Is the moon inside earth’s atmosphere? If so, what are the consequences?

It's long been known that Earth's atmosphere gradually reduces in density, and is detectable to very high altitude.

A 1976 atmospheric model has data up to 1000 km of altitude. At that altitude, atmospheric density is on the order of 10-15 kg/m3. This is already better than the best vacuum we can create on Earth's surface. So we can call this an atmosphere (because it contains gases also found on Earth at ground level), but its consequences are limited.

It's sparse enough that at this altitude, atmospheric drag on satellites is negligible for the lifetime of a satellite.

Earth's atmosphere loses some mass to space: at high altitude some atoms achieve escape velocity. This is replenished by evaporation and outgassing on Earth. For example, helium gas starts its life as alpha particles from uranium decay, the gas seeps through the ground, rises through the atmosphere (because it's much lighter than air) and eventually escapes.

We already know with very high accuracy that the Moon slowly moves away from us as it's slowed down. Most of that slowdown is caused by tidal effects, but there will also be an atmospheric component. Due to the very low density, this component will be small.


The largest catalogues, such as Hubble's Guide Star Catalog or the US Navy USNO-B1.0 catalogue have entries for about 1 billion object (mostly stars) with magnitudes down to 21.

The galaxy itself has about 100 billion stars, but there is considerable uncertainty about that value. The true value could be less or a lot more. Between 50 and 500 billion is the range of uncertainty

So we have catalogued between 0.2% and 2% of the galaxy's stars with best estimate of about 1% of stars have been catalogued.


Do we know the availability of any material suitable for solar cells on the moon?

Welcome to Astronomy.SE!

The moon, like the earth, consists of 25% silicon by mass. Silicon is one of the main elements used in producing solar cells.

So sourcing the material is not the problem. The manufacturing is.

I assume your interest in this matter doesn't just stop at finding the materials, but also includes in-situ manufacture. I won't go for a detailed answer, but I will outline a few considerations.

You need your own silicon industry. It's the sort of thing that requires a large infrastructure behind it. For instance, monocrystalline silicon solar cells require high purity silicon (on the order of a few ppm) and pristine growing conditions for the crystal to grow. That means you need to have the infrastructure in place to mine moon rock, reduce the silica to silicon, and separate out as much impurity as you can. With the sorts of purity levels we're talking about, you'd probably need at the very least a medium-sized moon base already in place.
Making any sort of solar panel is hard. You need an industry to make the silicon. You need another industry to make the solar panels. This sort of stuff requires precision engineering, and complex (and heavy) tooling which you'll have to import from earth. You need a 'pyramid' of support structures and expertise to maintain such expensive equipment.
Amorphous SI is probably your best bet. Amorphous silicon seems to be your best bet. It can be applied through some form of vapour deposition, and from a brief bit of reading around, appears to be one of the easiest ways of making solar cells, though the efficiencies tend to be low, with the best reported efficiency being 14% and realistically achievable efficiencies being likely much lower than 10%. (see the bottom green line of this diagram)


Do stars tend to fuse all hydrogen even when the mass is small?

The difference between a brown dwarf and a star is not a sharp boundary. A brown dwarf is simply a ball of gas where the (small) fusion rate is incapable of providing a significant fraction of the luminosity (which is mainly provided by gravitational contraction).

A star will contract and reach a minimum luminosity, whilst a brown dwarf's luminosity will monotonically decrease throughout its life. The core becomes degenerate and it is this that provides the pressure that supports a brown dwarf, even though it becomes colder.

If you add some mass to a brown dwarf, then the result depends when you add it. If you add it early during the strong contraction phase then you might get a low-mass star. If it is added after the brown dwarf is degenerate, then it could contract further, but do so without increasing the temperature enough to ignite hydrogen strongly. This is an interesting problem that deserves a model simulation!

If fusion can start to any degree, then convective mixing in these very low mass objects will very gradually (trillions of years), turn nearly all their H to He.


How many galaxies can be seen in a given patch of sky

Do galaxies ever appear as points of light similar to faint stars

sure, just google for photos

  Telescope specs

it depends on your exposure time

How many galaxies does Hubble see?

A hellova lot!!

https://en.wikipedia.org/wiki/Hubble_Ultra-Deep_Field


To answer the main part of your question: Yes, there do exist such systems. They're called visual binaries. We generally need a telescope to tell them apart. Most binary systems look like a single star when viewed with only the naked eye; many cannot be resolved without the aid of a telescope. But visual binaries can.

  also is there any binary stars doesn't linked together by gravity

No, there aren't. You were right when you said that the definition of a binary system is basically two stars orbiting a common center of mass. If the stars don't orbit each other, they aren't a binary system.

By the way, check out Proxima Centauri, the nearest star to the Sun. It's one of the three stars in the triple-star Alpha Centauri system. The reason I mention it is that even though it's seemed for years like it was gravitationally bound to the system, that idea is now under debate.

Let me address something that Mitch pointed out. The classification of star systems as "visual binaries" is based solely on our ability to observe them. If we had better telescope, this class could change. An analogy might be our naming certain wavelengths of light "visible". They only depend on our perceptions, not some objective characteristic that all observers in the universe could agree on.


who use the GADGET-2 pack for simulation? can I do it with my PC?

Gadget-2 is publicly available. Google works well to find it.
According to their website, it runs also on PCs. But don't get your hopes up, a single PC doesn't have enough computing power to simulate a whole universe. You'll need a bigger machine for that.

They have installation instructions on the webpage (see "Users Guide" under "Documentation"). You will possibly need a linux machine to run it, I don't know if the code compiles well under Windows.


Could dark matter form coherent bodies the size of stars or planets?

No, it's not really possible, because if it could it wouldn't be dark.
To see why this is, think about some some collection of gravitating particles which is going to to collapse to a smaller, denser collection of particles.  To do this they need to lose a lot of energy: you can see why this is by thinking about where the average particle lives in the gravitational potential of all the particles.
Ordinary matter – clouds of gas and dust say – do this by getting extremely hot as they collapse, and then radiating energy electromagnetically.  Ultimately collapsing clouds of gas can get hot and dense enough that nuclear fusion can start and we get stars.
Dark matter, by hypothesis, doesn't interact electromagnetically, at all: if it did it would not be dark!  This means that the only way it has of losing energy is to put it into the average kinetic energy of the particles, and that means that collapse doesn't happen, because if you have a lot of particles with lots of kinetic energy they don't stay collapsed.  The virial theorem is important to understanding this.
So dark matter simply lacks a mechanism to lose energy since it can't radiate it away electromagnetically, and so does not collapse.
(In fact I think there are some mechanisms: dark matter could lose energy by gravitational radiation for instance, but this is an absurdly slow process except in really extreme cases.)
To answer the second part: it certainly was (perhaps is) a possibility that there are very large numbers of compact, massive dark objects which make up dark matter.  These things are known as massive compact halo objects: MACHOs (the alternative theory is weakly interacting massive particles which are, of course, WIMPs, and there are I believe also RAMBOs although I'm not sure what they are.  If there are not ARNIEs there should be (perhaps one kind of these will be CONANs?)).  These might either be failed stars, isolated planets or (I like this idea) small black holes: there's a wide range of possible BH masses below stellar mass BHs and the point at which they start becoming bright due to Hawking radiation.  In the case of small BHs these would probably need to be primordial (made in the big bang) as there's no plausible mechanism for them being the end-point of stellar evolution.
But it turns out that such objects are observable.  Because they are massive they will deflect light, and they can therefore be detected by gravitational lensing as they pass in front of more distant luminous objects (stars, really).  Well, you can do the statistics of how many of these things there would need to be (which is really a lot), and how often you would expect to observe lensing events from them as a result.  And then you can do a survey and look for lensing events.  And the results of such surveys (some of these are pointed to by the above Wikipedia article) are that there are not nearly enough MACHOs to account for dark matter, although I think there is still a little wiggle-room, depending on who you ask.


Unlike "planet" the IAU hasn't attempted to precisely define "moon". General usage requires that a "moon" is a natural satellite of a planet (or dwarf planet, asteroid, or perhaps even of another moon?) and it is big enough for us to have seen it as an independent body.

This contrasts with the "moonlets" that have been detected in Saturn's rings by their graviational effects on the even smaller ring particles. We know that a body exists there, but it is too small to be directly imaged.

As such the question "How many moons does Jupiter have" is probably unanswerable, since it depends on the sensitivity of your telescope and how close you are to the planet.

The smallest moons are about 10km across, but moonlets in the rings exist that are only a few hundred metres across. So in current usage, 10k might be roughly where people stop saying "moon" and start saying "moonlet".

Language exists to serve us, the definition of planet was only made to solve a particular naming problem.  No such problem has arisen for moons. So there is no authoritative answer.


Other than relative temperatures, is there any difference between a red sun and a yellow sun?

No.

Small red dwarfs may be fully convective. Larger stars like the sun have a region in which the energy is transferred by radiation. So in small red dwarves, the star becomes fully mixed.

Red Giants are quite different and are large and old stars, swollen by their powerful cores.

Both red and yellow stars produce large amounts of neutrinos. Yellow stars produce more than red dwarfs because they are more powerful. Red Giants would produce more than yellow dwarfs, as they more powerful.

Nothing that could explain superman.


I think you mean that kappa = sigma / m.   The sigma is the "cross section," like the area you expose when you slice a watermelon in half.  The m is the mass of the particle that has the cross section in question (it's not literally the cross section of the physical particle, but an effective cross section for interaction with light), or more often, the mass of all the particles that come along with the cross section in question (often there is one particle contributing most of the cross section, but several particles that come with it which provide the mass).  For example, if you mix protons and electrons, the electrons have a larger cross section because their low mass makes them interact more easily with light, but that lower mass also means it is not the electrons, but the protons, that provide the "m" in the above formula.

