What is the spectral reflectance of starlight in a close binary?

Stars are far from perfect blackbodies due to scattering/reflection.  This is especially true for hotter stars, because of all the free electrons, but even cooler stars can reflect a significant amount.  For example, in aanda.org/articles/aa/pdf/2001/19/aa1009.pdf you will see that they use a reflection albedo of 0.30 for the K star and 1.00 for the F star, but the latter number is not meant to be taken seriously, they simply don't care if the light is reflected or absorbed and re-emitted because it isn't an important term. But the value of 0.30 for the K star might be meant more seriously, though it is still not regarded as a critical parameter because it only affects the color of the light that is reflected, not the total amount of light (given that stars are in radiative equilibrium, so must ultimately return all the incident light, whether it happens by reflection or heating).

Indeed, stellar emissions are often characterized by "effective temperature," to connect the surface flux of a star to the Stefan-Boltzmann formula for the emission of a blackbody by using a T parameter that is not necessarily the actual temperature.  When using this notion, as is quite common for dealing with stars, there is no essential difference between heating of and reflection from the surface of the star in question. The details of the difference have to do with the shape of the spectrum, but that shape is generally not a Planck function anyway, so as soon as one is using the "effective temperature" concept one has already parted company from a detailed understanding of the shape of the spectrum.  (When you do want the details of the spectrum, you will have to model the situation with some care.)

You are right that the albedo depends on the type of star (hotter stars having high scattering albedo due to all those free electrons), and many stars do show a lot of scattering from their surfaces.  Scattering/reflection elevates the effective temperature above the actual temperature of the layers being looked at, but again, if you stick to effective temperature, it makes no difference if there is scattering or heating of a blackbody, because effective temperature explicitly refers to the outgoing  flux with no claims on the actual temperature.  It is quite common for this distinction to be swept under the rug, and most stellar temperatures are effective temperatures, not actual temperatures.

Another consequence of radiative equilibrium is that the "reflection effect" cannot alter the total luminosity of the binary system seen from all angles, so if the effective temperature of the surface of the stars is elevated by reflection and/or heating, as seen from some directions, this must be exactly compensated by the reduced brightness due to eclipses seen from other directions.  So the reflection effect is part of the study of eclipse light curves, which are quite useful for understanding things like the sizes of the stars.

Added:  a nice paper on the polarization of reflected light was provided in the comments by uhoh, it is  nature.com/articles/s41550-019-0738-7 .  The idea is that if the stars are not too close to each other, there's not a lot of reflected light, but it is still noticeable by virtue of the fact that it is highly linearly polarized in the direction perpendicular to the line between the stars.  So it shows up much better in linearly polarized light, a potentially important new diagnostic of binary systems.


What parameters determine whether galaxies colliding will result in a merger or a hit and run?

The main factor is the velocity of the encounter. The higher the relative velocity between the two galaxies, the easier it is for them to pass through each other without being slowed down enough for a proper merger to take place, and without being strongly distorted by the encounter. (In a very high speed encounter, the two galaxies will spend almost no time close enough for tidal forces to be effective.)

In clusters of galaxies, where the typical velocities are around 1000 km/s, mergers are rare; in small groups, where the velocities are around 100 km/s, mergers are more common. So the Milky Way and Andromeda are pretty much destined to merge.

During a merger it's possible for stars in the outer parts of the galaxies to be ejected by tidal forces, but the central regions of both galaxies should definitely merge. Thus, the supermassive black holes (one in each central region of the original galaxies) will end up in the center, too. (Unless you have something exotic like a merger where one of the galaxies already has an unmarked binary SMBH; then you might get one of the three SMBHs ejected via a 3-body interaction. But I suspect that's pretty unlikely.)


How can you determine the initial volume of a planet's atmosphere?

In the Q on Earth Sciences SE What is the mass of our planet's atmosphere, hydrosphere, and cryosphere combined? this formula was derived: $$ M_{\mathrm{atm}_\bigoplus} \approx \frac{4\pi R^2_{\scriptscriptstyle \bigoplus} P_{\mathrm{sea\,level}}}{g_{_\bigoplus}} $$ which can be rearranged to derive pressure from the total mass of the atmosphere. $$  P_{\mathrm{sea\,level}} \approx \frac{g_{_\bigoplus}M_{\mathrm{atm}_\bigoplus}}{4\pi R^2_{\scriptscriptstyle \bigoplus}} $$


The sun is an excellent source of light for all the planets and moons, although it gets a little dodgy way out near Pluto. The moon reflects about 10% of the sunlight that hits it, that's why we see it. Venus' albedo is about 0.75, what with its clouds and all. That means 75% of the sunlight that hits it is reflected back into space for us to see.


Are we really seeing the asteroids from the past?

Most things in space move very predictably. For reasonably dense objects, like stars, planets or asteroids, gravity far outweighs all of the other forces acting on them, and it's not too hard to take account of gravity and predict their movements (if you have a powerful enough computer. For things like clouds of very thin gas it's a bit harder. So from the positions and movements of all the distant objects that we can see (at the time when the light we see them by left them), we can pretty accurately predict their movements a considerable way into the future, for instance up to possible collisions a few centuries out (for an asteroid) or a few billion years out (for Andromeda).

Of course there is a possibility of something unexpected happening -- between the light we see now being emitted and a possible future collision, a supermassive black hole travelling at close to the speed of light could pass through the Andromeda galaxy and scatter it, for instance, but we don't see things like that happening anywhere else in the universe, so it seems pretty unlikely to happen here.


How can the orbit of a Kuiper Belt Object be differentiated from the transit of a rogue planet?

I'm not speaking from an informed position here, but two things come to mind.

First is that a rogue planet is likely to be traveling very fast, relative to our solar system. An object from within our solar system (a Kuiper/Oort object) is going to have an orbital velocity. Something that isn't a part of the system at all could be traveling much, much faster.

Secondly, if a rogue planet somehow slowly drifted into our system at almost no speed, and then got pulled into a sort of orbital speed by the sun, the direction of its orbit might give it away. For example, it's orbital inclination could be way off--even perpendicular to the normal orbital plane. It also might be orbiting in the wrong direction--that is--clockwise, instead of ante-clockwise.

That being said, there's no reason that some disturbed Kuiper belt object might not have an eccentric orbital inclination or even an anti-clockwise orbit. But the larger the object, the less likely this is, because it would need a larger and larger disturbance to throw it so out of whack.


Are exoplanets at dwarf stars less likely to have super-rotating atmospheres or asynchronous tidal locking?

I'm not exactly sure what you're asking, cause you touch on a few related points.

