At what depth on Mars would the atmosphere have equal pressure of that on Earth?

Nasa has a atmospheric model of mars:
$$0.699 *e^{-0.00009 h} $$

A naive application of this model, solving for a pressure of 101 kPa, gives a depth of -55 km.

The Armstrong limit depth (at which water boils at body temperature) is -24km

The model assumes constant temperature, and gravity (it doesn't correct for the fact that at 55 km below the surface you would be well into the martian mantle and the temperature would be very much higher, and deep enough for a measurable difference in gravity). There's no "goldilocks depth" at which you would only need an oxygen supply.

These depths are not achievable with current technology. The deepest mines on Earth are about 4km deep, and even the Kola superdeep borehole only managed 12km


The Sun or Our Galaxy revolves around what?

As a previous comment states; it is complicated. But in principle the Sun is just one object in a large collection of objects which we call the Milky Way Galaxy. All the objects in this Galaxy orbit the barycentre (or barycenter, if you prefer the American spelling). The barycentre is, essentially, the common centre of mass of the whole galaxy. In the same way the planets in our solar system don't actually orbit the Sun, they orbit the Solar System Barycentre.

As for the Milky Way Galaxy, that too is a member of a larger group of galaxies, and they all orbit their common barycentre.


Has the Mars dust storm affected its naked-eye appearance?

Is it possible for an experienced naked-eye (or small telescope)
  astronomer to have noticed the change in appearance brought on by this
  dust storm?

Yes, it is definitely possible. Without dust storms it looks redder.

Source:

The opposition of Mars, 2001: Part I - Richard McKim


Say we were to bring Titan to Earth's orbit. How much would its atmospheric pressure change?

As Temperature rises, so would the pressure, this would work immediately, as heat from the sun reaches Titan's surface.

The atmospheric escape (and thus, pressure loss, as you describe it) would work on a much longer timescale.
We can make a few estimates to determine this timescale: On the most simplified level, the composition if Titan's atmosphere is that of Earth's - it's mostly molecular Nitrogen.
So the mean molecular weight $\mu$ would be the same. The temperature would be similar compared to Earth by assumption of your experiment, then the only important difference is the mass and radius of the two bodies, reducing the escape speed by a factor of ~10.

Titan has 2% of the mass of Earth, so the ratio of escape speed to thermal molecular speed at the exobase, from where molecules are lost, would approximately scale like $\sim 1/\sqrt{10}\approx 3.1$
To determine escape rates, those go roughly like $\sim \exp(v_{th}/v_{esc})$ and would thus increase by $\sim \exp(3.1)\approx 22 $, compared to today's level.

I don't know present estimates for the lifetime of Titan's atmosphere, but this factor would e.e. bring it down from 5 billion years to only 250 million years. That's really short, geologically speaking!

But of course it's not enough to affect you in your pressure suit on the surface. You'd rather be concerned about an overpressure than underpressure.


Calculating Radial Velocity from Hipparcos

You cannot calculate the radial velocity from a proper motion. The proper motion is the motion you see the star move on the sky, so perpendicular to your line of sight towards the star. The radial velocity is the velocity of the star along this line of sight. Perhaps this picture will help:

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

Radial velocities are actually calculated by measuring the Doppler shift of the spectra of stars, but this is not very easy, and you need the raw spectrum data. You are probably best off just taking the values of the radial velocity from that site for example.


Happy lunar new year! When will "sunset" happen for Chang'e-4 and Yutu-2?

tl;dr: To an accuracy of about a half of a day:

Third quarter is sunrise for Chang'e-4 and Yutu-2
First quarter is sunset for Chang'e-4 and Yutu-2

Moon seen from Earth:
From Wikipedia's article Selenographic coordinates

Longitude
Longitude on the Moon is measured both east and west from its prime meridian. When no direction is specified, east is positive and west is negative.
Roughly speaking, the Moon's prime meridian lies near the center of the Moon's disc as seen from Earth. For precise applications, many coordinate systems have been defined for the Moon, each with a slightly different prime meridian. The IAU recommends the "mean Earth/polar axis" system,2 in which the prime meridian is the average direction (from the Moon's center) of the Earth's center.3
2 https://astrogeology.usgs.gov/Projects/WGCCRE/WGCCRE2003preprint.pdf
3 http://lunar.gsfc.nasa.gov/library/451-SCI-000958.pdf

Simplifying further, seen from Earth, 0° longitude wobbles near the center of the lunar disk. (see also 1, 2, 3)
Chang'e-4:
Chang'e-4 is in Von Kármán crater (also here) at about 44.8° South, 175.9° East.
So Chang'e-4 and Yutu-2 sit almost diametrically opposed to the Earth longitude-wise, and so sunset on the Moon's prime meridian is close to sunrise for the pair, and vice-verse.
Answer:

What phases of the Moon as seen from Earth correspond to times of sunrise and sunset for those two?

First quarter is sunset for Chang'e-4 and Yutu-2
Third quarter is sunrise for Chang'e-4 and Yutu-2

Location Confirmation:
We know where Chang'e-4 landed because it returned video of it's own landing.
YouTube Video: https://www.youtube.com/watch?v=jyjyf4ANGIo
From How did Chang'e-4 hover, rotate, and then descend so gracefully? (Video).
Additional verification that Chang'e-4 and Yutu-2 are in Von Kármán crater:
From The Lunar Reconnaissance Orbiter Camera recent post First Look: Chang'e 4:

Arrows indicate position of Chang'e 4 lander on the floor of Von Kármán crater. The sharp crater behind and to the left of the landing site is 3900 meters across (12,800 feet) and 600 meters (1970 feet) deep. Image was shrunk by more than a factor of ten (full resolution available inside the post). LROC NAC M1303521387LR [NASA/GSFC/Arizona State University].

Area around lander enlarged by a factor of two relative to the native pixel scale (see zoomify below), bright speck between two arrows is the lander. The large crater in the center (just right and below arrows) is about 440 meters (1440 feet) across. LROC NAC M1303521387LR [NASA/GSFC/Arizona State University].


Photometric redshift - empirical methods (machine learning)

This was too long for a comment, but is not a real answer since I'm not completely sure, but:

My guess is that "representing the galaxies" refers to the "type" of galaxy that you intent to observe, where by "type" I mean e.g.
Lyman-break galaxies (Steidel et al. 1996),
Lyman $\alpha$ emitters (Partridge & Peebles 1967),
sub-millimeter galaxies (Blain et al. 2002),
(U)LIRGS, etc.

These terms all refer to selection methods (i.e. observational techniques), and hence also to physical differences. The closer the training data are to the observed sample, the better your algorithm will be at linking their redshifts to their photometric properties.

Mass is just one property, there's also e.g. dustiness, star-burstiness, stellar population, age, and others. "Region" is probably less important, but since the clustering of galaxies also affect their properties (e.g. their morphology, Dressler 1980), it could potentially influence the result.


very bright object (visible even after sun risen)

This is the planet Venus.  It is always bright (when visible) but exceptionally so at the moment.  Venus shows phases (similar to the moon) and is currently nearly half illuminated.  That, combined with its relative closeness (about 75million km) is why it is particularly bright at the moment.

It is often visible in broad daylight, but can be difficult to spot because of the bright sky background.


What should I read next after Neil deGrasse Tysons "Astrophysics for people in a hurry"?

I liked Black holes & time warps by Kip Thorne. Here's the link. While being easy to understand, he does go into some details regarding black holes and the warping of time, which I think are interesting to the casual reader. If you have a science degree, you might not learn much new informations, but it's a good read.


Does the lower mass of Uranus suggest it swapped position with Neptune?

The Nice model of solar system evolution proposes that both Uranus and Neptune migrated outward from their original orbits due to numerous encounters with planetesimals,
stimulated by a temporary 2:1 resonance between Jupiter and Saturn.
Simulations going back billions of years can put either Uranus or Neptune initially closer to the Sun.
Desch 2007, assuming that the Nice model is correct, argues for a particular density profile of the protoplanetary disk and notes that a closer proto-Neptune fits that profile better.
Some solar system scientists have problems with the Desch model, but Uranus's smaller mass is consistent with it.


Is the equinox absolutely true east everywhere?

The celestial equator's plane intersects the horizon due east and due west, no matter where you are on Earth (except at the poles where the horizon is the celestial equator).

Those that have the sunrise at the exact time of the equinox will see the sun rise in the east (90º). For the rest, it will be too late, the sun will have made the crossing, and it could be something like 90.00001º or 89.99999º, although it will still be perceived as due east.

The last equinox was March 20th, 03:50 (UTC). At that moment the sun was rising very close to Maputo, Mozambique (table), so everybody on that meridian saw the sun rise in the east, exactly at 90º.


I agree with the comment by StephenG. This is a stuck pixel. As the link suggests, they are a very common flaw in digital cameras, and not an astronomical feature.

I think this because it appears, on zooming in, to be a single pixel defect, most light sources even with the best focus, cover more than 1 pixel. The filter you are using is very powerful removing nearly all light. An object would have to be exceptionally bright to be seen through such a filter. If it was real then it would be so bright that when you removed the filter, that one spot would have totally overexposed your image. Since stuck pixels are pretty common, and you might not have noticed them in a picture taken at day, or of a star-field, it is a safe bet that that is the problem.


Your "GSC 2.3" entry is actually from GSC 2.4.1, as shown in the header lines containing the string "GSC241".
According to the record structure for GSC 2.4.1, Vmag is "V-band magnitude" while VpgMag is "Photographic V-band magnitude". The magnitudes are associated with a code that describes what bandpass/filter they are using. In your case, the VpgMagCode field is 1, which corresponds to the emulsion+filter IIaD+W12 (au used by the Palomar Quick-V survey). Confusingly the VMagCode field is set to 3, which corresponds to a Johnson B-magnitude. Perhaps this is an error in the catalogue and should be 4 (Johnson V-magnitude)? The source code of 19 indicates the magnitude is from APASS (the AAVSO Photometric All-Sky Survey). The APASS website provides some further information on the magnitudes used:

Johnson B and V were chosen to extend the Tycho calibration to fainter magnitudes and to match the many archival published datasets. These passbands can be easily used to create sequences for any VSX / GCVS star. Sloan passbands provide a direct link to the SDSS, SkyMapper, PanSTARRs etc. surveys, and provide a homogeneous catalog of brighter stars.  The Johnson B and V magnitudes are on the Vega system; the Sloan u',g',r',i',z_s and Z magnitudes are on the AB system.