So these quantities depend on the composition of the gas you have in mind.  It's easy for an electron-proton [edited] plasma, because if you don't worry about resonant interactions in the plasma, you can just consider the cross sections of the individual particles, and you get "Thomson opacity" of 0.4 square centimeters per gram (that's sigma/m for the sigma of an electron and the m of a proton, rounded to one significant figure because nothing is ever really a pure electron-proton plasma).  Purely ionized helium is also easy, because if you again ignore any resonant interactions between the electrons and the nuclei, it's just the same Thomson opacity except reduced by the extra neutrons [edited] in the heavy helium nuclei.  Since there is one neutron per proton, that reduces 0.4 by factor of 2, yielding 0.2.  [edited for forgetting that the 0.4 already includes the protons.]


Are any of the zodiac stars about to go supernova anytime soon?

I don't really know what you mean by "Zodiac stars". I suppose you mean the stars of the 12 traditional Zodiac constellations.
Alpha Scorpii, known as Antares, or Jyeshtha in Hindu astrology, is a red supergiant. It is in a similar stage of its lifecycle to Betelgeuse. It is expected to explode in a supernova in the next 10000-100000 years (the timing is not very well understood)
Other potential supernova progenitors are more distant and fainter, such as HD 168625 in Saggitarius, which is not visible with the naked eye. There is also the recurrant nova U Scorpii, a candidate for a type Ia supernova (which occurs when a white dwarf accretes a critical amount of matter) Again it is far to faint, even during an outburst, to be a naked eye star (but it would certainly be visible if it exploded in a supernova).


How was the distance from the Earth to the Sun determined

There are a bunch of ways.  The simplest is triangulation.

Consider a transit of Mercury or Venus, when the planet gets between the Earth and the Sun. If you observe the planet's path across the face Sun from two different places, measuring the chords across the Sun's disk and the timings, using no more than trigonometry, you can get an accurate distance to the Sun.

Here's a good discussion of the method on the Physics SE

(It's worth noting that it is easy to measure the distance to, say, Mars in terms of the ratio of the distance to Mars compared with the Earth-Sun distance, than it is to measure the Earth-Sun distance in miles.  Hence the Astronomical Unit.  We accurately knew many astronomical distances in AU before we knew the length of the AU equally accurately.)


How did they decide the length of one year even though it wasn't known that earth revolves around the sun?

Originally a year would be noticed as the time between two winters (or two harvests, two annual floods etc), although the exact number of days might not be known with any accuracy.

As people start to observe the sun more carefully, it would be noticed that the position of sunrise and sunset move regularly. Stonehenge and other ancient monuments seem aligned to midwinter sunset or midsummer sunrise. Since this would allow one to predict the seasons, it would have been very useful.

More careful observations led to the geocentric model, in which the sun takes just over 365 days to move around the Earth once relative to the background stars. The length of the year was known to be about 365.25 days since the times of the ancient Egyptians.


There are a number of flaws with your idea (despite the relativity breaking transatlantic journey times that would be possible).

Inflation is a popular theoretical model put forward to solve three main issues with the Big Bang cosmology: the horizon problem, the flatness problem & the magnetic monopole problem (see the wiki for details). It consists of a scalar field called the inflaton, a physical field that pervades all space. During inflation it undergoes a phase transition to a lower energy state, releasing huge amounts of energy that drive the expansion. To force the field back to a higher energy state would require inconceivable amounts of energy, and have to be applied across the entire universe. Not only is this impossible, but also annoyingly prevents its use for local galactic travel.

  Or maybe a technically very advanced civilization could keep its solar system inflated by the time of a deflation of the whole universe

I assume you mean contraction, as opposed to the current expansion that we see today due to dark energy. Again this would be unfeasible due to the reasons outlined above, and if contraction was to proceed to it's conclusion, i.e. a singularity like the Big Bang, then no information from this universe could be carried through that singularity, even if there was another universe on the other side.


How to measure distances to stars by means of spectroscopic parallaxes?

Actually the method described on Wikipedia is not the method that is meant by Spectroscopic Parallax. To determine the spectroscopic parallax, you'll need a spectrum of the star and measure the widths of the spectral lines. Compact stars have higher surface gravity, which means that their spectral lines are broader. This means that the radius of the star can be calculated from line widening. Using the observed effective temperature $T_{eff}$ and the radius R, you can get the Luminosity from Stefan's Law, which gives an absolute magnitude and together with the apparent magnitude, a distance modulus and therefore a distance.
This gives an accuracy of 10% in luminosity and 5% in distance for main sequence stars and 25-30% in distance for giant stars. (see de Grijs, 2011)

Wilson and Bappu (1957) proposed to use the width (W_0) of the emission core of the CaIII absorption line at 3933 Angstrom. Using data from Hipparcos the relation between the absolute magnitude in V and the width of the absorption line was calibrated to:

$M_V = 33.2-18.0 \log(W_0)$

by Pace er al. (2003).
So to use this method you would need to be able to measure the width of the CaIII absorption line, but other calibrations using the MgIIk line also exist.

As these distance measurements are calibrated using Hipparcos, they will never be able to be more accurate until better calibration data is available (i.e. with Gaia).

The much more simpler method as described on Wikipedia might also be used but that is even less accurate. NB. The relation between the colour (colour index B-V) and the spectral class is not linear. Many Hertzsprung-Russel diagrams show both colour and spectral type on the X axis but this is not really correct. A K0Ia supergiant has a different colour index than a K0V star on the main sequence.
On page 50 in de Grijs, 2011, this relation is shown in Fig 2.11. It should also be somewhere on Wikimedia (to which it is attributed) but I cannot find it.
EDIT: Here it is: http://en.wikipedia.org/wiki/File:H-R_diagram.svg


Where can I learn these Astrophysical techniques? [Read below]

If you are going to do data analysis, you need to understand how the fitting procedure works. This means a lot of statistics, and new terminology and so on, which is hard and takes much time.
If you want to start to fit, e.g., a spectrum, you should definitely read the XSPEC guide. You can find a pdf online as well.
In short, fitting is to take a model and measure how well this model fits to your data. To quantify this "how well", usually the chi-squared distribution is considered:
$\chi^2 =  \sum\limits_{i=1}^n(\frac{X_i - \mu_i}{\sigma_i})^2$
where $X$ is your data (observed value), $\mu$ is your expected value (which, in this case, corresponds with the prediction of the model), and $\sigma$ is the variance on your data point (the error).
This is the most general information that you need.
You can read more here, and especially on the Numerical Recipes (don't be scared by the huge format of the last document, the pages you need are only few around Chapter 15).
For fun, you can just try to play with XSPEC, to take confidence, and see at least how things change when you change your model or your data (especially if you know which model is the best-fit for your data). To understand everything will take years, but if you never start, you'll never arrive ;)


Does General Relativity really predict Black Holes?

Well, yes, but we must be careful with the meaning of "predict".
The Schwarzschild solution, developed by Karl Schwarzschild in 1916 [1], is the first closed-form, explicit solution of Einstein's field equations for gravitation. It describes a spherically symmetric, static, vacuum spacetime. The solution goes singular at a specific radius (the Schwarzschild radius). In the weak field limit, it correctly replicates the Newtonian gravitational field of a compact object.
Though Birkhoff's theorem (general relativity's version of the shell theorem) was not yet known in 1916, the Schwarzschild solution was nonetheless recognized as the general relativistic description of the gravitational field outside a compact gravitating body. The fact that it went singular at the Schwarzschild radius was either ignored or taken to imply that objects that are as small as, or smaller than, this radius cannot exist.
In any case, just because a solution exists in general relativity does not mean that objects described by that solution exist in Nature. (E.g., the field equations admit solutions that blatantly violate causality; yet I don't see any time machines out there.) For all they knew, the Schwarzschild solution was nothing more than a mathematical curiosity, an idealized case that does not describe reality.
For this reason, I'd suggest that a much more groundbreaking paper is that of Oppenheimer and Snyder from 1939 [2]. This paper demonstrates that a spherical cloud of dust initially at rest will undergo "continued gravitational contraction". An observer that is falling with the collapsing matter would see total collapse in a finite amount of time, but to a distant observer, the collapse will continue forever, the object asymptotically approaching, but never quite reaching, its Schwarzschild radius. And it was in 1957 I believe that Regge and Wheeler first demonstrated that a Schwarzschild singularity is stable under small perturbations [3], i.e., perfect symmetry is not required. (Wheeler, of course, was also the first to popularize the name, "black hole".)
Lastly, as I was reminded in a comment, we should not forget Penrose's 1965 singularity theorem [4], which introduces the concept of a trapped surface and shows that gravitational collapse is indeed a very generic feature of general relativity. After all, this is the result that earned Penrose the 2020 Nobel prize in physics.
In light of that, I think we can confidently state that general relativity predicts black holes, but only because we know not only that black hole solutions exist, but also that physically realizable configurations of matter can collapse into black holes (or, at the very least, to objects observationally indistinguishable from black holes, which of course may evaporate in stupendously long but finite timeframes due to Hawking radiation [5]) and that the solutions are stable under small perturbations, i.e., perfect symmetry is not a prerequisite.

[1] https://arxiv.org/abs/physics/9905030
[2] https://journals.aps.org/pr/abstract/10.1103/PhysRev.56.455
[3] https://journals.aps.org/pr/abstract/10.1103/PhysRev.108.1063
[4] https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.14.57
[5] https://www.nature.com/articles/248030a0


Why is there a matter/anti-matter asymmetry in the universe?

Currently, there's no single explanation for this.

The simplest explanation is that there isn't any imbalance — there are large antimatter bodies in the universe; just separated from us. However, these would have to be pretty far away (otherwise the boundary annihilation would be detectable). Theoretically, this requires antimatter to have clumped early on, which is rather unlikely/impossible.

A far more likely explanation is that charge-parity symmetry is violated. In other words, flipping the charges and parities in a reaction may not lead to the same dynamics — the outcome (when the charge/parity is flipped back) or distributions of outcomes may not be the same. Usually, we expect that if we flip the charges on both sides of a reaction, the reaction proceeds in the same manner (with same probabilities, etc). There is evidence that suggests that this may not be true. If this is indeed the case, then the matter-antimatter imbalance is not that hard to explain.


Where do asteroids impact upon the Earth most often?

The occur all over the world.  On this image each yellow circle is an recent impact of a small asteroid the size of the circle is related to the energy of the fireball (the sizes of the circles don't represent crater areas). You will note that there is no region that get more impacts, and similarly, nowhere is safe.