The reason planets around red-dwarf stars are thought to be tidally locked is because the tidal force is comparatively much greater for the habitable zone.   Take, for example, a star with half the mass of our sun.   It would have, roughly speaking, 1/16th the solar output, so a planet, to get the same amount of heat, would need to be 4 times closer or 1/4 AU.

The tidal force increases by the cube of the distance, or, 4^3, or 64, and the mass, 1/2, means that for just 1/2 the solar mass, the tidal force on a planet getting equal heat is 32 times greater.   That's a ballpark ratio to the 5th power, which is highly significant.    When you get into solar masses 25% the mass of the sun or less, the tidal force for a habitable zone planet can be thousands of times greater than the tidal force for the habitable zone in our solar system.   Because of the high tidal forces, planets in the habitable zone of small stars are very likely tidally locked, at least for any near-circular orbits.  That's pretty darn close to certain when the star is below a certain size.

There are some theories that thermal heating of a planet's atmosphere, which takes time, so it doesn't happen precisely at mid-day, but a couple hours after that, there's expected to be a tidal torque in the upper atmosphere and an equilibrium where the constant rotation of the atmosphere drags on the surface and keeps the planet rotating, so the planet never becomes tidally locked.    Venus might be an example of that.    But the articles below suggest that this is predicted to only happen in the larger red-dwarfs, 50% the mass of the sun or larger.  Article here and here for more details.

In summary, for smaller red dwarf stars, tidal locking is probably very common within the habitable zone.   For larger red dwarf stars, you might get a mix, growing less common for further out planets.

Rotations maintained by wind torque are probably, at least, to my thinking, not very rapidly rotating, similar to Venus, a slow rotation makes super-rotating trade winds easy.  (I say trade winds because I don't think weather-wind or local wind is considered super-rotating), so I think we should just consider relatively permanent wind speed and direction).

And for any slowly rotating planet, Super-rotating wind should be quite common.

For the very smallest stars, like Trappist 1, which at .08 solar masses isn't far off from the minimum size for a red dwarf, it's inner most planet has a period of just 1.5 days, it's corresponding rotation (Sidereal, not solar) has a 1.5 day period too.   That's probably a fast enough rotation to generate a significant Coriolis effect and some interesting weather (provided the planet hasn't lost it's atmosphere - which is also possible with close orbits around small stars).   That's an equatorial surface velocity of over 600 km/h, so for very close orbit planet around the smallest red dwarfs, super-rotating winds may not happen.

Looking at the planets orbital periods around Trappist 1.  The inner 2-3 planets might rotate too fast to have super-rotating winds.   The 5th one, for example, with an orbital period of 9 days, corresponding to a sidereal rotating speed at the equator of not much more than 100 km/h, it's probably likely to have super-rotating winds.  The longer the orbital period, the more likely super-rotating winds are to happen.  Any tidally locked planet with an orbital period over 10 days, should be statistically likely to have super-rotating winds (compared to sidereal rotation).   Solar rotation, well, by that measurement, all tidally locked planets with atmosphere have super-rotating winds.


How to calculate the orbital period and the density of a planetary body?

The formula for orbital period is given on Wikipedia:
$$T=2\pi \sqrt\frac{a^3}\mu$$
where:

$T$ is the orbital period in seconds
$a$ is the orbit's semi-major axis in meters
$\mu = GM$ is the standard gravitational parameter

$G$ is the gravitational constant
$M$ is the mass of the more massive body in kilograms

So $T = 2 \pi \sqrt \frac { (172 \cdot 10^9) ^ 3 } { 6.674 \cdot 10^{-11} \cdot 1.9884 \cdot 10^{30} }$. Can you take it from here?
As for density, the volume of a sphere is given by the formula $V = \frac43\pi r^3$; density ($\rho = \frac MV$, with $M$ the mass) is usually given in grams per cubic centimeter, so it makes sense to convert to those units. That gives the following calculation:
$$\frac { 6.15 \cdot 10^{27} } { \frac43\pi (6.743 \cdot 10^8)^3 }$$


Are black holes different sizes thanks to the different sized stars that created them?

The size of a black hole is not defined by the expanse of its matter content but (usually) its event horizon.  The radius of the event horizon  (of a non-rotating and uncharged black hole) depends on the mass:

$$r_S = \frac{2 G M}{c^2}\\ \small M : \text{black hole mass} \\ \small G: \text{gravitational constant} \\ \small c : \text{speed of light} \\ \small r_S : \text{Schwarzschild radius}$$

Yes, the size of black holes as final stage of stellar evolution stems from the original star's mass, though some mass is ejected/radiated away during the supernova.

Black holes do grow (increasing $M$ and $r_S$) when they accrete matter from disks or companion stars.

Not all black holes we (kind of) observe are from stars: for example supermassive black holes in galactic centers, which can have billions of solar masses.  However, it is not yet fully understood, how they form.

Note: The most common black hole descriptions actually don't include any matter.  They are defined in an empty spacetime except for one singular point that is usually interpreted to hold all the matter.  The mass of the black hole is then assigned by identifying the gravitational pull in the Newtonian approximation at some distant point.

How the interior of real, non-vacuum black holes looks like, is still up to discussion.


How would a flat mirror on the Moon reflect sunlight

The (weird for historical reasons) defintion of magnitude is that a difference of 5 magnitudes corresponds to a factor of 100 in the brightness of the source.

So a difference of 3-(-27) = 30 magnitudes is a difference of $10^{12}$ in brightness.

The surface brightness of the mirror and the sun is effectively equal, since the mirror reflects 100% of the sun's light, so the only reason for the sun to be brighter is that it is larger. The brightness is in direct proportion to the apparent area of the source.

This is because brightness is a simple linear scale. Two suns of the same size and brightness would be twice as bright as one. So if the sun is $10^{12}$ times brighter it must cover an area $10^{12}$ times larger.


What is the relationship between the focal length and f number of a telescope compared to a guide scope?

The aperture of the guide scope matters because it affects how faint the stars are which are available for guiding - the larger the aperture the fainter the stars it will pick up which gives you more choice when picking a good one as a guide star. This is especially important if you are imaging from an urban area where light pollution may otherwise limit the number of visible stars.
The length of the guide scope is not directly important, although generally larger diameter objective means longer anyway. The weight of the guide scope does however have to be also considered as your mount has to be able to deal with the extra load you are placing on it, lighter is better here of course.
When you talk about pixel size I presume you mean pixel size of the CCD camera you are attaching to the guide scope itself? I am not expert enough to go into great details there, if you need more in depth advice on that side of astrophotography I can thoroughly recommend the Cloudy Nights forum, there are some very experienced astrophotographers on there who are very willing to share their knowledge and experience with all no matter how much of a beginner you may be.


