Why are asteroids with zero orbital inclination rare?

Short answer: The scarcity of asteroids with an inclination near zero is an expected result of normally distributed inclinations in 3-dimensions about the normal vector to the reference plane, rather than selection bias or an orbit "clearing" around zero degree inclination.
Long answer: Orbital inclination is typically defined as the angle between a reference plane and the orbital plane:

An equivalent definition is the angle between a reference plane normal and the axis of revolution of the orbiting body overlay-ed here:

Notice that the axis of revolution can be mapped to a point on the Celestial Sphere. We can generate a Gaussian distribution of points on the unit sphere centered with mean at the reference normal using the von-Mises Fisher distribution.  The specific algorithm for generating sampling points is given on Stack Overflow.
Here is a group of 20 revolution axis intercept points with their corresponding circular orbits:

We can generate and plot a thousand points on a unit sphere (that I rotate to show the structure):

Note that only the very top points of this distribution correspond to inclination angles of nearly zero. As the inclination increases, the points get less dense, but the areas on the unit sphere are larger. The below plot is looking down from above the North Pole, with the orbits with inclination less than 2 degrees colored red, and the orbits with inclinations between 2 and 4 degrees colored green.  Note that even though the points are the most dense at the pole, the area covered is the smallest, so 13 points are red, while 50 points are green.  This is the explanation for why we see very few asteroid orbits with inclinations of less than 1/2 a degree from the original question.

Below is a plot with the y-axis as inclination and x-axis as longitude compared with the OP's post. I chose a standard deviation that was much too large, but the 1000 point Monte Carlo demonstrates the same statistical effect asked about by the OP.

Notes:

The above example is for circular ellipses all with the same SMA for convenience, but I don't think we lose any generality.

I assume the reference plane in the picture used by the OP is the Earth's ecliptic, but the above answer works for the Sun's equatorial plane, or Jupiter's orbital plane, or any other close reference plane.  We can assume that the asteroid inclination distribution 2D mean is actually going to be a bit offset from any choice of reference plane normal.

Orbital Mechanics can be very non-intuitive.  Inclination is a 1D value, but its representative underlying probability distribution is best modeled on a 2D spherical surface in 3D.

It's somewhat ironic that I claim this is strictly a math problem, but there are no equations in the above post.  The power of the Monte Carlo approach is in its simplicity and resultant appeal to mathematical intuition.

Statisticians/Probabilists should immediately recognize the distribution of orbit inclinations as a variant of the Rayleigh distribution which should lend a hint that a compression from 2D variables to 1D has occurred.

My Matlab code is available on request.


Why do PMS stars on the Hayashi track remain at a constant temperature while they contract?

The surface temperature of fully convective stars is set by their opacity mechanism, often emission when the H minus ion is created.  The rate the star emits light is not set by the contraction rate, it is the other way around-- the star loses heat at a rate set by its structure, and that's what sets the contraction rate.  (As said, the rate of gravitational energy release is always double the rate the star loses heat to starlight.)  The reason the surface temperature always has to be something like 3000 K is pretty complicated, but it seems to be due to the fact that this temperature is a kind of peak in the surface's ability to emit light.  Convection sets the temperature structure, but there are a range of possible solutions with different internal energy, so the history of heat loss picks out whichever solution has the right internal energy.  But just why the surface is always about the same T is still the trickiest part to understand, I am hoping to get a better handle on that eventually.  Ignore anything you see in introductory textbooks, they never get it right.


How do you go from transit times to Earth-Sun distance?

We know, precisely, the ratio between the Earth-Venus and Venus-Sun distances. Kepler's laws give us this.
We know, in terms of apparent size in the sky, the precise size of the Sun in degrees - just by looking at it.
Placing Venus between the Sun and the Earth, we can look at it from one point in the Earth (call it $N$) and see Venus against the Sun at a point I'll call $n$. Repeating from another place (call it $S$) at a different latitude, we get another point, $s$. If we know how far apart these two points are in the sky, we can use the relative distance of Venus to tell us the angle, at Venus, of the lines $N$-Venus-$n$ and $S$-Venus-$s$. Since we know the distance in kilometres between $N$ and $S$, that means we can solve the triangle and get the distance in kilometres between Venus and $N$ or $S$. (I am tacitly omitting a few steps here, such as the fact that the "distance" between $N$ and $S$ is a north-south distance and not a great-circle distance along the surface of the Earth). Having the distance between Venus and $N$ or $S$, we can multiply that up (since we know the ratio accurately) to the distance between the Sun and $N$ or $S$.
So it remains to find the distance in the sky between $n$ and $s$. Direct measurement cannot be done accurately enough. Accordingly we use transit time as a proxy. A line going through the middle of the Sun is longer than a line slicing across the top or bottom of it. So transit time tells us how far north or south of the Sun's diameter the point $s$ or $n$ is travelling.
Three transit times are actually needed, not two. To see this, suppose that the times for $s$ and $n$ were equal. All we would then know, strictly speaking, is that they were symmetrically disposed about the diameter - not how far north or south of the diameter they were. But since we know how fast Venus travels across the face of the Sun we can calculate the duration of a hypothetical transit along the diameter, and that figure can then be combined with just two actual observations.
To summarise:

Transit durations give latitudes on the Sun's disc.
Latitudes on the Sun's disc give an angular distance between two transit points.
– The ratio of orbit sizes transforms this into the angle of a long thin triangle based at Venus and joining the two observation points on the Earth.
– Trigonometry gives the Venus-Earth distance in terrestrial units.
– The ratio of orbit sizes, again, gives the Sun-Earth distance in terrestrial units.


Why can we still see 10 billion year old galaxies?

Those galaxies were 10 billion light years away from Earth. So light would take much more time to reach here and that light which is now 10 billion years older can be seen now. Even light from the Sun takes 8 minutes to reach to us. So if somehow sun disappears suddenly(very unlikely) we wouldn't know for 8 minutes.


What is the difference between precession and axial tilt?

The Earth orbits the sun. The plane in which it orbits is called the Ecliptic plane, and perpendicular to the plane is a direction, known as the ecliptic pole. The direction of the Ecliptic pole is towards the constellation "Draco". The Earth's North pole doesn't point in the same direction. It points towards the pole star "Polaris" in Ursa Minor. The difference between these two poles is the axial tilt (it is about 23.5 degrees)

Over time the direction of the North Pole changes, it moves in a circle around the Ecliptic pole, taking about 26000 years to complete one circuit. This is precession. The amount of tilt doesn't change much since the North Pole moves in roughly a circle. The tilt varies over a few degrees with a period of about 41000 years. The variation in tilt is not precession.

So axial tilt and precession are not the same thing. One is the angle between the ecliptic pole and north pole, the other is the motion of the north pole relative to the ecliptic.


Why don't stars move in the night-sky as the moon does?

This is referred to as diurnal motion, due to Earth's rotation on its axis, and it affects apparent motion of stars differently depending on their position on the skies relative to the axis of Earth's rotation. For example, on northern hemisphere, the star that appears not to move at all is positioned so that the earth's axis of rotation points directly towards it, and is called a Polaris due to its role as a pole star. Stars further away from true north (not to be confused with magnetic north), will appear to prescribe a circle around it as the Earth completes one rotation on its axis:

                                A long exposure photograph of the night skies, showing their apparent motion

So what you assert, that the stars don't seem to move isn't exactly true. Stars closer to celestial equator will move at equal apparent radial velocity as any object fixed on the night skies would (so at the speed of Earth's rotation), while the stars closer to celestial poles, while maintaining this radial velocity, will prescribe a smaller circle and appear to be more stationary. And the stars exactly aligned with the Earth's axis (like Polaris, but there isn't any such exact counterpart on the southern hemisphere), would appear not to move at all.


How long have the TRAPPIST-1 planets been in a habitable zone?

The luminosity of Trappist-1 is estimated to be $5.25\times 10^{-4}\ L_{\odot}$, but it has not always been like this.

The luminosity of a brown dwarf decreases with time and it this measured luminosity (along with the spectral type) that allows an estimate of the mass and a lower limit to the age using stellar evolutionary models.

If I look at the Baraffe et al. (2015) low-mass evolutionary models and look at the locus of luminosity versus time for a $0.08\ M_{\odot}$ star like Trappist-1, you can see that the current luminosity implies an age of $\sim 500$ million years. But if you go back in time, the star was more luminous and for that reason, planets that are currently in the habitable zone (said to be planets e,f,g) were not so in the past.

The details of a habitable zone (HZ) calculation can be complex, but basically the radius of the habitable zone scales as the square root of the luminosity. If planets d and h are not currently in the HZ then we can use these as a conservative definition of the HZ boundary.

From this (and using the published orbital radii of the planets), I can see that if the luminosity is increased by a factor of 9, then none of the planets b-g are in the HZ, it is larger than all of their orbits. Trappist-1 had a luminosity that was 9 times larger when it was younger than 27 million years old. On the other hand, if I want to move the HZ just outside the orbit of planet e (and simultaneously include planet h inside the HZ), then this would occur when Trappist-1 was at an age of 206 million years. As a final thought, you can see from this particular model that Trappist-1 may fade by a further factor of two as it gets older. This decreases the HZ radius by a factor of 1.41 and would mean that g (and possibly f) would fall outside the HZ, whilst d (and possibly c) would be brought into the HZ.

It should be noted however that: different models give slightly different results, these loci are mass dependent and the mass is not known, it is inferred from the same models using a temperature estimate (which is also uncertain). So whilst my qualitative conclusions about the past location of the HZ are likely to be correct (though the detailed age numbers are model-dependent), the future behaviour of the HZ is more uncertain because Trappist-1 may be slightly more massive than assumed and already reached its minimum luminosity.

The luminosity evolution of Trappist-1, assuming a mass of $0.08M_{\odot}$ and the models of Baraffe et al. (2015). The horizontal dashed line marks the best estimate of its current luminosity, for which planets e-g are said to be in the HZ. If we go back in time, the rightmost vertical dashed line marks the age below which the luminosity increases to the point that e becomes too hot to be habitable. Then further back, the leftmost vertical dashed line marks the point where all the currently known planets (b-h) become uninhabitable.