The very large yellow circle is the Chelyabinsk meteor in Russia. It was the biggest such impact over land in recent times.


visualisation of the universe's expansion

You were correct, you teacher was incorrect.

It is the space that expands - much as a surface of a balloon does.

An explosion is a poor analogy in contrast because - as you suggest - it implies something to expand into.

For another way, not as accurate as the balloon analogy but maybe helpful - imagine being trapped inside an expanding loaf of bread in an oven. You have no knowledge of the world outside the loaf - and you may assume the loaf is infinite in extent. But as the dough rises the gaps inside the loaf get larger without the mass of the loaf itself increasing. You could compare our position in the universe to being inside such an infinite loaf.


If getting shined on by a gamma ray is an extinction level event how do we detect them?

The danger from gamma-ray bursts (GRBs) is the potential to reduce atmospheric ozone drastically, allowing harmful ultraviolet radiation to pass through the atmosphere and make life on Earth much more difficult. However, you have to be fairly close for this to happen. Gehrels et al. (2003) calculated that a typical GRB from a supernova would have to occur less than 8 parsecs from Earth to cause serious damage to the ozone layer - or, as they put it, to cause levels to drop low enough that they would

  double the “biologically active” UV ﬂux received at the surface

Obviously, though, we can detect gamma-rays from distances farther than 8 parsecs! The closest GRB detected so far, GRB 980425, is likely the result of SN 1998bw, and thus originated 140 million light-years away - over 42 million parsecs away. The radiation from it was nowhere near high enough to harm life on Earth.

Your second explanation, therefore, is the correct one. The intensity of the signal decreases according to the inverse-square law - that is, $I(r)\propto r^{-2}$ - so at large distances, the flux is small enough that there is negligible impact on Earth's atmosphere.

Phiteros also raised a good point: To cause severe problems from Earth, the gamma-ray emission from a GRB - which is often along an axis from both poles of the progenitor - would have to be pointed nearly directly at Earth. For example, there was once a scare that a supernova by either of two stars in the WR 104 system could cause problems for Earth, but the odds that the beam would hit Earth full-on are slim.


The galactic coordinate system is usually used for galaxies. It is based on the galactic plane — approximately where the galactic disc is.

It uses galactic latitude (for the y-value of the objects) and galactic longitude (x and z values, based on the disc).

There is also a supergalactic coordinate system used for superclusters.

Its equator is an invisible line called the supergalactic plane, which is the preferred distribution of galaxies in a supercluster.


Why the light curve goes down when the planet is behind the star?

Think of when the planet is at the "side" - there's a little bit of light from the planet (ie, reflecting off the planet) shining towards us.

Could it be due to that little bit of light - when the planet is behind the star, it no longer reflects towards us?

Maybe that's the effect you have in mind?


How does 2015 RR245 compare to the other KBOs predicting "Planet 9"?

It was the first thing I thought too, but I would say it won't count in the statistics.  The earlier paper by Batygin & Brown was talking about bodies orbiting further out.

2015 RR245 orbits between 33.7 and 129 AU.  The "cluster" of bodies considered in the paper have aphelion distances from 150 AU upwards.  So I think it would be a mistake to try and fit this new body into the existing figures.


Difference in gravity on both sides of the moon

Very, very little

It's important to keep in mind that the Moon is constantly accelerating toward the Earth. The component of gravitation that one feels on the surface of the Moon is the difference between the gravitational acceleration toward the Earth at the point of interest and the Moon's gravitational acceleration toward thbe Earth as a whole.

At the sub-Earth point (the point on the Moon closest to the Earth), the gravitational acceleration toward the Earth is slight greater than is the gravitational acceleration of the Moon as a whole toward the Earth. The Earth very slightly reduces the gravitational acceleration toward the center of the Earth at the Moon's sub-Earth point.

At the antipode to the sub-Earth point (the point on the Moon furthest from the Earth; I'll call this the anti-sub-Earth point), the gravitational acceleration toward the Earth is slight less than is the gravitational acceleration of the Moon as a whole toward the Earth. The Earth once again very slightly reduces the gravitational acceleration toward the center of the Earth at the Moon's anti-sub-Earth point. The effect is nearly (but not exactly) identical to that at the sub-Earth point.

There are places on the Moon where the gravitational acceleration toward the Earth does slight increase the observed gravitational acceleration, which is the set of points where the Earth is more or less on the horizon. This effect here is about half that (in magnitude) of the effects at the sub-Earth and anti-sub-Earth points.

A qualitative discussion

Ignoring that the Moon is rather lumpy in a gravitational sense, the gravitational acceleration on the surface of the Moon due to the Moon itself is $$\vec a_M = -\frac{\mu_M}{{r_M}^2}\hat r_M$$
where
$\mu_M=G M_M$ is the Moon's gravitational parameter,
$r_M$ is the Moon's radius,
$\hat r_M$ is the vector from the center of the Moon to some point on the surface of the Moon, and
$\boldsymbol a_M$ is the gravitational acceleration due to the Moon at that point.

At that point, the gravitational acceleration toward the Earth is $$\vec a_E = \frac{\mu_E}{||R_E \hat r_E - r_M \hat r_M||^3}(R_E \hat r_E - r_M \hat r_M)$$
where
$\mu_E=G M_E$ is the Earth's gravitational parameter,
$R_E$ is the distance between the centers of mass of the Moon and the Earth,
$\hat r_E$ is the vector from the center of the Moon to the center of the Earth, and
$\boldsymbol a_E$ is the Newtonian gravitational acceleration toward the Earth at that point.

The Moon itself accelerates Earthward due to Earth gravity by $$\vec a_{M,E} = \frac{\mu_E}{||R_E \hat r_E||^2}\hat r_E$$
The specific weight (force measured by a spring scale divided by mass) at the point of interest is $$\begin{align}
\vec g &= \vec a_M + \vec a_E - \vec a_{M,E} \\
&= -\frac {\mu_M}{{r_M}^2}\left(\hat r_M-\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^2\left(\frac{\hat r_E - \frac{r_M}{R_E}\hat r_m}{||\hat r_E - \frac{r_M}{R_E}\hat r_m||^3}-\hat r_E\right)\right)
\end{align}\tag{1}$$
In the case where $\hat r_E = \hat r_M$ (the point on the Moon closest to the Earth) equation (1) simplifies to $$\begin{align}
\vec g
&= -\frac {\mu_M}{{r_M}^2}\left(1-\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^2\left(\frac{1}{||1 - \frac{r_M}{R_E}||^2}-1\right)\right)\hat r_m \\
&\approx -\frac {\mu_M}{{r_M}^2}\left(1-2\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^3\left(1+\frac32\frac{r_m}{R_E}+\cdots\right)\right)\hat r_m
\end{align}$$
In the case where $\hat r_E = -\hat r_M$ (the point on the Moon furthest from the Earth) equation (1) simplifies to $$\begin{align}
\vec g
&= -\frac {\mu_M}{{r_M}^2}\left(1-\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^2\left(1-\frac{1}{||1 + \frac{r_M}{R_E}||^2}\right)\right)\hat r_m \\
&\approx -\frac {\mu_M}{{r_M}^2}\left(1-2\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^3\left(1-\frac32\frac{r_m}{R_E}+\cdots\right)\right)\hat r_m
\end{align}$$
The dominant perturbing factor, $2\frac{M_E}{M_M}\left(\frac{r_M}{R_E}\right)^3$ is the same for both points and is rather small, about $1.5\times10^{-5}$. People at these two points would weigh a tiny bit smaller than they would if the Earth was not present. The difference between the two points is tinier still, a factor of about $2\times10^{-7}$.


Can there be elliptical rings around a planet?

Short answer: no.

Long answer:

There are many collisions within ring systems, and collisions always work, over time, to push orbits to a circular shape (or destroy the rings). Any deviations tend to be quickly corrected.

For the same reason, rings tend to be extremely flat and extremely thin. E.g. Saturn's rings are only dozens of meters thick; given their diameter of 300 thousand kilometers, if you cut a model of the rings from ordinary size printer paper, the paper model would be too thick. (would not represent the thickness of the rings to the correct scale)

Another characteristic of rings is that they are equatorial. If they are not in plane with the equator, precession will drive them towards that configuration.

You could apply a perturbation that makes a ring non-circular, or non-flat, or non-equatorial, but then the system will quickly evolve towards circular, flat, and equatorial (or will be destroyed).

https://www.uwgb.edu/dutchs/AstronPowerPoint/Rings.ppt


In the absence of any detailed information, such as @Uhoh asks for, this is just a guess, but here goes. I'm assuming you live at mid northern latitude.

The moon rises between south east and north east.  If we consider full moons, then during  the summer, the full moon will rise in the south east.  As we move into the autumn and winter, the full moon progressively rises further north.  If you are looking out of an east facing window, then through mid May to mid July the moon will appear to rise to the right side of your window, but will gradually move northwards, ie to the left of your window.

The reason behind this is that at full moon, the moon is diametrically opposite the sun.  At mid summer, the sun rises in the north east, and in mid winter it rises in the south east.


Almost all the time, the best eyepiece to use would be the 20 mm lens for seeing the moon. If you want a better view, then you can go for the 10mm. That's what I would do...


Gravitational lensing in Newtonian physics

A photon is an entity defined in the context of a relativistic field theory, and so it doesn't really make sense to talk about the Newtonian bending of a photon. Necessarily, we need to substitute an analogous question that's sensible in the Newtonian framework. To do so, we can imagine a classical corpuscle of light--appropriately enough, a theory of light advanced by Newton himself.

There are many problems with the Newtonian conception of light, but that's an issue for electromagnetism rather than gravity.

Critically, for a test particle, the trajectory depends on the initial velocity only, and not the mass. So we don't even have to address the mass of the light corpuscle at all to talk about its trajectory, as the acceleration is the gradient of the gravitational potential and if considering force (or gravitational potential energy), the mass cancels out anyway. If you wish to bring in the mass explicitly, we can still think of the trajectory of a massless particle as a limit of trajectories of particles with masses tending to zero but having the same velocity — a trivial limit because they're all the same trajectory in the Newtonian theory. On the other hand, if we recognize that light carries momentum, a Newtonian light corpuscle shouldn't have zero mass, so the question of what to do with a genuinely massless particle evaporates.

