gas giants, the number of "hot Jupiters" is about the same as the number of "hot Neptunes." A little under 9% of known gas giants are in the habitable zones.

An article by John Wenz has the discoveries thus far collated into a chart:

The claim, "most of them are "Hot Jupiters", orbiting so close to their stars that the heat causes them to swell up like a hot air balloon," is at best a bit of a stretch - the only reference that I found for this was a Gizmodo article about Kelt-9b. The Nature paper, A giant planet undergoing extreme ultraviolet irradiation by its hot massive-star host, that discusses Kelt-9b says,

  The planet is also extremely inflated relative to theoretical models, with a radius of $\sim 1.9\ R_J$. Poor redistribution of
  heat and radius inflation (both of which are also observed in WASP-33b), have been linked to high stellar insolation, although the exact physical mechanisms remain uncertain.

So they did observe that Kelt-9b had a larger radius than is normally expected from a planet of that mass, and it receives massive amounts of radiation from its star, but it cannot be concluded from that that it is the same as putting a balloon filled with air near a heat source.


When is the belt of Orion slanted from right to left, going up? (Egypt)

EDIT: entirely new answer based on comments

You may actually be on to something here.

The most obvious (and thus least interesting) answer is that they were
depicting Orion shortly after sunset, not shortly before sunrise (all
times UTC):

Here is the helical rising of Orion today, and it does indeed go right
to left:

If we go back 1000 years, it's much closer to being a straight line:

And if we go back to 1473 BCE (which is -1472), the rising is clearly
left to right:

Interestingly, if we go far enough south, the rising is right to left
even in -1472:

Note that position in which Orion rises doesn't change much from day
to day. In other words, the position of Orion's stars at helical
rising are the same as the position of Orion's stars at rising any
other day of the year. Because of precession, the position does
change over long periods of time, but not day to day.

The only problem with the "most obvious" answer above is that we know
the Egyptians were very interested in helical risings, so it would be
odd to see a depiction of a "helical setting".


Wormholes are a theoretical solution to Einstein's equations, but, unlike black holes, they have never been observed and probably don't exist.

If they did exist, they would almost immediately close due to the presence of matter in them. To stop them from closing you would need some "exotic matter" with remarkable properties, like "negative mass". Exotic matter has never been observed, and probably doesn't exist.

Trying to push something massive, like a black hole close to a wormhole would almost certainly cause the wormhole to close. Perhaps if the wormhole were sufficiently big with enough negative mass to stabalise it it would be possible for the black hole to traverse it. The black hole would pass through the worm hole just like anything else. At the other end you would see the black hole coming through the wormhole.

But remember, fundamentally, wormholes don't exist.


Could an impact have resurfaced Venus 300 million years ago?

Would an impact event leave visible traces like impact basins, or
  could the entire surface melt and reform as it is today, as I suppose
  Earth did when the Moon formed? Could Venus have been a very different
  planet up until 0.3 billion years ago? How could one find out, what
  kind of investigation would be needed?

Certainly giant impacts were fairly common, though I don't think they were common any more as recent as 300-500 million years ago.   That would have been a kind of unusual event.   Mostly they occurred when the solar-system was young.

Mercury probably had one

Venus may have too - and I recommend this article as very loosely related.

The Earth had one - as you know.

and Mars too, but not quite as "giant", though still very big and it left a visible mark that spans nearly 1/2 of Mars' surface.  See Here and Here.

The Moon even had one and there is evidence in the thickness of the Moon's crust and location of it's volcanic magma pools/dark spots.   See Here and Here.

The point I'm trying to make with all these examples is that unless the Impact is very very large, it would leave a impact footprint of sorts.   The impact has to be large enough to not only liquefy the surface but to also allow the surface enough time to reset in mostly uniform layers so you don't get 1/2 of the planet with a thicker crust than the other half.   Impacts of that size are statistically improbable, within the last 300 million years.  Most of the big impacts happened long before then.

Wikipedia suggests that there's other reasons for believing the global resurfacing though interior heat idea.  Venus Global resurfacing event - Wikipedia - Venus lack of a magnetic field, it's influx and loss of water and D to H ratio and it's Sulfur content.     Do these have had to happened through internal heat resurfacing?   Now, the article says there's no proof, only some pretty good evidence, so it it possible Venus resurfacing was due to an impact?    I see no reason why it wouldn't be possible.

