Trying to use Canon 60D DSLR with Skywatcher 130P. Can't focus image

As you suspected, the problem is probably that, as with most Newtonian telescopes, the image plane is too close to the outside of the tube, so you can't physically get your camera sensor close enough to the image plane.
There may or may not be adapters that will solve this problem for your specific telescope and camera. Barlow lenses help but aren't always enough. Luckily your telescope is collapsible, so you should be able to bring the image plane far enough out by collapsing the telescope a little. Attach your camera to the telescope as before and slowly collapse the telescope. You should see the image gradually come into focus.


What would happen if someone had a telescope and watched Betelgeuse when it goes supernova?

No, it would not be a problem. Supernovae are not at all like flashbulbs – they brighten over a period of many days and dim again even more slowly. Here are a number of different light curves taken from Wikipedia:

The rise is fast on an astronomical scale – several orders of magnitude over a period of roughly ten days – but very slow on a human scale. An amateur looking at it would not notice any significant change in brightness, but if the same person came back a few hours later or the next night, the change would be very evident.

As far as we can tell, the reason is that the light at peak brightness is caused by emissions from material blown off by the explosion. For example, in Type 1a SNe, most of the light is from the radioactive decay of the huge mass of ejected nickel-56 (half life 6 days).

The Wikipedia article on supernovae is quite good and covers this all in more detail.


Clear boundary between core and shell of a gas star

This depends on the initial mass of the star, but generally the gradient of the core is not sufficiently strong on the main sequence to justify a well defined core-envelope boundary, which is why if a star on the main sequence experiences Roche-lobe overflow the entire star is essentially destroyed, e.g., in a binary system.
The core of a main sequence star develops due to the nuclear fusion of hydrogen into helium. The main sequence ends when the core is exhausted of hydrogen, and then low mass stars enter the giant branch phase and high mass stars enter the Hurtzsprung-gap: in both cases the star begins shell burning of hydrogen around the core. Then, once the temperature in the core is sufficiently high, core helium burning begins which brings the star to the next stage of development - the supergiant phase.
The star chokes like an engine does when it switches from one form of nuclear fuel to another, resulting in contractions of the core and expansions of the envelope, resulting in the star's surface radius to generally increase as it ages.

If it does, then what is keeping a helium core distinct from its hydrogen shell, for a main-sequence star that transforming to a giant star?

So, once a star is on the supergiant phase, its core is sufficiently well defined to support itself under the envelope by the luminosty pressure from the core and the shells surrounding the core, and the core supports its own self gravity by the degenerate matter that develops composed mostly of oxygen and carbon (i.e., the electron degeneracy pressure).
There is also rotational support and its effects, which is an active area of research. The core-envelope boundary is (literally) a turbulent region where chemical mixing occurs which drives convective currents through the envelope and which can result in angular momentum transfer between the core and envelope resulting in spin coupling between core and envelope. This spin coupling can have implications for the late evolution of the star. I asked a question about this that's remained unanswered.
Enhanced rotational mixing of the chemical elements inside a main sequence star can lead to chemically homogeneous evolution, in which there is no longer a core-envelope boundary since the helium has been mixed throughout the star.


Venus presents same face toward Earth at closest approach. What do you make of that?

I wouldn't call the reference dubious so much as very much out-of-date. The paper in question was written in 1969, which was still pretty early in our understanding of the dynamics of the Solar System, including Venus. There would have been a lot of uncertainty in many of the measurements, such as rotation speed, which would have been determined by radar. Consequently, the possibility of resonance was there, at least until better measurements were available. However, the wikipedia article on Venus, and the attendant 1979 paper citation (#90) indicate the idea of a resonant Venus was rejected decades ago once sufficiently accurate radar observations of the planet had been made.

So, clarifying of the wikipedia article you mention is probably in order.


What is star composition after formation

The star reaches a critical mass and ignites, blowing away the surrounding material. (OP emphasis)

I think the misunderstanding is that it's not that explosive when the star "lights up". The process is in some sense continuous, so the nuclear reactions start slowly and gradually increase in strength as the star continues to contract, grow, and heat up inside.  Once the amount of radiation from the surface becomes sufficiently hot, we can speak of "blowing away" material, but even then it's not violent. The absorption of light in the gas creates a pressure that pushes the gas away. It picks up pace as the star settles down (and depending on how massive the star is), but it's broadly a consistent stream. (Astronomers are probably unwise to use the word "ignite" because all we mean is that the nuclear reaction rates become non-negligible. It's not like a rocket taking off or something.)

Generally, star formation is a poorly-understood process and a hot topic for research.  For stars larger than about 7 solar masses, we think that nuclear reactions begin before all the material has gathered in the star. i.e. while some is still flowing on. If this happens through a disk, then the light can stream out everywhere but the equator without much affecting the growth of the star.

That said, at very high masses, you do run into a problem of blowing mass away. For this reason, we don't really expect to see many stars over a few hundred solar masses, and those that are can be violently unstable (like Eta Carina, which threw off about 10 solar masses of material in the mid-19th century).

And as for stars genuinely "blowing up", this does happen in other circumstances. For example, in the core of a solar mass red giant, matter becomes nearly isothermal.  It eventually gets hot enough to ignite helium, and it nearly all lights up at nearly the same time in a helium flash.  This briefly produces as much light as a small galaxy! But only very briefly, and it doesn't make it to the surface for us to see. Incidentally, this is also the same mechanism as Type Ia supernovae: an isothermal carbon-oxygen white dwarf accretes matter until the carbon lights up, and the whole white dwarf basically ignites at the same time.


Why do transits of Earth across the Sun seen from Mars follow a pattern of occurring after 26, 79 then 100 years?

Such transit events occur when Earth and Mars are simultaneously at a node where their orbital planes intersect.

Between transits at the same node, both Earth and Mars must complete whole numbers of orbits, where 1 Mars year = 1.8808 Earth years.
Meeus and Goffin 1983 identify patterns of 79-year and 284-year intervals:

79.0 Earth years ≈ 42.0 Mars years
284.0 Earth years ≈ 151.0 Mars years

Their Table II arranges the May and November transit dates in separate "panoramas" with 284 years between rows and 79 years between columns.
These are somewhat analogous to a saros-inex panorama of solar eclipses.

Between transits at opposite nodes, both Earth and Mars must complete odd numbers of half-orbits:

Nov→May: 25.5 Earth years ≈ 13.56 Mars years
May→Nov: 100.5 Earth years ≈ 53.43 Mars years

These Mars year fractions differ from 0.5 because the half-orbit near perihelion takes significantly less time than the half near aphelion.
Meeus and Goffin Table I also shows some intervals of 21.5 and 53.5 years outside the present few centuries.


Structure of the torus of the Active Galactic Nucleus

This question was asked more roughly 8 years ago, and for 2 years there is a review article around which I found worth mentioning and summarizing. The manuscript is called Redefining the torus:  A unifying view of AGN in the infrared and sub mm and work of Sebastian Florian Hönig. Here an excerpt of the abstract:

The advent of high-angular resolution IR and sub-mm interferometry allows for spatially-resolvedobservations  of  the  parsec-scale  environment  of  active  galactic  nuclei  (AGN),  commonly  referred  toas the “torus.”  While molecular lines show the presence of large, massive disks, the IR observationsappear to be dominated by a strong polar component that has been interpreted.

The main result seems to be the following schematic picture (Fig. 4 of the manuscript) of the AGN which is an overlay of various observations:

The purple ring are observations in the near-IR.
The yellow torus and the gray dashed lines are from ${\rm H}_2$ emission.
The main brownish cone is from mid-IR.
The redish wobbly plane with the red (solid) arrows and the purple (dashed) arrows are sub-mm obseravtions (${\rm CO/HCN/HCO^+}$)
Bright green dots on the $x$-axis are ${\rm H_2 O}$ maser emissions.


Where would be the absolute most logical place in USA to observe the total solar eclipse on August 21, 2017?

I assume that when considering where to view the Solar Eclipse from, you're looking for the place with the best observance of totality. This elicits two factors, the degree of totality and the time elapsed during totality. The best observed eclipse will be a total eclipse, i.e., 100% totality, and for the longest time possible.

A map of the path of 100% totality is shown below. In it, you can see that grey band that passes across the US. Anywhere within that band, you'll be able to see a 100% total solar eclipse.

If you view this interactive map from NASA, you can see two specific points along the totality band that are called out, one labeled GE and one labeled GD.

The GE marker signifies the point of greatest eclipse which is defined as:

  The instant when the axis of the Moon's shadow cone passes closest to Earth's center.

The GD marker signifies the point of greatest duration, quite sensibly defined as:

  The instant when the length of the total (or annular) phase reaches a maximum along the central path of a solar eclipse.

So, if you're looking for the point of 100% totality that lasts for the longest amount of time, you'll want to be at the point of the greatest duration. Geographically, this is pretty close to Carbondale, IL.


Why was the July 27th 2018 lunar eclipse not visible in North America?

Nothing so romantic as the universe being fair.

The path of the moon is tilted by about 5 degrees from the path of the sun. but twice a month the moon must cross the path of the sun.  When this moment coincides exactly with full moon there is a lunar eclipse. Since the sun is on the opposite side of the Earth, lunar eclipses are only visible during the night, and this one happened to occur during the day for the USA. When it is day in the USA it is night in China, and the eclipse was visible from the other side of the world.

Of course, these things aren't random, but there is no simple formula like "if you see a solar eclipse, then the next lunar eclipse won't be visible."  You can just go to any of the eclipse prediction websites to find when the next one will be visible in your area.


What would be the practical consequences (on earth) if the Moon was not tidally locked?

If the Moon was not tidally locked, it would mean that the Moon is in the wrong orbital distance from Earth. The Moon is tidally locked because it is close to the Earth. If the moon were closer, it would approach the Earth's roche limit, be torn apart, and its debris would become a ring for ~100-200 million years. If it were too far, it would continue to drift outward until it was free of Earths' gravity.

By being tidally locked, the Moon has been extending the Earth day by slowing down the Earth's spin, to about ~6 hours from an 18 hour day to a 24 hour day. This process will continue into the future until the Earth is also tidally locked with the Moon. The Moon, while in this state, has also stabilized the Earth's seasons by stabilizing the Earths' orbital axis.

Lastly, a possible indirect consequence would be upon human history. Mankind would have probably abandoned the celestial sphere model in favor of the spinning Earth model much sooner had they seen proof that a celestial body could spin and rotate, much like when Galileo saw that moons were orbiting Jupiter.