When people talk about the Newtonian deflection of light, they are typically considering a hyperbolic trajectory of test particle at the speed of light under Newtonian gravity. If the angle between the asymptotes is $\theta$, then $\theta = \pi$ represents a completely straight trajectory unaffected by gravity, and the eccentricity is
$$e = 1 + \frac{v_\infty^2R_p}{GM} = \frac{1}{\cos\left(\frac{\theta}{2}\right)} = \frac{1}{\sin\psi}\text{,}$$
where $R_p$ is the periapsis distance and $2\psi = \pi-\theta$ is the measure of deflection. On the scale of the solar system, it doesn't matter whether we set $v_\infty = c$ or anywhere else along the trajectory. For example, if the velocity at peripasis is $c$ instead, then $e\mapsto e-2$, and so is negligible for light deflection due to the Sun, $e > 10^{5}$. The total deflection is approximately
$$2\psi \approx e^{-1} \approx \frac{2GM}{c^2R_p}\text{,}$$
which is half the correct general-relativistic prediction.

Note that here we did not assume that the speed is constant along the hyperbolic orbit. That would not be consistent with Newtonian gravity. Rather, what we have is a situation where if $v = c$ anywhere along the trajectory, then the speed along any other point is so close to $c$ that it doesn't practically matter for considering light deflection.


Does precession affect this measurement?

No special name.
The problem that I see here is that the distance will vary over the year.  At the moment of solstice, you can easily find the distance by basic trigonometry:

If the center of the Earth is C, then $AC/AB = \tan(23.5^\circ)$ so $AB = 6370/\tan(23.5^\circ)=13000\, \mathrm{km}$

At any other time the distance will be different, and at the equinox the axis of the earth is parallel to the tangent plane of the Earth, so the distance is infinite. So the distance AB is not a fixed value, and varies over the year.
In the longer term the precession of the equinoxes will affect when solstice occurs, but not the angle 23.5.  This angle also varies in the long term between 22.1° and 24.5°, over a cycle of about 41,000 years.


Why 225088 Gonggong is not a dwarf Planet yet?

As the comments already say, an object being a dwarf planet is a matter of convention. If the IAU says it's a dwarf planet, it's a dwarf planet. Otherwise, it's not. The requirements you are listing from Wikipedia are the IAU criteria for pronouncing objects as dwarf planets, but that does not mean that all objects fulfilling these criteria are dwarf planets. (don't ask me for the reasonings of the IAU, i'm not an expert on astronomer's brain twists)

There are a lot more examples of objects that could be dwarf planets, and are considered as being that by many researchers, despite these objects not having the status of dwarf planet. Wikipedia's List of possible dwarf planets has about 200 objects facing the same issue as 2007 OR10.

edit: as someone in the comments mentioned, not all objects on this list are actual candidates for dwarf planets (which is not what I meant, but I might indeed have implied this)


Is our supercluster part of a galaxy filament?

The Laniakea Supercluster contains several galaxy filaments, even just reading the wikipedia articles you provided you can see the difference in size of the two:

Galaxy Filament

  They are massive, thread-like formations, with a typical length of 50 to 80 megaparsecs, (163 to 261 million light years) that form the boundaries between large voids in the universe.

Laniakea Supercluster

  The Laniakea Supercluster encompasses 100,000 galaxies stretched out over 160 megaparsecs (520 million light-years).

The reason why Laniakea isn't the largest "structure" in the universe is because it inst a single object, rather an arbitrarily assigned group of galaxies. From this image below you can see how the supercluster is given an imaginary edge based on the peculiar velocities of the galaxies within it, there is no observable edge to the supercluster, unlike galaxy filaments that have visible edges.


Is there any hard evidence that rogue planets exist?

Rogue planets have been discovered by infra-red imaging because planets are hot when they form.  Here's a list of a few.

There's 2 types of Rogue planets.  One is failed stars.  condensing pockets of gas and dust that form similar to how our solar-system formed, but that are too small to form stars.   Source.

The 2nd type is planets that escape from a star's orbit.   This can happen by the star ejecting material and losing mass which causes the planets to expand in more distant orbits, some eventually escaping, or by gravitational assist either planet on planet or by two stars passing quite close to each other.    It's statistically impossible for Rogue Planets not to, from time to time, get ejected from solar systems, so they have to exist at least in reasonably large numbers throughout the galaxy, though I don't think it's well known how common there are.

Our solar-system might have ejected a planet - see here.   In general, larger planets can eject smaller ones but mostly not the other way around, but two stars that pass too close to each other can eject any planets, mostly ones with more distant orbits.  A star that loses a lot of it's mass or a star that goes nova can push planets out of it's orbit too.


How good is the HST compared to ground telescopes?

Adaptive optics techniques have only been successfully applied in the far red and near infrared. HST produces images with spatial resolution equivalent to ground based adaptive optics images, but it works at near infrared wavelengths and visible wavelengths.

The spatial resolution of HST at visible wavelengths has not really been matched with ground based telescopes. "Lucky imaging" has shown some promising results, but even here, this is usually applied at far red wavelengths.

Except now see this How did VLT's adaptive optics obtain this resolution for Neptune? Is it really working in visible wavelengths? Progress occurs! (Though it now seems my statement above should only be slightly modified, since the picture contains no information shortward of 550nm and is largely dominated by light at 600-920 nm).


What are reliable resources for weather and all other atmospheric obstructions?

You might try https://clearoutside.com/forecast/50.7/-3.52  It gives weather and visibility among other things.

They have an app for iPhone and android. Advertises site was developed by astronomers.


Size and mass of comet ATLAS (C/2019 Y4)

The confusion comes from the difference between the nucleus and the coma.

The nucleus is a small icy body, only a few km across.

The coma is the cloud of gas and dust released from the nucleus as it warms up.  With not much gravity, the coma spreads out into space, and it can be hard to say exactly where the edge of the coma lies, however, a coma "the size of Jupiter" would not be unusual. Active comets can have comas that are larger than the sun.

But the mass of the coma is tiny. It is made of a very low-pressure gas. It is nearly all empty space, most of the mass of the comet remains locked in the nucleus.


Planetary system orientation & the search for exoplanets

The multi-planet system HR 8799 is pretty close to face-on. Here's a video made by interpolating images taken over 7 years showing the orbital motion.
The planets of HR 8799 were detected by imaging. The majority of exoplanets have (at the time of writing) been detected by the radial velocity method, which requires radial motion and thus is most sensitive to planets close to edge-on, and the transit method which requires that the planet passes in front of the star as seen from Earth (thus will only spot planets extremely close to edge-on). The imaging method does not have this issue, so can find both face-on planets (like HR 8799) and edge-on ones, like Beta Pictoris b (video).
Astrometry is another exoplanet detection method that would allow detection of face-on systems. It has been plagued by a number of false starts (e.g. the infamous "detection" of planets around Barnard's Star by Peter van de Kamp) but hopefully once the full Gaia dataset is released there will be numerous astrometric exoplanet detections, which may include near-face-on systems.


How do we know that black holes are spinning?

The gravitational field of spinning matter, or a spinning black hole, causes matter around it to start spinning. This is called "frame dragging" or "gravitomagnetism", the latter name coming from the fact that it's closely analogous to the magnetic effect of moving electric charges. The existence of gravitomagnetism is tied to the finite speed of gravity, so it doesn't exist in Newtonian gravity where that speed is infinite, but it's present in general relativity, and for black holes it's large enough to be detectable.

Also, for purely theoretical reasons we expect that all black holes are spinning because a non-spinning black hole is the same as a spinning black hole with an angular velocity of exactly zero, and there's no reason why a black hole's angular velocity would be exactly zero. On the contrary, because they are so much smaller than the matter that collapses to produce them, even a small, random net angular momentum of the collapsing matter should lead to a rapidly spinning black hole. (The classic analogy for this is an ice skater spinning faster when they pull their arms in.)


Is it dangerous to look at an L-type brown dwarf from too close?

A typical L-type brown dwarf is about 1200-2200 K in surface temperature and is about the size of Jupiter. Using the Stefan-Boltzmann law, we can deduce that the hottest brown dwarfs have a luminosity of$$\Big(\dfrac{2200}{5778}\Big)^4 \cdot \Big(\dfrac{1R_J}{1R_\odot}\Big)^2 = 0.00021224 L_\odot$$
According to Wikipedia, the dimmest apparent magnitude that causes pain to look at is $-25$. We can use the luminosity formula for apparent magnitude and distance calculate how far away an observer must be to experience this magnitude: $$0.00021224 = 0.0813   \text{ ly}^{-2}d^2\cdot10^{-0.4\cdot(-25)}$$
Solving the equation gives $d=0.03232 \text{ AU}$, or $67.62 $ times the brown dwarf's radius. Using this useful HZ calculator and plugging in $T_{eff}=2200$ and $L=0.00021224$, we get the distance must be between $0.015$ and $0.032$ AU, which is inside that limit. This range is inside the limit for $-25$ apparent magnitude, so no, looking at brown dwarfs at their habitable zone is not guaranteed to be safe.


Why doesn't sunrise shift as we revolve around the sun (ignoring seasonal variations - so at the equator)

This shift is already included in a "day".
24 hours is already the time from sunrise to sunrise, meaning the Earth has performed a full rotation, plus a little bit more since it has moved along a degree in its orbit. That's solar time, the common usage of a "day".
The time the Earth needs to rotate exactly once,so that a star is in the same position in the sky, is a sideral day. It's slightly shorter at 23 hours 56 minutes and 4 seconds.
You can convert between them based on the fact that there's one less solar day than sideral days in a year:
$$d_{sideral} = \frac{year}{\frac{year}{d_{solar}} + 1}$$
Note that the difference between a solar and a sideral day is far from the only subtility arising from the Earth's orbit.


Is there a threshold on distance/size for a tidal locking?