So the answer to your question is quite uncertain and depends critically on the age of Trappist-1 now and of course whether the planets were always at the orbital radii they are now$^{*}$. As you can see from the plot above (note the logarithmic scale on the x-axis), the luminosity evolution referred to above takes place early-on. If Trappist-1 could be as young as 500 million years, then life on planet e may only have been possible for 300 million years. However, if the star is slightly more massive and is 10 billion years old, then life has had 9.8 billion years to get going.

If you are talking about planet f, then it has had slightly longer ($\sim +100$ million years) inside the HZ, and planet g slightly longer ($\sim +70$ million years) again. Planet h will have spent comparatively little time (in the past), less than a few hundred million years, inside the HZ.

$^{*}$ The abstract of the discovery paper by Gillon et al. (2017) briefly discusses the possibility that the planets migrated inwards after formation, through a process of "disc-driven migration". If so, then this will not alter the discussion above. Disc around very low-mass stars may be longer-lived than those around higher mass stars, but have essentially dispersed after $\sim 10-20$ million years (Kennedy & Kenyon 2009; Dawson et al. 2013; Binks & Jeffries 2017), and the planetary configuration would have to be settled where it is now by the time the disc has gone.


How could original astronomers have used the clouds to find distances?

No, because the size of the halo is determined by the properties of ice crystals, the distance to the Moon is irrelevant. So you get a 22° halo
 with both the Sun & the Moon, even though the Sun's much further away.


2SB is Dual Sideband, as opposed to DSB - Double Sideband. Here are a couple of papers you might find relevant and interesting:
a 2SB upgrade replacing a DSB
a paper that mentions the advantages of 2SB over DSB
Answers to What is the functional difference between DSB and 2SB mixers, and why exactly would 2SB be better for radio astronomy? in Signal Processing SE go into further detail.


Do most planetary systems have fewer planets than the solar system?

Perhaps the way to answer this is ask - could we detect the planets in our solar system if we were looking at the Sun, using current technology, from distances of many light years?

The short answer is that we could detect Jupiter using the Doppler radial velocity technique, if we observed for more than 10 years (at least one orbit is required). If we were lucky, and the orientation is right, we might then also be able to detect a transit of Venus or the Earth, using a satellite observatory like Kepler. Kepler could detect Earthlike planets by the transit technique, but the solar system is not "flat" enough that you would observe multiple transiting planets.

So the answer is that we would currently have seen Jupiter and maybe one other planet. Therefore we cannot at the moment conclude that 8 planets is an unusually high number; it may be quite typical. Although we do know that solar systems can be much more densely populated with planets (in their "terrestrial planet zones") than our own.


The black hole itself does not emit the radiation, but rather the space just around it. Because of quantum effects, small particles and anti-particles will spontaneously create everywhere (and later annihilate themselves in a collision). If this happens just outside the event horizon, one particle will inevitably fall into it, never to be seen again. The other particle is now 'free' and this is the radiation Hawking predicted.

  Can we detect a black hole's radiation?

Well, this radiation is way too small to be detected by itself, even if the black hole would be at our doorstep. However, the radiation makes the black hole less massive and smaller and finally it will vanish, producing a gamma ray flash which we hope is detectable, even from a very large distance.
As @JamesK notes in the comments, this evaporation will take extremely long for the 'common' black holes caused by supernovae and those found in galaxy centers, but smaller primordial black holes, formed in the very beginning of the universe, could be evaporating now and that's why we have a telescope looking for those gamma ray flashes.


SDSS Image FITS files have negative values. What do these negative values mean?

Without more information about the specific nature of the fits file, my immediate guess is that you're using a file that has already been corrected for things like bias and flat fielding.

It is true that for a simple exposure (what we might call the "science" image), the fits image should have only positive values as each pixel is a (proxy) count of how many photons hit it and how can you have a negative number of photons hit your detector? But often astronomers need to clean up their science image from various sources of noise. One such example is cleaning it with a bias frame. A bias frame is an exposure with the shutter closed and the exposure time set to zero. Essentially you just pull the signal from the CCD without taking a picture. In a perfect world, the CCD will be blank and empty and there will be nothing to read out if you never exposed, but in the real world things are messy and you have lingering electrons all over the place. Reading a bias frame allows you to get some estimate of this noise that can inherently exist in a CCD. An astronomer will then take this bias frame and subtract it from the science image to remove this noise, but of course this can result in your science image having negative pixel values. This usually isn't a problem though.

This resource looks pretty comprehensive about the basics of cleaning science frames.


gravitational time dilation multiple sources

In the static weak-field approximation, the metric is approximated in terms of the the Newtonian gravitational potential $\Phi$ as:
$$\mathrm{d}s^2 = -\left(1+2\Phi\right)\mathrm{d}t^2 + (1-2\Phi)\mathrm{d}S^2\text{,}$$
where $\mathrm{d}S^2$ is the metric for Euclidean $3$-space. A more general situation requires one to solve a multi-body problem in general relativity, although the parametrized post-Newtonian formalism provides a canonical approximation scheme. So the completely general answer to your question is "it's very difficult problem of numerical general relativity that doesn't have a neat analytic solution."

However, at large distances from slowly-moving objects (i.e., large compared to their Schwarzschild radii, we can approximate a multi-body time dilation relative to a stationary observer at infinity as just given by the sum of their gravitational potentials:
$$\frac{\mathrm{d}\tau}{\mathrm{d}t} = 1 - \frac{1}{c^2}\sum_k\frac{GM_k}{r_k} \text{.}$$
Note that this is consistent with the Schwarzschild formula for large radial coordinates, since by the Taylor-MacLaurin expansion,
$$\frac{\mathrm{d}\tau}{\mathrm{d}t} = \sqrt{1-\frac{2GM}{rc^2}} = 1 - \frac{GM}{rc^2} + \mathcal{O}\left(\frac{R^2}{r^2}\right)\text{,}$$
where $R = 2GM/c^2$ is the Schwarzschild radius. This also means that simply multiplying the factors is a pretty good approximation as long as those factors are close to $1$.


Icy moons: why hasn't the ice sublimated away?

I'll pick up on your example as it can be used very well for illustrating the main points:

First: The solar intensity in Jupiters orbit is less than 4 percent compared to Earths. So much less energy comes from the Sun which limits possible sublimation significantly
Second: Sure tidal flexing leads to fountains. These fountains blow material into space. But this does not mean that all the material will leave Europas surface forever. It can fall back or it stays in Europas orbit and returns to the surface due to the effects of Jupiters magnetic field or gravitational forces of the other moons.
Third: Sure the heat of the tidal flexing has to go somewhere. And it is largely irratiated away. As Europas surface is quite bright it reflects most sunlight away instead of absorbing it. This makes Europas surface quite cool (50-125K). Compare it with the darker neighboring moon Ganymede (70-152K). As you can see the albedo (which means the reflectivity of the surface) is far more important than the tidal heating.
Fourth: Jep some of the water is split up by the radiation. This leads to an oxygen 'atmosphere' on Europa with 0.1 µPa surface pressure. While this is detectable it still is practically a vacuum. There is very little oxygen there, probably not more than a few hundred or thousand tons on the whole moon. Not enough to threaten the existence of moon with a mass of 4.8 * 10^22 kg

In summary: Yes, icy moons lose some of their material. But for exactly this reason they only appear in the outer solar system where the solar radiation is low enough to make this effect negligble.


What is the large astronomical instrument shown in this 1918 eclipse expedition photo?

This construct was a telescope. Citing this article:

  Owing to the difficulty of erecting a double tower, most other astronomers follow the mechanically simpler plan of laying the camera tube horizontally and allowing the sunlight to be fed into it by means of a plane mirror driven by clockwork to counteract the westward motion of the sun.pg. 255

A Popular Astronomy article also discusses the eclipse observation in detail. They describe the primary telescopes used for photography. The first was a 65-foot telescope (that measurement being the length, not the aperture).

  This instrument was in charge of W. A. Conrad, who was assisted by W. L. Veale. The camera was horizontal and was directed toward the sunrise point. Light from the corona was reflected from a silvered glass mirror 10 inches in diameter attached to a coelostat controlled by clockwork. The lens was 7 1/2 inches in aperture and the plates were double coated special coronal, 17x20 inches, made by the Eastman Kodak Co. The diameter of the sun's image on the plates is about 7 1/4 inches. Five exposures were made of 5, 15, 35, 15, and 5 seconds duration respectively.

They also had a 105-inch telescope which was also laid horizontally and had the light directed into it via a coelostat. While I can't be completely sure which telescope is pictured in that image, given the size, my guess would be the 65-foot telescope.


Could you use luminosity changes over time to calculate distance?

It is an interesting question, but it won't work because the luminosity of astrophysical sources is not constant enough, and because measurements cannot be taken with the required precision on long timescales.

Let's assume that you have an astrophysical sources that has a constant luminosity (note, I think you have confused luminosity with apparent brightness) and which has a perfectly measured line of sight speed $v$.

At some particular moment
$$  F(t) = \frac{L}{4\pi D(t)^2}\ ,\tag*{(1)}$$
where $L$ is the luminosity, $D(t)$ is the (changing) distance and $F(t)$ is the observed flux.

We could observe how the brightness changed with time
$$ \frac{dF}{dt} = \dot{F} = -\frac{2L}{4\pi D^3}v\ . \tag*{(2)}$$

If the luminosity is unknown, we can eliminate it by combining equations (1) and (2) and rearranging to give $D$ in term of the observables, $F$, $\dot{F}$ and $v$.
$$ D = -\frac{2F}{\dot{F}}v\ , \tag*{(3)}$$
where a positive $v$ indicates movement away from us.

So it could work in principle, but let's put in some numbers. Let's say we have a source at $D=10$ pc and let's give it a big stellar velocity, say $v = 100$ km/s. That would mean (from equation 3), that $\dot{F}/F = 6\times 10^{-13}$ s$^{-1}$ or
$$\frac{\dot{F}}{F} \sim 2\times 10^{-5} \left(\frac{D}{10\ {\rm pc}}\right)^{-1} \left(\frac{v}{100\ {\rm km/s}}\right)\ {\rm year}^{-1}$$.

So you would have to have instrumentation capable of measuring a change in brightness of 2 parts in 100,000 every year, assuming that the luminosity didn't change at all. And the required precision gets more demanding for larger distances or for slower velocities.