Using the calsky.org website, this is a chart showing a view of the eastern night sky from a point near San Francisco, CA at 7pm this evening.  (The green tick mark on the horizon is due east).

Orion is located between the moon and the horizon, and will be rising quickly.


If space is negative energy and matter is positive energy then does that mean the universe is finite?

As everyone is saying, this is a very loose "plain English" version of some complex ideas, but if space was infinite, you could view this suggestion (that the overall total energy of the universe is zero) as applying to any sufficiently large region of space. So if you pick any 10 billion lightyear diameter sphere you would expect to find the energy of the matter in that sphere more or less cancelled out by the gravitational potential energy arising from the separation of that matter.


Why shouldn't you look at the Sun through a camera using eclipse glasses?

Electronic viewfinders (such as the ones using LCD or OLED screens) are excluded from this discussion. There's no danger of eye injury there, except perhaps danger of damaging the camera's sensor if exposure is very long and the lens aperture is big.

It's the classic or optical viewfinders that are targeted by the warning.

  I can see how the enlarged image of the Sun through a telescope or binoculars might overwhelm the eclipse glasses' filters.

A few observations.

It's really the larger aperture of the instrument that is the problem here. The pupil of your eye is only a few mm in diameter. The objective lens or mirror of a telescope could be anywhere between dozens of mm to hundreds of mm (or thousands for very large instruments). The ratio of areas between telescope aperture and eye pupil is even greater - it's the square of the ratio of diameters.

Let's say your eye's pupil is 2 mm in diameter. Let's say you use a 50 mm aperture telescope (small refractor or binoculars). The diameter ratio is 25x. The area ratio is 625x.

All the light captured by the very large area of the instrument is funneled into your eye through your pupil. With the instrument, now you're getting 625x more energy from the Sun, compared to the naked eye view. It's already dangerous to look at the Sun with the naked eye - with the instrument it's 625x more dangerous. And this is with a very small refractor.

There is a class of "solar filters" that are made to be mounted on the instrument's eyepiece, or after the eyepiece. THESE ARE VERY DANGEROUS! All the increased, focused energy of the Sun is now absorbed by the filter, which can warp, melt, crack, or burst into flames. A filter failure at this point is likely to injure the user. White light filtering for solar observations must always occur ahead of the instrument, not after it.

As for why classic, optical viewfinders were included in that list: it was out of an abundance of caution. Most of those viewfinders do not capture more light than your eye does, but some do. It's better to be safe than sorry. You can't expect everyone to be able to tell whether their viewfinder is dangerous or not. So, in a warning addressed to the general population, just tell them to stay away from it.

Now, if you install a full-aperture solar filter (like the Baader solar film) ahead of the viewfinder, then it's safe to use - provided the filter is attached firmly to the camera and cannot be blown away by some random gust of wind.


Need Simple equation for Rise, Transit, and Set time

I'm not sure it qualifies as "simple", but, using
http://idlastro.gsfc.nasa.gov/ftp/pro/astro/hadec2altaz.pro (and some
additional calculations/simplifications):

$
   \begin{array}{|c|c|c|c|}
\hline
    \text{Event} & \text{Time} & \phi  & Z \\
\hline
    \text{Any} & \text{t} & \tan ^{-1}(\cos (\lambda ) \sin (\delta )-\cos
      (\delta ) \cos (\alpha -t) \sin (\lambda ),\cos (\delta ) \sin (\alpha
      -t)) & \tan ^{-1}\left(\sqrt{(\cos (\lambda ) \sin (\delta )-\cos (\delta
      ) \cos (\alpha -t) \sin (\lambda ))^2+\cos ^2(\delta ) \sin ^2(\alpha
      -t)},\cos (\delta ) \cos (\lambda ) \cos (\alpha -t)+\sin (\delta ) \sin
      (\lambda )\right) \\
\hline
    \text{Rise} & \alpha -\cos ^{-1}(-\tan (\delta ) \tan (\lambda )) & \tan
      ^{-1}\left(\sec (\lambda ) \sin (\delta ),\cos (\delta ) \sqrt{1-\tan
      ^2(\delta ) \tan ^2(\lambda )}\right) & 0 \\
\hline
    \text{Transit} & \alpha  &
\begin{cases}
                   \delta >\lambda  & 0 \\
                   \delta =\lambda  & \text{Zenith} \\
                   \delta <\lambda  & \pi
                  \end{cases}
    & \frac{\pi }{2}-\left| \delta -\lambda  \right| \\
\hline
    \text{Set} & \alpha +\cos ^{-1}(-\tan (\delta ) \tan (\lambda )) & \tan
      ^{-1}\left(\sec (\lambda ) \sin (\delta ),-\cos (\delta ) \sqrt{1-\tan
      ^2(\delta ) \tan ^2(\lambda )}\right) & 0 \\
\hline
    \text{Lowest Point} & \alpha +\pi  &
\begin{cases}
                   \delta >-\lambda  & 0 \\
                   \delta =-\lambda  & \text{Nadir} \\
                   \delta <-\lambda  & \pi
                  \end{cases}
    & \left| \delta +\lambda  \right|-\frac{\pi }{2} \\
\hline
   \end{array}
$

where:

$\phi$ is the azimuth of the object
$Z$ is the altitude of the object above the horizon
$\alpha$ is the right ascension of the object
$\delta$ is the declination of the object
$\lambda$ is the latitude of the observer
$t$ is the current local sidereal time

Note the two-argument form of arctangent is required so that the
results are in the correct quadrant:
https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Two-argument_variant_of_arctangent

Additional caveats:

If $\left| \delta -\lambda \right|>\frac{\pi }{2}$, the object is
always below the horizon, and the equations for rising time and
setting time will not work.
If $\left| \delta +\lambda \right|>\frac{\pi }{2}$, the object is
always above the horizon (circumpolar), and the equations for rising
and setting time will also not work.
The measurements above are in radians. You can convert $\pi \to
  180 {}^{\circ}$ for degrees.
Because we use the local sidereal time, the longitude doesn't
appear in any of the formulas above. However, we do need it to find
the local sidereal time, as below.
To find the local sideral time $t$ in radians, we use
http://aa.usno.navy.mil/faq/docs/GAST.php and make some substitions
to get:

$t = 4.894961212735792 + 6.30038809898489 d + \psi$

where $\psi$ is your longitude in radians, and $d$ is the number of
days (including fractional days) since "2000-01-01 12:00:00 UTC". Traditionally, we use $\phi$ for longitude, but I'm already using it in the formulas above for azimuth.

If you combine the formula for local sidereal time and
azimuth/altitude and assume excessive precision, you get my answer to
https://astronomy.stackexchange.com/a/8415/21

Additional computations for these results at:
https://github.com/barrycarter/bcapps/blob/master/STACK/bc-rst.m

I was going to add some graphs to show how the altitude is NOT a sine wave and how the azimuth is NOT a straight line (although you might expect them to be), but they turned out not to be terribly instructive/helpful.