You can see an example of tidal locking and atmosphere simulation for a planet closely orbiting a dim star. They show a simulation of the atmosphere and some interesting theories about the movement of gasses due to tidal locking (convection) that occurs between the bright and dark side of the planet.  The link goes directly to the discussion of tidal locking:

KEPLER 186F - LIFE AFTER EARTH

A similar question was posted in the physics forum and it looks like that could be close to the answer you are looking for:

Tidal Lock Radius in Habitable Zones

I am also interested in tidal locking and hope someone may post a more concise answer.


What would happen if a person was out in space? would they get picked up by the pull of gravity from inbetween our planets, or sun?

There are no currents.

It all depends on the initial speed of the object, and the distance to the sun and planets.

If it's moving pretty slowly initially, it will eventually get captured in a loop orbit around some planet or maybe around the sun.

If it's moving pretty fast initially, it will described a trajectory through the system and eventually leave, never to return.

This is no different from what happens to asteroids, comets, the planets themselves, etc. This is all the play of gravity. This is called orbital mechanics. It's part of what is popularly known as "rocket science".

There is an app called Universe Sandbox. Install it on your computer and play with some solar system simulations. Setup a simulation with a central star, and just one body near it. Launch the body with various speeds. See what happens. The various scenarios unfolding from there are the answer to your question.


First Shooting Star Seen from Mars

  space.com  2005

  The background image shows the meteor near the top-left and the horizon at the bottom. A red arrow shows the direction of travel. The inset is a larger version of the meteor itself. The graph is a "light curve" that aided in tying the meteor to comet Wiseman–Skiff.
  (Image: © Nature/NASA/Spirit/F. Selsis et al.)

This was made by Spirit's panoramic camera, so it probably would be visible to the naked (helmeted) eye.


How to invert fits headers order (primary/secondary)?

EDIT:  I moved the last paragraph to the beginning because it's really a Python question, but I left my "manual" method answer in case it helps, too.
Forgot to add this link to Python code on Github.  It helped me get started with displaying images using Python and dealing with the headers without altering them.  On line 19, I changed it to "image_data = image_file[1].data" because the image was in the HDU [1] instead of HDU [0].
I've had this exact problem when dealing with TESS images - I can't open them in the "usual" software I use for all of my other fits images.  Inverting the headers won't solve it, either, because one of the keywords in the first header is needed.  If you are only trying to make the change for one image, it's easy to do manually.  You can edit the headers in Notepad or any other text editor.  Here's the general workaround I use if it's only one file (or a couple of files) - if you need to open a lot then this is too tedious.

Make a copy of the file and change the extension of the copied version to ".txt"
Open in a text editor like Notepad (I like Notepad++ better, but any text editor will work)
Turn on "word wrap" so you can see the header info, there aren't carriage returns or line breaks.  Each keyword line is exactly 80 characters long regardless of what info it holds.
Adjust the width of your text editor to get close to 80 characters wide and you'll see the headers "line up" pretty well but not perfectly.  There is almost always a situation where the entire 80 characters are used so there isn't a space between the last character of a keyword line and the first character of the next one which will throw off the word wrapping.  But that's ok, just frustrating.
Look through the header info until you see the beginning of the first extension (probably named "SCI").  It's fine to select and delete the keyword lines you don't need, the ones breaking it up into two HDUs, as long as you don't need the info.  Some of them will be entries such as "MISSION= Hubble" or something like that along with info about the pipeline that created it, etc.
When I've located NAXIS, I put the cursor in front of it and select everything up to the end of the first keyword line in the first HDU which should be "SIMPLE= 1" (or true). The first keyword line with SIMPLE= 1 is 80 characters long and MUST remain 80 characters long or it will break it, so I delte everything between NAXIS and the end of the SIMPLE line.  Look to see what the next line is after SIMPLE and delete to the first letter of that keyword. The SIMPLE keyword line is required and should be the first one. Make sure you don't add a carriage return or it'll break the file.
Save your changes and edit the extension to ".fit" or ".fits" and it should open fine.

As long as keep each chunk of keyword info 80 characters in length, you can manually force it to have only one header so you can view the image.  This is, of course, not practical for a large number of images and you could lose data that IS important to you such as the date, etc.
To view fits files without having to edit the header, check out SAOimageDS9.  It has never failed me in opening an image fits file with multiple HDUs.  If you are trying to work with large sets of images there are other workarounds depending on what you're trying to do.  Good luck, I hope this works for you!


What do gravitational waves allow us to understand?

I think the wikipedia page gives a reasonable overview of why gravitational waves give us a window on the universe that is either not observable or is complementary to the view we get from electromagnetic waves (or neutrinos).

But let's have a specific example. The potential of LIGO, as we have seen today, is to detect the gravitational wave signatures of merging black hole binaries.

The power radiated in gravitational waves strongly depends on the orbital separation of the black holes, it also depends on their masses. The frequency of the radiation will change as the black holes spiral together. The "noise" that this makes in the final few seconds as the black holes merge is the "chirp" that is detected by LIGO. Have a look (listen!) to what a chirp might "sound" like at this web page from University of Birmingham.

The exact characteristics of the chirp (duration, frequency etc.) can for instance yield the masses of the black holes and of their final configuration and may tell us about spin rates of the black holes. Yet these same mergers may produce zero electromagnetic wave signatures that could be seen by conventional telescopes.

Furthermore, because the amplitude and the frequency of the waves depend differently on the masses and distance to the GW source, these binary chirps acts as "standard candles" in a similar way to type Ia supernovae. In other words, measurements of the chirp independently gives the masses of the merging black holes and the distance to them.


There are elongated craters from grazing impacts, and these can sometimes get rather odd shapes like Schiller on the Moon and Orcus Patera on Mars (the later is a bit contested, it might have formed from something other than an impact).

Double impact craters is another possibility. Some craters occur simultaneously (more direct picture). Splitting impactors is likely the one way of getting truly complex topologies. Most of the time they tend to follow each other, producing crater chains.

If terrain shifts craters may be deformed, and depending on details in the rock they may also get odd shapes. Meteor Crater in Arizona is somewhat squarish.

Craters cannot generate that complex patterns since they are just explosions out from a center and impactors are usually single objects or random clouds. In principle one could arrange asteroids along a trajectory to spell out letters upon impact, but it is an exceedingly unlikely configuration so most impacts will only manage 'O'.


When were the spiders first observed on Mars?

I have the impression these were first observed in images taken by the Mars Orbiter Camera (MOC) of the Mars Global Surveyor, probably sometime between 1997 (arrival at Mars) and 1999 (when the main mapping program began) -- or possibly as late as 2000 (when they are first mentioned).
In Piqueux et al. (2003), a paper specifically focused on the spiders, they say

The term ‘‘spiders’’ was used by Kieffer [2000] (although the term may
have originally been coined by Andy Ingersoll [H. Kieffer, personal
communication, 2003]) and we use this term in this paper.

In Kieffer (2000) we find this statement

Dark radially converging dendritic patterns are visible in MOC images
of some portions of the spring polar cap [7]; these have been termed
"black spiders" by the MOC team [8].

Unfortunately, the published version of that short conference proceeding article is missing all of the references after number [4], so I don't know if there is more specific useful information available. The conference at which Kieffer (2000) was presented was in August, 2000, so I'd put mid-2000 as the latest possible date.


Which celestial object in the solar system has the fastest tangential velocity at its surface?

Hands down it's Jupiter.  It has both the largest radius and faster rotational velocity of the planets, and smaller objects will not be able to compete with that huge radius.  Jupiter's rotational velocity is approximately 13.4 km/s (with some variation due to wind currents, as it is a gas), which can be determined both from Doppler shifts of reflected sunlight, and also from watching the motion of equatorial cloud features.  Saturn is only 10.9 km/s, as it has a longer rotation period and a smaller radius.  None of the other planets can compete with Jupiter, and smaller objects would need prohibitively short rotation periods to have any chance.  For example, the Earth would need to have a day that was about half an hour long to compete with Jupiter.  Even a molecule that spins millions of times a second would need a size on the order of a cm to compete with Jupiter.


Why Milky Way and Andromeda are being drawn together if there was 'Big Bang'?

The evidence for expansion is that the redshift is proportional to distance.

The redshift of a galaxy can be divided into two components: that due to the cosmological expansion, which stretches the wavelength of light whilst it travels towards us; and a peculiar motion with respect to the cosmological expansion, which causes a straightforward doppler shift.

The former term increases with distance - this is known as Hubble's law. The redshift here is always positive (ie always a stretching of wavelength). It is this that tells us the universe is expanding. The latter term is caused by the gravitational effects of other nearby galaxies and clusters of galaxies on the galaxy in question. It is typically of order a few hundred km/s and can be positive or negative (ie it can cause a redshift or blueshift).

Redshift due to cosmological expansion completely dominates at distances greater than a few hundred million light years. Up till then (and Andromeda is only 2 million light years distant), peculiar motions can result in blueshifts for some galaxies.


How to map moon phase number + "lunation" to moon phase name?

The WeatherBit 16-day forecast documentation gives these output field descriptions:

moon_phase: Moon phase illumination fraction (0-1)
moon_phase_lunation: Moon lunation fraction (0 = New moon, 0.50 = Full Moon, 0.75 = Last quarter moon)

The illuminated fraction doesn't help you distinguish waxing and waning phases, so moon_phase_lunation is the value to use.
One possible mapping is:
0.0  0.06  0.19  0.31  0.44  0.56  0.69  0.81  0.94  1.0


The average surface temperature on Venus is 462 °C. You would need a temperature above 800 °C before you would notice any light.

Added:
Here are some calculated radiation outputs of the surface temperature of Venus, a 1000 °C black body and the sun.


What kind of star would allow for life like our Sun/Earth and then go nova?

In the real universe, stars that go supernova are very large stars that are short lived, or, they could be white dwarfs that accrue matter and overcome the Chandrasekhar limit.

Very large stars, 8 solar masses or more - about the minimum size for a star that goes supernova, the life span of that star is too short for a planet to form, cool and develop life.   The life of an 8 solar mass star is about 8 to the 3rd power or 1/512th the life of our sun.   On the order of 20 maybe 30 million years.   You can't get a planet to form and have advanced life that quickly.