A few more details on the Johnson magnitude system can be found on this web page: the system is set up so that the star Vega has a V-magnitude of 0.03, and the colors (e.g. B-V) are 0. Details of the Johnson-Cousins UBVRI filters can be found on the website of the Leibniz-Institut für Astrophysik Potsdam.
The details of the magnitude in SIMBAD can be obtained by going to the reference listed there: Norton et al. (2007). This paper indicates that the magnitude is from the SuperWASP photometric survey. The details of the filter the SuperWASP survey uses can be found on the SVO Filter Profile Service. Norton et al. (2007) describe the value in section 3 of the paper:

the mean SuperWASP magnitude,
defined as $-2.5 \log_{10} (F/10^6)$ where $F$ is the mean SuperWASP
flux in microVegas; it is a pseudo-V magnitude which is comparable to the Tycho V magnitude

A microVega is a millionth of the flux from Vega. Note that the above formula implies a V-magnitude for Vega of 0, in contrast to the Johnson system.
As for which one you should use, it depends on what you are doing. If you're comparing with other SuperWASP results (or perhaps objects in the Tycho survey), use the SuperWASP V-magnitude. If you're comparing with other magnitudes in the Johnson system, go with Vmag from GSC 2.4.1.


This may be a partial answer depending on what you do or don't know how to do at the moment, and the level of accuracy you require. Feel free to add some feedback.
If you'd like to use (or learn to use) Python then you can solve this problem trivially using Skyfield!
It depends on the level of accuracy that you need. The time difference will be of the order of 20 milliseconds for a 6000 km difference in light-path distance for example, but in that time the Earth moves (in some direction) about 0.02 * $\sqrt{GM/a}$ where GM is 1.327E+20 m^3/s^2 and $a$ is about 1.5E+11 meters, or about 600 meters.
If you don't want to worry about small corrections right now, just calculate the dot product between the normal vector pointing from the direction of the event and the vector drawn from the known observation site to the proposed site, that's the path length difference. Divide that by the speed of light to get the time difference.


Why is the Sun's atmosphere (the corona) so hot?

It is still an open question, even though it is clear that it is linked to the magnetic field of the Sun. The following hypothesis are the following:

Heating waves mechanism: propagation of magnetohydrodynamic waves can heat significantly the corona. These waves can be produced in the solar photosphere and propagate carrying energy through the solar atmosphere; they turn in the end into shock waves that dissipate energy as heat in the corona.
Magnetic reconnection mechanism: magnetic field lines are tied to the surface of the Sun and even when the plasma moves, they stay tied to their original position (their "foot points"); therefore, plasma movements can drag the magnetic field lines and entangled them in geometries such as loops, that can eventually "reconnect". "Reconnection" corresponds to a modification of the field line topology (see the image that pictures it better than my words). During this process, heat and energy are released quite efficiently.
Nanoflares mechanism: the dissipation of a lot of small scale structures (slow convective motions of magnetic foot points could create current sheets, coutinually dissipated and reformed, providing heating through Ohmic dissipation). You can see it as a kind of a lot of magnetic loops reconnecting together, instead of one big loop reconnecting alone.

Black lines represent the magnetic field lines. They are tied to the surface of the Sun (the foot points). One of the magnetic field lines is looping, and it is "crossing itself". When it reconnects, it forms, on the top, a magnetic loop and, on the bottom, an magnetic arch. As pointed out by this diagram, magnetic reconnection mechanism is also associated to Coronal Mass Ejection (CME), since the energy released by magnetic reconnection is powerful enough to eject a significant amount of matter.

Source:  A Contemporary View of Coronal Heating


Movement of the satellites of the planets . . .

Inner moon orbits tend to low inclinations with regard to planet's equatorial plane. Charon is an exception to this general rule. Nineplanets.org has a nice data base.


At what annual rate are new exoplanets being recognized? How does it compare to new asteroids?

Assuming exoplanets have been discovered at a constant rate since 1995 (not very true) then the rate is very approximately 117 exoplanets a year. Comparing to the Minor Planet Center (MPC) archive stats which are plotted below:

which shows that approx. 761,000 asteroids of all types have been discovered since 1995 April or roughly 22,000 a year. (The lower black line is for the smaller number of asteroids that have been seen enough times and the orbit is secure enough that they can be numbered).

Exoplanets suffer from magnitude and volume-limit problems in that to get a large enough number on the sky to discover large numbers of them, requires going faint. Since confirming an exoplanet and ruling out other explanations such as blended background stars and M dwarf or brown dwarf binaries (which give similar transit depths but different radial velocity amplitudes) requires a large additional effort and high precision spectroscopy to confirm them, this becomes the bottleneck requiring tens or more of hours of large (4m) telescope time to confirm each of the fainter exoplanet candidates. Additionally the most numerous exoplanets in the 2-15 times of the Earth range are very tough to confirm as the RV amplitude is very small (few m/s or smaller depending on the orbital period).

  I'm also wondering if there is any chance there will be a crossover in the rates in the next few years.

Because of the limitations discussed above, the chances of there being a cross-over in rates any time soon is slim to none.


For any given meteor shower in a given year (e.g. Quadrantids, now) where can I see a rough prediction of how many days it will last?

The International Meteor Organization publishes an annual Meteor Shower Calendar.
The PDF document for 2020 predicts a Quadrantid peak on January 4, 08:20 UT, and says:

  The peak is short-lived with an average duration (full width at half-maximum, FWHM – that is the period with ZHR above half of the peak level) of about four hours.

Unfortunately it only gives peak durations for a few showers.


What is the galaxy M87 doing these days?

The answer to your question is relatively straightforward: Not much. I'll delve into it in two parts: Star formation and the activity of the supermassive black hole at the center.
Star formation
M87 home to many Population II stars. They have little hydrogen and helium in them, and the clouds of gas and dust in the galaxy don't have much, either. In fact, many dust clouds can be destroyed by the intense radiation emitted by the accretion disk of the central black hole. This combines to make it seem unlikely that there will be much star formation in the galaxy.
That said, just because a galaxy doesn't have a lot of materials for star formation doesn't mean it can't form lots of new stars - at least, no under certain special circumstances. Galaxy collisions can induce high rates of star formation - starburst galaxies are good examples of this. While M87 might not be at a high risk for a collision, there is still the possibility of one. M84 is nearby, and may have had an encounter with M87 in the past. While it is a long shot (and M84 has a low star-forming rate), an interaction with the two could spawn new stars.
The supermassive black hole
You seemed particularly interested in the black hole at M87's center. It is a supermassive black hole, with a mass billions of times that of the Sun. Quite a monster, by many standards. Yet it is not doomed to swallow the galaxy whole anytime soon. It is sucking in gas and dust at a rate of $\sim10^{-1}M_{\odot}$ each year (see Di Matteo et al. (2002)) - and there are trillions of solar masses of gas, dust, and stars in the galaxy. For comparison, Sagittarius A* accretes matter at a rate of between $\sim10^{-8}M_{\odot}$ and $\sim10^{-4}M_{\odot}$ per year, depending on the exact model (see e.g. Quataert et al. (1999) and Shcherbakov et al. (2012). Still, the supermassive black hole in M87 couldn't accrete all of the matter in the galaxy in the age of the universe, if it continued at this rate.
In summary: Star formation in M87 is at a low rate, and while a galactic interaction could create a new wave of star formation, it is unlikely. The central black hole is sucking in lots of gas and dust, but it will take billions upon billions of years before it eats up a substantial amount of matter.


What would be the best inertial reference frame to describe objects in the solar system (other than the solar system's barycenter)?

Use the initial Center Of Mass (COM) as an origin.

It's trivial to show that the COM of a closed N-body system (in Newtonian physics) moves with a constant velocity.  This is a relatively easy system to work with and transforming to other coordinate systems, such as the heliocentric one you mention, is relatively trivial compared to computing the 3D N-body system with good accuracy.  A word of advice : if the math of converting between coordinate systems is a problem for you, the mathematics and algorithms of an accurate simulation will be likely much worse.  I'd suggest trying out such conversion mathematics as a warm up exercise and if you're comfortable with that then dive into the simulation.

COM can be shown to move at constant velocity because it's a closed system (or can be treated as such over the timescale you mention), and so net force on the COM is zero, meaning it has to move with constant velocity.

Adding a small body like an extrasolar asteroid makes next to no difference to the system as a whole.  It's a minute change.  You can treat such objects as special cases, ignoring their effect on the system and just calculating the effect of the system on them.  If you're planning to add very large bodies and see what happens then you need to include them in the general system all the way through the simulation.

Note there are a lot of existing simulations out there that implement a wide range of approaches.  This is a problem that has been extensively studied and a search online would yield a myriad of approaches.

The motion of the Barycenter is complex.  I'd avoid this complication in what is already gong to be a messy mathematical system to simulate.

Your main priorities should be gathering accurate data for initial starting values (non trivial) and choice of algorithms.  Choice or ordinate system is comparatively minor.


Be aware there are differing formats. SPK is the newer format - the older text format seems to be deprecated.

It's worth having a look at the documentation on this jplephem python library as it explains the format reasonably well. If you do use SPK, you'll need to read all of the documentation in NASA's SPK repository

I'm not sure if this is from the same author as this library - worth a read as it discusses the FITS format.


Our galaxy, the Milky Way, forms part of a galaxy group called The Local Group. What are the parameters that dictate what a galaxy group is?

A galaxy group is a set of galaxies that are close together and gravitationally bound, i.e. barring outside influences they will stay together indefinitely. That means a galaxy that is in the space occupied by the group but "passing through" it at a high velocity would not be considered part of the group.

The upper limit for calling it a group is roughly 50, bigger accumulations of galaxies are called clusters (which may consist of many groups). The borders between adjacent groups may be uncertain.

These aren't mathematical definitions and actual usage is often based on tradition more than on measurements. It's just like geography: you could debate endlessly whether a given piece of high ground should be called a hill or a mountain, and whether something is a separate mountain or a subpeak, but sensible people recognize it as a waste of time, make up a clear but arbitrary numeric distinction without strongly insisting on its usage.


How long would it take a Boeing 747 to go around Jupiter?

It will depend on the speed of the jet, however:

Jupiter has an equatorial circumference of 449,200 km, a current-generation 747 has a top speed of 988 km/h so it will take about 455 hours or nearly 19 days (18.9) (this doesn't allow for re-fueling).


How precise are the observational measurements for the speed of gravity?