Why would the side of the moon that faces earth be as dark as the far side of the moon?

Somebody suggested that the side of the moon that always faces the earth is as dark as the far side of the moon, but is that really the case?

That is not the case.
I'll look at two different wavelengths: visible and radio.

Visible
Doing astronomy from the surface of the Earth when the Moon is full is much more difficult than is doing astronomy from the surface of the Earth when the Moon is new. While much of this difficulty arises due to the Earth's atmosphere, some is the direct result of the presence of a largish, well-lit object that has same angular size as does the Sun. The full Earth as seen from the Moon is, on average, over 40 times brighter than is the full Moon as seen from the Earth. Reading a newspaper on the Moon at night under a full Earth would be a piece of cake compared to reading a newspaper on the Earth at night under a full Moon.

Radio (and microwave)
These are the wavelengths where an observatory on the far side of the Moon would truly shine. The Earth radiates significant amounts of electromagnetic radiation in the radio and microwave wavelengths. There are a few places on the Earth where radio towers and microwave towers are absolutely forbidden, such as the area around the Murchison Radio-Astronomy Observatory in Australia, the area around the Itapeting Radio Observatory in Brazil, and the area around the National Radio Astronomy Observatory (Green Banks, West Virginia) in the US. These Earth-bound radio quiet zones are still plagued with radio and microwave radiation reflected by the atmosphere and by radiation from satellites orbiting the Earth.
Radio telescopes on the far side of the Moon would be shielded from the large amounts of electromagnetic radiation in those wavelengths emitted by the Earth, and possibly even from the electromagnetic radiation in those wavelengths emitted by satellites orbiting the Earth. Moreover, radio telescopes on the far side of the Moon would not be hindered by the Earth's atmosphere. The transparency of the Earth's atmosphere at radio wavelengths ends at the very long wavelengths that would be useful in studying the early universe, and it ends at the upper microwave that would be useful in studying molecular gas clouds.


When could the Roche limit be larger than the Hill sphere?

The Hill sphere is approximately given by

$$R_\mathrm{Hill} \approx a_\mathrm{p} \left(\frac{M_\mathrm{p}}{3M_\ast}\right)^\frac{1}{3}$$

Where $a_\mathrm{p}$ is the radius of the planet's orbit, and $M_\mathrm{p}$ and $M_\ast$ are the masses of the planet and the star respectively. This is an approximation to the size of the Roche lobe around the secondary.

The Roche limit (not to be confused with the Roche lobe) is the limit at which tidal forces will disrupt an object held together by its own gravity. Roche derived the following formula:

$$R_\mathrm{Roche} \approx 2.44r_\mathrm{p} \left(\frac{\rho_\mathrm{p}}{\rho_\mathrm{s}}\right)^\frac{1}{3}$$

Where $r_\mathrm{p}$ is the radius of the planet, and $\rho_\mathrm{p}$ and $\rho_\mathrm{s}$ are the densities of the planet and the satellite respectively. Note that there are various different formulae for the Roche limit depending on the different assumptions being made, see the Wikipedia Roche limit article for details. For a spherical planet, this can be rewritten in terms of the planet's mass:

$$R_\mathrm{Roche}
\approx 2.44 \left(\frac{3M_\mathrm{p}}{4 \pi \rho_\mathrm{s}}\right)^\frac{1}{3}
\approx 1.51 \left(\frac{M_\mathrm{p}}{\rho_\mathrm{s}}\right)^\frac{1}{3}$$

So the condition you're interested in becomes:

$$1.51 \left(\frac{M_\mathrm{p}}{\rho_\mathrm{s}}\right)^\frac{1}{3} \gtrsim a_\mathrm{p} \left(\frac{M_\mathrm{p}}{3M_\ast}\right)^\frac{1}{3}$$

Cancelling and rearranging gives:

$$a_\mathrm{p} \lesssim 2.18 \left(\frac{M_\ast}{\rho_\mathrm{s}}\right)^\frac{1}{3}$$

For a satellite density of 3300 kg/m3 (similar to the Moon) with a host star the mass of the Sun, this corresponds to a planetary orbit of about 2.6 solar radii.

Needless to say, this is an extremely close star–planet separation that invalidates a lot of the approximations used to compute that limit. For example, the planet will be close to or within its Roche limit with respect to the star. If it is not itself being disrupted, it will likely be significantly non-spherical.


I am not very familiar with orbital dynamics (so please correct me if I'm wrong). I was told that, for instance in the case of the mean motion resonances that cause the majority of the Kirkwood gaps in the asteroid belt, not only the ratio of the periods, but also the timing is important.

Let's take Pluto as an example, which is in 2:3 resonance with Neptune. Although Pluto crosses Neptune's orbit, both bodies will never get closer to each other than a certain threshold. That is, because the timing between their orbits was preventing close encounters in the first place, which is then enforced by the resonance.

If the timing was different, close encounters would be possible, thus destabilizing the orbit of the smaller body. In the case of mean motion resonances, there is an algebraic expression that allows to investigate the stability of a resonance (see, e.g., here https://en.wikipedia.org/wiki/Resonant_trans-Neptunian_object [toward a formal definition]).


(Much of this echoes what antlersoft says in their answer)

For a phone photo through the eyepiece that looks about right to me!
The size... the brightness... both are as I expect.
What you could try is to use the manual mode of your phone's camera and set the ISO down to minimum (100) and the shutter speed down to something like 1/60s.
Take a few shots, pick the best one, and make a zoomed+cropped version.
You should be able to see a point of white on the south pole and a hint of dark patches.
If the seeing's bad then you'll get awful results no matter what, so wait until the view is nice and crisp.
Now, when you look through the eyepiece, how does it look to your eyes?
Is it too bright? If so then keep watching the planet and your eyes will adjust and the contrast will improve.
Is it too blurry? If so then back off the magnification a bit until it's not blurry any more. A smaller crisp image is much better than a larger blurry image - your eyes will be able to pick out small details in a tiny image with time and practice.

Here's a lucky snap I got a few weeks ago:

And here's a zoomed and cropped version:

And here's one from one minute beforehand with automatic camera settings:

These were taken through the eyepiece of a 10" F/5 Dobsonian - very similar to yours. (I can't remember which eyepiece I had in at the time, I'm afraid! I think it was a 4.5mm but it could have been an 8.8mm with 3x Barlow lens, or even a zoom lens plus the 3x Barlow...all I do remember is that the seeing was exceptionally good that night and I was making the most of it!)
The good one was ISO 100 and an exposure time of 1/60s. The bad one was ISO 100 and an exposure time of 1/17s.
Hopefully these photos demonstrate that everything has to come together to be able to get a halfway decent photo: the seeing, the collimation, a steady hand, the camera settings... :-)


There are actually a number of words / expressions related to this.

  Aureole

[Bright edge of object - sometimes with colour of light source vs object] This is when an object is silhouetted against a bright light source and it gives a ring of colour. This typically occurs when the light source is smaller than the sillhoutte. I think this is the closest meaning because its meaning implies the colouration caused by diffusion (or even refraction - in this case) of light around the sillhoutte. For example, a girls head in front of a candle: the fine granularity of the hair diffuses the light, bending it and causes a bright circumference of light around the extremities of her hair. This word is associated in optical physics with a number of other diffusion phenomina too, such as the aureole effect.

  Corona

[Colour of the light source, shape of light source] This is a ring of light when the sillhoutte is much smaller than the light source (or even no sillhoutte at all!), and there is practically no interaction between the light and the extremities of the sillhoutte, as a result most of the colour is from the star.

Noteable mentions:

  A diamond ring

[Angle of light source to object] is where the eclipse has passed either side of total and there's a flare on one side.

  Bailey's beads

[Shape of object to shape of light source] are when there is a total eclipse and there are flares all around the circumference, caused by imperfections in the planet's surface and / or the irregularity of the star's corona - entirely diffraction based.

  Glory

[Entirely colour of diffraction / refraction of light source around object] This is an optical phenomina which is less common, and is almost entirely based on refraction, giving colour bands in the vicinity of the object, typically in concentric colour bands. A bit like a rainbow. This only occurs in the context of where the observer and object are in an atmosphere so is least applicable to this context - although technically the whole solar system is within the atmosphere (or heliosphere) of the sun, right up to the helioshock, where the heliosphere interacts with either other stars or forms a boundary with outter space, but it's not inconceivable that this could occur in a nebula, or even give rise to a

  Brocken spectre

Incidently: the blue aura, in this case, is actually caused by a physics phenomina called Rayleigh Scattering which is where the atmosphere has the effect of appearing to change the wavelength of light as it passes through. Often this can be blue, however in some situations it might be red.

There is a whole family of words for these kinds of meanings, and a many of them are poetic. Some of the words relate to colour, some to shape, some to diffraction, some to refraction. This is one of those situations where knowing a little science could help you chose the right word.


Which quantum reality will be seen in a far away galaxy

So, the presenter explains that two observers who are not moving relative to one another will agree on the meaning of "now".  However if one observer is moving, then they will no longer agree, and what one considers to be now will be in the past or future of the other.

He observes that a sufficiently distant observer, in another galaxy 100 million light years away that is moving at "bicycle" speed relative to us would have a 200 year discrepancy in their understanding of "now".

But this alien would have absolutely no way of observing our universe now. They could not see us in our "now", nor in their "now" of 200 years ago, because they are 100 million light years away. They can only observe how our galaxy was 100 million years ago.  They can certainly not observe our galaxy in the future, even if they travel towards us at speeds approaching the speed of light.

So the explanation of different "now" is about agreement on clocks, not about actual observations.

I think then the program errs by treating Special Relativity as a full description of nature, rather than as a model. Relativity is a classical theory, in that in Relativity, if you know the initial state, you can predict the future state. There is no uncertainty principle in general relativity. It is correct to say that in Relativity the future is as real as the past. It is unclear if the same is true in Quantum Mechanics, it depends on how you interpret the maths. Whether the future exists or not cannot be answered by theory.

Since different interpretations of QM are based on the same maths, there is no way of testing which interpretation of quantum mechanics is "correct" or even whether calling an interpretation "correct" has any scientific meaning. In the "many worlds" interpretation, the notion of a single space-time which you can take slices of is lost completely.

I might note that I'm not a great fan of the explanations in that series, which seem to me to be confusing on a number of levels, and are more aimed at creating "wow" than being clear. For example, the mixed metaphor of "slices of bread and aliens riding bikes" seems to have confused you into thinking that there are aliens who are already seeing how our planet will be 200 years from now.