You might also be able to get simpler formulas if you set $t$ to be the "hour angle" (which is $\alpha-t$ in the current setup).


Which galaxy is receding from the Milky Way the fastest? What is known of the mechanism behind its recession?

This question may be a duplicate to: On what scale does the universe expand?

The mechanism for the expansion of the universe is gravity, manifested through what's known as Dark Energy. Dark Energy is a component to the universe which today, composes 68.3% of the energy density of the universe. You may be confused as to how gravity could be responsible for the expansion of the universe, but the equation of state of DE,

w = pressure/density

is negative (w_DE = -1), which means that it has anti-gravitational properties. Additionally, its density remains constant as the coordinates of the universe change, which is the reason why it dominates later on in the universe's history (while all other components, matter (dark or otherwise) and radiation/relativistic matter, become diluted with the expansion of the universe).

To answer your question, because the expansion of the universe is happening at all points in space, it looks like the "center" of expansion is at every point in space (see: Space expansion in layman terms), the galaxies furthest away (highest redshift), are the ones moving quickest away from the Milky Way Galaxy.


Explaining gravity assist to a 6-year old

I like the trowing a ball against a speeding train analogy.

The important factor is that Jupiter is moving, quite fast.  About 13 km/s.   The other important factor is that Jupiter is very massive, so it has a lot of gravity and it can draw objects towards it, causing their direction to bend.

The tricky concept for a 6 year old is that gravity is a zero-sum game.   The object that falls towards Jupiter, in this case it's a hyperbolic orbit but regular orbits work too, so as it falls towards it, it adds velocity then flying away from it, it loses the same amount of velocity, so the net velocity relative to Jupiter is zero.  No change.

But because Jupiter is moving, Jupiter can tug the spacecraft along with it.  The velocity relative to Jupiter is unchanged but the velocity relative to the sun changes, Jupiter is like the train that the ball bounces off of.

The opposite can happen too.   Jupiter can be used to slow a spacecraft down. It depends on the angle of approach, like throwing a ball against a train while it moves away from you.

You can also do this experiment below (with 2 balls not 3 to represent the gravity assist, though it's a cooler experiment with 3).  It's the same principal.   Bouncing off a moving object (Jupiter) is very different than bounding off a flat surface that isn't moving.

https://www.youtube.com/watch?v=2UHS883_P60

All that said, I'd still point out that this is a tricky concept and not to worry if they don't understand it.


Why does Nancy Grace Roman = 100 × Hubble? Why is the new space telescopes wide field camera so much wider than the old one's?

Hubble was designed in the late 1980s, when electronic detectors were small (the original Wide Field/Planetary Camera had two different 2x2 arrays of 800x800-pixel CCDs); it also had to carry multiple different instruments, each taking up nominal space in the focal plane. So it was optimized to use small fields of view to go with the small detectors.
WFIRST/Nancy Grace Roman is much more tightly focused on large-field surveys, so it can devote most of the field of view to the IR camera. The reason it is now possible to do this is probably a combination of three things:

Large, sensitive, low-noise IR detectors are now available; the current plan is to use a total of 18 individual 4k x 4k detectors -- a huge advance over the original mid-1990s 256 x 256 NICMOS detector installed in HST in 1997, and over the single 1k x 1k detector in the infrared channel of WFC3 (installed in 2009);
We have lots of experience in building array mosaics with lots of individual detectors (the planned $\sim 300$ megapixel mosaic is relatively modest compared with the gigapixel arrays in Pan-STARRS1 and LSST/Vera Rubin);
Computer technology has advanced to the point where we can now store all the data such an instrument would take and transmit it to the ground in a reasonable time. Hubble can currently take and transmit between 2 and 3 gigabytes of data per day; Roman is forecast to take and transmit about 500 times as much data.


How should I think about sunlight for the purposes of analyzing the angles it makes w/ earth?

This is a first-order approximation, and works well for simple models of irradiation of the atmosphere or the Earth's surface.

But, a simple experiment with a pinhole camera or a single lens will show that this model doesn't work in imaging situations.  The sun is an "extended object," meaning it is not a point source, so the image you get is not what you would expect from a point source.  Worse, even if the sun were a point source, the distance to the geometric focal point is different very slightly from the distance if all the sun's rays were perfectly parallel.  See the Lensmaker's Formula,  $ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} $ where p and q are the distances from the lens to the source and the image respectively.   (I'll warn that for a reasonable lens, say focal length $ f = 10 m$, the difference between focal location for rays that are parallel and rays from a point source $150\space million\space km$ away is ridiculously small)

Similarly, if you want to model the umbra and penumbra during a solar or lunar eclipse, you'll need to take the physical extent of the sun into consideration.


Is the duration of a sidereal year stable?

It changes some.   The solar energy pressing against the Earth from the sun, the kinetic energy from coronal mass ejections hitting the Earth, tidal forces from the Earth-Sun relation and even gravitational tugs form other planets and meteors and space dust that hit the earth all change the Earth's orbit a teeny tiny bit, also, changes in the Sun's mass either by the sun accumulating mass from space dust or comets or losing mass, which most suns do over time.   Each tiny orbital change changes the sidereal year (and all the other years) a tiny little bit.    Perhaps this only amounts to millionths of a second per year but some of these forces are consistent and can lead to predictable changes over time.  Others variable, leading to uncertainty.   The long term changes to the Earth's orbital distance are currently unpredictable.

It's generally believed that the Earth is slowly moving away from the sun, but I don't believe there's much agreement on how fast this is currently happening.   Articles here and here.   The 2nd one estimates 1.5 CM per year or 15 KM per million years which even over a billion years (15,000 KM) is quite a small distance compared to 150 million KM.   But estimates are difficult for a variety of reasons.

I asked a related question here, with a bounty and didn't get any definite answers.   It's possible to mathematically model the movement of some planets (Jupiter to an extent and Neptune/Uranus and perhaps Saturn based on the Jupiter model).  It's thought Jupiter moved both towards then away from the sun over the first few billion years of the solar system.  The Moon's distance from the Earth is also mostly predictible and can be modeled back into the past based on the Oceanic tidal bulge, but modeling the Earth;s distance from the sun over the past is much harder.  It's a tiny object compared to the Sun and Jupiter and N body systems have a lot of chaos and unpredictability to them.   4 billion years ago it's possible that the Earth was quite a bit closer to the sun, but I don't think anyone has a good model for how much closer.   The faint young sun paradox and the theory that young suns eject much more matter and have more active coronal mass ejections suggests to me that the Earth was likely closer to the Sun billions of years ago and not further away.