The Sun's (a normal star) brightness varies by 1-2 orders of magnitude more than this every year and there just isn't any astronomical instrumentation that has the capability or stability to make such measurements. Therefore it is unfeasible for at least two reasons.


Spinning black holes acting as a slingshot

Skyhooks or tethers are most useful for lifting mass from a surface into space, increasing velocity. Kerr blackholes have no appropriate surface for basing mass. Instead a gravity assist maneuver may take advantage of the high spin speed of a Kerr black hole, which induce frame-dragging where the space around the blackhole (within the ergosphere) is dragged at the speed of light in the same direction as the spin.
https://nasa.fandom.com/wiki/Gravity_assist#Limits_to_slingshot_use
The Penrose process also offers a means of extracting energy from Kerr blackholes, if the rotational energy of the black hole is located not inside the event horizon but inside this ergosphere.
https://en.wikipedia.org/wiki/Penrose_process


How was Earth's "quasi-satellite" 2016 HO3 "first spotted" and it's orbit determined?

Phil Plait's blog, Bad Astronomy, states (emphasis mine)

One of them was just discovered recently by astronomers: the asteroid 2016 HO3.
. . .
Here’s how that works. The asteroid was first seen in April 2016 in observations of the sky taken by the Pan-STARRS observatory, designed to look for asteroids and comets that get close to Earth. That’s recent enough that a really good orbit for it is hard to determine, but it turns out it was seen in older observations (those can be found by tracing the orbit backward and checking if any observations of it were archived), providing a much longer baseline and therefore a better orbit.

In other words, it seems like a basic approximation of the orbit was created, and the astronomers looked at images taken of areas near where the asteroid should have been, according to the calculations. The calculations could then be refined, and further previous sightings could be investigated.


I have been an amateur astronomer for over 30 years so I can give you some practical advice on this one.

A Barlow lens is an attachment which fits into the eyepiece holder. It's main purpose is to give increased magnification with your existing eyepieces. They are usually marked as 2x or 3x which indicates how many times they will multiply the magnifying power of an eyepiece. They can be useful for large telescopes which can comfortably handle high powers but for a smaller scope they can be of limited use.

The two problems you would encounter if you used one are, first, that the more you magnify an image the fainter it gets, with a smallish diameter scope like yours this means that you will find it very difficult to see faint objects. The other one is the fact that locating and tracking objects as they move (due to the rotation of the earth) becomes much harder as you increase magnification, the field of view of the eyepiece gets much smaller as you increase the power.

I would personally suggest that you get used to using your new scope for a while as it is, then, see whether you might need one in the future. If your main area of interest turns out to be the moon and planets then a Barlow might be useful at some point, but if you enjoy wide field views of nebulae and star clusters then that is less likely.
Hope this is of some help.


How far into the future can we go by traveling close to a black hole?

The way that you have specified the question, the answer is as far as you like. You simply put your spaceship into any orbit around the black hole and wait.

A more sensible question is what is the largest time dilation factor that can be accomplished - i.e. that maximises your travel time into the future for a given amount of proper time experienced on the spaceship.

This in turn is governed by how close to the black hole you can come and still tolerate the tidal forces. If you don't put a limit on this (your first case), then the answer is again infinite; you can hover as close to the event horizon as you like, using an enormous amount of rocket fuel, and the time dilation (see below) can be arbitrarily large.

Your second case is more realistic. Roughly we can say that the tidal acceleration across a body of length $l$ is given by $GMl/4r^3$, where $M$ is the black hole mass and $r$ is the distance from the black hole. If we make this acceleration equal to say $1 g$, and your body length $l \sim 1$m, then for a $5M_{\odot}$ black hole $r \simeq 2500$ km (well outside the Schwarzschild radius of 15 km).

If you could "hover" at this radius, then the time dilation factor would be
$$\frac{\tau}{\tau_0} = \left( 1 - \frac{2GM}{rc^2}\right)^{1/2},$$
where $\tau$ is the time interval on a clock on the spaceship and $\tau_0$ is the time interval well away from the black hole.

For $M=5M_{\odot}$ and $r = 2500$ km, this factor is 0.9970.

If the spaceship is in a circular orbit, the factor is $(1 - 3GM/rc^2)^{1/2} = 0.9955$.

These factors are perhaps not as big as you might have imagined!


How to calculate the thickness of the ice crust on a frozen ocean planet/moon?

Based on assumptions, if you are ok with approx solution then you can solve heat transfer equation in steady state.

Heat transfer equation

Ignoring the time - long time has passed to attain equilibrium.

Heat flux assuming outward and hence internal ocean would be sum of all internal heating mechanism.

Easiest would be if somehow you can come up with surface temperature value. Then you can solve conduction equation through ice for temperature and  weight for pressure and just see at what temperature and pressure ice will go into liquid. Simply asking what is ice layer thickness to maintain a internal heat flux and given surface temperature.

You need one boundary condition. Other way would be little harder, starting at centre with outward heatflux. You need to know both phase and composition since density and cp both changes with these conditions.


Determining the percentage area of the moon's disk shaded with respect to time

If I interpret your question correctly, you want to calculate the area of the moon disk as seen from Earth which is shaded, correct? You might then be looking for a way to calculate the position of the moon's terminator line. For this, you might have to dive into celestrial mechanics, more specifically about the ephemerides of moon and Earth. This is necessary as you  have to calculate the position of Earth and moon in an appropriate sun-centered coordinate system.
In a basic course on celestial mechanics this does not necessarily occur since the priority there is to calculate the stability of the orbit of objects.
Related questions and answers

Get a table of Moon phase angle per day
How to calculate the moon phase angle?

References

Klaus Hempe, Jürgen Molt: Sterne im Computer. Berechnungsprogramme für den Hobby- Astronomen (1999). IMHO an accessible explanation of the numerics, but in German.

http://www.internetsv.info/MoonCalc.html seems helpful.

Wolframalpha's moon phase calculation always give you the illuminated percentage of the moon disk.


Solid-body tides on the Moon
There are various good models for solid-body tides on the Earth. Are there any recommendations for tides on the Moon?

This answer starts with simplistic explanations and links with trivial calculations and progresses to links that, along with their references, provide exacting and lengthy approaches. @scb, you might want to skip halfway through the answer.
To make accurate calculations one needs to know about various parameters of the moon: composition of the crust and internal structure of the moon, Love numbers, precession, tidal acceleration, even time dilation - each results in a tiny error if omitted, a larger error if you skip them all.
An explanation of the anomalies of the Moon's crust is offered in Space.com's article: "Mystery of Moon's Lumpy Gravity Explained".
A particularly poor calculation is offered on this NASA webpage "SP-345 Evolution of the Solar System - 9. SPIN AND TIDES". It leaves out a lot (like Love numbers, precession, tidal acceleration, etc.) providing a quick and easy calculation along with a bit of unknown inaccuracies.
A short and quick explanation, less than you asked for, is provided by Space.com's article: "Moon Bumps: Earth's Gravity Creates Lunar Bulges" which explains:
"Earth's gravitational pull is so powerful that it creates a small bulge on the surface of the moon. ... The corresponding distorting effect on the moon, called the lunar body tide, is more difficult to see, because the moon is solid except for a molten core. But Earth's pull raises a small bulge about 20 inches (50 centimeters) from the surface on the near side of the moon and a matching bulge on the far side. [Rob's additional comment: they won't be equal.]
The same side of the moon always faces Earth, but the bulge does move around a few inches over time, wobbling and following Earth's pull like a magnet, as the moon shifts slightly during its orbit.
Scientists observed the lunar body tide using NASA's Lunar Reconnaissance Orbiter satellite, which is mapping the height of features on the moon's surface, and NASA's Gravity Recovery and Interior Laboratory satellite, which is mapping the moon's gravitational field. The satellites measured the height of 350,000 locations spread across areas of the moon closest to Earth and areas of the moon on the opposite side from Earth. The satellites passed over each location several times, so scientists could compare the height of each spot from one satellite pass to the next. By identifying which spots changed height, the researchers could track the lunar tide.".
A detailed calculation of the shape of the moon with plenty of math is covered in: "Evidence for lunar true polar wander and a past low‐eccentricity, synchronous lunar orbit". While covering a vast number of factors affecting the shape of the moon today body tides is amongst the points covered:
"Abstract
As first noted 200 years ago by Laplace, the Moon's rotational and tidal bulges are significantly larger than expected, given the Moon's present orbital and rotational state. This excess deformation has been ascribed to a fossil figure, frozen in when the Moon was closer to the Earth. However, the observed figure is only consistent with an eccentric and non‐synchronous orbit, contrary to our understanding of the Moon's formation and evolution. Here, we show that lunar mascons and impact basins have a significant contribution to the observed lunar figure. Removing their contribution reveals a misaligned fossil figure consistent with an early epoch of true polar wander (driven by the formation of the South Pole‐Aitken Basin) and an early low‐eccentricity, synchronous lunar orbit. This new self‐consistent model solves a long‐standing problem in planetary science and will inform future studies of the Moon's dynamical evolution and early dynamo.
1 Introduction
The long‐wavelength lunar figure is triaxial, due to the combination of rotational deformation (which creates a rotational bulge) and tidal deformation (which creates a tidal bulge along the Earth‐Moon axis). However, it has been known since Laplace [1878] that the observed deformation is significantly larger than the predicted deformation assuming hydrostatic equilibrium and the Moon's current orbital and rotational state (Figure S1). This difference has been ascribed to the presence of a fossil figure, frozen in when the Moon was closer to the Earth and the rotational and tidal potentials were larger [Sedgwick, 1898; Jeffreys, 1915; Lambeck and Pullan, 1980]. More recent work by Garrick‐Bethell et al. [2006] has shown that the observed deformation not only requires freezing the fossil figure while on an orbit with a smaller semimajor axis, but also with significant eccentricity. Matsuyama [2013] extended this work by taking into account the effect of finite lithospheric rigidity; however, the required initial orbital eccentricity remains large.
...".
That article includes many references which explain how the numerous equations were derived. As this illustration shows the article covers the development of the Moon's shape throughout history and attempts to calculate the minute bulging more exactly than the links referred to at the start of this answer.