What was the orbital path of earth before collision with a Mars sized planet?

The total angular momentum would always be conserved about the Sun (assuming it at rest), as there is no torque about the Sun. Apart from that, I think it is hardly possible to say anything about the orbits. Also, it may be true that the pre-collision orbital plane and the post-collision orbital planes changed due to the collision, as small transverse perturbations must also have resulted from such a collision. However, I would also like to see what other members suggest.


Are we already moving faster than light?

Short Answer

NO.

Long Answer -

The Earth "wobbles" on its polar axis.  This motion is not relevant (IMO) to the calculations needed to return to earth from interstellar space as it is an "in situ" motion, but it exists.
The Earth revolves {spins} on its axis. For these calculations we will use the equator with a circumference of approx. 25,000 miles. One rotation of Earth is approx. 24 hours. Again this is "in situ" motion so not relevant to space travel, but adds to our "stationary chair" model
25000/24 =1041.7 MPH
The Earth is orbiting the Sun once a year. The circumference of the Earth's orbit is approx.  607.6  million miles {or 940 million kilometers}. One year is approx. 365 days
365 days X 24 = 8760 hours
607,600,000/8760 = 69,360.73 MPH

  So far we have basically 3 Directions of Motion {Wobble, Spin and
  Orbit} and a combined speed of 1041.70 + 69,360.73 = 70,402.43 MPH for
  a person sitting in a chair at the Equator.

The sun {and hence the solar system} is moving towards the constellation Hercules, namely to the star Lambda Herculis at 12 miles per second {or 20 kilometers per second} which is 43,200 MPH
The Solar system is also moving upwards, at 90 degrees to the plane of the Milky Way, at 4.34 miles per second or 15,624 MPH. But we are actually leaving the Galaxy, out about 50 light years now and will be moving out to 250 light years before it reverses. Details of the mechanics of this are explained in the link below. We also crossed the Galactic plane 2 million years ago.
The Solar system is orbiting around the Galaxy at an "estimated" speed of 124 miles per second {or 200 kilometers per second} which is 446,400 MPH. The way that figure has been calculated can be found at the link below.
So our "armchair Astronaut" is now moving through 6 different directions and a combined speed of approximately 574,585 MPH

  69,361 MPH Spin and Orbit    43,200 MPH Towards Lambda Herculis
  15,624 MPH Perpendicular to Galactic Plane  446,400 MPH Orbiting the
  Galactic Center {or Galactic Spin Rate}  574,585 MPH  or 256862.478 m/s Speed of Earth
  within Our Galaxy

light's= velocity = 3*10^8 m/s
So the ratio = 256862.478 / 3*10^8 = 0.00085620826


Could a pulsar in orbit around a black hole feed the black hole with its radiation?

Having the pulsar emiting radiation without decaying, the black hole not decaying via Hawking radiation and also the orbits of both objects not decaying and in a particular dynamic, I would say yes, veeeery slowly.

Thing is the radiation from the pulsar comes from particles that fall in it (dust, etc) and are "reoriented and shot" (roughly speaking) as high-energy radiation near the poles because the magnetic field.
The black hole would attract even more particles than the pulsar so considering both objects in a space with same density of particles, the black hole by itself should suck in more matter/energy compared with matter/energy from the pulsar beam.

Also consider orbital implications. A black hole (unless supermassive at center of galaxies) and pulsars have equivalent masses. Not the same, but not different in orders of magnitude if the black hole is a stellar remnant as the pulsar.
That means that the orbit will be, probably, more like a double-star system, with the black hole moving also around the center of masses (you can thing about it as Pluto-Charon orbit system).
Now. Consider how stars are formed (from a dust cloud, it starts to collapse, gain angular momentum, etc). Chances are that the orbits and the spins of both objects are in a very similar planes. Think that the beams of the pulsar are shot mostly from the poles (with some inclination though).
I think is almost impossible for a naturally formed stars that one becomes pulsar, the other black hole, and also they orbit in a way that the beams of the pulsar are going to feed the black hole. The beams would just scape from the system, forming ~90º relative to the line from the pulsar-black hole.

Last, about the "feeding". Lets consider that all the above is possible, just very unlikely. We have a scenario where the orbit of a pulsar is ~90º compared with its spin, like Sun-Uranus. Then a fraction of the time it feeds the black hole (twice a "year").
Black holes eat stars sometimes. The size of the black hole grows when it eats, and is one of the theories about the mass of supermassive black holes in the center of most galaxies (just eating... the other is direct formation via a sudden collapse of super-massive dust cloud even avoiding a star formation).
I try o point that this feeding via a pulsar beam would be super slow compared with direct feeding of surroinding matter like the supermassive black holes do.

I would say to the black hole: Sir, eat the pulsar :)

And maybe it would become true: considering very long periods of time, orbits decay. The black hole would maybe collide/absorb the pulsar (as recently detected via gravitational waves)
Finally, depending on the period of time we are talking about, we have to consider black hole evaporation throught Hawking ratiation (it would make it smaller).
On the other hand, I considered pulsars powered by matter orbiting/falling in it. there other types, so a rotation-powered one, after trillions of years, could maybe decay in an almost neutron star.
I don't know about the numbers so maybe these effects are almost zero, not sure.

PD. This is my first answer in StackExchange and I'm not a native english speaker, feedback appreciated ;)


Could a space radio telescope fill multiple purposes?

See http://en.wikipedia.org/wiki/HALCA and http://en.wikipedia.org/wiki/Spektr-R.

HALCA was a Japanese 8 meter radio telescope satellite used for Very Long Baseline Interferometry (VLBI) from 1997 to 2003.  Spektr-R is a Russian 10 m dish launched into space in 2011 with apogee nearly as far away as the moon, and still going.
So, yes a radio telescope in space is quite useful for long baseline interferometry.  You probably do not want to also use it for deep space communications because that would interfere with the scientific use.  It is also not very useful for asteroid work because the odds of one going by is not greater than the odds of one going by the Earth and there are more powerful radio dishes on the ground.  Typically, one would not bother having a strong source for bouncing off asteroids on a radio telescope meant for extragalactic work.  And you sure would not want to risk sending it into the asteroid belt!


Why can't we see the supermassive black holes supposedly at the center of galaxies?

The main reason supermassive black holes aren't visible in images of our galaxy is because of all the dust and stuff in between the center of the galaxy and us.  The Milky Way's bulge is ~16000 light-years thick.  There are millions of stars, and huge clouds of dust in between us and the black hole, which block visible light.  Most of our measurements of the center of our galaxy are made in radio wavelengths.

In addition, compared to galaxies, black holes are small.  A supermassive black hole might have a radius of up to 400 AU (for the really large ones).  The nearest spiral galaxy to us, Andromeda, is 778,000 parsecs away.  This corresponds to an angular size of 0.0005 arcseconds.  Even the Hubble Space Telescope only has a resolution of 0.043 arcseconds in the optical.  That means that Hubble can only see things that are 100 times this size.  So we wouldn't even be able to see any black holes in another galaxy.

That being said, we can make observations of black holes using indirect means.  For example, one of the ways we are able to precisely measure the black hole in our galaxy is by watching stars orbit around it.  We take pictures of these stars in the infrared and radio wavelengths, which pass through the dust.  By measuring how fast these stars are moving, we can calculate the mass of the black hole.

In other galaxies, we can observe other effects of black holes, such as active galactic nuclei, and quasars.

So despite being unable to get an optical picture of a black hole, we still know that they're there.


Is it possible to create a traditional clock solar system?

Is it possible to create the proposed system of satellites?

Yes (in theorey). Hierarchical multiple systems (like the one proposed in the OP) tend to be stable if the periods differ by $\sim 5$ or more and the orbits are near-circular. So, such a system could be stable. A slight problem may be the commensurability of the periods (the fact that their ratios are rational numbers), which implies orbital resonances. However, these resonances are rather weak (the lowest ratio is 1:24), but they may mess up the system in the very long term (a detailed analysis would be required to find out). In practice, it will be extremely hard to set-up such a system, even with advanced space engineering.

What are some values for the weights, rotational speed, and radii that would work?

Since this is a strongly hierarchical system, the masses $m$ and semi-major axes $r$ (=radius if the orbit is circular) can be worked out from Kepler's third law
$$
    \frac{2\pi}{T} = \Omega = \sqrt{\frac{G(m_1+m_2)}{r^3_{1-2}}}
$$
for each sub-system. Giving
\begin{align}
   \frac{4\pi^2}{1\mathrm{sec}^2} &= \frac{G(m_S+m_M)}{r^3_{S-M}} \\
   \frac{4\pi^2}{1\mathrm{min}^2} &= \frac{G(m_S+m_M+m_H)}{r^3_{SM-H}} \\
   \frac{4\pi^2}{1\mathrm{hour}^2} &= \frac{G(m_S+m_M+m_H+m_D)}{r^3_{SMH-D}} \\
   \frac{4\pi^2}{1\mathrm{day}^2} &= \frac{G(m_S+m_M+m_H+m_D+m_X)}{r^3_{SMHD-X}},
\end{align}
where $r_{ABC-D}$ denotes the distance (more precisely: the orbital semi-major axis) between object $D$ and the centre of mass of objects $A$, $B$, and $C$. Since these are 4 equations for 8 unknowns (4 masses and 4 distances), there is some freedom in the design of your 'clock'.

Does the universe have a limit on how many layers of satellite systems it permits?

So far, our formula only considered gravitational forces and neglected all else. Moreover, post-Newtonian effects (general relativity) have been ignored. As long as these assumptions remain valid (orbital speeds $\ll c$, orbital distances $\gtrsim\,$cm, solid objects with sizes $\ll$ distances so that tides are neglible, no electric or magnetic fields, vacuum), there is no limit. However, in practice, there is always some elctro-magnetic field and never perfect vacuum...


Would alcohol wiping damage multi-coated eyepiece coating? How else can I take COVID precautions on the eyepiece?