The earth's orbital eccentricity also changes with time, on a cycle of roughly every 100,000 years but according to Wikipedia, that doesn't change the length of the year, only the shape of the orbit.

Finally, and just for fun, it makes logical sense to me that the Moon orbiting the Earth creates small apparent fluctuations at least from the point of view of the Earth, because the Earth in effect orbits the Earth-Moon barycenter in an ellipse roughly 1/81st the size of the distance to the moon, or about 4,480 KM in radius every 29 days.     Depending on whether the Moon is ahead of the earth or behind it, the Earth year can change by as much as 5 minutes.    I suspect that this is largely ignored when measuring Earth years though and the Earth-Moon barycenter is used.   If we measured the Earth's center we should get much larger variations year to year.  (see picture).

Source.

(if anyone can improve on this, feel free, I have the feeling it's only a partial answer, cause logically it makes sense that the Earth-Jupiter positions or Earth-Venus should change, if only slightly, the Sidereal year).*


How was Cruithne's orbital time calculated?

It's a good question!

I went to the JPL Horizons web page and looked up 3753 Cruithne and saw that the current orbital solution is based on observations from 1973 to 2018, about 45 years. That's a fraction of an orbit, but very careful measurements of it's position on the celestial sphere (RA, Dec) over time will allow astronomers to calculate its position and speed by fitting those observations to many possible trajectories and finding the one that's the best fit.

JPL/HORIZONS               3753 Cruithne (1986 TO)         2019-Feb-14 23:48:43
Rec #:    3753 (+COV) Soln.date: 2018-Sep-27_06:26:23    # obs: 677 (1973-2018)

The orbit is elliptical but has a nearly-identical semimajor axis as Earth, so it's period around the Sun is one year just like Earth. That means that it in 45 years it has been possible to observe about 45 orbits around the Sun, allowing astronomers to nail down its orbit with great precision.

I plotted the distance from Earth to Cruithne from 1973 to 2018 below from the Horizons data. At a half of an AU and 0.1 arcsec astrometric position precision (averaging over many observation) that gives the position to tens of kilometers over a span of decades.

Then using accurate orbital modeling using the gravity of all the solar system's bodies as described here it is possible to project the position with great accuracy, as long as non-gravitational forces like unpredictable outgassing aren't too strong, and predictable non-gravitational forces like solar radiation pressure can be modeled as described here.

Over time the orientation of the elliptical orbit slowly precesses around the Sun due to the gravitational effects of Earth. This is the ~770 year period you've mentioned.

In the video Near-Earth Asteroid 3753 Cruithne they hold the position of the Earth fixed and you can see how Cruithne's orbit sloshes back and forth.

This GIF is too large to embed here, but it demonstrates the sloshing nicely: https://en.wikipedia.org/wiki/File:Animation_of_3753_Cruithne_orbit.gif


How much could Earth's orbit shift without causing an ecological catastrophe?

Depending on how one defines "quickly" and "significantly," I think a gravitational influence strong enough to do that would tear apart the Earth through tidal effects long before the change in orbit would make the planet unsuitable for life.


What visible differences do Geminids have to other meteors

Well done on your observations, I've been completely clouded out.

You are quite right, the Geminids are noted for a variety of colours, and a good number of fireballs, though most observers note a predominiance of yellow meteors This is partly due to their composition and density (coming from a peculiar "rock comet") and partly to their velocity (35km/s) which is fairly fast.

There is a good chance of seeing some good trails with this shower. Brighter meteors and most fireballs will leave a trail, and the Geminids have more of these than most other showers.


Why does the Event Horizon Telescope (EHT) not include telescopes from Africa, Asia or Australia?

Would it increase the diameter if they would include some from there?

No. Not by much, at least. The telescopes are already ~20,000 km apart, so you can't create a longer baseline that still has a simultaneous view of the target.

Don't forget: Earth is a sphere. Only one half of that sphere can observe M87 at the same time.

Telescopes in the Eastern hemisphere would enable more observations, but I don't know if that would bring any improvement over what they already had.


If the Sun disappears right now, when the absence of its gravity affects on Earth?

As we know by General Relativity, gravity propagates by the speed of light c = 299,792,458 m/s in a vacuuum (like space).
The average distance between the earth and sun is about 149.6 million kilometers, so it takes light about 8.3 minutes from the sun to earth.
This means, if the sun were to disappear right now, we would have to wait 8 minutes and 20 seconds until we would realize. Luckily, by all known laws of physics, this is not possible :)


As the planets evolve during their protoplanetary stage and accrete materials from the protoplanetary disks, which are gravitationally collapsing interstellar dust and gases, these accreted particles retain some of the angular momentum from the materials they form from and being in constant motion.

      Generated image (virtual fly-by) from a simulation of the accretion period of the protoplanetary disk, showing preservation of      angular momentum in the orbit around a Jupiter-size planet, as it clears its neighborhood. (Source: Frédéric Masset)

One nice description for this angular momentum preservation, and why the planets appear to rotate faster than their surrounding protoplanetary disk goes like this:

  Conservation of angular momentum explains why an ice skater spins more
  rapidly as she pulls her arms in. As her arms come closer to her axis
  of rotation, her [rotation] speed increases and her angular momentum remains the
  same. Similarly, her rotation slows when she extends her arms at the
  conclusion of the spin.

  Source: Scientific American article on Why and how do planets rotate? (George Spagna)

So it could be described as this axial rotation of planets resulting in conservation of the angular momentum of the materials in the protoplanetary disk, forming during the accretion period of the planetary system as the protoplanets gain in weight, and preserve this angular momentum due to inertia of their radial velocity.


Why don't we see the gas behind the black hole?

The reason is that your animated gif shows a geometrically thin, optically thick accretion disk, whereas the disk around the M87 black hole is a geometrically thick, but optically thin disk.

The main difference is that you do not see any evidence of the structure of the disk in the picture - it is nearly transparent. You do however see plenty of gravitational lensing - that is what the bright ring is. Many of our sightlines to hot plasma run through a small annulus at the photon sphere (at 1.5 times the Schwarzschild radius) making it bright. Because of lensing, the apparent size of this ring, as viewed from infinity, is enlarged to a radius of 2.6 times the Schwarzschild radius. Contrary to your question, we see photons emitted from all around the black hole in that bright ring.

There are many GRMHD (general relativistic magneto-hydrodynamic) simulations that do agree with what is observed.


What is Copernicus referring to when he states that "circles have poles different [from the earth's]"

Remember that Copernicus was espousing a heliocentric Solar System.  What he means by "pole" is the object's axis of revolution and its axis of rotation.  The Earth rotates around its poles: i.e. the North and South pole.  A line between those poles is our axis of rotation.  The axis of revolution would be the axis about which the Earth orbits.  This would correspond to an imaginary line at the Sun.
 Copernicus is presenting the idea that the axis of rotation is not the axis of revolution.