"Figure 3. Our proposed model for the formation of the Moon's fossil figure. (a) The Moon coalesced from the debris of giant impact. (b) Under the action of tides, the Moon migrated outward and cooled from a magma ocean, perhaps experiencing periods of non‐synchronous rotation or high eccentricity. (c) As the moon cools, it forms an elastic lithosphere capable of supporting long‐term deformation, resulting in the fossil figure. (d–e) The South Pole‐Aitken (SPA) basin forming impact occurs, resulting in ~15° of reorientation, placing SPA closer to the south pole. (f) Subsequent large impacts and mare volcanism occur but do not significantly alter the lunar figure. (g) Ultimately, the Moon migrates to its current orbital configuration.".
Additional references:
"Lunar Laser Ranging Science" (18 Nov 2004), by Williams, Boggs, Turyshev, and Ratcliff.
"Lunar Geophysics, Geodesy, and Dynamics" (Homepage, direct .PDF link) (2003), by Williams and Dicky. That paper contains this useful text:
"The present lunar rotation model involves elastic solid body and fluid core. It will become more complicated in
the future. Numerically integrated physical librations (no fluid core) are available with DE403 at the JPL ephemeris
web site (http://ssd.jpl.nasa.gov/horizons.html).".
That website has moved, it's current address is: https://ssd.jpl.nasa.gov/?horizons - a bit of searching there turns up this URL: Lunar Constants and Models which says:
"Title:     Lunar Constants and Models Document
Author:     Ralph B. Roncoli
Date:   September 23, 2005
Reference:      JPL Technical Document D-32296
Summary: The primary purpose of this document is to provide a single source for the constants and models to be used in the trajectory and navigation design of missions whose objective is to orbit or land on the Moon. A secondary objective is to provide the mission analyst with some basic background information about the Moon, its orbit, and the previous missions that have explored the Moon. As a result, this document contains more information than the typical constants and models document. Some of the data are required for mission studies while other data are simply provided for "educational purposes". this document provides only brief descriptions of the constants and models. The user should consult the references if more detailed information is desired.
Format:     PDF
Link:   download here".
So that last list of references is "build your own rocket and land on the moon" accurate.


What can a lunar-based telescope do?

First, you would have no atmospheric distortions to interfere with the viewing.
Second, there would be no atmospheric absorption of ultraviolet or infrared light.
Third, there would be no cloud cover, so viewing can be almost continuous (the sun may interfere).

  However, the star should be bright and its distance should not be large.

I don't know what you reason for saying that, but it is unimportant since the Hubble telescope looks at all the stars.

Disadvantages:
The telescope will need to be self cleaning. There is a lot of dust kicked up when as meteoroid strikes.
An extended manned mission to the moon to set up a large array.


Fortunately it is not something that we have yet had to face.

Diverting the impact of a comet would not be easy, especially if it was a newly discovered comet on a nearly parabolic orbit. We would probably have less than 2 years to prepare.

Remember that the Earth is orbiting the sun at 30 km/s. So if we can delay the comet by three or four minutes, that turns a direct hit into a miss. But most of the plans to divert objects apply to asteroids whose orbit is well known and for which we could have tens of years in which to act. For a new comet there may not be time.


Determine the moons of Jupiter through a telescope

Use an orrery that will let you specify a specific epoch and vantage point. There are some quite fancy ones online, for example this Solar System Scope:

What you do in this particular tool is click on the calendar bar below and enter date and time of your observation (if you forgot that, there's a good chance your photograph has a time stamp, either of the file or in its EXIF data). You can adjust epoch later on, too. Then double-click on Jupiter, then its Orbit button, then click on link As seen from Earth below, then click Go back button and use mouse wheel to zoom out so all four Galilean moons display. Then compare with the photograph that you took. Note that it might take a bit of rotation for you image to align perfectly. I use the old neck twisting and turning technique, but photo editing tools or even rotatable screens will do just as well, if not better.

Note that this mentioned tool does offer settings to switch between orrery and realistic model and large and realistic sizes of displayed celestial bodies. But there are many such tools online or downloadable for both computers and smartphones. I think even this one is available as an app for smartphones for free, but last time I tried to install it, it didn't work for me. Maybe they fixed that. Do check because it's a lot easier if you can check orbital alignment of Galilean moons as seen from Earth directly on the field, as you're doing your observations.

Have clear skies and I wish you happy hunting for planets and moons of our Solar system and beyond!


Absolute/apparent magnitude and distance for HIP31978 inconsistent?

The absolute magnitude you quote appears to correspond to the Hipparcos parallax distance estimate, which is $\sim 1000 \mbox{ ly}$, not the $\sim 100 \mbox{ ly}$ distance you have used.


Can a black hole formation be stopped or interrupted

Your black hole is quite massive. It has a mass that is much greater than the mass of the Earth, (about 1000 times the mass of the Earth. And the rest of the Earth isn't moving very fast relative to the black hole.  So in a short amount of time everything falls into the black hole. This will be a quite energetic process with a lot of high intensity radiation being given off. But in short "everybody dies" and quite quickly too.


How are "parallel fields" implemented on the Hubble Space Telescope?

As the question Instrument aperture sizes on Hubble Telescope shows, the focal plane area is large enough to focus on several instruments at the same time (but with each capturing a different area).
If two objects of interest are separated by a certain angle (the instruments are fixed within the focal plane), the telescope can be rotated so that two instruments are exposed, one to each object.  If desired, the telescope can later be rotated to swap the instruments on each object, so that both objects get a wider spectral coverage, for example.

No optical components point the instruments (cameras/spectrometers) at different objects; rather, the whole telescope orients on one direction in space, and by rotating around the main mirror axis to a particular angle (matching the positions of the fixed instruments) can choose to expose multiple instruments to multiple objects in that Field-Of-View.  All instruments (except the WFC UV/IR sensors) are always exposed to something, but data collection/transmission stresses the systems, so 'parallel' viewing is used selectively.

Only the Wide Field Camera uses a pick-off mirror (steerable for UV or IR sensors). The other instruments are aligned with their patches of the focal plane and used COSTAR lenses for correction until Service Mission 4 removed COSTAR and replaced the actual instruments (which now have built in corrections). The Wide Field Camera was replaced several times and had its own corrective mirrors instead of a lens via COSTAR.


What happens if the Moon moves twice as fast?

No.  The orbital speed is related to the distance of the moon by the equation
$$v=\sqrt{\frac{GM}{r}}$$
Where G is a constant, M is the mass of the Earth (adjusted for the mass of the moon) and r is the average distance of the moon(wikipedia).
So there are two interpretations of your question: firstly what would happen if we gave the moon a big push that doubled its speed? Well this would cause the moon to change its orbit from a roughly circular to an open orbit.  The moon would sail off into space (slowing down as it is dragged back by the Earth, and end up orbiting the sun). The escape velocity is $\sqrt2$ times the orbital velocity, so if you are in orbit, and double the speed, you are sure to escape.
If you gave it a more gentle push it would enter an elliptical orbit.  The speed would then be non-constant. If you increased its speed by 10% it would move 10% faster at perigee, but slower at apogee and on average its speed would actually be less than before.
The only way to increase the speed is to decrease the distance to the moon.  If you reduced the distance by a factor of 4 (to about 95000km).  This would cause the moon to double its speed.


What would be the maximum theoretically possible angular resolution?

The answer that you are looking for rests on two main pillars. Are you limited in your ability to resolve objects because of the atmosphere like most ground-based telescopes, or because of the fundamentals of the wave nature of light (as per space telescopes)? I am assuming the latter in your question.

One of the biggest rule-of-thumb equations when it comes to astronomy is the equation that you are looking for in your answer. Astronomers define angular resolution with regard to the Rayleigh criterion: two point sources are regarded as just resolved when the principal diffraction maximum of one image coincides with the first minimum of the other (Wikipedia). See Airy Disk functions and the Wikipedia article on Angular Resolution for a better idea as to what this means.

The accurate, but still approximate, equation for the resolution of an object $\theta$ (radians), for a given wavelength of light $\lambda$, over some telescope's diameter $D$, is given below so long as the wavelength of light and the diameter is measured as the same length unit (i.e meters, feet, angstroms).

$$ \theta = \frac{1.22 \lambda}{D} $$

Obviously, the atmosphere is a pain, so, adaptive optics in most famous telescopes try and correct to this limit. To exceed this limit, major breakthroughs would need to happen in the wave nature of photons themselves. If your "technology far beyond" includes this, then, I don't think your question can be answered as you would hope aside from the standard "who knows". Currently, only space-based telescopes operate at this limit.

For the most part, this is really all you need. It does not matter what object you are observing, this still always applies.

Edited: Included unit specifications.


The Big Rip happens if the equation of state for the dark energy has $p/\rho = w<-1$, and all empirical data give us $w\approx -1$. A Big Crunch requires a pretty high value of $w$ (it must go above -1/3 to just stop the acceleration), but of course if dark energy is changing over time it might do that.

So tentatively ruling out Big Crunches is the reasonable thing to do given observational evidence and the assumption that the future will not be wildly different. If one thinks we should be more open to dark energy changing (which is going to be rather unconstrained by observation for the foreseeable future) then we should also be more open to $w$ moving across the Big Rip boundary.


During what time of the year can Centaurus be seen from Tokyo?

You can use an application that shows the stars at any time and location. Stellarium is free, has nice features and isn't too complicated. You can select a star/constellation and flip through the time of day and through the days to see how its elevation changes over time.

In general, everyone with the same latitude sees the same sky at a certain date (assuming an equal elevation, which counts if you're standing on Mt. Fuji). So those dates for New Sealand are irrelevant for you.

Using the StarmapPro application, I made the following chart for your location (Tokyo) showing the elevation of Centaur at midnight for each month.

As you can see, it's visible from halfway march until the first half of may, with its maximum around the first week of april. It comes barely above the horizon, so you'll need a clear view to the horizon and try to avoid light pollution (good luck with that in Tokyo ;-)


Spotting the booster rocket of the Chinese space station?