After calling Orion (they were nice to answer the phone on a Saturday), they said they can't say anything about using alcohol but they do have an optics cleaning kit to sell.
So I tried to find the ingredient of this cleaning kit, but instead found this, which is from the skyandtelescope web page:

For tougher dirt or stains, various lens-cleaning solutions are
available. Good ones are pure isopropyl alcohol or methyl alcohol
(methanol), available in drug stores and hardware stores,
respectively. Standard, diluted isopropyl rubbing alcohol works well
too and is easier to find, but avoid alcohol preparations with other
ingredients that may leave stains. Camera shops sell lens-cleaning
fluids such as Crystal Clear, which is pure methanol, but you can get
methanol much cheaper in a hardware store. Also available are “lens
pens” with a soft, retractable, solvent-impregnated cleaning pad.
When cleaning lenses, apply fluid to your cleaning pad, not the lens
itself. You don't want any fluid to seep in around the lens's edges;
it may carry dissolved gunk to glass surfaces inside and leave stains
where you can't reach them.

And this cloudynights web page supports this information:

The cleaning solution I use is 91% pure isopropyl alcohol. A pint
bottle is less than two dollars, and is useful to have around the
house anyway. Q-Tips are the 100% cotton type. They are used to apply
the cleaning solution, and remove the excess.
A word on the alcohol. The 91% pure isopropyl alcohol contains two
ingredients on the label, alcohol and purified water. Alcohol
evaporate easily at room temperature, and leaves no residue. Purified
water will not leave a residue, but will take longer to evaporate.
Some rubbing alcohols may contain lanolin or other oils to prevent the
alcohol from excessively drying the skin. These should be avoided.
Read the label. We want alcohol and water as the only ingredients.

So in conclusion, 90%+ isopropyl alcohol (the other 10% being water) is actually recommended to use on eyepieces after dust is removed and if the alcohol is applied to a cleaning material to wipe on the eyepiece, not directly onto the eyepiece so that it does not seep through the lens. This is wonderful because alcohol kills any virus and bacteria as a bonus!
What I plan to do will be taping a big piece of Saran wrap that covers the eyepiece, and use a new piece each time. It'll be a big hassle, but I don't have enough time to get the 90% or pure isopropyl alcohol now.

Update: The little gathering was a success! I initially had a rubber band to hold the Saran wrap down, but after 2 minutes it got dropped into the grass and couldn't be found. Since I wanted to reduce any wrinkles in the Saran wrap and hold it taut, using tape was not practical, so I ended up using my hands to hold a large sheet of wrap for each family. We saw Jupiter, Saturn, and Mars!

Update #2: This is the response from Orion, and they discourage using alcohol, so be cautious, maybe test it on a tiny spot first with a tiny amount.

Original Inquiry: Can I use alcohol
wipes to disinfect the multi coated Orion E Series 7-21mm Zoom
eyepiece?
Thank you for contacting Orion Telescopes & Binoculars. We do not
recommend using alcohol wipes on any optics , as this could damage the
coatings on the lenses. You can on the outside of the lens but not
anywhere near the glass lens. We would recommend cleaning fluid for
the lens.
https://www.telescope.com/Orion/Orion-Optics-Cleaning-Fluid-for-Coated-Lenses/rc/2160/p/5810.uts?keyword=fluid
Please do not hesitate to contact us if you have any further questions
or concerns!
Thank you, Leanne Customer Service Orion Telescopes & Binoculars


The key to this kind of comparisons is to get a ratio. How much bigger is one number than the other?

From Buenos Aires to Los Angeles, there are nearly 10,000km. The size of a grain of sand varies a lot, but let's say it's a millimetre in diameter.

That's 4 orders of magnitude in the numbers themselves, then three orders of magnitude from there being a thousand metres in a kilometre, and three more from there being a thousand millimetres in a metre.

$=10^{10}$

This is how much larger the LA - Buenos Aires distance is than the diameter of that grain of sand. It is also how much larger the distance out in the universe is compared to the diameter of the Earth.

The Earth's diameter is a bit more than ten thousand kilometres, so after multiplication, we will end up with about $\approx 10^{14}km$. A light year is about $10^{13}km$, so we get out to the handful of closest stars.


Will there be any more planets come into our solar system and make it their "home"?

Short answer: Yes, but it is very unlikely.
The probability that a random object drifts into the Sun's sphere of influence (SOI) is low (excluding tiny stuff like space dust and small rocks). But the chance that a full-blown rogue planet enters our Solar System is many orders of magnitude lower. Let's set up a Drake equation:
Let $R_n$ be the number of rogue planets in the galaxy, $R_{ns}$ the fraction of rogue planets that may enter the solar system, $R_e$ be the probability that one enters the Sun's SOI, $R_i$ be the probability that the planet will survive an encounter with a Solar System planet, $R_s$ be the percentage of all possible orbits for the rogue planet after the interaction that are long-term stable, and $R_o$ be the probability that the rogue planet gets captured into an orbit. Therefore, $$R_o=R_nR_cR_iR_s$$


According to the Astronomical Society of the Pacific's website "How Fast Are You Moving When You Are Sitting Still?", we are moving:

Rotation of the Earth: 1600 km/h

Orbital rotation of the Earth around the sun: 107,000 km/h

The sun around the centre of the galaxy: 792,000 km/h

Going further, according to the article "Milky Way Galaxy: Facts About Our Galactic Home" (Redd, 2013) quotes the sun's speed around the centre of the galaxy as 828,000 km/h

The Andromeda Galaxy and the Milky Way are headed for a collision at 112 km/sec

According to "Milky Way moving away from void" (2007), the Milky Way galaxy is moving from the local void at 600,000 mph (960,000km/h)

According to the article "The Universe is expanding at 74.2 km/sec/Mpc" (Plait, 2009), the Universe is expanding at a rate of 74.2km/sec per megaparsec (3.26 million light years).

There seems to be some divergence of values in the articles, but these give a good estimate.


If we pumped an large EMP into High Earth orbit and it detinated would it act as a shield against solar radiation?

I can think of 2 ways to create an EMP and while both involve explosives in both cases the EMP is an effect of the blast and not a detonation per se. Would it create a radiation free "bubble "? Yes but only briefly and hardly worth the trouble. In fact the first way an EMP is created is as a side effect of a nuclear explosion. Hardly worth the trouble then if the goal is to clear out a patch of space to then fill it full of fallout. Rapidly collapsing a strong magnetic field essentially by crushing the magnet in an explosion may also produce an EMP although less strong and your space cleared of radiation would be but for a fleeting moment and only potentially cleared of radiation susceptible to electromagnetic influence. If you think of the charged particles that you are trying to deflect as bullets and the electromagnetic lines of force as walls then your walls have to be strong enough to withstand the bullets or the particles  will blithely rip through the EMP like it wasn't even there.


On the Equator, why are the shortest shadows seen at 11:32, and not at Noon?

If you look at the position of the sun at the same time each day, over the course of a year, you will find it traces a "figure of 8" pattern in the sky called an analemma. This may seem an extremely obscure thing to have a name, but it is important for sundial makers, and comes from sundial cant.

Now this means that the time that the sun will cross the meridian will vary from day to day. At the equator, the sun will travel directly overhead only on the equinoxes. In June, the sun will be in the North at midday, In December it will be in the South.

Each country is in a time-zone. It is a matter of convenience that the time in the country is all the same, rather than each location using their local solar time. At your location in Costa Rica you have found that solar noon occurs at 11:32. The Costa Rican's could choose to put their clocks forward an hour, but if they did, then local noon would be at 12:32.

The effect of time-zones is most extreme in the far west of China, where Beijing time is in use. There, local noon occurs at about 15:00. China chooses to have a single time zone, in contrast to other large countries like the USA or Russia.


There is an explanation for why rings flatten out here. The general mechanism is that particles collide, and gets a very uniform momentum. Thus, any set-up giving unusually thick rings is in essence "cheating".

Here are some ways:

Moons can cause spiral waves in the rings, giving them more of a structure in the z direction. The ones known in Saturn's rings has a modes amplitude of just 10-100 m, but larger Moons can easily increase that.

Another way is simply having massive rings. Then they can not get more flattened, as there are no more empty space to remove.

A tilted ring relative to the Planets orbit around the star is going to experience tidal forces, as long as the radius of the rings is some notable fraction of the planet's orbital radius. From the context that sparked the question, that is not a suitable mechanism though, along with the possibility of having a so low density that particle collisions are rare.

However, more promising:

The halo ring of Jupiter is estimated to be around 12500 km thick (about the same as the diameter of the Earth), and are very fine dust kept from condensing into a disc by both the magnetic fields of Jupiter, and by iterations with the Galilean Moons.

We have four planets with rings in the solar system, so the sample size is quite small. Applying some small-sample-size statistical methodology, in this case an unusual application of the German Tank Problem, we can give a rough but realistic maximum thickness of a ring:

$$N \approx m+\frac{m}{k}-1$$

Where $m$ is the highest observed value, and $k$ the sample size.

Modified slightly to get a non-integer version that makes some sense, we get:

$$\text{max}_{\text{thickness}} \approx 12500 \; \mathrm{km}+\frac{12500 \; \mathrm{km}}{4} \approx 16000 \; \mathrm{km}$$

By no means a very certain limit, but at least about what can obtain from what we know.


What is the magnitude of variation of Earth's orbital inclination?

In a solar system model by Varadi Runnegar and Ghil Earth's inclination varies between 0 and about 0.05 radians (or about 3 degrees)

The variation is rather chaotic, but you may note that there seems to have been a qualitative change in the pattern about 60-70 million years ago


Plausible reason for the Hot Neptune gap

An explanation could be provided by irradiation and evaporation of planetary envelopes.

The introduction in the article by Bourrier et al. 2018, which is the source of the NASA press release, explains that as exoplanets migrate inwards, they are increasingly challenged to hang on to their envelopes due to irradiation and heating (insolation) by their star.

The Neptune-sized planets are less capable of retaining their envelopes because of lower densities and lower "surface" gravities. The idea is that many of them get stripped of their outer envelopes and end up as "Cthonian" planets that are of "super-Earth" size.

A hypothesis is that many hot Jupiters have enough mass and density to maintain a large size for billions of years and hence it is only the hot Neptunes that are depleted.

The recent work on GJ 3470b (cited in the NASA press release) supports this hypothesis by finding a very large atmospheric loss rate from a Neptune-sized object at the edge of the "hot Neptune desert".


Why are most planetary orbits nearly circular

The combination of the conservation of angular momentum and gravity give you the inclination and direction.  Gravity will condense material in the axis direction of the angular momentum, and collisions and gravitational interactions will damp oscillations in that direction, forming a disk of material.  The angular momentum resists gravity in the plane perpendicular to the axis, maintaining the extent of the disk.  Over time, the direction of rotation of the angular momentum is the direction of the vast majority of the material, due to collisions and interactions with that smaller population going the wrong way on a one-way street.