The accuracy is extremely good thanks to LIGO and a binary neutron star merger.  It's accurate to about 3 parts in 1 quadrillion!
See the paper Gravitational Waves and Gamma-rays from a Binary Neutron Star Merger: GW170817 and GRB 170817A
The paper constrains "the difference between the speed of gravity and the speed of light to be between $−3×10^{-15}$ and $+7×10^{−16}$ times the speed of light."
Older response before the LIGO observation:
One of the more recent of gravity was first measured in 2003 using Jupiter and a quasar.  The measurements were for the purpose of proving a finite speed and weren't very precise.  See the paper for more.
More recently in 2013 a Chinese group built a model using Earth's tides that helped them narrow it down.

... [T]he speeds of gravity are from 0.93 to 1.05 times the speed of light with a relative error of about 5%. This provides first set of strong evidences to show that the speed of gravity is the same as the speed of light.

This is so far the most accurate measurement I've seen.  See the paper for more.
In the near future, LIGO may be able to provide more accurate measurements by comparing the distance among detectors and the delay of observation.
Also related, see the question "How is the most accurate value of  G  measured?"


How is it possible for a collision to be responsible for Uranus's axial tilt?

For objects on the scale of a planet, the state of matter doesn't really matter much.  A colliding planet or planetesimal would not "just pass through"  The amount and density of the gas would prevent that.
Some easy experiments:  Put your hand outside a car as you are driving along, you'll feel that "gas" (ie air) has real substance and pushes against your hand quite hard.  Now imagine that you are driving at 20 km/s.... What happens when a piece of rock from space hits the atmosphere? It doesn't just pass through, it burns up in the air as a meteor.
Moreover the pressure and density in Uranus is huge.  Not so far below the surface are layers of water,ice, ammonia, methane.  And the pressure is so great that the density is more that that of water.
The collision of two gas planets could result in a combined planet that had a rotation axis inclinded at 90 degrees or more, and tidal effects and equatorial bulge effects would pull the moons into orbits around the equator.


What telescope might this be a model of, with Sir Fred Hoyle?

The photo looks like it could be from the late 1950s/early 1960s. Jodrell Bank's Lovell telescope had been commissioned and people were looking for the next big thing in radio astronomy.

That next big thing was the fixed Arecibo dish, with a diameter of 305m, commissioned in 1963.

It is clear though that the picture is not a (very accurate) model of Arecibo and the location is wrong. I would guess instead that this picture purports to be the design for a radio telescope on the Moon or is an upscaled model of the Dover Heights "sandpit" telescope (see below) built in the 1950s in Australia.


What percent of all stars that have lived (up to now) are stellar remnants?

Most of the remnants are white dwarfs, because their progenitors can be much shorter lived than the lifetime of the galaxy and are much more numerous than the higher mass progenitors of neutron stars and black holes.
A census of the local neighborhood suggests that about 10% of "stars" are white dwarfs. This number is low because (a) only stars more massive than about $1M_\odot$ have had time to become white dwarfs and (b) the birth stellar mass distribution is heavily weighted to stars less massive than this.
Extrapolation of this birth stellar mass function to estimate the numbers of stars born with $M>8M_\odot$ (almost none of which are still on the main sequence now) suggests there are a further $\sim 1$% of neutron stars and $\sim 0.1$% black holes, though these fractions remain to be observationally confirmed and the latter number is very uncertain even as a theoretical calculation.


Does a week represent something in astronomy?

The concept of a month derived from the amount of time it takes for the moon to cycle from new moon to new moon (which is roughly 29.5 days). The modern month has experienced changes from this original concept due to trying to fit a standard number of months within a solar year.

If you divide the lunar month into quarters, each quarter is approximately 7 days long. This is believed to be the origin of the seven-day week.


Do comet orbits lie in the same plane as the other planets?

Those two comets happen to have large orbital inclinations to the Ecliptic: 117$^\circ$ for
C/1858 L1 (Donati)
and 162$^\circ$ for
1P/Halley.
Long-period comet orbits are oriented essentially at random, suggesting that they originate in the
Oort cloud,
a sphere several thousand AU from the Sun.
Many short-period
comet orbits are inclined only a few degrees to the Ecliptic,
suggesting origins in the
scattered disc,
a torus several tens to a few hundred AU from the Sun.


Does this black hole magnetohydrodynamics equation even superficially make sense?

I believe your assumption of the partial derivatives with respect to time being identically zero is incorrect.  When you are talking about working in phase space, i.e., Liouville's theorem with functions of positions and velocities, then you are correct that the partial derivatives of one phase space variable with respect to the others is zero.  But here (I presume), they're looking at the temporal dynamics of the black hole magnetosphere.

It is a little dense (but if you're working through Equation 4.24 above then you'll have no trouble with this), but you might find the stuff in this paper as filling in a lot of the background equations.


Flares in space are known for coming from stars. Sunlike stars, red dwarfs, etc, almost any kind of star has flares.

Hot Jupiters can be the size of Jupiter,(hence the name). They can be bigger; however, if they get too big, they become brown dwarfs. This means that hot Jupiters are too small to have flares.

In addition, many hot Jupiters have unstable atmospheres. They are so close to their parent stars, that they lose their atmospheres. No atmosphere, no flares.


periodic inclination change circumbinary planet

The period of the stars' orbits is likely so much shorter than the planet's orbit that from the planet's perspective the pair looks like two donuts, or toroidal-shaped masses, possibly overlapping depending on the masses and sizes.

In orbital mechanics you can think of this roughly as a single star with a really huge J2 term (gravitational quadrupole moment).

In addition to apsidal and nodal precession (which you would have seen) J2 is well known to induce oscillations in inclination. For example see J2 long-period perturbations in the inclination.

Also from Two-Body, J2 Perturbation & J4 Perturbation Propagators:

  The solutions produced by the J2 Perturbation and J4 Perturbation propagators are approximate, based upon Keplerian mean elements. In general, forces on a satellite cause the Keplerian mean elements to drift over time (secular changes) and oscillate (usually with small amplitude). In particular, the J2 and J4 terms cause only periodic oscillations to semi-major axis, eccentricity and inclination, while producing drift in argument of perigee, right ascension and mean anomaly.


What is the Milne-Eddington Approximation?

I don't have time to write up a full answer with derivation, but you might want to have a look into the standard radiation transport literature (e.g. the book by Mihalas&Mihalas) or the radiation transport in stellar interiors (e.g. Kippenhahn) or literature for irradiated exoplanets (e.g. Guillot (2010)). The ME-approximation is a standard result for grey radiation transport at optical depths >1, when the radiation flux is diffusive, i.e. $\rm F_{rad} = c/(3\rho \kappa_R)\nabla E_{rad}$.
Then you can couple the equations for $\rm E_{rad}$ and $\rm E_{gas}$, integrate and solve for $\rm E_{gas}$, which is $\rm \rho c_vT$. This gives you the ME relation, which is valid for the diffusive radiative transport in gases, and precludes interesting things like the greenhouse-effect, temperature inversions, etc.


Why can we trust Hubble Time if the rate of expansion is not constant?

$H_0^{-1}$ is only a rough estimate for the age of the universe and you have correctly identified the reasons why not.

A correct age estimation relies on knowing $H_0$ and the densities of matter and dark energy, so that the past expansion history of the universe can be correctly modelled. Even this relies on an assumption about how dark energy behaves.

A more interesting question is why a value of $H_0 \simeq 70$ km s$^{-1}$/ Mpc gives an $H_{0}^{-1}$ of 14 billion years, which is within a few per cent of the current best estimate for the age of the universe of 13.8 billion years. In the $\Lambda$CDM cosmology model, the reason for this cosmic coincidence is that the universal expansion underwent a period of decelaraton up until about 4 billion years ago, when it started to accelerate again (the red curve on the plot below).

As a result, a tangent to the red curve, showing the size of the universe versus time, at the present epoch almost goes to ${\rm size}=0$ about 14 billion years ago. If we were to go backwards or forwards in time by 5 billion years, the agreement between $H_0^{-1}$ and the age of the universe would not be so good. At earlier times $H_0^{-1}$ would have overestimated the age of the universe, whereas at later times $H_0^{-1}$ will underestimate the age of the universe.


What are the velocity, mass, and charge distribution of the solar wind.

Velocity
The solar wind speed has a large range of variation, between ~250–820 km/s [e.g., Chen et al., 2014; Gopalswamy, 2006; Jian et al., 2011, 2014; Kasper et al., 2012; Maksimovic et al., 1998; Marsch, 1983; McComas et al., 2013; Schwenn, 1983; Stverak et al., 2008, 2009] near the ecliptic plane.  These values are not including interplanetary shocks, which can have speeds exceeding 2000 km/s.

The speed is generally higher at higher latitudes out of the ecliptic plane, tending to be over 650 km/s [e.g., McComas et al., 2008; 2013].

Number Density
The number density also has a large range of values, from ~2–90 $cm^{-3}$ [e.g., Chen et al., 2014; Gopalswamy, 2006; Jian et al., 2011, 2014; Kasper et al., 2012; Maksimovic et al., 1998; Marsch, 1983; McComas et al., 2013; Schwenn, 1983; Stverak et al., 2008, 2009].  Again, these do not include interplanetary shocks or coronal mass ejections (CMEs).

Charge State
The alpha particle to proton number density ratio varies between ~1-5%, depending on solar cycle and solar wind speed [e.g., Kasper et al., 2012; Schwadron et al., 2014].

We have also measured the ratio of $O^{7+}/O^{6+}$ and $C^{6+}/C^{5+}$, finding ~1-30% and ~20-200%, respectively [e.g., Schwadron et al., 2014].

  Near the earth within the magnetosphere in the ecliptic

The properties of the terrestrial magnetosphere vary so widely, you would need to narrow down this question.  For instance, the charge states are completely different (e.g., we observe $O^{1+}$ but not $O^{7+}$) but the number densities range from ~$10^{-2}-10^{3} \ cm^{-3}$.

  Near the earth but outside the magnetosphere in the ecliptic

See responses above to first part.

  Outside the ecliptic at 1 AU

We don't have any measurements near 1 AU that are at high latitudes.  Some spacecraft have done out of ecliptic polar orbits with high apogees, but the heliocentric latitudes were still within ~$10^{\circ}$ of the ecliptic plane.  The notes above discuss our only real measurements out of the ecliptic by the Ulysses spacecraft.

  I would like to understand the energy content of the solar wind, and how it compares with the solar radiation(solar constant is 1360 W $m^{-2}$?