The name of the module is Tianhe (天和), and the name of the rocket is Long March 5B, but it got a special designation of 2021-035B. It is not really official, but I think it is good enough to get some data. I found this website for current position (live on globe). Your userpage says you live in Europe, so the rocket will fly over Greece at 00:30 CEST (22:30 UTC) tomorrow (in 1.5 hour if you have time to be awake for that long). I also calculated the maximum angular speed (on May 6) of around 1.85 °/s (on May 7, it is 2.7 °/s).
There are already some images; I found only that one. The booster's visual magnitude is around –1, so pretty bright to spot.
So, if you have time, try to look to the South and spot the booster at 22:30 UTC today. Maybe you are going to have luck tonight : )

There are some equations I derived with basic trigonometry about the rocket's visibility:
Distance $D$ at which the rocket can be seen right on horizon:
$$D = R \ \arccos\left(\frac{R}{R+h}\right)$$
Where the calculator calculates radians, $R$ is Earth's radius and $h$ is the altitude of the booster (you can get that on previously given website; at the moment it is 170 km).
The angular height of the booster which is $D$ kilometers away from us (on Earth) is given by:
$$\alpha = \frac{\pi}{2}-\arcsin \, \left(\frac{(R+h)\sin{\frac{D}{R}}}{\sqrt{2R^2+2Rh+h^2-2R(R+h)\cos{\frac{D}{R}}}}\right)$$
Again, the calculator must be set in radians.

Update: The rocket fell on May 9 at 02:14 UTC into the Indian Ocean.


Is a supernova's core temperature absolute zero just before collapse?

No, absolutely not. The core of a core-collapse supernova is one of the hottest places in the present-day universe. The temperature as the star runs out of nuclear fuel in its core is around 6-10 billion Kelvin. As it collapses, the core gets even hotter, perhaps as high as 100 billion Kelvin for a few seconds, before neutrino cooling starts to become effective.

We know that the temperatures are getting as high as this because at least two neutrino instruments on Earth detected roughly "thermal neutrinos" with energies of $\sim 10$ MeV coinciding with SN1987A in the Large Magellanic Cloud (see for instance Mirizzi et al. 2015).


Why are the ALMA receivers' ADCs only 3-bits?

It is wasteful to sample with many bits because the signal to noise ratio at the ADC of a radio telescope is typically << 1, so using many bits would just be resolving noise. (An exception to this is when there is strong radio-frequency interference that needs to be resolved, but this is not a big problem for ALMA due to its location and observing frequencies).

High dynamic range measurements arise after averaging together many samples (or correlations of samples), which boosts the SNR to a meaningful level.

Using very few bits at the ADC does introduce quantization noise that reduces the efficiency of the instrument, but 3 bits is enough to achieve 96% efficiency [1].

[1] Convenient formulas for quantization efficiency


Is there a retrograde star that passes closer to the galactic center than our Sun?

There's a whole stream of them: https://iopscience.iop.org/article/10.3847/1538-4357/aadba3
Named Phlegethon, this is a stellar stream that covers about 75⁰ of the sky, and is probably the remnants of a disrupted globular cluster.
It orbits the galaxy retrograde on orbits that take it between 4.9 and 19.8 kpc from the galactic centre. The orbits are highly non-Keplarian. The image shows the orbits with the sun marked as a yellow dot.

Other streams exist and may have similar shaped orbits.


How do I adjust the sunrise equation to account for elevation?

Higher elevations by themselves do not change the time of sunrise.

The horizon at sea level is 90 degrees from the zenith. The horizon at Denver is still an angle of 90 deg from the zenith if you assume that the ground is "flat". In other words, even though you are 1 mile high in Denver, you cannot see the horizon at 0 mile elevation.

If you are on a hill or mountain and looking down so that the distant horizon is more than 90 degrees from the zenith, then sunrise would occur early. If you are in a valley and the sun rises behind a hill or mountain, the sun rises later.

So you need to know the elevation of the observer and horizon (or obstacles). From that you can calculate the altitude of the sky in that direction. From that you can calculate the time of rising.

(edit Jan 24) For points that are close to each other, simple trigonometry can be used.
$$\tan(\theta)=y/x$$ where y is the difference in elevation between points 1 and 2, and x is the distance. Sunrise occurs when the altitude of the Sun is $\theta-sun's\;radius-refraction$.

For more distant points with latitude (lat) and longitude (long), solving for the sides and angles of a triangles gives an approximate solution. (I am sure there are more precise formulas that account for the Earth's shape.) From the figure below,
$$a=radius\;of\;Earth+Elevation1$$
$$b=radius\;of\;Earth+Elevation2$$
$$\cos(\gamma)=\sin(lat1)\sin(lat2)+\cos(lat1)\cos(lat2)\cos(long1-long2)$$
$$c=\sqrt{a^2+b^2-2ab\cos(\gamma)}$$
$$\cos(\alpha)=\frac{b^2+c^2-a^2}{2bc}$$
$$\theta=\alpha-90$$
and sunrise occurs when the altitude of the sun is $\theta-sun's\;radius-refraction$. Naturally, you need to check all locations along the line of sight between the observer at point 1 and the obstruction at point 2 to find the highest obstruction.


There is a "South Star" called Sigma Octanis located in the constellation Octans, but it is so dim that virtually nobody calls it the south star.

Just another side note: Polaris doesn't always point North. The reason is that our Earth has a precession effect which make the North pole circulate around the sky. This picture illustrates my point.

EDIT: Just to clarify, the picture is showing the circulation of the south celestial pole around the south ecliptic pole.

EDIT2 : References: http://en.wikipedia.org/wiki/Pole_star#mediaviewer/File:Precession_S.gif
http://en.wikipedia.org/wiki/Pole_star


Binary pairings that haven't been discovered yet?

As far as I am aware, there are no known examples of black hole+neutron star (BHNS) binaries (e.g. this paper from 2018 notes that "No BHNS binaries (BHNSs) have been discovered so far by radio surveys"). BHNS mergers should be observable with gravitational wave astronomy, so this situation may not last for too much longer. There is a report of a possible BHNS merger detected by LIGO and Virgo (event S190426c). This event has yet to be confirmed, follow-up searches by Hosseinzadeh et al. (2019) and Goldstein et al. (2019) for transients have failed to find a counterpart to the event.

Binaries containing an sdB star and a neutron star have also not been found, though theoretical models predict they should exist in our galaxy (Wu et al., 2019). Sufficiently close sdB+NS binaries should be strong gravitational wave sources, so LISA may discover them.

Depending on how you want to define various "types" of star, you can probably come up with more examples. For example, there is the concept of singleton stars which have never been members of binary systems or had close encounters with other stars, which by definition have not been found in binary pairings.

There are also some weird categories of objects with few known members. For example there are the LP 40-365 stars, which are runaway stars with atmospheres dominated by oxygen and neon, likely the partially-burnt remnants of peculiar type Iax supernovae. None of the four known examples of this class are binary systems, plus the circumstances that lead to their creation makes it seem fairly unlikely that we'll find one in a binary.


Description of Henyey tracks on wikipedia incorrect?

The wording provides a contrast with the Hayashi track phase that immediately precedes it, where the luminosity decreases by orders of magnitude with little change in effective temperature.

You are correct though, the Henyey track for low-mass stars is not horizontal in the HR diagram, however it is a critical point that the luminosity does not change very much (less than a factor of 3) as the effective temperature increases, since it implies that the star slowly shrinks, and by the virial theorem, the core temperature gradually rises and is ultimately able to sustain nuclear fusion.


When will solar eclipses become lunar transits?

When looking at definitions of [transit] and [eclipse] or more general [occultations] it is noteworthy that the degree of covering is not defined, so technically speaking each eclipse is also a transit and an annular eclipse even more so.

Details for next eclipses can be found at [this NASA page]. The next annular eclipse, err, transit happens next Sunday 2017-02-26 and is visible in South America, south Atlantic and Afrika.

As you were referring to the [increasing distance between earth and moon] (+3.8 cm /yr), it then depends on an exact definition on when an eclipse is not regarded as an eclipse anymore. Assume for simplicity that this is the case, when the diameter of the moon has halved (one quarter the area), this means that the distance had to be doubled.
This would take [380 Gm] / 3.8 cm/yr = 10 Tyr (using SI prefixes).

Note that one might need to take other effects into account, too, for example the [increasing distance of earth and sun] (~15 cm / a) and that the [sun is becoming a red giant] in a couple of Gyr.

Sorry, would have loved to post more links, but are not allowed to currently.


Where's the Prime Meridian of celestial coordinates

The equatorial plane and the ecliptic plane cross each other, like shown in this illustration:

The "Vernal Equinox" is simply the direction of the line where they cross.

You can not obtain the Earth latitude from the ecliptic latitude alone, as the planes are not parallel. You must know the ecliptic longitude as well.

From equatorial to Earth latitude it is much simpler, the latitude is equal to the declination.


Could I create a staircase to the moon and walk to it?

A space elevator has been proposed. With currently available technology it would be very expensive to build one, but not necessarily impossible.

Assuming the staircase or elevator being fixed at some point on Earth near the equator, the Earth's gravitational pull is compensated by centrifugal force at a height of about 36,000 km, in the geostationary orbit.

That's the height you would have to climb to, before you could slide "down" to the Moon along an even more expensive construction stepwise adjusting your velocity to that of the Moon. The total trip would be at least about 400,000 km, that's ten times around the Earth.

Your weight would decrease, the higher you climb from ground to the 36,000 km height. Assuming you make 3 km of height each day, it would take you about 34 years to reach geostationary orbit.

Then sliding down with 10 m/s would take at least one more year to reach the Moon. Hence, it's possible with a one-way trip of about 35 years, assumed the technical infrastructure is provided.

But it looks easier to take an appropriate vehicle.


Does the axial tilt of Mercury and Venus vary?

There is almost no data available on the internet as such.
Though, the theory that they (Mercury and Venus) are stabilised by the Sun's Tidal Force seems very likely as they are much closer to the sun than any other planet. The Sun very easily dwarfs these planets in both size and gravitational and tidal attraction.
I found this Wikipedia page that I think might be useful : https://en.wikipedia.org/wiki/Axial_tilt
It says that it 'might' be that they are stabilised due to the Sun.
But this article I found changed my mind : http://hosting.astro.cornell.edu/~jlm/publications/2005I09374.pdf
It clearly says that Mercury's obliquity is not consistent with time. It also contains other info that makes it must-read if you are interested in this topic.
And although I couldn't find anything convincing on the subject of the obliquity of Venus,  I found this one article that seemed quite interesting :
http://articles.adsabs.harvard.edu/cgi-bin/t2png?bg=%23FFFFFF&/seri/AJ.../0075/600/0000273.000&db_key=AST&bits=4&res=100&filetype=.gif
Venus' obliquity is 177.4 degrees, to be exact, according to this : http://solarviews.com/cap/misc/obliquity.htm
It also displays the approximated obliquity of other planets in the Solar System.