The circularity of the orbits is the result of a more dynamical process.  Hopefully someone else here can explain it better than the following.  My simplistic understanding is that a bunch of orbits crossing each other is not stable.  Planets and planetesimals have their orbits changed continuously until such time as they settle into orbits that have fewer interactions with other bodies.  Sort of a natural selection.  In the long haul, this results in a relatively stable configuration with circular zones that tend to not interfere with the other circular zones.  In fact, this is now part of the definition of the word "planet", in that to be called such, a planet needs to clear the neighborhood of its orbit.  Though its not clear how it could occur, a bunch of elliptical zones that happen to be co-aligned would also not be stable since orbits precess due to various influences.

Where the material was less dense than where the planets are currently must certainly have many objects in eccentric, retrograde, and/or highly inclined orbits.  Pluto is the first hint at that, with a relatively high inclination and eccentricity compared to the planets.  Sedna is far more eccentric.

As for other systems, I'm sure there must be some oddballs out there.  However the Kepler mission has observed mostly circular orbits for planets around other stars.


Why is the Color scheme of natural satellites in our solar system based of light shades of grey?

The colour of a natural satellite is almost entirely based off the atmosphere and surface of the satellite. The moon Io, of Jupiter is not grey and is coloured because of sulfur on its surface that create a colour. Io has a very thin atmosphere.

So that brings us to your question, why is the majority of natural satellites grey? We know that moon's colours are dependent on what they are made up of and if they have an atmosphere, so the basic answer is that the majority of moons are grey because they do not hold an atmosphere, and are made up of rock. Our moon is made out of basalts which are a grey rock. You can probably go out and research what certain moons are made of, and find out more about the rocks on them and how they affect the colour of the planet.


To quote from wikipedia

  There are several distinct phenomena which can be used to measure mass. Although some theorists have speculated that some of these phenomena could be independent of each other,[2] current experiments have found no difference in results regardless of how it is measured:

  Inertial mass measures an object's resistance to being accelerated by a force (represented by the relationship F = ma).
  Active gravitational mass measures the gravitational force exerted by an object.
  Passive gravitational mass measures the gravitational force exerted on an object in a known gravitational field.

  The mass of an object determines its acceleration in the presence of an applied force. The inertia and the inertial mass describe the same properties of physical bodies at the qualitative and quantitative level respectively, by other words, the mass quantitatively describes the inertia. According to Newton's second law of motion, if a body of fixed mass m is subjected to a single force F, its acceleration a is given by F/m. A body's mass also determines the degree to which it generates or is affected by a gravitational field. If a first body of mass $m_A$ is placed at a distance $r$ (center of mass to center of mass) from a second body of mass $m_B$, each body is subject to an attractive force $$F_g = Gm_Am_B/r^2$$ where $G = 6.67\times 10^{−11} N kg^{−2} m^2$ is the "universal gravitational constant". This is sometimes referred to as gravitational mass.[note 1] Repeated experiments since the 17th century have demonstrated that inertial and gravitational mass are identical; since 1915, this observation has been entailed a priori in the equivalence principle of general relativity.

In space the third phenomenon (passive gravitational mass) is not terribly useful, because we may not have a known gravitational field to hand, or both the object being measured, and the instruments doing the measuring may be accelerating under the influence of the same field (this is what happens in orbit).

The others are fine. For measurement purposes it can be helpful to avoid having to measure forces exactly, and this is where conservation of momentum comes in. Suppose you have an object of unknown mass $m$ and an object of known mass, which, for simplicity of calculation we'll assume to be 1kg. Initially they are at rest relative to one another. Now we let some interaction happen between them (it doesn't matter what, so long as no other masses are involved) as a result of which they move apart. We measure the speeds of movement (in a frame in which they were both initially at rest), say $u$ for the unknown mass and $v$ for the 1kg mass (these must be in opposite directions). So conservation of momentum tells us that $mu - 1.v = 0$ or $m = v/u$.


The life course for a massive star from birth to death using the HR Diagram

I won't primarily explain the H-R diagrams, because I think focussing on some of the underlying physics is essential for understanding, especially stellar nucleosytheses.
For simplification let's assume the star initially consists of nothing else than ordinary hydrogen and traces of carbon and nitrogen.

Nuclear fusion of hydrogen forms helium; fusion of helium forms carbon; fusion of carbon leads to heavier elements like neon, also oxygen, sodium, magnesium; neon decays to oxygen; oxygen fuses to silicon and others; silicon fuses stepwise with helium to iron. Each of these phases of burning needs higher temperature, and it releases energy. The phases can be subdivided in first core burning, then shell burning.

Earlier phases last long enough to be be visible from outside, resulting in motion within the HR diagram. Heating from phase to phase generally results in an overall expansion of the star.
The last phases last only a short time, too short to propagate effects to outside.

Fusion of iron consumes energy. Therefore the star gets a problem after the core consists of iron. Outer pressure cannot be stopped by fusion and further heating: The core collapses to a neutron star or a black hole, while the outer part is blown off by a supernova.
Neutron stars usually rotate rapidly, and cool down over very long periods of time.

After this overall framework you may go deeper into details of the phases.


How to approximate the distance at which the thin disk peaks along a given LOS from Earth for a given galactic coordinate per 1 sq deg?

If you have a ray from an origin point $\mathbf{x}_0$  with unit direction $|\mathbf{v}|=1$,  $\mathbf{x}(s)=[x(s),y(s),z(s)]=\mathbf{x}_0+\mathbf{v}s$ and look at the density times distance $\rho(s)=\rho_0 e^{-R(\mathbf{x})/h_R}e^{-z(\mathbf{x})/h_z}s^2$ the derivative is
$\rho'(s) =\rho_0\left\{ [-e^{-R(\mathbf{x})/h_R}R'(\mathbf{x})/h_R]e^{-z(\mathbf{x})/h_z}s^2 + [-e^{-z(\mathbf{x})/h_z}z'(\mathbf{x})/h_z]e^{-R(\mathbf{x})/h_R}s^2 +[2s]e^{-R(\mathbf{x})/h_R}e^{-z(\mathbf{x})/h_z} \right\}$
$=\rho(s)\left\{[-R'(\mathbf{x})/h_R]+[-z'(\mathbf{x})/h_z]+[2s]/s^2\right\}$.
Now, $R'(s) = (x(s)v_x+y(s)v_y)/R(s)$ and $z'(s)=v_z$ so the criterion for a maximum or minium is $0 = -(x(s)v_x+y(s)v_y)/h_RR(s)-v_z/h_z + 2/s$ or $$\frac{2}{s} = \frac{[x_0v_x+y_0v_y] +[v_x^2 + v_y^2]s}{h_R\sqrt{(x_0+v_xs)^2+(y_0+v_ys)^2}}+\frac{v_z}{h_z}$$
I can't see how to simplify it further.
I think the right approach here is just to do a numeric solution using golden section search rather than do a lot of algebra that anyway ends with a root finding algorithm. Doing that also allows you to use any stellar density model.


Calculating distance from center of mass of two stars in a binary system

It actually doesn't really matter: the angular separation is sufficiently small that you can use the small angle approximations $\sin \theta \approx \theta$ and $\tan \theta \approx \theta$ (where $\theta$ is in radians). In both cases the next term in the Taylor series is $\mathcal{O}(\theta^3)$ and the effect on the answer of dropping this and all subsequent terms of the Taylor series is in most realistic cases going to be utterly dwarfed by the uncertainty in the distance.


North node, South node correct understanding

The first few diagrams show the lunar nodes coinciding with the first and last quarter phases, which is only the case every 173 days or so.
When the nodes are aligned with the new and full phases, eclipses can occur:

In this diagram from timeanddate.com,
our point of view is north of the ecliptic, the ascending node ☊ is in the 5 o'clock direction, and the descending node ☋ is in the 11 o'clock direction.
Nodal precession moves them retrograde (clockwise from north) about 19.4° per year, an 18.6 year cycle mostly due to the Sun's gravity.


Lightkurve nan values after bin light curve

Thanks to all for your help. Because of the planetmarker's comment I found out that it is also necessary to set the time_bin_size parameter. After upgrade to lightkurve 2.0.6, there is also parameter bins, which sets time_bin_size automatically.
bin = fold.bin(bins=101)  # bins instead of n_bins


Why is space and time spoken of as one thing

There are different words for different aspects of space. For example, consider: length, width, and height. Other words include depth and breadth. We can speak of them as different things if we choose to, but we generally consider them to be part of unified concept of space. Why?

It's because we understand that these words just pick out measurements along specific directions in space set by context or the speaker's viewpoint, and that there is no intrinsic difference between those directions. Depending on which way they are facing, what's length to one person is width to another.

More formally, there is a symmetry between directions in space. We can rotate freely and the intrinsic properties of space look just the same, e.g. the Euclidean metric (infinitesimal distance formula)
$$\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2$$
is invariant under the orthogonal group of rotations, $\mathrm{O}(3)$. Together with spatial translations (which are like rotations about a point at infinity), this forms the Euclidean group of isometries, $\mathrm{ISO}(3)$. Because of this symmetry that "intermixes" directions of space, we think of those different directions as being unified one into one thing: space.

In special relativity, spacetime has a Minkowski metric (in units of $c=1$)
$$\mathrm{d}s^2 = -\mathrm{d}t^2 + \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2\text{,}$$
and it is invariant under the Lorentz group $\mathrm{O}(1,3)$ that act like rotations in spacetime; together with the translations (in space and in time), the full isometry group is the Poincaré group. In fact, the usual Lorentz transformation (say, boosting along the $x$-axis) is exactly a rotation with by a hyperbolic angle in the $tx$ plane.

That's why people began to think of spacetime as one unified whole. The reasons are essentially the same as why people think of space as one unified whole despite it having different directions: there is a symmetry that "intermixes" those directions, albeit slightly differently in spacetime because of the different sign of the temporal and spatial directions in the metric.

In general relativity, this is generalized further, but the details aren't immediately relevant here.