The solar wind ram(dynamic) pressure is typically only ~1 nPa or $10^{-9} \ J \ m^{-3}$.  This is highly variable and can change in milliseconds (e.g., interplanetary shocks), but that would still only be $10^{-6} \ W \ m^{-3}$.  If we make a hand-wavy argument that this drops to zero in ~3 $R_{E}$ (i.e., upper bound on thickness of magnetosheath), then the power per unit area can be up to ~20 W $m^{-2}$.  However, I would not read too much into that number as the actual power dissipated per unit area is different for numerous reasons.

References

C.H.K. Chen et al., Geophys. Res. Lett. 41, pp. 8081, 2014.
N. Gopalswamy, Space Sci. Rev. 124, pp. 145, 2006.
L.K. Jian et al., Solar Phys. 274, pp. 321, 2011.
L.K. Jian et al., Astrophys. J. 786, pp. 123, 2014.
J. C. Kasper et al., Astrophys. J. 745, pp. 162, 2012.
M. Maksimovic et al., Geophys. Res. Lett. 25, pp. 1265, 1998.
E. Marsch, Fifth International Solar Wind Conference 228, pp. 355, 1983.
D.J. McComas et al., Geophys. Res. Lett. 35, pp. L18103, 2008.
D.J. McComas et al., Astrophys. J. 779, pp. 2, 2013.
N.A. Schwadron et al., J. Geophys. Res. 119, pp. 1486-1492, 2014.
R. Schwenn, Fifth International Solar Wind Conference 228, pp. 489, 1983.
v. Stverak et al., J. Geophys. Res. 113, pp. 3103, 2008.
v. Stverak et al., J. Geophys. Res. 114, pp. 5104, 2009.


Can I see Uranus and Neptune with a household telescope?

Yes, you can.

I use a Newtonian reflector, 200mm objective, 1000mm focal length.

Uranus is sometimes right on the fringe of naked eye but looks a lot clearer as a small but recognisable blueish disc.

Neptune, sadly, just looks like a star to me ... but you can find it.


How can I calculate the right ascension of an observer on Earth's surface in the TEME frame with python.skyfield?

An Earth-centered observer looking at that point on Earth’s surface would also be looking toward the point that is the zenith for that location on Earth.
Thus, you just need to find the local sidereal time (LST) for that Earth location.  That gives you the right ascension on the meridian for that place and time.  The meridian passes through the zenith, so that would give the R.A. of the desired point.
It looks like Skyfield has a GMST (Greenwich Mean Sidereal Time) attribute for any time object you create.  To go from that to LST, convert your longitude to hours (15 degrees per hour) and subtract if it’s west, add if it’s east.
So something like this:

from skyfield.api import load
ts = load.timescale()     # Timescale object
t = ts.now()   # Time object

RA_hours = t.gmst - west_longitude_degrees/15.

where you could change to a different time if you want, and you’d have to define that longitude variable.
You’d also want to check for negative values and make sure your answer falls into the 0-24 range, adding 24 if needed.


Delay of Moon passing a defined meridian each day

The Moon is moving around the Earth in the same direction it is rotating. This means that it would be moving across the sky in the opposite direction the stars are, in case the Earth stopped rotating, however, the Moon is moving in the same direction as the stars due to diurnal motion (only slower, because it also has velocity in the opposite direction). Its angular speed across the celestial sphere will be

$\omega = \frac{2*\pi}{T_{stars}} - \frac{2*\pi}{T_M} = \frac{2*\pi}{T}$

where $T_M$ and $T_{stars}$ are the sidereal period of the Moon and the sidereal day length. Substituting the values above, we get that the Moon takes about 51 minute over 24h to go around the celestial sphere, crossing the same meridian. Note that this is a rough approximation because the Moon's orbit is not circular, but it explains the value you mentioned.


What makes protoplanetary disks start rotating? (Initial energy needed to rotate)

Two rocks placed in space with no relative motion are going to be attracted by gravity, and hit. 3 rocks, placed in space with no carefully rigged symmetry, will likely miss each other, as the gravitational attraction of the additional rock changes their course. Those near misses are the beginning of rotation. Multiply that effect by trillions, and you have a protoplanetary disk. Of course, a lot of those molecules, dust and rock won't even start out with zero relative velocity, so near misses are inevitable.


Is it fruitful to search for a correlation between rates of in-falling matter in black holes and the expansion rate of our universe?

As a layman, I can tell you that there are some issues with your proposal.

The biggest one is that White holes probably don't exist.  They're purely theoretical and, even worse than that, they're mostly not even well defined theoretical objects.  For a physical hypothesis to work, every element of the theory needs to be precisely defined.  White holes aren't well defined.  (now, let me add that, perhaps White holes are well defined by Einstein's field equations, I'm not sure, but the specifics go above my paygrade).

And running time in reverse, a black hole really doesn't look anything like a big bang, and, granted, we don't know what the/a big bang looked like, but we know what a black hole swallowing matter looks like.   It absorbs matter by accretion disk and it shoots out jets along the accretion disks axis.   Running time backwards you can see streams of matter and the accretion disk reforming into stars and dust that flies away from the black hole but that gravitational consumption in reverse That looks NOTHING like the big bang is thought to have looked.

The moments after the big bang are believed to have been a very hot, dense and rapidly expanding quark/photon/fundamental particle soup kind of thing.   Nothing in today's universe, backwards or forwards in time, looks anything close to that, so running time backwards doesn't really help.

The very early universe and the ratio of hydrogen to helium and the presence and near perfect uniformity of the background radiation all imply a mostly uniform, very hot, very dense, and only the slightest bit lumpy rapidly expanding origin, that only works under modern physical laws if space expands faster than the speed of light.

We can re-create those conditions, but only for the tinyest fractions of a second, and with only a handful of particles, in particle accelerators. You might think that inside a black hole, similar conditions exist, but that's not entirely true.  Inside a black hole, gravity is dominant.   Due to faster than light expansion and uniformity, it's not believed that gravity was a dominant force in the moments after the big bang.  It was a force, but not a dominant force.   The space of space-time, for example, is very different inside a black hole than in the moments after the big bang.

A final issue with your question is whether the big bang was an explosion, or, more specifically that it originated from a source-point-origin at all, (as it's often presented), or whether it was an expansion that took place everywhere (everywhere stretch, so to speak).

A theoretical white hole ejecting matter would be a source point and a source point is subject to the inverse square rule and we'd expect to see a variation in density and evidence of a direction towards the origin.   We don't see that.  The observable universe appears remarkably homogeneous, which is only possible, from a source-point, if the size is enormously large.   The rate that distant galaxies move away from us is also very difficult to explain without some kind of dark energy expansion.

The variations in the image are exaggerated.  The cosmic background radiation is very homogeneous, implying we were inside the expansion, not part of some material that was ejected.

Source.

I realize that you didn't say explosion specifically, but asked about a source object ejecting material, but both have a similar effect.  Ejected material spreads out from it's source.   The universe doesn't appear to have been formed in that way.   Nice summary of the expansion theory here.

That's my layman's understanding anyway and I'm not sure I should have gone there.  Origin of the universe questions are kind of out there.


Looking for methods to preserve night vision

Your problems with shielding your eyes from light probably come because it is easy to shield them from direct light from the car lights, but hard to shield them from reflected car light coming from every angle.

I suggest that you experiment with closeing your eyelids while shining lights at your face.

I just tried shutting my eyes while looking at a light about a foot and a half from my face.  Everything looked a very dark grey  or black color, and it didn't look any brighter when I looked at the light bulb.
I often close my eyes when a passenger in a car and facing toward bright sunlight.  The sunlight shines red through my eyelids and I tend to see geometric patterns - when the car passes into shade from trees the color changes to blue.  So the geometric patterns sometimes flash between a red and a blue background rapidly.
But looking at the light bulb right now I don't see any light shining through my eyelids, because sunlight is many times as intense as the light bulb.
So possibly you  might want to experiment with staring at lights of varying intensity with you eyes closed to see how much light, if any, penetrates your eyelids. And yu can try calculating how bright a car's lightswill be at your distance from the road, and comparing it to light sources which doo or do not penetrate your eyelids.
Maybe simply closing your eyes is enough to prevent loosing your night vision.

You could wear a helmet with a hinged opaque face plate that you can slam down when you hear a car coming.  You could have opague cloth handing down several feet from the bottom of the helmet to prevent light from getting in through the bottom, or maybe tape the bottom of the helmet to your clothing.

If you own the place you observe from, you could built a tiny windowless shed or shack there and make certain it is pitch black inside with the door closed, and run into it and close the door when you hear a car coming.

You could set up a tent made of totally opague fabric or plastic sheets, and when you hear a car coming jump into it and knock down the tent pole to collapse it on you.  When the car passes, struggle to get out of the tent, perhaps by finding the tentpole and putting back up.

I hope you might find some of my suggestions useful.


In an area with lots of light pollution, which clear sky clock parameter is most important?

I would agree that transparency is the more important item, but it does depend on what you are observing. Deep sky objects such as clusters, nebulae, and galaxies will benefit from the darker skies by the better transparency. The Sun, Moon, planets and double stars will benefit more from better seeing than better transparency (but my guess is that you are not using a large dob for the Moon and planets.)


Is there a star simulation software that can handle mass ejections and supernovae?

Regarding Terminology:
'Mass ejections' (at least semantically) are very different from supernovae.  A "mass ejection" would generally refer to something like what the sun does as part of its normal activity --- ejecting very small amounts of plasma, or on the other end of the spectrum, the ejection of massive shells of material by very massive (e.g. $M \gtrsim 40 M_\odot$) stars$^{[1]}$.  Supernovae, on the other hand, are the explosion and destruction of an entire star (or at least everything outside of the core).

Simulations
The sad fact of the matter is that there are no simulations which can fully self-consistently explode a star.  In other words, there is no code that can start with a star and have it naturally explode$^{[2]}$.  There are many specialized codes, however, which deal with particular aspects of the explosions.  The initial collapse and explosion is usually done with very specialized, complex hydrodynamic codes by people like Christian Ott, Adam Burrows, Chris Fryer, etc.  These codes are not public.

More general hydrodynamic codes are usually used to study the effects of supernovae, i.e. how the blast waves evolve under different circumstances.  In these simulations, the user artificially deposits a huge amount of energy in the center of a stellar model (mimicking the energy produced by an explosion), and then sees how the system evolves.  One of the most popular and advanced codes for this is called FLASH, produced and managed at the U of Chicago.