Do triple or more times reflecting telescopes exist?

Cool question!
If you allow for flat mirrors, I've seen five myself. Answer(s) to How did I flip some mirrors around in the dark at 3 AM and change the focal length of a 24 inch Boller and Chivens? show four mirrors and there was one more on the floor below to make the beam horizontal. If you count the aluminized reflective grating as a mirror it was six.
 ALL IMAGES: click for larger size
For more info, see Nasmyth and coudé focus

For more examples of three mirrors built into the telescope itself (as opposed to an eyepiece diagonal for viewing, see answer(s) to

How does this telescope work, and what is it for?
What telescope is this, and how does it work?

For more little mirrors that send light to different instruments, see answers to

What is a pupil slicer, and how does it work with anamorphic optics in VLT's ESPRESSO Echelle spectrograph?
Where exactly is the modification that first corrected the spherical aberration in Hubble's primary mirror?

But three or more curved mirrors is interesting
One example is the Hubble Space Telescope using the Wide Field and Planetary Camera 2 or WFPC2.
Basically it is another whole telescope inside the Hubble, re-imaging the primary focal plane with a second purely reflective optical Cassegrain system to avoid chromatic aberration. (There would be no way to correct a fused silica lens system from 115 to 1050 nm.)
From what-when-how Hubble: The Big Telescope

Figure 3.36 WFPC2 optics. Light enters the optical train from the main telescope at left.


What exactly is a fortney grid transmission spectra modeling

Planets with atmospheres are not discs with hard edges, when viewed in transit across the star. The atmosphere allows light to pass through the limb of the planetary disc.
The radius within which the planet is effectively opaque to light from the star will depend on the density and composition of the atmosphere as a function of height and also the wavelength of light that it is illuminated by. This is because the opacity of the atmosphere will be much greater at wavelengths where some molecule or atom that is present has an absorption feature.
Therefore if you observe a transit at different wavelengths, you will see different transit depths, because the planet has a different effective area at different wavelengths.
For example, if the atmosphere contains lots of atomic sodium, then because sodium strongly absorbs light at 589nm, the planet is opaque to a greater radius at 589nm, and the transit depth will be bigger at 589nm than at wavelengths either side of this.
Ps. (Jonathan) Fortney is the name of the scientist who developed the grid of models.


Identifying cluster of exoplanets between Deneb and Vega

That's the field the Kepler telescope was observing during its primary mission (before the second reaction wheel failure). The shape is due to the detector array.
I.e. this is not a real pattern in the actual distribution of exoplanets, it's a consequence of how the observations were done.


Is it practical to build a giant telescope in Moon? (Considering the fact that the atmosphere of moon is very rare)

It's very unlikely that large optical telescopes will ever be built on the Moon, because the Moon is almost the worst possible place to build them. (The surfaces any of the  planets other than Earth are worse.)  It has no particular advantages over orbit and costs a lot more to build there.

The Moon looked like a good location when observatory technology was film-based, because observatories required people and people needed a base to live in and work best under gravity.  But we now have the technology to shade an orbiting telescope from the Sun, point it exquisitely accurately, and take endless photos without changing the film -- all in orbit.  Orbit is easier to get to than the Lunar surface -- even the Webb (should it ever launch) which will be parked a million miles from Earth, is easier to get to than the surface of the moon.

If we had the space capabilities to build a giant reflector on the Moon, we could build an even bigger and better one in orbit for the same cost.

You spoke of FAST. FAST is a radio telescope. It's likewise much cheaper to build radio telescopes in orbit than on the Lunar surface. The one possible advantage of the Moon is that the far side of the Moon might have advantages of being especially radio quiet.

But even there, it seems likely that any space radio telescope would be a large array of radio telescopes joined to form a giant synthetic aperture, and that would almost certainly want to be much larger than earth-based instruments -- for which the Moon is already too small.


If the fifth gas giant in the early Solar System was completely ejected, where would it be now?

It could be anywhere in the Galaxy or even (unlikely) have left the Galaxy.
The kinematics of freely orbiting objects in the Galaxy are heated - that is, the velocity dispersion of objects increases with time.
For objects of age of 4 billion years we might expect a dispersion of about 10 -15 km/s in each velocity coordinate. That's about 10-15 pc/Myr. So in 4000 Myr, an object could have scattered to most parts of the Galaxy, or at least to a reasonably wide and thick annulus around the Galactic centre.
It is for similar reasons that, even though the Sun was probably born in a cluster of $\sim 10^4$ stars, none of those siblings have been firmly identified.


Are we moving away or towards each other?

At the present time this doesn't hold for the observable universe, since the observable universe is only a tiny fraction of the surface of the 3-sphere (assumed it's a sphere at all). The observable universe is flat within a 0.4% error.

Based on this, we are moving away from each other. And the universe is expanding. So even if it would be curved, we would move away from each other. Only in a future Big Crunch scenario, we would eventually move towards each other.

Finite flat cyclic spaces can be constructed, but there is no significant hint, that the universe is of such a shape, or any other non-trivial topology.


What is the origin of the OBAFGKM classification system?

The Wikipedia article on stellar classification has the full list of letters:

Class
Surface Temperature
Notes

O
> 33,000 K

B
10,000 - 33,000 K

A
7500 - 10,000 K

F
6000 - 7500 K

G
5200 - 6000 K

K
3700 - 5200 K

M
2000 - 3700 K

L
1300 - 2000 K
L T and Y are those new letters

T
700 - 1300 K

Y
< 700 K

History
This Cornell doc explains much of the history behind the system:

In the 1890s, many scientists were interested in developing a classification scheme for the stars. Edward C. Pickering at Harvard University, together with his assistant Williamina P. Fleming, assigned stars a letter according to how much Hydrogen could be observed in their spectra: stars labeled A had the most Hydrogen, B the next most, and so on through the alphabet. There were 22 types in all: this scheme was rather cumbersome, and it wasn't clear what its physical significance was.

In 1901, another of Pickering's assistants, Annie Jump Cannon, also begun to work on the classification sequence. Her meticulous observations led her to simplify the 22-type scheme into a simple sequence of temperature: OBAFGKM. Not only was the scheme much simpler than the previous, but it related the amount of hydrogen observed to a physical property of the stars for the first time.


How can one determine the angle between the line of nodes and the major axis of the orbit of Mars?

The name for this angle is the Argument of Periapsis.
The Argument of Periapsis (or in this case, Perihelion, since Mars orbits the Sun) is the angle, in the plane of the orbit, measured in the direction of travel around the Orbit, from the Ascending Node, through the body being orbited, to the periapsis of the orbiting object.
Princeton's Planetary Systems page lists the Argument of Perihelion of Mars as $286.5°$ in 2006.
This would make the interior angle between the major axis of Mars and the line of nodes to be $360°-286.5° = 73.5°$, as of 2006.


Is there a theoretical limit on telescope's resolution?

The absolute limit of a telescope resolution is given by diffraction.
No matter how perfectly built and aligned is a telescope, you cannot resolve angles smaller than
$$\theta \propto \frac{\lambda}{D}$$
where $\lambda$ is the wavelength of interest and $D$ is the diameter of the telescope. This is why a number of millimeters and radio telescope (and also some -antennas) are huge.

The first figure here shows the basic principle. Imagine that you can divide what you observe in tiny squares and consider each one as a point like source. Each one will generate a diffraction pattern when passing into the telescope and if two points are too near, you cannot distinguish between them.

Wisely you put your telescope in space: turbulence in the atmosphere degrade the signal, and the best/biggest telescopes have hard times to go below $0.5 arc seconds$

Interferometry comes to the rescue increasing $D$ from the telescope size to the distance of two (or more) telescopes (called baseline). Interferometry has been used in radio astronomy since decades.
From what I hear optical interferometry is much much more complicated and as far as I know the only large scale attempt is the VLTI project.

So you could imagine to have a constellation of relatively small telescopes spread over hundred thousands or millions of kilometers. But this have the huge problem that you have to know the position and timing of every one of them with an impressive precision (my guess is that the precision in position is of the order of $\lambda$).

The other problem is light collection. If you want to see something very faint you have two options: 1)you observe for a lot of time or 2) you build a bigger telescope (6 to 40 meters in diameter).
And here also the largest baseline interferometers cannot do much, as the amount of light that they collect is just the sum of the light collected by the single telescopes.

To conclude: to observe Pluto or Ceres with high enough accuracy you would need a large number of large space telescopes very far away one from the other with perfect telemetry. It's far easier and cheaper to got there to take pictures.


Are there galaxies with 2 or more super massive black holes orbiting each other?

Yes, there are galaxies with two supermassive black holes in the center, see for instance 4C +37.11

Most likely such galaxies are formed by collision and merger of two galaxies, and their cores have not yet merged. Source


What do these statements about interstellar magnetic fields really mean?

What does the phrase "...magnetic fields tap into every free energy source they can hold onto and grow..." refer to in the astronomical context of interstellar space?

Interstellar space is filled with charged particles/dust which have their own magnetic fields (e.g., their own dipole moment or any field induced by the charge moving). When you apply a larger, more global magnetic field to these small charges, very often an alignment occurs (e.g., the particles dipole will align with the larger field, or the motion will suddenly change to move along the field line). This causes these smaller fields to add to the larger field, making it even larger. This is a foundational assumption for academics which study light polarity in terms of dust grain alignment via Radiative Torque (RAT) - namely that the background global magnetic field of interstellar space aligns charged grains with that field, thus further contributing to the overall field.

What does the phrase "...only magnetism and gravity can “reach out to you” across vast distances. Electricity, by contrast, is local and short-lived, since the positive and negative charge in any region will neutralize overall. But you can’t cancel out magnetic fields; they tend to add up and survive." refer to in the context of interstellar space? Is it the lack of free magnetic monopoles, or the dearth of polarizable ferromagnetic material?