Given Venus' Heat trapping atmosphere, higher temperature and greater gravity, which also means, greater impact velocity, at least escape velocity of about 10 km/s where as the Mars impact that covers a hemisphere was thought to be about 5 km/s and the impacting object into Mars, at least 1,600 km in radius which is nearly the size of the Moon.   You might not need that big to impact Venus and create a complete resurfacing, but I wouldn't think you could go too much smaller.

I love the Venus had a moon that crashed into it idea cause that seems to work on a few levels.  A giant impact that created Venus could have happened 4 and change billion years ago and created a Venusian Moon, which over time, could both slow Venus' rotation and work it's way towards Venus in the process, eventually crashing into it.   It seems at least, possible, though I'm not sure what evidence could be gathered to prove it.    I like your idea a lot and I think it's much more likely than this one:  Earth stole the Moon from Venus.

As to Venus being very different prior to it's resurfacing.    The sun was only about 3%-5% less luminous when Venus resurfaced and given Venus' proximity to the sun, I have a very hard time seeing Venus as not already in a run-away greenhouse effect even then.   I love the Venus once had oceans and maybe life hypothesis, but with a CO2 rich atmosphere, it's hard for me to see it as likely.   My guess is that Venus was hot and virtually without water 300 million years ago too.   Unless it somehow stored a huge amount of it's CO2 underground, I have a hard time not seeing it as too hot for oceans 300-500 million years ago.   Maybe, but I'm skeptical.    It would be very neat to have a picture of what Venus was like prior to it's resurfacing and over the hundreds of millions and billions of years before that.

Edit

In this article it mentions that it's improbable that Venus once had a Moon based on angular momentum.   A planet-Moon system maintains it's angular momentum and if the moon spirals in towards the planet, it should carry significant angular momentum with it.   Such an impact should give Venus a pretty rapid rotation which would be difficult to explain given that it rotates so slowly now.  Unless perhaps, the Moon was in a retrograde orbit, Moon going in one direction the planet in another and the combined angular momentum largely cancelled out, but adding more specific conditions tends to decrease the probability.   I still like the hypothesis, but the article above says it's unlikely.


If Earth is tilted, why is Polaris always above the same spot?

You are correct that the axis of the Earth's rotation is tilted with respect to the plane of its orbit by 23 degrees. But it is incorrect that the direction that the axis points changes by a large amount (it should be 2*23 degrees) over a 6 month time span.

Your assumption: If axis it pointed at Polaris at (1), then it should be pointed at a different star at (2).

The axis remains pointed in the same direction throughout the entire year because the laws of physics are that the axis of a spinning object remains pointed in the same direction unless a torque acts on the body to change its orientation. There are torques acting on the Earth (namely the Sun and Moon), but it takes 13000 years to change the direction from "one way to the opposite way", not 6 months.

Correct: The orientation of the Earth's spin axis remains pointed at the same star throughout the year (ignoring the slow, 26000 year precession cycle).


What would the sun look like from aphelion of The Goblin's orbit?

Yes, easily.

1955 AU is a long way, but it is only 0.03 of a light year. The sun would still be less than 1/100 of the distance to the next star.

The brightness of the sun would be much less: about 4000000 times less bright, but that is about 16 magnitudes. The sun from Earth is -27 magnitude, so the sun would appear as a magnitude -10 star. It would be much brighter than any other star, and would easily outshine Venus as seen from Earth. In fact it would be 100 times brighter than Venus. It would be as bright as a half-moon (strong enough to cast shadows, but not to read by)


Are there any astronomical phenomena that could emit strong radio waves with multiples of a discrete frequency?