The size of a black hole is determined by its energy and angular momentum. For a non-spinning black hole, there is a simple linear relationship between the black hole's mass and its Schwarzschild radius:

$$r_s=\frac{2GM}{c^2}$$

where $r_s$ is the Schwarzschild radius, $G$ is the universal gravitational constant, $c$ is the speed of light, and $M$ is the black hole's mass. The mass of the black hole comes from all the matter and energy that went into creating the black hole, as well as any additional matter & energy that's fallen into it afterwards.

Strictly speaking, the Schwarzschild solution is for a universe containing nothing but a single eternal black hole. But although it's unphysical it's still a useful solution when we want to model spacetime with spherical spatial symmetry (including planets & stars, not just black holes).

For a rotating black hole, the situation is more complex, and there are several horizons. Please see Wikipedia's article on the Kerr metric for details.

We cannot say exactly what happens at the core of a black hole. As discussed in this answer by Florin, we need a theory that unites General Relativity and Quantum mechanics to answer such questions.

A pure GR black hole has a mathematical singularity at its core, but most astrophysicists believe that's unphysical, and that a proper quantum gravity theory will eliminate that singularity. However, it's likely that the core of a black hole is still very small, since the quantum gravity corrections probably don't kick in until the size gets smaller than an atom. And of course even with a quantum gravity we still won't ever be able to observe black hole cores to validate the theory.

So we cannot currently say exactly what happens to energy and matter once it reaches the core of a black hole. But we do know that once anything crosses the event horizon of a black hole it can no longer affect the universe outside the event horizon, and that anything inside the event horizon must rapidly (in its own proper time) fall towards the core of the black hole, and that no known force can prevent that process.

The gravitational field of a black hole is sometimes described as a "fossil field". All matter & energy falling into the black hole modifies the spacetime curvature as it approaches the event horizon. And once it crosses the event horizon it can no longer change the spacetime curvature outside the horizon, so those curvature changes are preserved (until something else comes along to add its own curvature changes).

So it doesn't really matter what happens at the core of the black hole, since the exact nature of the black hole core doesn't affect what's happening outside the event horizon.


How does the number in stellar classification indicate properties within the range of that class?


As you see from the app, there are no hazy objects there, and in the light polluted Mumbai skies, deep sky objects would not be easily visible. There are no bright naked eye comets visible at the moment.

Satellites appear like moving stars, not hazy. The most likely hazy object would be a high cloud or part of a contrail, perhaps lit by the light of the city, or by the first light of dawn.


Mass-magnitude relation of evolved stars

A star with absolute magnitude of -7 is big and powerful, It would be among the most powerful stars in the galaxy.

The star would be near the end of its life, so comparing with other aging stars:

Aldebaran is slightly more massive than the sun. It is a red giant (or orange giant according to some sources) It has an absolute magnitude of about -0.6. Aldebaran is too small to undergo a supernova.

Betelgeuse is much bigger. Estimates of its mass put it at about 11 solar masses, but it could be bigger. It has a magnitude of about -6 (it is variable and there is some uncertainty on its distance).

Eta Carinae is one of the most powerful stars in this part of the galaxy. It is also variable, but has an absolute magnitude of -8—-9, and mass of 100-200 suns

AG Carinae is a LBV with a mass of 50 suns and a minimum absolute magnitude of about -8, but becoming much brighter during outbursts. (as bright as -11)

So an evolved star with a magnitude of -7 could be supposed to be between the LBV stars and Betelgeuse, with a mass of around 15-20 suns. Stars do get very unstable this late in life, and there could be a great deal of uncertainty.

Others here know more about supernovae mechanics, and may be able to explain any discrepancy between the supposed stellar mass, and the mass of the ejecta.


See this post about binary stars, which contains relevant information:

  Stars are far from perfect blackbodies due to scattering/reflection.
  This is especially true for hotter stars, because of all the free
  electrons, but even cooler stars can reflect a significant amount. For
  example, in aanda.org/articles/aa/pdf/2001/19/aa1009.pdf you will see
  that they use a reflection albedo of 0.30 for the K star and 1.00 for
  the F star, but the latter number is not meant to be taken seriously,
  they simply don't care if the light is reflected or absorbed and
  re-emitted because it isn't an important term. But the value of 0.30
  for the K star might be meant more seriously, though it is still not
  regarded as a critical parameter because it only affects the color of
  the light that is reflected, not the total amount of light (given that
  stars are in radiative equilibrium, so must ultimately return all the
  incident light, whether it happens by reflection or heating).

  Indeed, stellar emissions are often characterized by "effective
  temperature," to connect the surface flux of a star to the
  Stefan-Boltzmann formula for the emission of a blackbody by using a T
  parameter that is not necessarily the actual temperature. When using
  this notion, as is quite common for dealing with stars, there is no
  essential difference between heating of and reflection from the
  surface of the star in question. The details of the difference have to
  do with the shape of the spectrum, but that shape is generally not a
  Planck function anyway, so as soon as one is using the "effective
  temperature" concept one has already parted company from a detailed
  understanding of the shape of the spectrum. (When you do want the
  details of the spectrum, you will have to model the situation with
  some care.)

Is it possible that there is a star which acts as a mirror and reflects back the light almost as it was received?

That depends on what you mean by "as it was received". That post shows that the concept of albedo applies to stars as well; they can be reflective.


What type of energy is escaping from black-hole's poles?

The article you've read is not quite accurate/correct. A more correct pictue is as follows:

A star may approach a super-massive black hole (SMBH) so closely that the tidal forces of the SMBH tear it appart. The distance to the SMBH at which this happens is often referred to as the tidal radius. For a (non-rotating) SMBH with a mass in excess of about $10^8$M$_\odot$, the tidal radius is within the event horizon and hence this process is invisible.
About half (depending on the initial orbit of the star) of the stellar material is caught by the SMBH onto a elliptic orbits (the other half escapes on hyperbolic orbits), but due to gas interactions and dissipation (shocks transfer orbital energy into heat which is then radiated away) it soon circularises to form a (or contribute to an existing) accretion disc. (Note that feeding from stars is not thought to be the main mechanism for the formation of accretion discs; a more likely process is the direct infall of gas clouds).
Material in the accretion disc orbits the BH on nearly circular orbits. It cannot simply fall into the SMBH because its angular momentum is conserved. However, gas at adjacent radii orbits at different velocities (differential rotation), i.e. the disc is shearing. Viscosity in the gas then causes angular momentum to be transported outwards, mass being transported inwards, and the material to heat up. The material in the disc is a hot (presumably) magnetized plasma, and the viscosity is thought to be provided by the so-called magneto rotational instability. The time scale for this process is, however, very long and only becomes efficient ($\sim10^6$yr) rather close ($\sim100$AU) to the SMBH.
The temperature of the disc material is extremely high, such that it radiates in the X-ray and UV, visible as quasars. This radiation is the main energy output from the accreting BH and constitutes about 10% of the gravitational energy gain $mc^2$ ($m$ being the mass accreted). This radiation interacts with any surrounding material (including the outer accretion disc itself, in particular if it is warped), generating an outflowing wind
(by its radiation pressure) and possibly preventing further accretion (this corresponds to the Eddington limit).
The rotating magnetized disc may also generate collimated bi-polar outflows, commonly referred to as jets. However, IMHO, only a small fraction of the energy can be emitted in these jets, already because the jets have low entropy (they are dominated by ordered motion), while the hot disc has high entropy.