Orbital decay via gravitational emission in a hypothetical, two-body universe

Using this equation from Wikipedia for the approximate merger time:
$$\frac{5}{256}\frac{c^5}{G^3}\frac{r^4}{m_1m_2(m_1+m_2)},$$
and Wikipedia's values for the Moon's equatorial radius and mass, and the launch mass of the HST, we can use this query in Google:
((5/256) * (c^5/G^3) * ((100 miles) + (1738.1 km))^4 / ((11110 kg) * (7.347631E22 kg)^2))
to perform the calculation.
The result is $\approx 3.4497 \times 10^{46}$ seconds, or around $1.093 \times 10^{39}$ years, which is about $7.93 \times 10^{28}$ times the current age of the universe.
That equation is for a circular orbit, and it's only a crude estimate based on orbital energy lost via gravitational radiation. It mostly ignores the change in the orbital radius, since that's fairly small until the final stages of the merger.
There's some information about the effect of orbit eccentricity  on changes to orbital angular momentum and energy due to gravitational radiation at Two-body problem in general relativity.


I would start by doing a search on VizieR, select "Globular_Clusters" from the right-hand list box (this tells me there are around 900 catalogs in this category) and click "Find Catalogs".

Some of the returned results will be catalogues of stars within a given cluster, the density map on the right-hand side of the list will show these as a single point. Other catalogues will be lists of multiple globular clusters, the density map will show there is data from all over the sky.

I hope that's useful for you!


Dataset for machine learning MK stellar classification

That I know there is the XHIP catalog via VizieR and you can enter a range in UMag (or B- and V-band) for example of -20 .. 20. Check the box SpType and Tc to get also the spectral type and temperature (double check the literature for the quantities you need)
On the column on the left titled Preferences, set the quantity max to unlimited (it's the number of data) and the box just below indicates the format to download the data, if you want *.csv select CDS Portal and click on Submit.
You will be redirected on the CDS portal and simply click on Save and then on MyData, it will show you a list of the dataset you saved and you can select the format of your file (csv, fits, etc.) and then download.
With the simple range of the V absolute magnitude between -20 and 20 you get more then 100k stars
Edited: I noticed that downloading the data through the CDS portal does not give you the quantities that you check, but some fixed one. Maybe I can suggest to select ascii text/plain and then download the page.


How can cosmic inflation make an infinite universe homogeneous?

Inflation is used to explain why the observable universe is extremely homogeneous.

Without inflation, we can do the following crude calculation. The cosmic microwave background was formed about 300,000 years after the big bang, at a redshift of about 1100. Thus causally connected regions at the epoch of CMB formation would have a radius of $\sim 300,000$ light years, which has now expanded by a factor of 1100 to be $3.3\times 10^{8}$ light years in radius.

This can be compared with the radius of the observable universe, which is currently around 46 billion light years. This means that causally connected regions should only be $\sim 4 \times 10^{-7}$ of the observable universe, or equivalently, patches of CMB of $\sim 2$ degrees radius on the sky are causally connected. This is clearly not the case as the variations in the CMB are no more than about 1 part in $10^{5}$ across the whole sky.

Inflation solves this by allowing previously causally connected regions to inflate to become larger than the entire observable universe.

You appear to understand this quite well, so I am not entirely clear what your question is. We cannot know whether the entire universe is homogeneous, since we cannot measure it. The cosmological principle is an assumption that appears to hold approximately true in the observable universe, but need not apply to the universe as a whole. Indeed it is not absolutely true in the observable universe otherwise it would be quite uninteresting, containing no galaxies, clusters, or other structure. I think the only requirement on inflation is that it blows up a patch of causally connected universe so that it becomes much bigger than the observable universe at the current epoch.


Galactic mass estimation from apparent magnitude and distance

First you must estimate the luminosity. That may be difficult unless the magnitude you have is a bolometric magnitude.

Then you assume some mass to light ratio (actually mass to luminosity ratio, where in solar units the M/L of the Sun is 1).

The M/L ratios of galaxies depend on galaxy type (and age, metallicity, star forming history) and vary from about 2-10. However, this is just the stellar mass. In most cases these ratios would need to be increased by a factor of 10 to give the total mass of a galaxy, which may be dominated by dark matter. The total M/L ratio of dwarf spheroidals like Tucana is likely to be $\sim 100$.

In conclusion, there is no single value of $c$ that can convert from a magnitude (luminosity) to a mass. Further reading: https://courses.lumenlearning.com/astronomy/chapter/properties-of-galaxies/


Is it possible to calculate how much a star's brightness dims when a planet transits in front of it from a viewer's perspective?

You answer your question basically yourself with the calculation - that's the right way to approach it, your math is correct. Stars and planets are big enough, that simple geometric optics work well to calculate the degree of occulatation when one body passes in front of another. This is the method used to estimate the size of exoplanets pretty accurately (we know stelar sizes from the mass-luminosity-radius relations).

Typical drop in luminosity from observed exoplanets is 1% or less - which happens to be exactly the square ratio of solar radius to jupiter radius. You correctly get bigger luminosity drop as you don't put both objects in infinity, thus don't directly compare radii but their apparant angular diameter.

The only not exactly straight-forward part is the influence of the atmosphere of the transiting body as it contributes to absorption depending on density. That is one of two reason that for any brightness curve of exoplanet transits you don't see an immediate drop in brightness (the other is the result of the limb darkening of the star). (See e.g. this light curve) This is used to infer exoplanet atmospheric properties. It can even be taken so far as to make atmospheric spectroscopy by comparing transition spectrum and normal spectrum of the star.

You, however, seem implicitly to ask (also) about the human factor, how we, as humans, perceive the brightness variation. That is an entirely different story as our eyes cover a MUCH larger brightness span as they work logarithmically - thus small changes in linear brightness (eclipsing part of the sun's cross-section is simply that) don't make a big impression. CCDs or CMOS work linearily, thus detect such changes much easier - but are bad when it comes to covering large brightness variation.

Anyone who observed a solar eclipse can tell the story that you don't notice much of a brightness drop until the very last moments when the sun becomes a very tiny sickle just moments before totality. The most notable difference is behaviour of the shadows (they change shape) and the sharpness of shadows. It gets (slightly) colder, as less light arrives - but the remaining light source is as bright per unit area as the already eclipsed part is so that the perceived brightness remains nearly the same.
It's similar when you switch on two identical lamps - the room does not seem twice as bright compared to only one lamp.


Is CMB Polarization simply the temperature gradient of the CMB?

Polarization of an electromagnetic wave refers to the extent to which the electric field (which is always perpendicular to the direction of wave motion) oscillates in one particular direction, rather than randomly in the plane perpendicular to the wave motion.

Complete linear polarization means the electric field oscillates to and fro along one particular axis.

Polarization of the CMB is produced by Thomson scattering from free electrons just before they combine with protons and the Universe becomes transparent to the CMB radiation.
The polarization occurs because the radiation field seen by the electrons is not uniform in space (not isotropic), it varies according to which direction the radiation is coming from. These anisotropies are the fingerprints of what went on in the early Universe and what the values of the cosmological parameters are.

There is an excellent primer on this topic - http://background.uchicago.edu/~whu/intermediate/polar.html

The CMB is a blackbody spectrum at a temperature of 2.7K. Differences in this temperature are seen on a variety of scales at levels of 30-60 microKelvin ($\mu K$). These are the "ripples" first seen by COBE, then WMAP and Planck with increasing precision and sensitivity to smaller angular scales.

Whereas expressing these spatial variations as differences in temperature seems straightforward, it is not so obvious why you would express polarization signals in the same terms.

In brief, what is going on here is that when a polarization signal is measured it is actually finding the difference between two perpendicular polarization intensities. Because these intensities are measured as changes of temperature in the bolometer arrays used as detectors, this is how the results of the polarization are expressed.

The polarization anisotropies are expected to be an order of magnitude smaller than the CMB temperature variations. That's because repeated scattering from free electrons can dilute the signal, so it is only light that has been scattered once that contributes.


How to tell between annular and total eclipse (directly)?

There is a massive difference in the brightness of the "ring". During a total eclipse, the photosphere of the sun is completely hidden and the corona comes into sight. The corona is much much less bright than the photosphere of the sun, so during a total eclipse, the sky becomes dark.

During an annular eclipse the photosphere is never completely obscured. The bright 'surface' of the sun is still visible, so the sky never becomes completely dark, the sun remains dangerously bright, and the corona is not visible. An annular eclipse is like the partial phase of a total eclipse, because part of the sun is still visible.


What would a gas giant consisting of only earths atmosphere look like?

Jupiter's atmosphere, if you could see through it, the inside of the planet would begin to glow visibly red from heat perhaps a couple/few thousand miles below it's upper clouds.

Source

The problem is, some two thousand miles of atmosphere above that visible glow is almost certainly well past the point of transparency.   Add to that, mixing and cloud formation would make seeing through all that impossible - no matter what the atmosphere was made of.

Hot gas rises and as it rises, some gas condenses from gas into liquid droplets (clouds), but these clouds form in updrafts so even though they start falling back towards the planet, the thick atmosphere and the updraft movement maintains the clouds for some time.   There's different layers of clouds too.   Jupiter's upper most clouds are Ammonia, but below that there's water clouds (and perhaps others).   go deep enough you might get sulfur clouds (I'm not sure) or more dense hydrocarbon (gasoline like) clouds.

An atmosphere that thick maintaining transparency is impossible.   Hydrogen and Helium are (to my knowledge) just as transparent as Oxygen and Nitrogen, maybe more, but the mixing and condensation gives Jupiter a constant cloud cover, different colors mostly represent different temperatures and depths from which that gas came from.

If you gave Jupiter an Oxygen-Nitrogen atmosphere but maintained it's inner core, the inner hydrogen would rise and mix with the Oxygen-Nitrogen.   Chemistry would inevitably happen and the Oxygen would bind with the hydrogen and make water.   Some of the Nitrogen would probably form Ammonia and in the end, Jupiter might not look that different.   Initially it might have a bluish tint, but only temporarily.  It would likely always have clouds and mixing some heavier/darker gases from below.   You'd never see it's core, no matter what.

Now if you had a planet with only one gas in it's atmosphere and entire makeup - avoiding any chemistry, say a planet made entirely of hydrogen or entirely of Argon or Oxygen - then you might get specific colors relating to the specific element.   But such planets are pure fantasy.

My best guess is that Jupiter's appearance wouldn't change much if you gave it an Earth-like atmosphere, but Jupiter is basically made of gas, so if you did that, it would just mix and it's inner hydrogen would rise to the top and Oxygen-nitrogen would sink.   It's the trace gases that give Jupiter it's color anyway, not it's hydrogen.

Layman's answer.  :-)


Gravitational pull needed to keep a gas in atmosphere

Yes, the formula you quote applies to some extend.