[1] For a technical article see, e.g.: http://arxiv.org/abs/1010.3718
[2] There are numerous reasons for this, first and foremost that it is an incredibly difficult computational task.  Stars are generally about $10^6$ km in size, while their cores are only a few km --- to understand the explosion, the code needs to model this entire dynamic range which is incredibly challening.  Additionally, the densities and temperatures involved in the cores of supernovae are way outside of the range of whats ever been explored in a laboratory --- so information about that materials behavior is still largely uncertain.  There are lots of other big challenges (e.g. incorporating general relativity, advanced nuclear reactions, etc) but these are some of the key issues people are exploring.


orbital angular momentum and the longitude of ascending node

My naive question is: is this precession rate exactly the same as the precession rate of the longitude of the ascending node or are there subtle differences between the two?

They're the same, assuming the star's equator is the principal plane of the coordinate system. One way to calculate the longitude of ascending node as $\Omega = \arctan(h_x, -h_y)$, where $\arctan(a,b)$ is the two-argument inverse tangent function and $h_x$ and $h_y$ are the x and y components of angular momentum vector.


Is there a strict chronological order that determines the shape of galaxies?

Galaxies evolve from spiral to elliptical.  The spirals are formed by patterns of new star formation in the disk surrounding the bulge, which contains mostly old stars.

As galaxies run out of hydrogen gas clouds, which is the raw material from which stars are formed, then no new stars form, and with no new star formation, the complex structures of the arms are lost. Elliptical galaxies are made of old stars and have little new star formations. They were formed when spiral galaxies merged, causing a burst of star formation, and leaving a bulge of old stars and little gas to make new ones. The ultimate fate of the milky way and the andromeda galaxy is to merge and form a giant elliptical galaxy.


What should I improve my telescope kit to see more details of Moon?

The theoretical resolution of a telescope - and the maximum useful magnification that you can use - increases with aperture (That's the diameter of the main mirror or lens, and not the f-ratio (which is what photographers usually mean by aperture)). As a rough guide, the typical everyday maximum magnification is around the same as the aperture in mm (and is achieved with an eyepiece of focal length equal to the scope's f ratio).

Given good seeing conditions (steady atmosphere, which can be rare) you can push the magnification to around twice that (with half the eyepiece focal length, or by using a barlow) - which is what manufacturers usually quote as the maximum magnification.

As magnification goes up, the brightness of the view goes down, since you're spreading the same amount of light over a larger area. (Increasing aperture increases the light gathering area; doubling the aperture catches four times as much light, so increasing the aperture allows brighter views at the same magnification. This is important for fainter objects, but the moon is bright enough that you don't usually have problems even with small scopes)

So the best way to see more detail is to switch to a good quality, larger aperture scope.


Why does a hot cloud need more mass to collapse?

It's more do do with having a higher pressure gradient than a higher pressure, though for a cloud of a set size, the two are equivalent.

For a cloud to be in equilibrium requires
$$ \frac{dP}{dr} = - G\frac{M\rho}{R^2},$$
where $dP/dr$ is the pressure gradient and $M$ and $R$ are the mass and radius of the cloud and $\rho$ is its density.

It we just make a rough approximation of a linear pressure gradient and that the pressure on the outside is zero, then
$$ P_c \simeq G\frac{M\rho}{R},$$
where $P_c$ is the central pressure, which is in turn, for a perfect gas $P_c = \rho k_B T_c/\mu$ where $T_c$ is the temperature and $\mu$ is the average mass of a particle in the cloud.

In order to collapse, the RHS has to get bigger than the LHS.  For a given mass, radius and density, then this means the temperature of the cloud must be less than some critical value.  i.e.
$$ G \frac{M\rho}{R} > \frac{\rho k_B Tc}{\mu}$$
$$ T_c < \left(\frac{G\mu}{k_B}\right)\left( \frac{M}{R}\right)$$

Thus a hot cloud won't collapse and a cold one will. But one way to get the hotter cloud to collapse would be to increase $M$ (while keeping $R$ constant).


Could the barycenter orbit of our sun be greatly underestimated?

This is basically equivalent to asking whether there's an undetected massive object in the outer Solar System.

Starting from the basics, the location of the centre-of-mass of a system of point masses is given by

$$\mathbf{X} = \frac{\sum_{i} m_{i} \mathbf{x}_{i}}{\sum_{i}m_{i}}$$

Note that the quantities in bold are vector quantities. Without loss of generality, we can use a reference frame where $\mathbf{X} = \mathbf{0}$, i.e. the origin of the coordinate system is the centre-of-mass. For a two-body system, this gives:

$$m_{1} \mathbf{x}_{1} = -m_{2} \mathbf{x}_{2}$$

So the second particle is located in the opposite direction of the first from the centre-of-mass, with the ratio of the distances $r_{1} = \left\| \mathbf{x}_{1} \right\|$, $r_{2} = \left\| \mathbf{x}_{2} \right\|$ depending on the mass ratio $q = m_{2} / m_{1}$:

$$r_{1} = qr_{2} = \left( \frac{q}{1+q} \right) r$$

Where $r = r_{1} + r_{2}$ is the total separation between the two objects. This means that the trajectories of the two objects around the centre-of-mass are scaled mirror-images of each other. So the semimajor axis of the reflex orbit of the Sun $a_{1}$ for a planet of mass $m_{2}$ with orbital semimajor axis $a$ is given by:

$$a_{1} = qa_{2} = \left( \frac{q}{1+q} \right) a$$

This orbit is larger for more massive planets (higher $q$) and for wider planetary orbits (higher $a$). Working this out for the major planets, the largest reflex orbit is due to Jupiter (1.06 solar radii), followed by Saturn (0.59 solar radii), Neptune (0.33 solar radii) and Uranus (0.18 solar radii).

If Planet Nine exists with the predicted properties given by Batygin et al. (2019), a ~5–10 Earth mass planet in a ~400–800 AU orbit would give a reflex orbit of ~1 to ~5 solar radii, which would likely be the largest contribution to the Sun's orbit around the Solar System barycentre.

The hypothesis that the Sun may have a distant binary companion or a giant planet in the Oort cloud (which would both be more massive than the proposed Planet Nine, and much further out), which has variously been termed "Nemesis" or "Tyche" would result in an even larger orbit around the Solar System barycentre. So far, these hypotheses have not held up to observational evidence.


Does a star fuse helium to beryllium on the main sequence?

Stars don't fuse helium to beryllium except as a very, very short intermediate step toward carbon. Helium-helium fusion to form beryllium is endothermic: It consumes energy. To make matters worse, the beryllium-8 that results has an extremely short half-life, less than $10^{-16}$ seconds. Helium would be the end of fusion in stars (and there would be no us) if not for a fluke: The beryllium-8 formed by helium-helium fusion has almost exactly the same energy as does an excited state of carbon-12.

This greatly increases the probability of a third helium-4 nucleus combining with a short-lived beryllium-8 nucleus to form carbon-12. This is stable. The next stage after hydrogen burning is thus triple helium burning (the triple alpha process), essentially bypassing beryllium except as an intermediary.

  When the star is fusing to beryllium, will the star still be in the main sequence phase and will it at that point start to grow into the red giant phase, or is there no given rule for when it will start growing?

A star leaves the main sequence well before it starts fusing helium. It leaves the main sequence when the star can no longer sustain hydrogen fusion in the core. This happens when the core becomes void of hydrogen. At this point, the helium left behind by hydrogen fusion is essentially ash. Hydrogen fusion proceeds on the edge of the core (shell burning), but the hydrogen-depleted core at this point is far too cold to fuse helium to carbon (not beryllium). So it collapses, and gradually becomes hotter.

The star starts fusing helium to carbon (and also oxygen) if the post main sequence star's mass is large enough. At this point, the red giant collapses and behaves almost like a main sequence star with a second life. That second life doesn't last very long, however.


What planet is better than earth to infer solar system configuration?

I will argue Mars.

A smaller diameter makes it easier to determine the round shape of the planet, and determine the diameter more exactly.
A thinner atmosphere and almost no clouds are better conditions for sky observation.
The small orbital radius of the Mars moons makes it easy to figure out the distance to them. (For example, Phobos is only in the sky about a third of the time seen from equator). Also, the fact that there are two of them makes two data points to discover Kepler's third law.
Closer vicinity of the asteroid belt calls for an earlier discovery of objects with an irregular orbit.
The Moon is a much closer example of something orbiting another object than Galileo’s discovery of the Jovian moons. "Mars is the centre of the universe" disproved.


Why isn't interstellar medium pulled in towards nearest massive object?

It is not true that the particles in the interstellar medium (ISM) are only acted upon by gravity. For instance,

In many cases a significant part of the ISM is ionized, in which case it interacts with magnetic field which permeates the gas and may in some cases be quite strong.
In the vicinity of massive and hence luminous stars, radiation pressure may exert a strong force on the ISM. They also emit copious amounts of cosmic rays (i.e. relativistic particles) that transfer momentum to the surrounding gas.
Supernova explosions create hot bubbles that expand and sweep through the ISM, resulting in shock waves and galactic outflows.

In most cases, however, what may prevent a gas cloud from collapsing is simply its temperature.
Despite all the above processes, and despite gravity being the weakest force, gas clouds do sometimes collapse to form stars. The criterion for doing so is that the gas is dense enough, and that its internal pressure (or thermal energy) is weak enough. This is described by Jeans instability, which formulates the criterion for a cloud of gas to collapse through equating pressure forces, or thermal energy, to gravity. One way to express this is the Jeans mass $M_J$ (Jeans 1902) which is the critical mass of a cloud where thermal energy is exactly balanced by gravitational forces:
$$
M_J = \rho \left( \frac{\pi k_B T}{4 \mu m_\mathrm{u} G \rho} \right)^{3/2} \\
\propto \frac{T^{3/2}}{\rho^{1/2}}.
$$
Here, $k_B$, $G$, and $m_u$ are Boltzmannn's constant, the gravitational constant, and the atomic mass unit, while $T$, $\mu$, and $\rho$ are the temperature, the mean molecular mass, and the density of the gas.

In the second line of the equation it is emphasized that $M_J$ increases with temperature, and decreases with density. In other words, if the gas is too hot, or too dilute, the total mass needed to collapse must be higher.

In general, gas will not collapse to form stars if the temperature is above some $10^4\,\mathrm{K}$. If the temperature is higher, the particles simply move too fast. Since various processes may easily heat the ISM to millions of degrees, the gas has to cool before it can collapse. On way to do this is by cooling radiation: Fast-moving atoms collide (either with each other or, more often, with electrons). Some of the kinetic energy of the atoms is spent exciting their electrons to higher levels. When the atoms de-excite, photons are emitted which can leave the system. The net result is that thermal energy is removed from the cloud, until at some point it has cooled enough to collapse.