If you take a large, galactic magnetic field and it passes through an interstellar dust cloud that is full of charged particles, those charged particles will align with the field and overall add to it, making the galactic field stronger in that area. The only thing that could weaken the magnetic field is something which could cancel it.
Electric fields can easily be cancelled because an electric field being radiated out by a proton can easily be cancelled by the "opposite" field being radiated out by a bunch of nearby electrons (hence the concept of the Debye length). Magnetic fields are harder to screen and, as you say, because the things that could screen them just don't exist in space.
I would also mention that if monopoles even existed, it would require that you could have two separate flavors of monopoles such that they could cancel each other out much like the positive and negative charges of electrons. A logical pick for flavors would be a "north" monopole and a "south" monopole, but I could imagine one constructing a more complicated, yet valid system. Since we've never seen a monopole, any monopole-based models and purely hypothetical at this point.


What is the physics of a gas mass subject to gravity in space?

The formula for the change of pressure with altitude is the same-you need hydrostatic equilibrium.  This means that at constant temperature and over small height differences (so gravity can be considered constant) the pressure falls exponentially with the scale height.  You have $\frac {dp}{dz}=-{\rho g}$  You can compute the local $g$ by taking everything closer to the center as a point mass and ignoring all the mass outside the current radius.  You can consider the compression adiabatic, but need to use the appropriate value of $\gamma$ in $PV^\gamma=k$ for the local temperature.  Calculating the radiation is hard unless there is enough dust around, in which case the grains can be black bodies.  If you assume rotational symmetry, you have a 2D problem, which you can model on a grid and step through.


How did I flip some mirrors around in the dark at 3 AM and change the focal length of a 24 inch Boller and Chivens?

What you did was to switch in a Coudé mirror, which sends the light down the equatorial mount as you described, to other instruments (besides the original focal point of the secondary mirror)
There are several designs to a Coudé path. The simplest uses a flat mirror and then places a second camera at the re-positioned focal plane, or re-images that focal plane using relay lenses.

Other designs can use a convex mirror as the first Coudé mirror to extend the focal length and send the light down into a "Coudé room" where other instruments are located.


As I understand it, you should not be using pip with Anaconda or you will get into a mess. Unfortunately, that's not the main issue here.
The workbook you linked has a banner at the top that mentions an initialization workbook, which is on this page. On that page it states that "Users unfamiliar with Anaconda/Astroconda should see the documentation."
I am unfamiliar with Astroconda, so I went to the documentation page which lists the following requirements:

System Requirements

64-bit Intel/AMD processor (x86_64)
64-bit Linux (glibc ≥ 2.12) or Mac OS X (≥ 10.7)
BASH or ZSH as your default shell environment (T/CSH is NOT supported)

I see from your command line that you are using this on Windows. That is not listed as a supported operating system: the software is Linux/Mac only.


Can Newton's gravity equation explain why black holes are so strong?

No you can't and the behaviour of bodies with mass and of light is completely different near a compact, massive object if you use Newtonian physics rather than General Relativity.
In no particular order; features that GR predicts (and which in some cases have now been observationally confirmed) but which Newtonian physics cannot:

An event horizon. In Newtonian physics there is a misleading numerical coincidence that the escape velocity reaches the speed of light at the Schwarzschild radius. But in Newtonian physics you could still escape by applying a constant thrust. GR predicts that no escape is possible in any circumstances.

Further; this numerical coincidence only applies to light travelling radially. In Newtonian physics the escape "speed" is independent of which direction you fire a body, but in GR light cannot escape from (just above) the Schwarzschild unless it if fired radially outwards. For other directions, the radius at which light can escape is larger.

GR predicts an innermost stable circular orbit. A stable circular orbit is possible at any radius in Newtonian physics.

In GR a particle with some angular momentum and lots of kinetic energy will end up falling into the black hole. In Newtonian physics it will scatter to infinity.

Newtonian physics predicts no precession of a two-body elliptical orbit. GR predicts orbital precession.

Newtonian physics predicts that light travelling close to a massive body has a trajectory that is curved by about half the amount predicted by GR. Even stranger effects are predicted close to the black hole including that light can orbit at 1.5 times the Schwarzschild radius.

The GR approach to gravity is fundamentally and philosophically different to Newtonian gravity. For Newton, gravity is a universal force. In GR, gravity is not a force at all. Freefalling bodies are said to be "inertial". They accelerate, not because a force acts upon them, but because spacetime is curved by the presence of mass (and energy).
In most cases, where Newtonian gravitational fields are weak, the consequences of this difference are small (but measureable - e.g. the orbit precession of Mercury or gravitational time dilation in GPS clocks), but near large, compact masses, like black holes and neutron stars, the differences become stark and unavoidable.


Have any co-orbital exoplanet pairs been discovered (and not subsequently retracted)?

One example is Kepler-91b; early observations included strange light curve behavior. While follow-up observations seem to confirm that it is in fact a planet, rather than some other object, Placek et al. (2015) suggested that there is a Trojan "planet" ($M\sim.025\pm.019M_J$) at one of Kepler-91b's Lagrange points - either L4 or L5.

The authors believe that this explanation explains the light curve data much better than the single-planet model, as the unexplained dimming could be the result of a "secondary transit" by the Trojan planet, rather than a "primary transit". However, predictions of the daytime temperature of the secondary planet fall squarely about 5,000 K, which is close to the temperature of the surface of the Sun. Albedo calculations are also higher than expected (0.49).

Therefore, the authors state that the light curve may therefore be a false positive, and Kepler-91b does not exist - and nor does the hypothetical Trojan planet.

Trojan co-orbital planets have been the main focus of theoretical and experimental research so far; see Giuppone (2012) for more information.


Do Satelites travel in train like formation?

Yes they can. It depends on how the satellites are deployed from the launch vehicle.


How bad could we reasonably expect a solar flare to impact earth, and what can be done to mitigate the impact?

Space Weather is the field within heliophysics that tries to understand this Sun-Earth relationship. When a Solar Flare occurs multiple things may happen and how it interacts with the Earth and our technology depend on different factors.

The first effect, the X-ray radiation produced by an Solar Flare affects the ionosphere and therefore the radio communications. This effect is almost immediate and there's no way to attenuate or prevent it. We can forecast the likelihood of a flare to happen on the day but still we don't have the knowledge enough to exactly predict when it will happen. The radiation from the flare it takes around 8 minutes to get to Earth, and at the time we see it the radio communications have already been affected. This effects mostly to VLF bands.

After a solar flare high energetic particles and coronal mass ejection could be produced, these two events have different effects on Earth and whether interacts with Earth or not depends on where it happens on the Sun and the properties of the solar wind at that time. High energetic particles (also known as Solar Energetic Particle events - SEPs) could arrive around 30 minutes after the flare happened, and Coronal Mass Ejections (CMEs) between 2 or 3 days. These times depends on the properties of the events, not all are the same.

CMEs could be somehow predicted, and are these the ones that could affect, between others, the power grid. It won't affect your computer. The reason of why it affects the power grids is due to a induced electric current happened by the interaction of the CME's magnetic field produces in the Earth's magnetic field. The electric current finds it easier to travel through the power grid than through the Earth itself (the cables are better conductors). In simple words, when that happens the electric grid "thinks" that there is an increase of demand and tries to compensate it generating more electricity and that can burn them - as it happened in Canada in 1989.

To reduce it effects there are a lot of engineering work around these systems and I'm not an expert in such topic to be able to explain them properly. However, a warning to the power grids of a couple of days in advance helps. The difficult bit is how to be exact in the forecasting. CMEs sometime are deflected by the solar wind and just a flank of it "touches" the Earth, other times small ones get accelerated and have a larger impact than expected. So, to be able to reduce the effects we need to have a better understanding of their origin and surroundings, understand their physics and be able to model what they are doing. There are few models out there (e.g., CCMC) but at the moment we need to keep researching to improve these further.

Answering to "how bad" a solar storm (not just a flare) can affect Earth is hard to quantified, in UK it has been classified as one of the highest priority natural hazards in the National Risk Register, and similarly in other countries. An event like Carrington's (the one you linked) could have a lot of impact in our current technological civilization and it's been estimated in many years and trillions of US dollars to restore the damage (see Royal Academy of Engineer report on Space weather).


Looking for help in understanding how black holes can move

I saw Dr. Kip Thorne in a documentary on black holes state that there is no longer any matter in a black hole once it forms. He said something like, the matter was there, but it has been crushed out of existence and basically has been transformed into the energy that produces such a great curvature of spacetime.

Isolated black holes are indeed vacuum solutions to general relativity. Thus in particular, the mass and energy density is identically zero everywhere in spacetime. Because of some technical issues, this does not imply that black holes have no energy or mass; rather, it means that we can't directly think of them as an integral of mass or energy density. See also this question.

  I also saw Dr. Andrew Hamilton in another documentary state that within the event horizon of a black hole (again paraphrasing), space is falling toward the singularity so fast that it effectively drags everything with it (including light).

For an isolated black hole in a particular frame field, this is a valid picture. For example, the Schwarzschild spacetime in Gullstrand–Painlevé coordinates can be interpreted as Euclidean space falling into the black hole. However, the spacetime is itself is stationary (even static) in the geometrical sense—there is a timelike field representing a direction in which spacetime geometry is left unchanged (this corresponds to Schwarzschild time, in fact). Similarly, rotating black hole spacetimes are stationary.

But generally speaking, you really can't think of black holes as some sort of suck-holes for space. It's just an analogy that applies to simple situations (isolated black holes, with nothing else in spacetime) and then only in a particular frame or coordinates. That's not to say it's useless--e.g., acoustic horizons in fluids are interesting analogies to event horizon--but don't take it too literally. See also this question.

  Based on 1 (no matter remains in the black hole) and 2 (space is being pulled towards the gravitational singularity from all directions), I try to picture the black hole moving, and it seems to me that if, say, the black hole moves in one direction, I can see how its event horizon is also moving, bringing that bit of space into it.

So what's the problem?

  But how can the other end of the black hole let space "escape" in order to complete the motion?