It can also be written in terms of Boltzmann's constant $k_\mathrm{B}$ as
$$v_\mathrm{rms} = \sqrt{\frac{3k_\mathrm{B}T}{m_\mathrm{m}}},$$
with $m_\mathrm{m}$ the mass of the molecule in question,
gives the mean ("root-mean-square") velocity of the gas molecules as a function of temperature $T$. Comparing this to the escape velocity
$$v_\mathrm{esc}=\sqrt{\frac{2GM_\mathrm{p}}{R_\mathrm{p}}},$$
where $M_\mathrm{p}$ and $R_\mathrm{p}$ are the mass and radius of the planet, respectively, gives an order-of-magnitude estimate of whether or not the planet will be able to maintain its atmosphere.

However, if the two velocities are equal, it would still mean that half of the atmosphere evaporated instantly, and that over time, a large fraction would disappear, although the time scale for this to happen may be very long. In order for the planet to really sustain its atmosphere, $v_\mathrm{esc}$ needs to be roughly 6 times larger than $v_\mathrm{rms}$.

Gravity is not the only factor determining the stability of the atmosphere, though. Radiation pressure from the star ("Solar wind") can easily rip a planet of its atmosphere. On the other hand, if the planet has an efficient magnetic field, this will shield it against the wind.

Other factors are also involved. But you're right in that at the very least, the planet needs a certain gravitational field to keep its atmosphere.


Is there a MATLAB equivalent of PyEphem?

Disclaimer: I've never used this toolbox and so cannot speak to the comparison of this toolbox with PyEphem.

There are some ephemeral and the like functions within Matlab, for example planetEphemeris, earthNutation, or moonLibration.

Unfortunately these functions are only accessible within the Aerospace Toolbox and do not come with a standard Matlab license. You'll need an entirely separate license for the Aerospace Toolbox to use these functions in your Matlab code.

If it is easy for you to obtain this license (be it from some university or organization), then by all means go for it. Otherwise, your best bet may be free packages like PyEphem.

I would also like to echo the comment made by David in that you should definitely learn Python. It is an amazingly simple, yet robust language that can do just about anything. And its free!

If you're really dead-set on using Matlab, but the PyEphem package is the only thing that does what you need, you can always consider porting between the two languages. Matlab has several tools and capabilities to interface with other languages. You can read more about using python packages natively in Matlab right here. I will say, this would probably be much harder than just going with Python outright though.


What is the current accepted hypothesis of what caused Triton's retrograde orbit around Neptune?

According to this source from SOA/NASA Astrophysics Data System on Tidal evolution in the Neptune-Triton system (Chyba, C. F., Jankowski, D. G., & Nicholson, P. D.),

  Transcript: Many investigators have speculated on the reason for
  Triton's retrograde orbit. Lyttleton (1936) suggested that both Pluto
  and Triton originated as prograde satellites of Neptune, only to
  experience a catastrophic gravitational interaction. McCord (1966)
  showed that tidal evolution could have brought Triton inward from a
  near-parabolic orbit, lending plausibility to a capture origin for the
  satellite. Jarrington and Van Flandern (1979) suggested Triton's
  peculiar orbit and Pluto's supposed "escape" were caused by ab
  encounter with a massive "rogue" body, a conclusion contested ny
  Farinella et al. (1980). Finally, McKinnon (1984) combined momentum
  and energy constraints with an improved knowledge of Pluto's mass to
  show that all Pluto-Triton interaction scenarios for the origin of
  Triton's retrograde orbit are untenable, leaving the capture
  hypothesis as the most likely possibility.

Now, that summarizes all the theories one might have about it, at least all I could find.

As far as evolution of the system is concerned, the same document follows up to it and demonstrates how tidal effects are responsible for the evolution of the system. "Degrading" is a strange term. Let us go with evolution.

One of the possiblities stated is that Triton will reach a perfectly retrograde state and stay in it for ever! For a detailed analysis, you may read up section 6 in the link provided, "The future of Triton"


Galaxies are a large organised collection of stars (& nebulae) outside of our own galaxy, very distant. They shine by their own light.

Nebulae are clouds of gas & debris from, usually, a stellar explosion (Nova, supernova, etc), within our own galaxy, lit up by nearby stars (maybe internal ones) - but the gas is not glowing with it's own light, generally.

So a galaxy is going to be a bright clump of glowing stars, regular in shape - be that elliptical, barred or other, generally they will appear symmetrical & regular. A nebula will be much more amorphous & patchy, possibly with shapes from the clouds of gas (c.f. Horsehead, Orion), and you will be able to see individual stars in it - unlike a galaxy where they are too tightly packed & distant to make out individual stars.


How did the reionization epoch make the universe "transparent"?

Nowadays, the universe is mostly ionized, and transparent.

Transparent means that photons are able to travel long distances.

So there comes the confusion: How could it be that recombination made the Universe transparent, and reionization keep it transparent?

After the Big Bang we had a very small Universe, filled with ionized matter. That amount of ionized matter caused the photons to be short-lived, and thus the Universe were opaque.

Then it came the recombination, and most baryonic matter became neutral, the photons were able to travel long distances, and the Universe thus became transparent.

And it was expanding. And at the same time clusters of matter were collapsing.

The combination of these two factors made the intergalactic medium density to fall a lot. Not only were there much more space for the same matter, but also the matter was concentrated in small bunchs making almost all the space almost empty.

  The second phase change occurred once objects started to condense in the early universe that were energetic enough to ionize neutral hydrogen. As these objects formed and radiated energy, the universe reverted from being neutral, to once again being an ionized plasma. This occurred between 150 million and one billion years after the Big Bang (at a redshift 6 < z < 20). At that time, however, matter had been diffused by the expansion of the universe, and the scattering interactions of photons and electrons were much less frequent than before electron-proton recombination. Thus a universe full of low density ionized hydrogen will remain transparent, as is the case today. [1]

So when the clusters of matter collapsed enough to become galaxies with stars and these stars reionized the intergallactic medium, there were almost no matter there, so photons continued being free to travel without obstacles, and so the Universe remained transparent.

So, it has been transparent for two different reasons: neutrality of matter, after the recombination, and scarcity of matter, after the reionization.

1: http://en.wikipedia.org/wiki/Reionization#Background


Why is the Warm-Hot Intergalactic Medium (WHIM) so hot, and what is "collisionless shock heating"?

It should probably be added that the article includes a glaring error of the type you often see when the science writer apparently did not take an elementary astronomy class (this is why we have such classes!).  When the article states that the "lost matter exists as filaments of oxygen gas", you can be sure that Michael Shull never said any such thing, because that language clearly asserts that the "lost" matter is oxygen.  Of course what is meant is that the previously undetected baryons exist primarily as free protons (and electrons), which is merely evidenced by the presence of detectable oxygen that is simply assumed to be in the normal ratio we find between oxygen and hydrogen.  We could also quibble that calling it "lost" is hyperbole, a bit like saying the brain matter of the writer was heretofore "lost" because it couldn't be seen by looking into his/her ears, and was only recently "found" in some recent MRI.   Astronomers have long expected the WHIM to be there, and knew it would be hard to see, so it was hardly "lost."  But it is certainly nice to have greater certainty that it is there.

And not only is there a crucial assumption of the oxygen-to-hydrogen ratio, the OP is astute to ask how well we can even know the oxygen content itself.  All we see is the light from the oxygen, so we have to model how emissive it is to know how much oxygen is there.  That's not always easy at low densities where the emissivity can differ from what you'd get in local thermodynamic equilibrium, as was pointed out in the question and the first answer also.  So not all the uncertainties have been resolved, but it's certainly a nice step forward.

One final point regards why the gas is so hot.  This is related to the fact that low density gas cools itself very inefficiently, because cooling generally requires collisions between the particles (which generates the light).  Heating, on the other hand, can occur in various ways that might involve interactions with something other than the particles, say with cosmic rays or with magnetic fields.  Those latter elements can be present even if the density is very low, leading to high temperatures.  There is also a radiative cooling instability above about 10,000 K, where gas that is heated above that starts to cool less efficiently (as hydrogen gets all ionized), and responds by being compressed by the ambient pressure, which drives the temperature up even more.


The basic tidal acceleration felt by a test mass near the surface of the Sun, due to a body of mass $m$ at a distance $r$ is given by
$$a_{\rm tidal} = 2\frac{Gm R}{r^3},$$
where $R$ is the radius of the Sun.

You can work out that Jupiter has the largest tidal effect on the Sun, very closely followed by Venus, and both produce tidal forces that are only a factor of $\sim 2.2$ times greater than that produced by the Earth on the Sun.

The distortion of the Sun could be modelled as following the consequent equipotentials (well inside the Roche lobe) in a binary system, but this is not analytically solvable and I could not locate any straightforward approximations. One way to judge the size of the effect would be to compare the tidal acceleration with the centrifugal acceleration at the Sun's surface $R\omega_{\rm Sun}^2$.

So $a_{\rm tidal}$ for Jupiter is $3.7\times10^{-10}$ m/s$^2$, whereas the centrifugal acceleration (at the solar equator) is 7 orders of magnitude larger at 0.006 m/s$^2$. Given that the Sun is almost spherical to 1 part in 100,000, because the centrifugal acceleration is only 20 millionths of the Sun's surface gravity, we can safely assume that tides due to planets that are 7 orders of magnitude weaker do not distort the shape of the Sun to any degree worth worrying about.

I was surprised to find that the potential influence of planetary tides on solar activity is still an area of recent research. Suggestions that planetary tides might be responsible for sunspots or the sunspot cycle on the Sun go back a long way. Grandpierre (1996) suggested that planetary tides influence flows in the Sun and could be responsible for the 11-year solar cycle. A NASA report by Hung (2007), although not peer-reviewed, does suggest a link between solar activity and planetary positions. The claim is made that major solar flares happen "underneath" the locations of the major tide-inducing planets (Jupiter, Venus, Earth, Mercury [in that order]).

I very quickly found a peer-reviewed paper in "Solar Physics" by Seker (2013), which discusses all this and reviews the evidence on sunspots. They conclude that the appearance of new sunspots is not correlated at all with the positions of the planets. However, they do raise the caveat that because the magnetic fields producing the spots are produced at the tachocline between the radiative core and convective envelope of the Sun and take several days to buoyantly rise. Thus the longitude at which the spots appear and where they are generated, with respect to the positions of the planets, can change significantly.