Secondary maximum in SNIa lightcurves for near infrared

The main power driving a supernova (of type Ia) light curve is the explosive detonation of a carbon/oxygen white dwarf. This produces a large amount of iron-peak elements (particularly Ni) via nuclear fusion of carbon, oxygen and alpha particles via magnesium, neon, sulphur and silicon.

The decay of the light curve is then largely governed by the radioactive decay of $^{56}$Ni via $^{56}$Co to $^{56}$Fe, which releases gamma rays that "reheat" the ejecta. The half-lives of these decays are about 6 and 77 days respectively, and this is the basic reason for the decay of the light curve happening on timescales of tens of days.

However, as you have noticed, in the (rest-frame) infrared part of the spectrum, there is a secondary peak.

Now, whilst the emission in the visible part of the spectrum is largely of a thermal continuum nature (the ejecta are typically at temperatures of $\sim$10,000K during the early parts of the decay), the infrared portion can be dominated (and certainly is after $>20$ days) by line emission (that is then smeared out by turbulent doppler broadening).

The species that emit this radiation are ionised iron-peak elements (mainly cobalt and iron). As the ejecta cools, the species that are present recapture electrons, becoming less ionised. At around 7000K there is a fairly abrupt transition between the gas consisting of mainly doubly ionised species to mainly singly ionised species. This results in an order of magnitude increase in emissivity of the gas due to radiative recombination of the ions and electrons. It turns out that this radiation emerges mainly at near infrared wavelengths.

As the gas cools further and becomes wholly singly ionised, or even neutral, then the emissivity falls drastically again. There is therefore a "sweet spot" where the infrared emissivity of the ejecta is maximised, that occurs about 20-30 days after the explosion, that manifests itself as a secondary bump in the infrared light curves.

Much more detail can be found in Kasen (2006).


How did Biden become VP? (naming of 2012 VP113)

This has to do with how minor planets obtain provisional names (see also Wikipedia). The year is divided into 24 half months, with a particular letter associated with each one. Each of these half-months has a number of cycles of length 25 depending on how many minor planets are discovered; a minor planet is assigned a letter corresponding to its order in the cycle, and the number of the cycle it is in. This means that the provisional name for a minor planet is, in general,

year+half-month+position in cycle+cycle number

2012 VP113 was discovered in 2012, in the first half-month of November (corresponding to the letter V) and was the 15th minor planet discovered in the 113th cycle that half-month.


First up, the tilt is exactly 23.45 degrees.
The reason for Earth's tilt is still not yet really proven, but scientists at Princeton stated on August 25, 2006 that planet Earth may have 'tilted' to keep its balance. Quote:

By analyzing the magnetic composition of ancient sediments found in the remote Norwegian archipelago of Svalbard, Princeton University's Adam Maloof has lent credence to a 140-year-old theory regarding the way the Earth might restore its own balance if an unequal distribution of weight ever developed in its interior or on its surface.
The theory, known as true polar wander, postulates that if an object of sufficient weight -- such as a supersized volcano -- ever formed far from the equator, the force of the planet's rotation would gradually pull the heavy object away from the axis the Earth spins around. If the volcanoes, land and other masses that exist within the spinning Earth ever became sufficiently imbalanced, the planet would tilt and rotate itself until this extra weight was relocated to a point along the equator.

Same goes for the lack of proof about the exact consequences on the planet or if it'll happen again. All that is still being researched and debated, but those Princeton scientists have thrown in some interesting perspectives which you'll discover when you visit the link I provided and completely read what they wrote.
Besides "consequences" for Earth as a planet, it should be noted that the tilt of Earth is the reason why we have seasons. So even when we yet have to find out what the consequences for Earth (as a planet) have been and/or will be, we do know that the tilt surely has consequences for all beings that live on planet Earth… the seasons that influence us all: summer, fall, winter, and spring.
And - last but not least - as for your question how they know where 0 degrees would be, you just have to look at the Earth's rotation around the sun. Take a 90° angle of the planet's orbit and you know where 0° would be. But I guess a picture explains more than a thousand words, so I quickly created the following graphic:


The Sun at solar noon (meaning that it is on the meridian) will be in one of these directions:

Due south if the declination of the Sun is less that your latitude. (For Brasilia, that occurs from Nov 6 to Feb 5, approximately)
Directly overhead (at the zenith) if the declination of the Sun is the same as your latitude. (For Brasilia, that occurs Nov 5 and Feb 6, approximately)
Due north if the declination of the Sun is more than your latitude. (For Brasilia, that occurs from Feb 7 to Nov 4, approximately)


Can this picture of the moon directly above the sun from the horizon be real?

No, it's not real.  Given that the moon appears to be nearly full, it would be near to the horizon opposite the sun, which is clearly not the case.  Also the framing of the sun and moon between the trees means that the sun and moon are on the same side of the trees, which is not possible when the moon is full.


How to determine the orbital period of an asteroid using Kepler laws given its orbit's radius?

Here's how I would do it. I'd convert everything to a single, standard set of units as recommended in the comments, and also stick to one digit before the decimal in scientific notation:

$$ T^2=\frac{4\pi^2r^3}{G\cdot M_{Sun}}$$

Using all numbers in the same units:

$r \ 4.5 \times 10^{11} \ (m) $

$G = 6.674 \times 10^{-11} \ (m^3 \ kg^{-1}  s^{-2}) $

$M = 1.989 \times 10^{30} \ (kg)$

$$ T^2=\frac{4 \cdot \pi^2 \cdot 9.11 \times 10^{34}}{6.674 \times 10^{-11} \cdot 1.989 \times 10^{30}} = \frac{3.60 \times 10^{36}}{1.33 \times 10^{20}} = 2.71 \times 10^{16} (s)$$

$T = 1.65 \times 10^8 \ (s) $

$\frac{1.65 \times 10^8}{365.25 \times 24 \times 3600} = 5.21 \ (years)$

Just fyi you can also call the product $GM$ the standard gravitational parameter and those numbers are more accurate than $G$ and $M$ are separately, as discussed in this answer.