Now if there's a 2 star system, like lets say, Sirius A and B, the larger star reaches it's red giant stage first prior to becoming a white dwarf and the smaller star keeps burning and in time, clears out much of the red-giant solar nebula left by the first star, then you can have a threat of a supernova binary system where the remaining star feeds the white dwarf, and that star lives long enough for planets to develop life.   Binary systems may have some issues, like the planets need to be fairly distant and/or the stars quite close and you have the white dwarf stealing matter from the star, so the star loses mass over time and the white dwarf likely emits intense UV and perhaps gamma rays, but in this way, it's theoretically possible to have a super-nova on the way and have a planet with life on it in orbit around the binary system.

Sirius A & B are actually not a good example.  They're too far apart and Sirius A is in the direction of too large and short lived and Sirius B is not expected to go supernova (maybe after Sirius A goes red giant - just maybe - that would put on quite a show if we're still around to see it, but I digress).

More realistic, much more common and just as deadly would be a star going red giant.


It appears (and I am no expert) that Saturn's ring evolution is governed mainly by "viscous spreading" - collisions between ring particles; and also by interactions with Saturn's moons (resonances); and bombardment with meteoritic material.

There appears to be no consensus on how old the ring system is. Most theories of their formation have this taking place early in the life of the solar system, but there are observations that suggest little contamination by meteoritic material and hence that the rings are young.

According to Salmon et al. (2010)  the viscosity in the ring is critical to its evolution and therefore estimates of ages and lifetimes. As the disc spreads due to viscosity, the viscosity itself will reduce, slowing the evolution and thus the ring system may not empty for many billions of years. On the other hand, comparing viscous spreading models with the current density profile of Saturn's rings suggests they are much less than a billion years old. However, it should be noted that other authors have used similar models to argue that Saturn's rings did indeed form 4.5 billion years ago and were much more massive (by a factor $>100$) then they are today! (Charnoz et al. 2011)


At a certain size, huge asteroids get classified as dwarf planets. Pluto has an atmosphere 100,000 times thinner than Earth, and Pluto is already one of the two largest dwarf planets known.

Asteroids (like everything) do have gravity, so nearby gas would be drawn to them. But it would take just very tiny distrubances for that gas to drift away, so what little there is would probably be close to undetectable.


The Lagrangian point $L_2$ is very close to the most distant point from Earth with an umbra.
$L_2$ is like the radius of the Hill sphere at $r=a\sqrt[3]{\frac{m}{3M}}$ for circular orbits, with $m$ the mass of Earth, $M$ the mass of the Sun, and $a$ the distance Earth-Sun. The ratio $\frac{m}{3M}$ of the Earth and the triple mass of the Sun is almost exactly $10^{-6}$, the cubic root hence $0.01$.

The diameter ratio of Earth and Sun is about $1/109$. Therefore the umbra of Earth ends near $92\%$ the distance to $L_2$.

The answer to another bonus question would then be: If Earth would be $9\%$ larger in diameter, but with the same mass, its umbra would end almost exactly at $L_2$.

Earth's orbit isn't perfectly circular, but the aphel/perihel ratio of about $1.04$ is insufficient to question the result qualitatively.
The error of the implicite assumptions $\tan x=x=\sin x$ is negligible at the considered level of accuracy.


How do we calculate heat flow or cooling rate for no fusion brown dwarf star?

Whether or not the object is just above or below the threshold (about 13 Jupiter masses) at which the core will become hot enough to burn deuterium is not really relevant to the calculation you wish to understand. The object has cooled beyond the point at which D burning has ceased. It has also pretty much contracted to a minimum sized configuration where electron degeneracy pressure supports it. Therefore very little gravitational potential energy can contribute to subsequent thermal evolution.

All that is left is to calculate the thermal energy content (roughly $3kT/2 \times N_i$ - the number of hydrogen and helium ions, where $T$ is the internal temperature; the degenerate electrons contribute very little to the heat capacity) and then divide this by the luminosity radiated from the surface. The calculation is a little more complex than that because the luminosity fades as the brown dwarf cools.

Just for kicks - here is a back of the envelope calculation. The central temperature of a contracting ball of gas is given by the virial theorem. The brown dwarf you are talking about will contract until D burning (nearly) starts at about $10^{6}$ K - let's assume that is an upper limit to the current core temperature. The average temperature in the interior will be a bit lower than that (let's divide by 2), so $\langle T \rangle \sim 5\times 10^{5}$ K.

If the mass is about $0.012 M_{\odot}$ and it is made almost totally of hydrogen, then the number of protons is about $N_i \sim 1.4 \times 10^{55}$.

Thus the total thermal energy is $E \sim 3N_i k\langle T \rangle/2 \sim 1.5\times 10^{38}$ J.

The current luminosity $L \sim 4 \pi \sigma R^2 T_s^{4}$, where $T_s \simeq 1100$ K is the surface temperature and $R$ is the radius, which will be about the same as Jupiter, i.e. $R\sim 70,000$ km. Thus $L \sim 5 \times 10^{21}$ W.

The current cooling timescale is $E/L \sim 10^{9}$ years - but this is a lower limit.

I am not surprised that this is somewhat lower than your estimate because the cooling time gets much longer as the object cools. This is clearly seen when looking at a proper model calculation; for example see the picture below taken from Burrows et al. (1997).

The lines show the cooling tracks of surface temperature versus (log) time. The red curves are "planets", the green are "brown dwarfs" - the dividing line is the D-burning threshold - seen as a plateau in the the green brown dwarf curves at early times. Your object is around this threshold. I have marked two points on the plot - the first corresponding to a 13 Jupiter mass object with a surface temperature of 1100K (you can see it must already be at least a hundred million years old) , the second marks the same object when it has cooled to about 273 K (zero Celsius). It does indeed occur after about 10 billion years.


How does the concept of a universe with no center work?

When we talk about the universe, we are really talking about one of two things:

The observable universe, which is everything we can possibly see.
The Universe, which is everything that has ever existed, currently exists, and will exist.

The observable universe has its own center, usually the Earth. It is a spherical region of everything that we can see, essentially anything whose light has reached us. We usually refer to this when we say things like "there are $10^{86}$ atoms in the universe."

In reality, everyone has their own observable universe, and it can change depending on where you are. An exoplanet far away has its own observable universe, and can receive light from different places. Essentially, you are the center of your own observable universe.

I assume you're talking about the latter, though. The Universe (notice the capital "U") is all of space and time and its contents. Anything that has existed, will exist, and currently exists is part of it.

The Universe is thought to be infinitely large, so it can't have a center. The center of something is the point equidistance from the edges, but if something spans infinitely long, it would just keep going. It wouldn't have an edge, and thus it wouldn't have a center. You couldn't find the point equidistant from the edges if it just spans infinitely.

You might ask, "then where did the Big Bang start? Surely it must've been the center of the Universe, right?" Well, you can say the Big Bang happened everywhere. Before the Big Bang, matter filled most of the Universe's emptiness. It was essentially dense, and so got extremely hot to the point where no hadrons could form.

It is thought that this temperature caused space itself to expand. Essentially, more space was created in between all the matter, until everything was able to cool down. This is what we call the Big Bang. It didn't happen at a certain point, but rather it happened everywhere.

EDIT: I understand your confusion. Let me just clear some common misconceptions:

The Universe is not like a ball. Rather, you can think of it like a flat grid, and its "expansion" just means that the distances between objects on the grid are getting larger. In essence, more space is being created between the objects. That's what we mean by expansion — that objects are moving away from each other, since more space is being created between them.

Here's an easy analogy: imagine you are walking your dog. Suddenly, the ground begins expanding between you. You and your dog will separated and continue receding away from each other.

That's essentially happening everywhere: space is expanding between everything, so we are drifting away from other galaxies. The Universe is infinite, and we can constantly drift apart from other objects because space is being created in between us. Here's a GIF I made that might help you get it:

You can see how the galaxies drift apart as the space between them increases. And this happens everywhere in the Universe. So let's keep that in mind and clear up the Big Bang.

The Big Bang did not happen at a single point, nor did the Universe begin at a single point. The Universe is, and has always been, infinite. The Big Bang was just when the Universe's expansion really began — that is, when objects started drifting away from each other. The Universe was still infinite, but there was less space between the matter.

This density caused the Universe to get extremely hot and expand. So here, space itself was distorted and began to expand. More and more space was created between the matter, and still is now (although now it is mainly due to dark energy instead of heat).


New NEOs are found daily. Is there any evidence gravity is pulling them to a centralised location?

No, they are being discovered where they are in their (usually fairly stable) orbits. They are not discovered because they are arriving or going to some particular location.

The one exception is long-periodic comets that have fallen in from the Oort cloud and now pass through the inner parts of the solar system, maybe to be captured by the gravity of Jupiter in a closer orbit.

(Strictly speaking all these objects are all pulled towards a central location, the Sun, but because they already have a sideways velocity that just means an orbiting motion where they are constantly falling around it.)

NEOs are too light and small to interact much with each other - the gravitational influence between them is microscopic compared to the influence from the planets which is usually microscopic compared to the sun's influence.

The reason we are seeing a lot more NEOs now is that new instruments are better and more searches are being conducted. LINEAR for example found  231,082 small Solar System bodies (of which 2,423 are NEOs) in the early 2000s and the Catalina Sky Survey has found 6,515 NEOs since 2009.


Basically you need to have the following equipment list
1.- A mechanical system track objects: you can build with 3d printer like:this prototype and/or python tracking mechanical controller
2.- A cmos/ccd sensor camera with over exposed function and color filters ,like this usb computer or use dslr camera telescope or like
3.- Automatic comparative histogram  (Software), you can use python OpenCV
enter link description here
I hope this links can help you, basically you need compare all histograms of same object


When is optical refraction important in astronomy?

Optical refraction is related to the change in direction of a light ray when the refractive index changes. Excluding Earth's atmospheres and instruments, I think that refraction has little/no impact in astronomy.

The only cases that comes to my mind where we can (probably) have some important refraction is in eclipsing star binaries or near edge on planetary systems.
Let's imagine a planet transiting behind his star. Some of it's light passes through the stellar atmosphere and gets refracted. As the atmosphere is curved and likely changes the refractive index with height, it acts like a lens dispersing (intuition says so) the planet light.