I think you've misunderstood the article - the quantity that seemed to be occurring at integer multiples of some number isn't the frequency of the radio emission but rather the dispersion measure (DM) of the source. As photons travel through the interstellar medium, interactions with free electrons mean that lower frequency photons take longer to reach the observer than higher frequency photons. The time delay of between photons with frequency $\nu_1$ and $\nu_2$ is
$$t_1-t_2=4.15\left(\frac{\text{DM}}{\text{pc cm}^{-3}}\right)\left[\left(\frac{\nu_1}{\text{GHz}}\right)^{-2}-\left(\frac{\nu_2}{\text{GHz}}\right)^{-2}\right]\;\text{ms}$$
with the dispersion measure defined as the integral of the free electron number density $n_e$ over the path between the source and the observer:
$$\text{DM}\equiv\int n_e\;d\text{l}$$
In a sense, the dispersion measure is a distance proxy. If we had some uniform $n_e$, then the DM would scale perfectly linearly with the distance to a source. That said, the Galactic electron density is not uniform in all directions; it's obviously much higher in the disk, meaning that a source inside the disk some distance from us will have a lower DM than a source in the halo the same distance away. Note that the DM isn't an intrinsic property of the source - it depends on the observer's location.
The point of the paper cited in the article (Hippke et al. 2015) is that the dispersion measures of the 11 fast radio bursts (FRBs) appeared to fall near integer multiples of 187.5 pc cm$^{-3}$. There's certainly variation, although I don't know how the deviations compare to the measurement uncertainties. The authors argue that this could be explained if FRBs are actually artificial - in other words, if there's some sort of human-made source emitting at different frequencies with each frequency emitted at a different time mimicking the effects of interstellar dispersion.
Our knowledge of FRBs has since grown as the sample size has expanded dramatically, and most known DMs vary quite a bit from the trend. I don't know if anyone has looked at this small sample again, but at this point, the extremely broad consensus is that FRBs are astrophysical (mostly extragalactic) sources. A broad range of mechanisms have been proposed over the years, with some since ruled out and others supported by further observations.
A note on pulsars: The article isn't stating the pulsars aren't FRB sources - the sample size back then was far too small to rule them out - but rather that fast radio bursts differ significantly from typical pulsar emission, which tends to be uniform. There are still viable FRB mechanisms involving pulsars, such as giant pulses of young pulsars, and neutron stars in general play a role in some currently-favored theories, such as magnetar flares.
All that said, I think it's worth answering the title question. It is indeed true that some radio sources emit at integer multiples of some frequency - namely, spectral lines from rotational transitions in molecules. In a molecule with moment of inertia $I$, the energy of the state with the angular momentum number $J$ is
$$E_{\text{rot}}=\left(\frac{\hbar^2}{2I}\right)J(J+1),\quad J=0,1,2,\cdots$$
A transition from $J$ to $J-1$ then releases an energy
$$\Delta E_{\text{rot}}=\frac{\hbar^2J}{I}$$
and leads to the emission of a photon with frequency
$$\nu=\frac{\hbar J}{2\pi I}$$
As a notable example, the $\text{CO}$ molecule$^{\dagger}$ emits a photon at $\nu\approx115\;\text{GHz}$ during the $J=1\to0$ transition, at $\nu\approx230\;\text{GHz}$ during the $J=2\to1$ transition, and so on. Keep in mind, though, that fast radio bursts are broadband sources, and while spectral radio lines will exhibit some broadening through certain mechanisms, it's simply not comparable - the emission profiles are completely different.

$^{\dagger}$Specifically, the common isotope $^{12}\text{C}^{16}\text{O}$. Other isotopes like $^{13}\text{C}^{16}\text{O}$ have corresponding lines at slightly different frequencies.


How can tidal heating lower Io's orbit?

It doesn't, at least not to first order. The first order effect is that tidal heating acts to circularize Io's orbit. Counter to that, orbital resonances with Europa and Ganymede act to make Io's orbit more elliptical. This leads to a nice hysteresis loop.

Suppose Io is in a fairly circular orbit. This results in reduced tidal stresses, thereby making Io cool down. A cooler and hence more rigid Io is less susceptible to tidal deformations than is a warmer and hence more plastic Io. Given two bodies in the same orbit, one warm and plastic, the other cool and rigid, the warmer body will suffer more tidal deformations than will the cooler one. This is captured by the object's $k_2$ Love number. The inevitable lag in response means that the response will not be symmetric about periapsis/apoapsis for an elliptical orbit, and the greater the plasticity, the greater the greater asymmetry. This is captured by the object's tidal quality factor $Q$.

This cooling of Io as its orbit becomes close to circular enables the ever-present resonance effects to now dominate over the circularizing effects. Io's orbit slowly becomes more elliptical. That elliptical orbit increases tidal stresses on the cold, rigid Io, eventually making it start to warm up and become more plastic. The circularization effects grow as the orbit grows more elliptical and as Io's interior becomes more flexible and more plastic. Eventually the circularization effects dominate over the orbital resonance effect, making Io's orbit become more circular -- until the cycle repeats.

Rinse and repeat, at least as long as that three-way orbital resonance between Io, Europa, and Ganymede holds up. How long that three-way tidal resonance has existed and how long it will last is, as far as I know, unknown.