For realistic black holes, there is no "other end". If you really want to tie everything together, then interpret Dr. Thorne's comment about stuff falling into the singularity as crushed out of existence as also applying to space. It doesn't come out of any "other end". It stops existing.

But really, that's past the point where the fluid analogy make sense anyway. I think Dr. Hamilton might note that the full maximally-extended Schwarzschild spacetime is a black hole, so one may be able to think as it "coming out" of the corresponding white hole, but he would also tell you that this white hole is a mathematical artifact that doesn't have anything to do with actual astrophysical black holes.

  But I want to understand why it is that space itself can cross the event horizon when other things can't.

Of course they can. In the infalling-fluid analogy, they're carried along by the space that's being sucked in. Perhaps you're thinking of the fact that it takes an infinite Schwarzschild time for an object to reach the event horizon. But that doesn't imply that things can't cross the horizon; rather, it's just a symptom of Schwarzschild coordinate chart not covering the horizon.

Intuitively, think of a coordinate chart as a "grid" drawn on a patch of spacetime. That patch may be the entire spacetime, or it might be just a piece of it. In the case of the usual Schwarzschild coordinates, it's just a piece... one that simply doesn't cover the horizon.

  Regarding the last part, I meant space crossing the event horizon the other way - out of the sphere demarcating the event horizon - whereas material objects go one way - in but not out, spacetime itself does not seem so restricted, at least not in the sense of "pinning" a black hole to one part of itself.

OK, your edited question provides some more context. But the answer is completely the same: what Dr. Hamilton is talking about in the video is the maximal analytic extensions of the rotating black hole solution, which does contain a passage into a completely different region (sometimes called different 'universe'). This is very much analogous to the maximally extended Schwarzschild spacetime, which contains a wormhole that comes back out of a white hole in another spacetime region, except that the extension of the rotating Kerr black hole spacetime contains an infinite chain of such connected regions.

I haven't watched the entire video, but it's clear that's what they're talking about when they say "in a ship propelled by pure mathematics", because the causal structure of such spacetimes is well-known. Once again, I direct your attention to Dr. Hamilton's own page, linked above, that explains that this dual waterfall picture is an artifact of mathematical idealization and not something that actually happens in reality.

However, even if you take the maximal analytic extension of such black hole spacetimes overly seriously, it's important to emphasize that you don't come back out into the same region, but rather a different one connected by a wormhole inside the horizon. If you're up to it, I also recommend Dr. Hamilton's conformal diagrams of said black hole spacetimes, which make it quite clear that's what going on is a whole "chain" of black holes and white holes.


The merged broadcast ephemeris is in the YYn directory where (YY is the 2 digit year, n means navigation e.g. 19n) and you are looking for the brdcDDD0.YYn.Z where DDD is the day number (of the year) and YY is the 2 digit year as before. These are compressed RINEX files; gunzip etc will be able to uncompress them. The RINEX format is pretty well documented and I've written Python code to read them.

The other files in those directories are the received GPS signals from the various GNSS monitoring sites. You may be able to make progress by using the map on the Archive Explorer to find a site near you, noting its code and then downloading its data file. For me the nearest is VNDP (Vandenberg AFB) and so for 2019-Jan-01 I would download vndp0010.19n.Z from ftp://cddis.gsfc.nasa.gov/gnss/data/daily/2019/001/19n/. Otherwise I think you would be looking at having to do your own visibility code to work out what satellites were visible from your location at the specific time (since the satellites have a ~12 hour orbital period and come and go)

You may want to look at RTKlib (binaries available there at github also) which has a lot of tools for processing these sorts of data and has rtkget which can automate the download of the various types of files from CDDIS etc.


The mass will pull into the shape of the sphere if the body is massive enough. The body will then basically be in hydrostatic equilibrium. A planet like Naboo couldn't exist b/c the denser materials would move to the centre, and the lighter elements would then form an atmosphere around the solid core.

Plasmas require really high temps and a terestrial plantet couldn't compress gases to the temps required to completely ionize the gas. That and what I said about the way the planet would form. There's just no way that a shell of rocky material could be supported by gas pressure in that way. The whole thing would be really unstable.


How do we know that an exoplanet is tidally locked to the parent star?

They are probably tidally locked, given some assumptions about age of the planetary system.

There's a formula to calculate how long it'll take for an orbiting object to tidally lock, and one of the outcomes of that formula is that large objects in close orbits don't take very long to tidally lock.

Since many of the exoplanets we've found are large objects in close orbits of old stars, tidal locking is a likely outcome.


Do gravitational waves really travel in "nothing"?

Well, space isn't really nothing.

Empty space (properly spacetime), also called the vacuum, is pretty complicated in modern physics. All of the fields, like the electromagnetic field and the electron/Dirac field, are there whether or not there are any particles (photons or electrons). The particles are vibrations of the fields – or to put it another way, they're vibrations of the vacuum.

Although people tend to say that gravity is different from everything else, it isn't really very different. The vacuum has a "shape", and it can vibrate in various ways. Some of the vibrational modes are gravitational waves. Others correspond to electrons, photons, etc. The details are complicated but the idea is simple.

The vacuum is different from an ordinary wave medium in some ways. Most importantly, it doesn't have a state of motion: you can't detect a "vacuum wind" or "vacuum current" resulting from moving through the vacuum. But if you keep that imporant caveat in mind, then the analogies of a "fabric of spacetime" or of gravitational waves as waves on the surface of the ocean are not so inaccurate.


Why the comets can cross the orbits of planets?

There are several reasons, the main ones are:

The orbit of some comets may appear to 'cross' a planetary orbit when viewing the solar system in 2-dimensions, but their orbit does not lie in the ecliptic plane (that the planet orbits follow).  An example is Comet Hale-Bopp:

Analogous to why Pluto won't hit Neptune despite their orbits appear to 'cross' in 2D representations.

Image source: http://spiff.rit.edu/classes/phys235/comets/comets.html

Some comets do have orbits on the ecliptic, particularly some short-period comets, such as Comet Encke:

Same image source as above

However, it is important to note that the chances for the comet and planet to be in enough of a vicinity to each other to result in the comet being 'captured' by the planet is low - so if the planet is not nearby, the comet can just keep on going on its normal orbit.

Though we have witnessed if a comet gets too close to a planet, when Comet Shoemaker-Levy 9 got caught in Jupiter's gravitational field, tidally fragmented and eventually collided in 1994.

in terms of whether a comet could be captured in orbit around a planet, this has been addressed as answers to the question "Can a comet orbit a planet?".


Let me put it this way.  If the Moon's orbital plane where exactly aligned with the Earth's orbital plane (which, by the way, includes the sun, not surprisingly) then there would be a solar eclipse on Earth during each New Moon.  But, in fact, the Moon's orbital plane moves about by many degrees in a very complicated fashion.  Figuring out why this was and how the plane moves kept Isaac Newton busy for a few years.

Both the Moon and the Sun subtend about 1/2 of a degree on the sky.  The distance to the Moon also changes considerably, so its angular extent varies around this 1/2 degree. So, to get a total solar eclipse, they need to be aligned to a very high degree and the Moon needs to not be near the far edge of its distance range.


The "stair-step" pattern visible in the first image is characteristic of the the Hubble's Wide Field and Planetary Camera 2 (WFPC2). This pattern is created by the arrangement of the four CCD chips. This image shows how the three larger "wide field" chips and the smaller, single "planetary camera" chip are arranged to create the pattern.

The bottom picture appears to be a slice of the original Hubble Deep Field, which was also captured with WFPC2.

So the answer, for these two images, is that the shape of the image is an artifact of image capture.

However, in general, finished images such as these could take up any arbitrary shape. Many of the modern images contain data from multiple views, multiple cameras, even multiple telescopes all drizzled together using specialized photo processing software.


Measuring declination in negative hours of arc?

It's not hours, minutes, and seconds, its degrees, arcminutes and arcseconds.

There are 360 degrees in a circle, 60 arcminutes in a degree and 60 arcseconds in an arcminute.

Since the planets lie close to the ecliptic, and the ecliptic tilted at 23 degrees relative to the equator, the declension of the planets rarely exceeds +-23 degrees, giving an appearance that their dec. is like a time.


Are we less safe now that Arecibo is no more?

The linked article is copied from a university press release.
Arecibo's article is more matter-of-fact but naturally also emphasizes the value of their own work.
NASA Goldstone can do planetary radar and remains operational.
As this page says:

Arecibo has twice the range and can see three times the volume of Goldstone,
while Goldstone, whose greater steerability provides twice the sky coverage and much longer tracking times, serves a complementary role.

FAST has no transmitter but presumably could take part in bistatic radar observations as GBT does.
Radar can measure distance and radial velocity ($r$, $\dot{r}$) in a single observation, especially useful when combined with astrometry ($\alpha$, $\delta$) since the two different kinds of data are orthogonal.
Given enough observations over a long enough time, astronomers can determine asteroid orbits rather precisely through optical astrometry alone.


It's angular diameter (a.k.a angular size or apparent size). This XKCD comic strip shows in an interesting way the angular size of some of the most remarkable objects in the sky:


Potential paradox with the Big Rip and the Multiverse?

I don't know whether this is a real paradox. Infinities are tricky :) For example, we don't have just one infinity. Also, our knowledge about the universe is too narrow too understand the system of our universe (or multiverse) and we can only imagine the universes as some balls which are expanding into some other space = multiverse. If multiverse exists, it is probably not some space with some balls in it. There are a lot of dimensions that are present in here.
So, this isn't a paradox: we shouldn't think about the universes as some physical objects expanding. But we also don't know what to imagine. We just haven't figured it out. But it is probably a lot different from our current physics, math and imagination.


Can you see Pleiades and Sirius at the same time on the sky, from Greece?

Stellarium is a free planetarium software, and it allows you view the position of stars in the sky at any date.

Sirius and the Pleiades are actually rather close in the sky. On the line of Orion's belt, on opposite sides. So for much of the the winter months, both are visible in the sky. For example in mid-December, they will be visible at midnight, in south-western skies. At the time of writing (8th October) they will be visible in the early morning.

It is a well known starscape, with Orion central, the Pleiades and the V of the Hyades on one side, and Sirus, scorching on the other.

Of course, the Pleiades and Sirius are "fixed stars", so the separation between them doesn't change. There is no time when they approach each other.