IAU Dwarf planet definition Point 2.b - is not a satellite

Or am I missing something here?

Yes. You are missing the word "Sun". The IAU did not want to tackle the problem of how to categorize exoplanets. There are only eight planets in the entire universe, plus a good number of (five and counting) dwarf planets. An expoplanet orbiting Alpha Centauri is not a planet or a dwarf planet, per the IAU definition. Some hypothesize that the early solar system contained five giant planets, but one of them was expelled. That rogue giant planet, if it exists, is not a planet per the IAU definition.

You are treating the word "orbit" as if it's a mutually exclusive term. It's not. The Earth's Moon, for example, orbits the Earth and it orbits the Sun. But since the Moon is a satellite of the Earth, it's not a planet or a dwarf planet. Side note: the concept of what distinguishes a "satellite" from a non-satellite is not defined.


Yes, you are indeed correct. We always see what was happening at a given location when the light was emitted there for us to be detected. Let's assume we observe a oscillating star 100 light years away by measuring the shift in frequency of the detected light. That means that the star was actually oscillating a hundred years earlier and we just happen to observe it now; But we (that means nobody) have no way of knowing what is happening now due to the fact that information can't travel faster than light (especially when the information is transmitted via light as in our case).


Scattered Intensity of Light vs Angles of Incidence and Emission

It is easiest to understand if you fix the incident angle and explain why the emergent intensity is largest at highly oblique angles.  Since the light comes from the outside, it only penetrates so far, and this in turn causes the atmosphere to act like a source of scattered light.  But the source of scattered light is brightest near the top of the atmosphere, since that's where more of the external light penetrates.  Whenever the sources are brighter near the top, it produces what is called "limb brightening," where if you look from highly oblique angles, you mostly probe those higher, brighter regions.  Looking down the normal is where you see the deepest into the atmosphere, where the external light does not penetrate as well.

You can also think about what the individual photons are doing, and ask what is their distribution over emergent angle.  If cos(i)=0, all the photons scatter right at the surface, so it is tantamount to introducing an isotropic radiation field right at the surface.  The photons that go outward will of course have an isotropic distribution, and an isotropic incident radiation field must scatter isotropically (that is a consequence of the principle of reciprocity).  So the emergent distribution is isotropic, but intensity is also per solid angle, so accounts for the foreshortening, and that's where the 1/cos(e) in the intensity comes from.  If cos(i)=1, on the other hand, the incident photons tend to penetrate more, and must diffuse their way out, which gives less of an advantage to low cos(e) after the foreshortening is included.

As for fixing the angle at which you are looking and altering i, here your result says that the intensity always peaks as cos(i) rises.  Your expression claims that is true at every e, so to me this suggests a normalization error.  You want to keep the incident F fixed, but that requires you must get the same outgoing F if you integrate over all e.  Your result says the I is higher at all e if cos(i) is higher, but that contradicts the idea that you are keeping the incident F the same.  Maybe your result is actually comparing the incident and emergent intensities, not the emergent intensity to incident flux.  Then at higher cos(i), for the same incident I, the incident F is falling, explaining your rise in I/F.


After choosing an arbitrary time t, you have the two equations you reference and two unknowns: $ λ$ and $φ$. The problem is that the equations cannot be solved directly, so an iterative approach is required. For example,

assume that $(λ−λ_1)=θ_1$. Note that this also lets you calculate $λ$
solve for $φ$ from the first equation
use that value of $φ$ in the second equation to solve for $λ$
If the values of $λ$ from equation 1 and 2 are different, try another value for $λ$ in the first equation.
repeat until the assumed value for $λ$ in equation 1 and the calculated value from equation 2 are equal.

It has been a long time since I did spherical trig, but I think the solution on the Physics SE is inefficient. It looks like they are calculating the sun's altitude at multiple point (multiple times of t) along the path, so the two equations two unknowns needs to be solved multiple times -- that is inefficient. It should be possible to determine the equation of the great circle through the departure and destination point. Essentially you need to calculate the longitude of where it cross the equator and the inclination. That probably needs to be solved iteratively, but it only needs to be solved once. Then for any time t, the position $λ$ and $φ$ can be calculated directly, and the sun's altitude can be calculated directly.


A 50mm aperture with 1000mm focal length is an f/20 ratio - which is a very very small aperture for the focal length and will result in a very faint image. The small opening angle will result in vignetting (that's what you see with the 'small image'). With those pre-conditions you won't get a bright image with a wide FOV.


Use this (http://www.ast.cam.ac.uk/~mjp/calc_parallax.html) but substitute the exoplanet for the Earth, you don't need to involve the Sun in your calculations, just assume the star is 3pc (parsecs) from Earth.

  From Figure 2, the distance between the Sun and the star is :
  d = r / tan P
  If P is 1 second of arc:
  d = 150 000 000 / tan 1" = 30 million million km
  This distance is called one parsec and is a basic unit for measuring astronomical distances.
  Distance in parsecs = 1 / P in seconds of arc


What is the longest observable wavelength of light using an optical telescope?

I guess it depends what you mean by an optical telescope.

However, if you just mean any telescope that can record images at visible light wavelengths, then it is possible to use these (with the appropriately cooled instruments) to make observation at so-called mid-infrared wavelengths of $\sim 20$ $\mu$m.

See for example on the Gemini telescopes
https://www.gemini.edu/sciops/instruments/midir-resources

The background noise at such wavelengths is extremely high - the telescope and sky look hot at these wavelengths - so these instruments have largely been superseded by space observatories like Spitzer and WISE.

Almost all large "optical" telescopes are routinely used to make observations out to $2.4\ \mu$m (the K band).


Why is the Earth's center still hot after millions of years?

There is a nice article from Scientific American, but the main point is:

  There are three main sources of heat in the deep earth: (1) heat from when the planet formed and accreted, which has not yet been lost; (2) frictional heating, caused by denser core material sinking to the center of the planet; and (3) heat from the decay of radioactive elements.

Other planets may have more or less radioactive material and this will make a difference over time as to whether they continue to have a molten core, or become volcanically and tectonically inactive.


Using emission lines to determine redshift of a quasar

So the two observed lines must be two of the three suggested lines, red-shifted by the same amount. That means the ratio of their wavelengths will be unchanged, so we need two of the suggested lines whose wavelength is close to the ratio $404.7/317.6$.  It's easy to check that only the first two are close to that ratio, so they must be those two. So the redshift should be equal to $317.7/121.6 = 2.61$  and to $404.7/154.9$ which is also $2.61$ (such close agreement in real data would be a bit unlikely, and lead one to check the authenticity of the data.


How can I calculate solar intensity at Saturn

Saturn is about 10.1 AU from Sun, so by the inverse-square law, its solar illumination is a little less than one-hundredth that of Earth, which is also called the solar constant with a value of 1362 W/m².

$$I = I_0 \frac{1\ \mathrm{AU}^2}{R^2} \approx 1362\ \mathrm{W\ m^{-2}}\frac{1^2}{9.583^2}\approx14.83\ \mathrm{W/m^2}$$

Since Saturn's relatively large eccentricity of 0.0565 puts it anywhere between 9.041 and 10.124 AU from the Sun, $I$ varies from 16.66 and 13.29 W/m^2 during its 29.5 year orbital period.


Is there a way to estimate or calculate the tidal range induced on a water-bearing planet?

Wikipeida (quoting Icarus ) gives
$$\frac{15}{8}\frac{mA^4}{Mr^3}$$
Where $m$ and $M$ are the masses of the moon and planet, respectively; $r$ is the orbit radius of the moon and $A$ is the radius of the planet. For Earth this is a little less than a metre.
Systems with multiple moons (or moon and sun, like on Earth) will have multiple bulges which can add up.
This tells you very little about the sizes of actual tides, which are strongly magnified by topography. It predicts tides of 0.7 m, whereas real tides have ranges of between 0 and 16 m.


Is it at all possible for the sun to revolve around as many barycenters as we have planets in our solar system?

The short answer is no; there is only one barycenter. Yes, you can count the Sun/Jupiter barycenter or the Sun/Saturn barycenter, or whichever barycenter you want, but the net effect of all Solar System bodies is to be considered when you calculate the actual barycenter of the Solar System. (And yes, that would include counting all the small asteroids and moons, even those yet unknown to humans, even though their combined effect is negligible.)
One could see it in such a way that yes, there are many barycenters, but the movement of the bodies is around the “average” barycenter. Somehow. But that’s not a good way to describe the system.


How are sulfuric acid clouds able to completely cover Venus?

The Venusian cloud deck is highly acidic. How much water there is in Venus's atmosphere above and below the cloud deck is irrelevant. What matters is how much sulfuric acid there is within that cloud deck.
And it is a lot, with the cloud deck droplets being between 75% to nearly 100%  sulfuric acid by weight. Since the atomic weight of sulfuric acid is about 5.44 times that of water, this weight range corresponds to between 35.5% to nearly 100% sulfuric acid by volume, or by molecule. It's only in the polar regions where the sulfuric acid content of the cloud droplets is at the low value of 75% by weight.
The top of Venus's thick cloud deck is marked by where incoming solar radiation can dissociate the sulfuric acid in the cloud deck. The bottom of Venus's thick cloud deck is where the high temperature (temperature increases with increasing depth in the cloud deck) dissociates sulfuric acid. Sulfuric acid is stable within these bounds.
The dissociation at the top and bottom of the cloud deck means there has to be a source of sulfuric acid to counterbalance the dissociation. The source is the dissociation of other molecules at the top of the cloud deck, plus the large amounts of sulfur dioxide ("large" compared to other trace components) in Venus's atmosphere. The dissociation above the cloud deck creates atomic oxygen. This atomic oxygen is very volatile and readily combines with sulfur dioxide to form sulfur trioxide. This in turn readily combines with water to form sulfuric acid.
In short, conditions just above the cloud deck replenish the sulfuric acid in the cloud deck.
References:
Barstow, J. K., et al. "Models of the global cloud structure on Venus derived from Venus Express observations." Icarus 217.2 (2012): 542-560.
Lee, Yeon Joo, et al. "Vertical structure of the Venus cloud top from the VeRa and VIRTIS observations onboard Venus Express." Icarus 217.2 (2012): 599-609.
Young, A. T. "Are the clouds of Venus sulfuric acid?." Icarus 18.4 (1973): 564-582.