Edit A similar description holds in general for any object passing behind an other one that have atmosphere.

And there is Gravitational Lensing (if you allow me), which has much larger impact on observations. This is caused by gravity bending light rays when passing near galaxies/cluster of galaxies(/stars/...). One of the differences of gravitational lensing with respect to standard lenses is that there is no change in refractive index, so it's achromatic (all wavelengths get bent by the same angle).

The effective index of refraction can be described as ( source: Narayan and Bartelmann(pdf) ):
$$ n = 1 + \frac{2}{c^{2}} |\Phi|$$
where $\Phi$ is the gravitational potential and is generally a function of position of the object.

Gravitational lensing is canonically divided in three groups:

Strong lensing, usually observed in galaxy clusters or around massive galaxies. The gravitational potential is so strong that the the image of a background galaxy is heavily distorted into arcs and rings, like in this striking image of Abell 2218 from HST:

(source: hubblesite.org)
Weak lensing. The light of a galaxy encounters matter (and a lot of dark matter) travelling to us and gets refracted. This doesn't have a dramatic effect as in strong lensing, but distorts the shape of the galaxy. And this distortion can be used to study, e.g., the dark matter distribution around some object or the content of the universe.
Micro lensing. Imagine to observe a star and somehow know that a blob of dark matter is going to pass in front of the star. The blob is not big enough to distort the star shape, but for sure it will increase by a small amount the luminosity of the star.


Why does an irregular moon point its longest axis towards its parent planet?

Rotation of a rigid body around the major axis (the principal axis with the highest moment of inertia, which will tend to be close to the longest axis of the satellite) minimises the kinetic energy for a given angular momentum. As a quick way to visualise this, write the equation for rotational kinetic energy $T$ in terms of angular momentum $L$ and moment of inertia $I$:
$$T = \frac{L^2}{2I}$$
At constant $L$, $T \propto I^{-1}$, therefore picking the axis with the highest moment of inertia will give the lowest kinetic energy.
As a satellite is not perfectly rigid nor perfectly elastic, deformations will tend to dissipate energy but angular momentum is conserved, thus leading it to evolve into this state.
Some lecture notes on the relevant dynamics can be found here, with the relevant result (the "major axis theorem") given in §4.11.

Any freely rotating body that is not perfectly rigid, will lose kinetic energy and whist its
angular momentum remains constant in space, it moves with respect to the body until the
body is rotating about its major axis.

This explains the rotation, likewise orienting the long axis towards the planet minimises the gravitational potential energy. As an example to demonstrate this, consider a rod of length $2r$ of linear density $\rho$ with its centre located at a distance $R$ from a mass $M$, with the rod at angle $\theta$ to the radius vector. Neglecting variations in the gravitational field perpendicular to the radius vector, the potential energy is
$$\begin{aligned}V &= \int_{-r}^{r} \frac{-GM\rho}{R+x \cos \theta} dx \\
&= -\frac{GM\rho}{\cos \theta} \log \left( \frac{R+r \cos \theta}{R-r \cos \theta} \right) \\
&= -2GM\rho\left( \frac{r}{R} + \frac{1}{3} \left( \frac{r}{R} \right)^3 \cos^2 \theta + \ldots \right)
\end{aligned}$$
where the last line uses the MacLaurin series for $\log (1+x)$.
All the terms in the sequence in brackets are positive and proportional to increasing even powers of $\cos \theta$, so the potential energy decreases as $\cos \theta$ increases. The minimum potential energy is therefore when $\cos \theta = 1$, i.e. when the rod is aligned with the radius vector.
The state of rotating around the major axis with the major axis aligned with the radius vector (long axis pointing towards the planet) is therefore the minimum energy state.


I am studying ways to make an anti-gravity machine (!!).  I believe it can be done, or theoretically be done using gravity itself as a "force" that pushes outwards instead of inwards.

It doesn't work that way.  No one has found a repulsive version of gravity and there is no evidence this is even possible.

And in science, we go by the evidence.

  A similar question was asked here, which speaks about why objects in the universe are round, which as stated, is due to Gravity.

Which the question's answer explains by the attractive nature of gravity.  No evidence (that word again) of repulsion.

  I understand Gravity is the strongest, most consistent "force" in the universe

You understand wrong.

It is considered the weakest of the fundamental forces.  This Physics SE answer explains why.

It appears to be strong only because on average, over a large scale, matter is generally neutrally charged, whereas mass (and energy) just gets bigger the more space you have (a little simplistic, but that's the general idea).

As for "consistent", well all the forces behave consistently or we would not be able to model them with mathematics.

  and to traverse it we have to push against it - jets, propulsion, etc.

More evidence for no repulsion !

  Never the less, I am perplexed why gravity's forces make objects round - eventually or over time. Why can't it turn a planet into a cube for example? Why can't Gravity make objects flat, and perhaps round at the perimeter ?

Because these are unstable configurations.  Even if they could form from natural processes (which is unlikely as I'll explain) the resulting forces and stresses (for large objects) would be too great to prevent collapse into a more stable shape.

Now smaller objects can form less rounded shapes - small asteroids can be all sorts of shapes.  Only their relatively small mass (and hence small gravity) allows them to resist compression into "nicer" round shapes.

But larger objects have lots of gravity simply because the have lots of mass.  The materials they are made of will not withstand compressive forces on them and they will form roughly round shapes, because that's the most resistive configuration.

So why more-or-less spherical ?

Because these larger objects tend to form in a way that builds up material equally all over the surface, so there is some tendency to be spherically symmetrical just from the way material accretes as it grows.

However once they get really big (planets, larger moons and so on) the forces on the materials will be so great that they will as a general rule, behave a little like fluids.  And fluids will end up in that nice spherical shape as they lack the rigidity to resist compression.  Again a little simplistic, but this gets really complicated quite quickly.  They key here is to stop thinking of e.g. rocky planets as rigid objects.  Maybe on a timescale of seconds they are, but on astronomical timescales they're not so rigid.

So for larger objects, more-or-less spherical is inevitable simply because they're large.

  If I can answer this question, I am sure I can find a way to "bend" Gravity for my invention.

No anti-gravity to be seen here.

Really, learn the basics of physics if your interested and hopefully you'll find that not only is it interesting, but it can explain things without the need to invent other explanations.  Usually. :-)


How has the average distance between asteroids in the asteroid belt changed over time?

The asteroids formed roughly where they are now, from matter in the sun's protoplanetary disc that didn't form into a larger body (at least partly under the influence of Jupiter's gravity)
Since early in the solar system's story, the asteroids have been stable. Most asteroids haven't changed significantly in 4.5 billion years.  And 4.5 billion years ago the separation between asteroids was comparable to how it is now.


Will the Sun's solar wind push a permanent magnet out of our solar system?

No. The magnetic field has two poles, the force on the two pole of the magnet is equal and opposite. This is why a compass needle will align in the North South direction, but is not pulled towards the North or the South.

Two magnets will attract or each other because the North pole of one is nearer the South pole of the other. You can get a net force on a magnet when the field-lines are getting bunched together. But the magnetic field in the plasma is not like this. To a first approximation they are parallel to each other and perpendicular to the equator of the sun. In fact the magnetic field is complex as it is carried with the flow of solar wind.

There would be no net force on the magnet, but it may tend to align with the magnetic field. This might cause a turning force, but no acceleration.

There would be some force due to solar wind particles impinging on the object, but since there is only something like 5 atoms per cm3, the force isn't very big, and it has nothing to do with the object being a magnet.


Can we simulate Earth's gravity in space?

Simulating gravity in space basically means simulating weight, which requires acceleration. So basically, the question is, how do we create acceleration in space.

The easiest method for simulating gravity in space is by spinning the space station. In this case, the reaction force to centripetal force substitutes the force of gravity, which pushes the inner contents to be pushed against the outer edge, giving a sensation of weight. This method is theoretically sound and can be used to simulate gravity.

The problem with doing this in ISS boils down to two things- size and shape.

The best shape for a space station having a simulated gravity is something like a doughnut spinning about its axis, so that the occupants are pushed against the outer wall. The shape of ISS is nowhere near this.

If one excludes the solar panels and the truss structure, the ISS is around 45m along the axis of the habitable modules, if we spin along the axis of the truss structure, this gives a radial distance of ~22.5m.

For simulating earth's gravity in ISS, we will require,

$$\omega^{2} r = g$$

$$\omega = \sqrt{\frac{g}{r}}$$

For g = 9.8 $ms^{-2}$,

$$\omega = \sqrt{\frac{9.8}{22.5}} = 0.66 rad/sec = 6.3 rpm$$

The problem is that at this rotational speed, the head and foot of the astronomer will have different linear velocities. The head of a 1.5m tall astronaut will be moving at 13.86 $ms^{-1}$, while the foot will be moving at 14.85 $ms^{-1}$, never mind they are standing at the bottom of a barrel. This will lead to Coriolis effect, which will be uncomfortable.

So, while earth's gravity cannot be simulated in ISS due to various constraints, we can settle for reduced gravity (with reduced, ~2 rpm) in an ISS sized spacecraft.


Adjective for things outside our solar system

As @AtmosphericPrisonEscape noted in comments, above, "Extrasolar" is the correct term.  E.g., we refer to "extrasolar planets", not "interstellar planets."

"Interstellar" refers to things among the stars (Merriam-Webster: "located, taking place, or traveling among the stars especially of the Milky Way galaxy")

"Extrasolar" means "outside the Solar System." (Merriam-Webster: "originating or existing outside the solar system")

One distinction is that "interstellar" doesn't include all space outside the Solar System -- "Interstellar space" is usually spoken of as distinct from, and much smaller than, "Intergalactic space".

The original question asked for "a general adjective for objects outside our solar system" and that would be "extrasolar".


How does the ITRS frame related to ECEF frame?

The simple explanation:
The International Terrestrial Reference (ITR) Frame is an Earth-centered, Earth-fixed (ECEF) frame, but with the axes of the ITR frame precisely specified taking into consideration the movements of the Earth's crust due to plate tectonics.