I guess the jury is still out, but this is quite "fringe" material. The solar cycle is certainly thought to be a product of the dynamo mechanism that produces the magnetic field. However, the location of sunspots definitely cannot be predicted with such models at the moment. Interest in this topic has been revived recently by claims of correlations between magnetic activity and the orbital positions of hot Jupiter-type exoplanets and even the suggestion that tidal effects or interactions between stellar and planetary magnetic fields might be responsible for increased magnetic activity (see Lanza 2014 for a recent review). For these close-in exoplanets of course the tides are many orders of magnitude larger than in the solar system (see equation above).


Is this object Mars?

I'm pretty sure this is simple lens flare.
While lens surfaces in cameras have antireflection coatings that strongly reduce reflections, they are not perfect. A very bright small spot like the Sun or Moon or a bright streetlight will result in near-mirror images opposite the center of the image.
Analysis from this answer:

Here I used this effect on purpose to reduce the brightness of the partially eclipsed Sun before it became annular. Solar eclipse of December 14, 2020):

complete image:
 click for larger


Different full moon face on different places on Earth same time

Yes, they do see a slightly different face of the Moon.

It's easy to calculate how different. If they're on nearly opposite sides of the Earth so that one sees the Moon just rising and the other sees the Moon just setting, they are seeing it from about 1.9 degrees apart. (A triangle with a base of 8000 miles (the distance they are apart) with sides of 240,000 miles (the distance to the Moon) yields an angle of 1.9 degrees (8/240 radians).

So they'll be viewing the Moon from 1.9 degrees apart and will see a full moon displaced by 1.9 degrees, or about 36 miles on the surface. So one will see 36 miles of Moon more on the Lunar east side than the  other and the other will see about 32 miles more on the Lunar west side.


How close would an LED have to be to your eyes to be as bright as the looking at the sun during midday?

Typical sunlight has a power of a few hundred watt per square meter. A typical laptop screen has a power of about 4-5 watt. So obviously there is a vast difference in how bright they are.
To turn this into an astronomy question we can use magnitudes to answer it. The sun has apparent magnitude −26.7. A laptop at an arms length is about 50 lux, which would be about -18.4 apparent magnitude. So to get a solar apparent magnitude I need it to be 8.25 magnitudes brighter. That corresponds to a brightness increase of $100^{8.25/5}\approx 1995$. So if each pixel shone about 2000 times brighter, you would feel like the screen produced sunlight.
You could of course try looking at the screen more closely, making it brighter that way. However, the $1/r^2$ law for how brightness changes with distance stops being valid when your eye get closer to the pixel than the pixel lateral size. Worse, the amount of photons your eye can receive will also be limited by your pupil aperture: you cannot see all of the screen light. So this is why pushing your face against a screen doesn't make it feel like the sun (trust me, I tried this right now with my screen and some sample sunlight).
If it is a smaller region that is very bright surrounded by black it may feel brighter: good visual design can often fake a bit of extra brightness that is not physically there.
I don't see a physical reason for true sunlight intensity screens being impossible, but they are not practical. They would drain batteries like crazy. A visually convincing spectrum may still be doable with less power than a full blackbody spectrum (that would also emit a lot of infrared - nice feeling on the skin, but would heat up the screen a lot) just like LEDs faking white light with just a few frequencies.


Stellarium shows the Moon and Mars very close together in the sky tonight (Saturday, 3rd October 2020), so yes, it was probably Mars that you saw.
Moon and Mars on 2020/10/03 (Stellarium)
Stellarium is a great tool for identifying astronomical objects (and satellites), and is absolutely free.
Very rough rules of thumb for identifying planets by eye:

Very bright, white, and low in the sky (early morning or evening): probably Venus (but might be Jupiter).
Bright, white, and high up in the sky: probably Jupiter (unlikely to be Venus, which stays close to the Sun).
Bright and red: probably Mars.
Moderately bright and pinkish: probably Saturn.
Faint, white, and low in the sky (early morning or evening): might be Mercury.
If it scintillates (twinkles), then it's a star.

Jupiter and Venus are easily confused, since they can both appear to be bright and white. You really need to consider the time of day and the elevation (how high it is in the sky). Since Venus is closer to the Sun than the Earth, you are more likely to see it in the morning or evening (before sunrise or after sunset). Venus can climb quite high in the sky, but it will never be more than 45° in elevation.


Why can I see the dark side of the moon?

You can generally see the unlit side of the moon when a considerable amount of sunlight is reflected off the earth. This reflected sunlight illuminates the unlit side of the moon. This is referred to as earthshine, and a decent explanation can be found at timeanddate.com. I seem to recall reading somewhere (but now can't find a reference) that it is more prominent when it is very cloudy on earth near where the moon is overhead.


Trigger Warning:  Major hacking going on here.
[edited from 1 ly to 1000 ly diameter]
Scaling from the image, I'll approximate the jet diameter as 1000 ly . So there's a volume of roughly 2e4*1e6 * pi/4 cubic ly.   For absolutely no good reason, I'll propose an electron ejection density of 1E6 per cubic km -- at least those are round numbers and easily scalable if anyone can find references to actual calculations/measurements --, so using 1 cubic ly = 8.468e+38 cubic km, I get 1.33e+56 electrons in the jet.


Ensemble average VS spatial average for a cosmological density field

I can't see the contradiction. One version says average over the regions, the other says integrate over one realisation (containing many regions). I suppose you could argue that one is performing a summation,whilst the other is an integration, but if the regions are small compared with the size of your observable universe then they amount to the same thing.
$$\frac{1}{\sum_i V_i} \sum_i \lambda_i  V_i \simeq \frac{1}{V}\int \lambda(r)\ dV$$


In which countries are there organised Messier Marathons?

As far as I have researched, there are really many countries that organize Messier Marathon, but also many astronomers organize it locally only for them and don't publish it on web, so I could write here nearly every country.


What's special about Saturn that it has distinctive rings?

Saturn's rings are unique in that a lot of the rings compositions contain new and recycled material. Much of the uniqueness is complementary from Saturn's moons, which play a critical role in shaping and recycling the rings materials.

A great example is the moon Enceladus, which contributes largely to the E Ring. Icy material is ejected from Enceladus, a process called cryovolcanism (spewing plumes of icy materials instead of silicate rock); due to the lack of atmospheric pressure and low surface gravity, Enceladus doesn't hold the material that it ejects, but rather feeds material into the E Ring. Discovered by Cassini, the material is comprised of simple hydrocarbons such as methane, propane, acetylene and formaldehyde.

A contributing factor for the unique shaping of the rings and much of their orbital stability stems from the moons also. Another well known example is the Cassini Division and a process called orbital resonance. Essentially the moon Mimas orbits once for every two orbits for material inside the Cassini Division. As a result of that orbital resonance the material receives accumulation of radial momentum, the effect of such accumulation forces a tug of war with the particles. Mimas attracts the material pulling it into the A Ring as the material increases in velocity, resulting in the division.

Other moons have also been documented to have this effect, such as the moons Pan and Daphnis in the A Ring, however it is thought that the effect from these smaller moons only create a division around 10km in width. Stability can be exaggerated from the F Ring, with the moons Pandora and Prometheus. These moons twist the material into a spiraling vortex, like a horizontal tornado encircling the planet. Processes of stability arise from moon ringlets, such as the Titan Ringlet. Titan has orbital resonance with material in the C Ring and provides radial momentum with the Colombo Gap, the gap is slightly elliptical rather than circular (just as Titan is) and the material is attracted towards Titan in the same pattern, in a process called apsidal precession. This creates a compression and expansion of the rings giving them stability while accumulating velocity due to orbital resonance.

Distortions from the moons on an independent inclination to the rings as they come close to the rings, also pull material from the rings and spread it around in chaos, additionally recycling material.

Each ringlet has an individual story as to the forces which make that ringlet unique, comprising of the orbital parameters of the moons and the unique positioning of the forces around them. Along with material that is recycled in a unique fashion not evident with other planetary rings system.

The uniqueness stems from these factors.


How are the elevations of the poles proportionate to distances of earth that have been traversed according to Copernicus

Here, I believe he is referring to the altitude of the celestial poles at different latitudes on the Earth.

If the Earth is indeed a sphere then the altitude of the celestial poles above the horizon should be the same as your latitude on the Earth. That is, as you travel along a line of latitude, the altitude of the pole should change accordingly.

So Copernicus was noting that this is indeed what we observe. At the North pole the North celestial pole is directly overhead, and on the equator it is on the horizon. This observation, therefore, is direct evidence that the Earth is spherical (or at least spheroidal).

You can see the evidence of this in star trail photographs. The stars will always appear to rotate about the celestial poles. This image shows the star trails above Ecuador, at a latitude of about 1$^{\circ}$. You can clearly see that the stars are rolling directly over head; with the North and South celestial poles located on the horizon to the right and left, respectively.


When Betelgeuse goes supernova, will the explosion hit any neighboring stars?

There have been 2 proposed spectroscopic companions, though never confirmed, traveling with Betelgeuse. These are thought to orbit the red supergiant. So when Betelgeuse eventually explodes, they would  certainly be blasted by radiation, atmospheres could be damaged or completely destroyed, and would lose their host star sending them on a new trajectory through space.  But with Betelgeuse having been ejected from its birth place and  traveling with no verified companion, its more likely that the surrounding galaxy would be left unchanged, aside from  supernova remnant left behind.
https://en.m.wikipedia.org/wiki/Betelgeuse has more information if you still have questions.


The moon orbits close to the plane of the ecliptic, which is tilted at 23.5 degrees to the equator, the moon is slightly off the ecliptic, and can be about 5 degrees above the ecliptic.

So, on the winter solstice (June 21) if full moon happens to be furthest from crossing the ecliptic, the moon would be 28.5 degrees above the horizon: roughly a third of the way between the horizon and the zenith. It can get the same distance below the horizon, and during summer the full moon would not be seen at all.

Our moon is unique in being close to the plane of the ecliptic, and not in the plane of the equator, which suggests its formation was not like that of other moons in the solar system.


The grey areas are known as "maria" (singular:"mare") and they were formed from flood  basalts. These are a type of volcano and they have formed on earth too, examples include the lake eruption on Iceland, and the Deccan traps, that played a role in the mass extinction 65 million years ago.

On the moon, the flood basalt is old. About 3-4 billion years old, with some regions probably younger, but nothing less than 1 billion. Volcanism is driven by the heat of radioactive decay. On the moon there are no longer enough radioactive elements to melt rock, and so the moon is no longer a volcanic active body.

There are no active volcanoes on the moon, and there is no prospect of any in the future.