The basic criterion is that compact ellipticals live up to their name by being compact -- small in size but with a high density of stars -- while dwarf ellipticals arguably fail to live up to their name because they are more extended and diffuse. (Given that "dwarf" generally means "faint" or "low in stellar mass" when applied to galaxies, you could argue that a better set of terms would be "compact dwarf ellipticals" and "diffuse dwarf ellipticals", respectively.)

The figure below (from Janz et al. 2016) shows size (effective radius "$R_{e}$", which is the radius inside of which half the galaxy's light is found) versus stellar mass for a variety of objects. The relevant part for this discussion is the upper right, where you can find regular elliptical galaxies (green plus signs), dwarf ellipticals (green pentagons), and compact ellipticals (red open circles, plus the black open circle labeled "M32", plus the red plus signs in the same region of the plot). From this you can see the dwarf and compact ellipticals have similar stellar masses (compact ellipticals can sometimes be more massive), but the dwarfs are physically larger. This means that, since they have similar stellar masses and thus similar numbers of stars, they are more spread out and diffuse than compact ellipticals, which are small and dense. Compact ellipticals seem to be an extension of the size-mass correlation of normal (higher-mass) ellipticals, so some astronomers argue they are a continuation of the normal elliptical population, while dwarf ellipticals are a deviation.


Why is iron responsible for causing a supernova?

Your question is a bit oversimplified because there are many types of supernovae based on the size and configuration of the star. But I can answer your question about "why iron" by considering what keeps a star from exploding in the first place.

In the simplest terms of star formation, when material from an interstellar nebula starts to collapse under its own gravity, the pressure and temperatures involved will become great enough to eventually start fusing hydrogen into helium (it's a bit more complicated than that, but I'm speaking in generalities). If you were to consider the helium atoms created by that process, you'll notice that each helium atom weighs just a bit less than the two hydrogen atoms that formed it. That bit of extra mass is given off as energy which is produced in great quantities as the hydrogen continues to be fused into helium.

During the star's "main sequence", the release of energy by the hydrogen-helium fusion helps counteract the weight of the star's gasses pushing inward. Material presses in; energy pushes out in perfect balance. This balance of gravity and energy output continues until the star uses up most of its hydrogen.

It's at this this point (when there is no hydrogen left in the core of the star to fuse into helium) that the fusion reaction stops and gravity will resume to collapse the star further. As this star collapses, it will quickly become denser and hotter until the temperature and pressures of the interior are great enough to start fusing helium into heavier elements… and the process continues.

That is, until the star starts fusing elements into iron…

The fusion into iron is the first element that does not create more energy than it takes to produce. The effect is that there is no net energy being produced to counteract the gravity pushing inward. So the outer layers will quickly collapse into a much denser and smaller ball causing the remaining star material to undergo fusion all at once, causing the supernova.

So, in that sense, iron is not the cause of the supernova, but its presence marks the inevitable end of this star's life cycle… in this particular scenario.

But understand that this is an oversimplification to illustrate the process you asked about. There are many other sequences of a star's life cycle. Our sun, for example, does not have sufficient mass to keep collapsing down with sufficient pressures to fuse heavier elements into iron. Without getting into other pathways for the production of heavy elements (even in smaller stars like our sun) — once our sun starts creating carbon and oxygen, the fuel starts to run out and the core will simply collapse and rebound as it swells up into a red giant, before losing its outer layers as a planetary nebula while the core shrinks to become a white dwarf (and eventually cooling into a black dwarf).


What Types of Radiation Emanate in the Future and are Perceived in the Present?

Nothing that is known to exist

Relativity does not allow for a massive particle to travel at the speed of light, but it doesn't prevent a particle from travelling faster than light. Such a particle has been called a Tachyon. No such particle has ever been observed. There are good reasons for believing that they don't exist.

Such a particle would be extremely strange. The mass of such a particle would not be positive, it would be imaginary! However you could not use such a particle to send a message to the past. The particles can't be localised, which means that you can't detect them as being "at" a particular place at a particular time.

In conclusion, such particles have never been detected, probably don't exist, and couldn't be used to send messages.

read more at http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/tachyons.html


How frequently does a comet drastically change its orbit in the Solar System (like 67P did in 1959)?

How many known comets are known to have had their orbits substantially
  changed by a single flyby?

Probably nearly all if not all of the (traditional) short period comets were deflected in closer to the sun by one of the gas giant planets.  The reason for that is simple.  Comets can't form in orbits with a near-sun fly-by so nearly all of them had to be deflected inwards by a gas giant planet.   From the Wikipedia article quoted above.

  Based on their orbital characteristics, short-period comets are
  thought to originate from the centaurs and the Kuiper belt/scattered
  disc

Short Period Comets also have a limited lifespan.  Halley's for example, with a perihelion of 0.6 AU, which is closer to the sun than Venus, it loses a significant amount of ice every time it flies past the sun and Halley's may lose all of it's ice within a hundred or so more fly-bys.  Source

Most Short Period comets are also planet crossers because they have highly elliptical orbits and planet crossing orbits aren't long term stable either.   So of the 586 known short period comets, it's safe to say they have an lifespan of (probably) less than a million years when averaged out.  Also, if you consider that Jupiter can work both ways, it can pull comets out of the inner solar system as well as toss them in, then you can work out that they would need to be replaced at least fairly often, perhaps on the scale of every 100 years or so, perhaps on the scale of decades.  (there's probably more accurate estimates out there, my estimate is on the rough side)

There's a great deal of variation with individual comets life spans though.   For example, a Halley-type orbit (see picture) doesn't cross Jupiter's orbit due to the 18 degree inclination, but Halley type comets are more rare (75 known per Wikipedia article above).   Jupiter family short period comets are more common (511 known), and those are also more likely to get jostled by Jupiter since they cross Jupiter's orbit.

Source of picture here.

Wikipedia also mentions Main-belt comets, but those aren't what I think of as traditional comets as they orbit within the asteroid belt and don't have highly elliptical orbits.   Because the Asteroid belt is mostly stable in a kind of gravitational zone between Jupiter and Mars, those "comets" may have been there for billions of years, and are only beginning to show comet like features as the sun has increased in luminosity.

The most common type of comet, long period comets, per the article, some 4,600 of them and about 10 new ones discovered every year and many many more exist but have not been discovered.  The Oort cloud has perhaps more than a trillion "potential" comets, or cosmic icy objects 1 KM or greater in diameter.   Source.

I don't know enough to say how Oort cloud objects are directed toward the inner solar system on average.   With so many objects, it's possible that some Oort cloud objects naturally formed with high enough eccentricity to pass through the inner solar-system once an orbit.  Some might be deflected by objects like Planet Nine (assuming it exists) or by repeated passes of dwarf planets or by the occasional pass of a rogue planet or the very rare close pass of another star, but it's way above my paygrade to try to estimate by which method Oort cloud comets are usually directed into the inner solar-system.  A percentage of those probably pass close enough to Jupiter or one of the other gas giants for a measurable deflection but I couldn't even guess what percentage.   Since these are the most common comets, it's very hard to give an estimate to your question in relation to these.

Loosely related, but Shoemaker-Levy was captured into Jupiter's orbit around 1960 or 1970.  It's hard to know what it's orbit was before then because the calculations are difficult, but it had to be a Jupiter crosser of some kind.  It's apogee (.33 AU) extended outside of Jupiter's Sphere of influence, which works out to about .22 AU, but it was a stable enough orbit to make a few cycles before crashing into Jupiter in 1994.

Jupiter's sphere of influence (.44 AU in diameter) is barely over 1% of it's orbital journey around the sun, but over hundreds or thousands of orbits, 1% is enough to clean out or deflect most of the objects that cross it's path.

Comet Ison, a long period comet was also (virtually) destroyed when it passed around the sun close enough to break apart in November 2013.   Comets seem to die fairly frequently.   Fortunately (or perhaps unfortunately), there's trillions more where they came from.   :-)

Hope that wasn't too vague an answer, but I think the short answer, based on deduction is that events like the one you mention with Jupiter deflecting Comet 67P are reasonably common, maybe every few decades, maybe once a century.   They're not rare events.   Earth will measurably deflect a small asteroid in 2029.   Asteroids are much more common than comets, but Earth is also much smaller than Jupiter.  Icy bodes are likely to pass close enough to Jupiter to have a significant deflection every so often and there's about a 50/50 chance on whether that deflection will push them closer to the sun or further away with a small chance that they will be caught in Jupiter's orbit.

Hope that wasn't too general.   Since this had gone unanswered, I thought I'd give it a shot.


How far can you see in solar plasma at just under 1 solar radius?

If you have a look at the top-left panel of Fig.11 in these lecture notes by Rob Rutten, you will see that the continuum opacity at optical wavelengths at the photosphere is about $10^{-6.7}$ cm$^{-1}$.

The inverse of this is the optical depth. You can see stuff through about 2-3 optical depths, so your "horizon", looking horizontally in the solar atmosphere is about 100 km.

However, the same plot shows that, because of the temperature and density variations with height, the photosphere becomes an order of magnitude more or less opaque at only 50km below or above the "photosphere" respectively.

What this means is that you can see about 100 km horizontally, but less than this (guesstimate 30km) looking downwards and more than this (guesstimate infinity - that is how the photosphere is defined!) looking upwards.

These numbers would all be revised downwards at the wavelengths of prominent absorption lines.

If you could look all around you, you would see a very asymmetric brightness/colour distribution. Overhead would be dimmer and cooler (redder); underneath, much brighter and hotter (more or less the same colour we see the photosphere).


Which Chandrasekhar Limit do I use? 1.39 or 1.44?

The "classic" Chandrasekhar mass is given by
$$ M_{\rm Ch} = 1.445 \left(\frac{\mu_e}{2}\right)^{-2}\ M_{\odot},$$
where $\mu_e$ is the number of mass units per electron ($\mu_e =2$ for ionised carbon, oxygen or helium; $\mu_e = 1$ for hydrogen, $\mu_e= 56/26$ for iron (56)).

This assumes a white dwarf star of uniform, ionised composition and an equation of state given by that for ideal fermions (electrons in this case). It also ignores rotation and it uses Newtonian gravity. When defined in this way, the Chandrasekhar mass is obviously composition dependent - but most white dwarfs are expected to be formed from something with $\mu_e =2$.

In practice, a white dwarf can never become this massive. There are (at least) 4 effects that can mean the de facto Chandrasekhar mass (if it is taken to mean the maximum mass of a stable white dwarf) and at least one effect that can increase it.