This website may be useful.


How many nested stable (1 Mio years) orbits are theoretically possible?

Yes, it's possible to have many nested levels, all stable. Each nested orbit needs to be inside the Hill sphere of its central body in order to be stable.

You can't nest any further when the Hill sphere gets too close to the Roche limit. Below the Roche limit the satellite disintegrates.

The exact number of nested levels depends very greatly on the characteristics of all bodies involved.


If outer space was a perfect vacuum, then there would be no planets or other objects occupying it. How much space are we talking about, and where is this space you refer to? for practical purposes, outer space is defined to as starting at a location past the Kármán line heading outwards from planet Earth. That does not however mean that it is a perfect vacuum. There are still particles, atoms, molecules and other stuff there.
I have also read that Astronauts detect the smell of slight traces of something burnt or burning. This can be caused by a physical effect to do with the way the brain reacts to certain conditions. It may not actually be due to the human olfactory system.
https://www.nbcnews.com/health/health-news/phantom-smells-may-be-sign-trouble-n890271
Even though such things may affect Astronauts, NASA or other space agency medical staff may not take it as a long term effect that cannot be corrected or stabilized. And as such still allow Astronauts to space walk with periods of recovery in between. Thus allowing the astronauts to effectively work in outer space, even though they experience such olfactory sensations.
Yeh, I also have seen that video with Astronaut Chris Hadfield. One of the biggest issues with Astronaut habitats in outer space is that they accumulate large amounts of dead skin cells. I mean, huge amounts! So if some of those dead skin cells deposited themselves on the space suits, and radiation outside the craft frazzled the cells, it would smell like burnt steak, when you come back inside the habitat/craft.


Why could a gas planet not get much larger than Jupiter?

Usually they can get is Almost twice.  Here's the evidence:
 It is difficult for them to become even bigger than that as once they reach 15 Jupiter masses or so there is enough pressure in the core to ignite deuterium fusion, so those are considered “brown dwarfs” rather than planets.
An example is TrES-4, which is located 1,400 light years away in the constellation Hercules. The planet has been measured to be 1.4 times the size of Jupiter, but it only has 0.84 times Jupiter’s mass. With such a low density, it is dubbed the puffy planet.
It was discovered in 2006 orbiting a brightish star 1500 light years from Earth. Its diameter of about 1.8 times Jupiter’s makes it the largest accurately measured planet. Strangely, though, TrES-4 is very light for its size. It has only 88 per cent of Jupiter’s mass, giving it a density of roughly 0.2 grams per cubic centimetre, less than that of cork. Just how a planet can be as fluffy as TrES-4 remains a mystery.


Would a red dwarf star resemble our own Sun at sunset to an observer on a nearby planet?

Your question may ulitmately be about the physiology of the eye, which is off-topic here.
The spectrum of the Sun seen low on the horizon is quite different to the spectrum of an M-type red dwarf. The reason that a red dwarf is red, is not just that it is cool, but that there are great chunks of the spectrum that are absorbed by molecules in the photosphere of the star. These molecules start to form when the temperature of the photosphere falls below 4000 K.
In contrast, the Sun has an almost continuous spectrum all the way into the ultraviolet, with relatively narrow absorption lines caused by atomic species in the solar photosphere. This is then modified, mainly, by Rayleigh scattering, which smoothly scatters blue light out of the line of sight towards the Sun.
Whilst any sort of spectrograph or prism could distinguish these differences, whether your eyes could is another matter. So whilst it is conceivable that the colours would appear similar for some particular angle above the horizon, the detailed spectrum would be quite different.


What is the temperature inside a Black Hole?

The temperature of a black hole as seen from the outside is the temperature of the Hawking radiation, which @jyoti has discussed.

The temperature you would experience as you fell freely through the event horizon, is a different question. I think the answer depends on what else is falling into the black hole at the same time. From its own viewpoint, everything that falls into a black hole reaches the singularity in a fixed maximum time (a few days for the very largest supermassive black holes) so you could only exchange energy with things that fell in around the same time as you did. If there was very little such stuff, then I don't think you'd notice anything out of the ordinary -- you'd be exposed to the cold of space, just as you were before (assuming the black hole is big enough that you haven't been heated up the tides of the black hole distorting you. On the other hand if lots of stuff was falling into the black hole with you then the whole lot would get churned up, and you'd be falling in as part of a mass of very hot plasma. Eventually, at least in a non-spinning black hole, you'd approach the singularity and be ripped apart by tides.


How does a Super-Massive Black Hole 'flare'?

The flare happens just outside the black hole.  Matter which aproaches a black hole but not perfectly straight on doesn't fall into it immediately. It gets pulled around the black hole, just as a satellite is pulled around the Earth.

If the object is a single small particle, that's almost the end of the story. Gravitational radiation and the precise effects of GR will cause its orbit to decay until it eventually falls in, but if it's small enough not much of anything is emitted in the process.

If there are lots of objects in the neighbourhood, or if your obect is something relatively large and diffuse (like a gas cloud, or a planet or star) on the other hand, different things happen. As gravity pulls on different parts of it with different strengths (tides) the object gets stretched and then pulled to pieces, and multiple objects collide and interact in all sorts of ways. Magnetic fields also start to play a role. Some of that matter escapes completely, but much of it ends up in a disk (a little like Saturn's rings, but less well organised) around the hole. By this point all the stretching and collisions have made it very hot, so it shines.

Now even within the disk there are collisions and friction, and they have the overall effect of heating the matter in the disk and pushing some of it out of orbit, so that it falls into the hole, getting even hotter in the process.

So finally we come to a flare. There are probably many kinds of flare-up in this basically unstable setup, but the most obvious is when a particularly large object or cloud of gas falls into the accretion disk, or when some instability in the disk causes a large amount of matter to fall out of the disk. Either way, the larger amount of hotter matter shines more, and we see a flare.


Is there a recent update on the mysterious M82 radio source reported in 2010?

The SIMBAD page for the source in question: EQ J095552.5+694045.4

From the list of references, the most relevant ones to this question appear to be as follows:

Joseph et al. (2011) suggest that the object could be a microquasar similar to SS 433, which has a high radio to X-ray luminosity ratio.
Gendre et al. (2013) argue against a supernova interpretation for the source:

  Muxlow et al. (2010) suggested that this
  object might either be due to accretion around a massive collapsed
  object or be a faint and unusual supernova (Joseph, Maccarone &
  Fender 2010). In the latter case, we would expect the flux density
  to steadily decrease over time, as the source would be too young to
  have reached the remnant phase. However, in the observations presented here, it shows
  a flux density at 5 GHz of ∼0.90 mJy, which
  does not seem to vary significantly over the period 2009 May–2010.
  The supernova hypothesis is thus unlikely.

Mattila et al. (2013) state that the nature of the source is still unclear but they favour the microquasar interpretation:

  The nature of the 43.78+59.3 transient remains elusive, and on
  the basis of the limited data available, we regard an extremely bright
  extragalactic microquasar as the most plausible scenario, perhaps
  from a high-mass X-ray binary such as LS 5039 (Clark et al. 2001),
  but with a high ratio of radio to X-ray flux. A bright extragalactic microquasar was
  proposed as the most likely explanation by
  Muxlow et al. (2010) and Joseph et al. (2011). We concur with that
  conclusion although it would mean that the NIR luminosity is a
  factor of about 30 higher than Galactic microquasars.

Varenius et al. (2015) note that the source was not detected with LOFAR.

So from these, the nature of the source hasn't yet been conclusively shown, though the most plausible scenario is that it is a somewhat extreme example of a microquasar.


How close are we to having the technology to measure planetary obliquity for exoplanets?

Carter & Winn (2010) suggest that the most promising means of detecting exoplanet obliquity would be through miniscule signatures imprinted on the transit light signal at ingress and egress (~ 200 parts per million for a planet as oblate as Saturn). Zhu et al. (2014) use this technique to make the first tentative detection of exoplanet obliquity from a Kepler object, the 18 Jupiter mass brown dwarf Kepler 39b (KOI-423.01). They measure an oblateness of 0.22±0.11. They also place some upper constraints on the oblateness of other planets in the Kepler catalogue.

Transit signal for KOI-423.01 over 12 orbits. The residuals of two models, one with oblateness and one without, are plotted at bottom. The Oblateness model is a better fit to the data.

Oblateness variations are thought to be conducive to exoplanet habitability by regulating temperature modulations, so I expect that measurements of this property over time will be a priority in future exoplanet observations for astrobiology and SETI studies.


What resolution would earth-moon telescope achieve?

According to Wikipedia, the angular resolution is inversely proportional to the baseline (the separation between the telescopes). The order of magnitude of the baseline of the Event Horizon Telescope is 10,000 km; the Earth-Moon distance is about 40 times larger, and Earth-Mars at its maximum 25,000 times larger. That means we could theoretically reach 1.5 arcmicroseconds (at this wavelength) with an additional telescope at the Moon, and 2.4 arcnanoseconds with Mars.

However, to obtain proper interferometry we need (regardless of the baseline size) to know the telescopes' precise location with the precision of a quarter of the wavelength. For 230 GHz radio waves, the wavelength is 1.3 mm, so the precision needs to be 0.3 mm. I suppose that this is incredibly difficult to reach for bodies moving through space.

I am not well-versed enough in the matter to say what the effect of arranging the telescopes in a line (on Earth-Moon or Earth-Mars scale, it's a line, unless you add space telescopes at e.g. L4 and L5 Lagrange points) vs. a grid like the Event Horizon Telescope is. But interferometry does work for two telescopes already.

Note that angular resolution is not the only metric you should consider; objects become fainter when you enlarge them, so you'll need more telescopes (and/or a longer exposure time) to get the same level of brightness in the pictures.


What you said :

  monoliths increase the mass of Jupiter

What you quoted :

    these monoliths increase Jupiter's density

Increasing density is not the same as increasing mass.

By squashing stuff together you can increase it's density without affecting it's mass at all.

  But if Jupiter were by scifi magic made as massive as a star, would its orbit remain the same regardless?

No, nor would any body in the solar system, including the Sun, remain undisturbed.  It would be a massively complex change.