The effects that reduce it:

Electrostatic interactions The electrons and ions do not form an ideal Fermi gas because of Coulomb interactions. The net result is to make a white dwarf slightly more compressible and the maximum mass about 2% lower. The correction is composition dependent; it is stronger for white dwarfs made of material with a larger atomic number.

General Relativity Massive white dwarfs are strongly affected by General Relativity. White dwarfs will become unstable at a finite density when the hydrostatic equilibrium is treated with GR. This finite density is reached at about $1.38-1.39 M_{\odot}$.

Inverse beta decay At high densities the electron Fermi energy becomes high enough to initiate inverse beta decay reactions. Electrons are captured by protons in the nuclei and this renders the star unstable. For a Carbon white dwarfs this will also happen at about $1.39 M_{\odot}$, for oxygen the threshold is lower at $\sim 1.37M_{\odot}$.

Pyconuclear reactions At high densities, even at low temperatures, fusion reactions can be initiated by quantum tunnelling. This changes the composition of the white dwarf and can change $\mu_e$ or lower the density threshold for inverse beta decay making the star unstable.

The effect that can increase the maximum mass for stability is rotation. Some authors have claimed that the limit can be increased to as high as $2.6M_{\odot}$ in certain circumstances, though these are usually referred to as Super-Chandrasekhar mass (e.g. Das & Mukhopadhyay 2013) and are not stable. A stability analysis of rotating white dwarfs in GR found that the maximum stable mass was only increased to around $1.47 M_{\odot}$ for a Carbon white dwarf (Boshkayev et al. 2012), but these would be rotating faster than any observed white dwarfs.


Yet another Question about the expansion of universe

Neither the amount of time it took for light to reach us emitted long ago from a distant galaxy nor the distance to that galaxy are directly observable. What is observable is the redshift, how much light has been shifted toward longer wavelengths.

The amount of time it took for that redshifted light to reach us can be calculated from the redshift $z$ by assuming various parameters regarding the universe (regular matter vs dark matter, the Hubble constant, whether the universe is open, flat, or closed, etc.), and by assuming that that galaxy was and our galaxy is more or less moving with the local Hubble flow. A distance value can be calculated simply by multiplying the calculated time by the speed of light.

Even though the second calculation is much simpler than the first, it is this second calculation that is rather misleading. There's an underlying assumption in this calculation that the universe is not expanding. The universe is expanding, so it's not quite valid to multiple very long times such as 13 billion years by the speed of light to get 13 billion light years. In fact, that distant galaxy is about 29 billion light years from us now and was only about 3.5 billion light years away when the light was emitted.


How to calculate the point on the globe with maximum magnitude with given Besselian elements and given time for a partial solar eclipse?

Jean Meeus, Morsels 2, chapter 69 talks about the Nelder-Mead simplex algorithm for minimising a function of two variables - in this instance magnitude as a function of latitude and longitude. Maximising the magnitude equates to minimising the negative of the magnitude. Apparently the method first appeared in magazine Byte in the 1960s, there are various references on-line.
Maximum magnitude always occurs near the polar circle, so try starting with three points equally spaced round the 60th parallel. They would have to be converted to Cartesian X Y with centre at the pole (giving an equilateral triangle), and the result converted back to latitude/longitude at the end. The sign of the element Y will make it obvious which polar circle to use.
Because the shadow is conic, not cylindrical, the resulting point will not (in general) be exactly on the horizon.


UK groups that investigate extraterrestrial life

Extraterrestial intelligence is a hard research topic as the search for evidence is not straight forward in any way. The search for intelligent life as far as I know is limited to observations in radio and for that you want to see if any SETI members are UK based.

However if you're more into astrobiology in general you might wanna look up the UK Centre for Astrobiology which is based in Edinburgh. I don't know any of the faculty but I do know that they are hosting a meeting for early career scientists in 2017. There are also some smaller research groups based in the UK but you might wanna check out the specific research aim for these (e.g. the group at Imperial College in London is focused on Mars) as it might not be the area within astrobiology that interests you the most.

There is also Mark Burchell who is a professor at the University of Kent. His group did some exciting work on Panspermia (the idea of life travelling through space) that I happened to read earlier this year. Note that this is focused on microorganisms so this is probably as far from "intelligence" that it gets.

But you can try to look up departments at different universities. You probably want to look under astronomy or sometimes physical or Earth science as the research aims of some groups put them more in e.g. Earth science than actual astronomy. Sometimes you can also find smaller groups under biology or life science.


The tidal locking problem concerning Earth sized planets in habitable zone around Red Dwarfs

There is quite simple formula that will give you the tidal-locking half-time

$$T = C\, \frac{a^6 R \mu}{  M_s M_p^2}$$

$a$ - semi-major axis, or simply the radius of the planet circular orbital trajectory in meters
$R$ - planet radius in meters
$M_s$ - planet mass in kg
$M_p$ - parent planet/star mass in kg
$\mu$ - rigidity, approximately $3×10^{10}$ for rocky objects and $4×10^9$ for icy ones.
$C$ - a prefactor dependent on units used. On Wikipedia, they write $C = 6 × 10^{10}$ if the time is to be in years, but this number is probably wrong. (See the discussion.)

Solving the problem of interaction between the parent star and between the planets will be difficult. Generally, if tidal-locking half-time between the planets is much shorter than the tidal-locking half-time between the star and the planet, the planets will tidal-lock to each other rather than to the star. So we ask, whether

$$ \frac{A^6}{  M_s^3} \ll \frac{a^6}{  M_s M_p^2}\;,$$

where $A$ is the semi-major axis of the orbit of the double-planet. For typical planet in the habitable zone of the red dwarf star, $a$ = 9,000,000,000 m (0.06 AU), while the distance of the double-planet could be for example $A$ = 30,000,000 m. $M_s = 6\times 10^{24}\;\mathrm{kg}$ and $M_p = 2.9\times 10^{29}\;\mathrm{kg}$. We get

$$2\cdot 10^{-41}\ll 6.3\cdot 10^{-36}\;,$$

which is well-satisfied and the planets will tidal lock to each other rather than to the parent star. Were the distance between the planets $A$ = 300,000,000 m, the inequality would yield

$$2\cdot 10^{-35}\ll 6.3\cdot 10^{-36}\;,$$

which is not satisfied and the tidal forces of the star would probably disrupt the double-planet. (The planets would not even be in the Hill sphere, which approximately gives the stable orbits.)

So the answer is: a very close double planets can tidal-lock to each other rather than to the parent red-dwarf star, but they would have to be much closer than the Earth to the Moon.


How are all the intermediate images between the "lensed" and "unlensed" endpoints of this video generated?

For me, it looks indeed very much like an interpolation, both from what I see and how the caption is formulated. Also: If it would be scientific simulation results, I would have expected whatever the value of the variable displayed on each frame - much like most journals would not accept a 2D plot without axis-labels and units.
Another hint: Martin Kornmesser is a graphic designer for public outreach not a scientist. I guess we would only know if we asked him - but I did not figure out his email yet.


Should I focus more on Aperture or Focal Length for a telescope?

There is one rule that is generally true for all deep sky objects (nebulae, stars, galaxies,...): Aperture matters!

For solar system objects, aperture is not that important.

The second most important thing is: What size are the objects you want to look at: Small objects need long focal lengths and high magnifications, large objects need short aperture for low magnifications.

With 400mm you could watch objects like:

Andromeda galaxy core
Orion nebula, other large emission or reflective nebulae (e.g. Pleiades)
large star clusters
low magnification lunar observations

With 900mm you could watch objects like

Planets (Saturn, Jupiter, Mars, ...)
high magnification lunar observations
planetary nebulae (e.g. ring nebula)

Note that 60 and 70mm aperture are still very small for telescopes! The aperture influences two things:

Light sensitvity: The larger the aperture, the more light you can collect. Very important if you live in a city!
Maximum resolution: Rule of thumb is that you can do aperture in mm times two as maximum magnification. I.e. for 60mm a 120x magnification is the absolute maximum which still is feasible.

The magnification is created by the eyepiece. E.g. when you have a 400mm focal length telescope and use a 10mm eyepiece, you get 400mm/10mm = 40x magnification.

Note: the shorter the eyepiece focal length, the more difficult it is to build. Good 5mm eyepieces can cost 100 USD and up. I personally started with a 750mm Newtonian with 150mm aperture and 25mm and 10mm eyepieces. That's a good allrounder, even though planets will appear rather small with the 10mm eyepiece. But you can later invest more money in good eyepieces, which you can re-use on better telescopes which you may buy later on.

Edit: One more thing -- the telescope mount is equally important as are the eyepieces and the telescope itself. A mount that fits the telescope easily is as expensive as the optical tube assembly itself. Hence many beginners start out with a Dobson telescope, which uses a very, very simple yet sturdy mount.


How close would the sun have to get to a supermassive black hole to have Jupiter stripped and slingshot into interstellar space?

Closer than about 10000AU, 1.5 trillion km or about 2000 times the orbital radius of Jupiter.

Three body systems are peculiar beasts, and but I'm going to change the question to "Would Jupiter be in a stable solar orbit?" This makes it a question of Hall radius.

The Hall radius is $r_H = a\sqrt[3]{m/3M}$ where $m/M=1/4000000$ is the ratio of solar mass to black hole mass (the mass of Jupiter is treated as negligible!) and $a$ is the orbital distance of the sun about the black hole

If we set $r_H$ to 5 (AU) and solve for $a$ we get a value of about 10000  (I'm only working at 1 s.f. accuracy here. In other words, if the sun is orbiting the black hole at 10000AU then Jupiter would be in a marginally stable orbit.  If the sun got closer than 10000AU then Jupiter would tend to orbit the black hole and not the sun.

However this is only an upper bound. On a hyperbolic orbit the sun would spend little time near the black hole and so even if Jupiter had drifted in the black hole's gravity as the sun passed by, as the sun and jupiter moved away, it would tend to fall back into orbit around the sun. To model whether Jupiter would actually be lost would require knowledge of the initial conditions and numerically solving the differential equations of motion.


Black hole darkness a result of gravity or temporal distortion?

As an object falls towards a black hole, light emitted by the object will become red-shifted by the gravity. As photons rise out of the gravity well of the black hole they lose energy, they can't slow down but their wavelength is stretched. The object will go dark as it approaches the event horizon due to gravitational redshifting.

Photons starting from the event horizon would be redshifted to zero energy, and no photons can travel from inside the event horizon.

We also will observe gravitational time dilation, the last second before the object reaches the event horizon will be stretched out indefinitely. The two effects are all part of the same relativistic theory of gravity. You can't unpick the two effects.