The comments under your question also mention that it is sometimes easier to use AU and years as units and use Earth's orbit for scaling to the 3/2 power, but only for things orbiting our sun.  With $r = 3$ AU:

$T^2 = 3^3$

$T = 3^{3/2} = 5.2 \ (years)$


Supernovae increase in brightness over several days and decrease over months.  Thus, whatever you saw, was not a supernova, sorry.


How does the appearance of the analemma vary with latitude

Analemma is a diagram showing the deviation of the Sun from its mean motion in the sky, as viewed from a fixed location on the Earth. Note that it says nothing about time- it can be any fixed time. Analemmas created at different times of the day have slightly different shapes. For example, see the solar analemmas taken at the same place at different times.

Solar analemma at 0700 UTC by Anthony Ayiomamitis, from solar-center.stanford.edu

Solar analemma at 1400 UTC by Anthony Ayiomamitis, from solar-center.stanford.edu
The best way to understand the effect of latitude on the shape of the analemma is to consider the effect of earth's axial tilt on the shape of the analemma.

If the Earth’s orbital path was elliptical, but its axis was not tilted, the analemma curve would be oval shaped. At the Equator, this line would be a straight line spanning from left to right or West to East.
The 23.5-degree axial tilt of the Earth affects the Sun’s apparent position in the sky – as the year progresses and the Earth continues to spin at an tilted axis and orbit around the Sun, the Sun seems to move up and down (North-South) in the sky. This has the effect of generating the two loops of the figure 8.

You're correct that the tilt of the analemma varies with the latitude. Observers in the Northern hemisphere will obtain an analemma curve with the broader loop at the bottom. This reverses for observers in the Southern hemisphere, where the broader loop is on the top of the curve. At equator, the anlaemma lies on its side. At north pole, only the top of the analemma would be visible, while it is the opposite at the south pole.


Where can I get J2000 epoch initial mean longitudes for all planets and celestial bodies?

Take a look at this link. It included Program Codes Listing also (Appendix C). But if you only want the initial mean longitude and the angular veclocity... the appendix C.1 does have them listed at the beginning... in it, the values are L0 and n for each planet... (The n in the listing is in arcsecond, you have to divide it bye 3600 to get your desired degree value.)


A plot of the Moon-rise/sets on the Horizon Location

We are wondering if the moon wobbles back and forth along the horizon the same way the sun wobbles.

Yes, but over a period of 27 days compared to a full year for the Sun.
If you visualize a compass, the angle that a star rises and sets is the same angle measured from due south (for those in the northern hemisphere). This figure shows a star rising 60 degrees from due south (at an azimuth of 120 degrees). The same star sets 60 degrees from due south (at an azimuth of 240 degrees).

The rise angle and set angle can be calculated from
$$cos(angle) = -\frac{sin(dec)}{cos(lat)}$$
where $dec$ is the declination of the star and $lat$ is the observer's latitude.
The same calculation applies reasonably well to the Sun and Moon. (The apparent diameters makes a small difference, as does atmospheric refraction. Let's ignore those effects for now.) Due to the motion of the Moon during the day (and to a smaller extent the Sun), the change in declination changes the rise and set angles by a small amount - a few degrees at most. For the following graphs, I have ignored this motion during the day.
For the year 2021, you can see how the rise/set angle and azimuth of the Moon changes from "north of east/west" to "south of east/west" during one month (approximately). Of course, the Sun requires an entire year to go through the same cycle.

The azimuth changes because the declination changes, and the declination changes because the Earth's axis is tilted 23.5 degrees from perpendicular to the orbital plane. The Moon's orbit is inclined another 5 to 6 degrees from that. As Uhoh has mentioned, the inclination of the Moon's orbit sometimes adds to the declination and sometimes subtracts from it. This longer cycle takes 18.6 years to complete. The rise/set angle for a full cycle looks as follows:

Here is another factor that may be affecting your observations. Between New Moon and Full, the Moon rises during the day. It may be difficult to see the Moonrise until it gets close to full. During this time span, the Moonset occurs at night. (So "measure" the Moonset and calculate the set angle from due south.) The opposite occurs from Full Moon to New: it rises during the night and sets during the day.


The luminosity is $4\pi d^2$ multiplied by the integral under your SED.
i.e. you cannot find the luminosity without knowing the distance $d$ to the source.
NB This assumes no extinction (unless that features in your SED model), that the radiation from the source is isotropic and that the source is not significantly redshifted or moving relativistically.


How does gravity have an effect from the inside the event horizon of a black hole with the rest of the universe?

As is the case even in Newtonian gravity, the gravitational attraction from a spherical body exterted on a far-away object is identical the that exterted by a point mass located at the the center of the body. This is the so-called shell theorem.

To an outside observer, due to the infinite time dilation at the event horizon of a black hole (BH), all mass that has ever fallen into the BH has actually never reached the event horizon, but is instead placed in a shell around the BH, still falling ever-so-slowly toward the BH.

Thus, what we feel is in fact the gravitation not from a point mass, but from a shell of matter, but there is no way for us to tell the difference.

Now you may ask, "What if we drop a massive object, e.g. a planet, into a symmetric black hole? Since it never reaches the BH, won't the effective gravitational potential be asymmetric?" What happens is that the way that space is "warped" around the black, the gravitational field is bent along with it in such a way that sufficiently far from the BH+planet, the gravitational field looks symmetric. Sort of like the following drawing shows close to (left) and far from (right) the BH.


It's probably impossible to prevent chromatic aberration, since there is no such thing as a perfectly achromatic lens. but you can certainly minimize it.

As you have mentioned, for a given spectral range, the best way is to choose a lens that is said to be "achromatic" even though the correction is never perfect.

But if you want to use a lens that is not achromatic, then you can try the following to reduce the chromatic aberration:

Use a lens that has a very long focal length, for more on that see How does making a refracting telescope very long reduce the chromatic aberration of an uncorrected lens? I figured out the answer there and will try to post an answer soon!
Minimize your spectral range with a filter, or at least cut out the short wavelengths (e.g. blue). The dispersion of glass is stronger at the shorter wavelengths.
If you are doing color astrophotography with a color sensor, then there may be a way to shoot the three color images separately, bringing each color into focus separately. One problem is that for color sensors used to reproduce human color perception, the red channel often has a blue bandpass to replicate the way color receptors in our eye work. See answers to Filter for RGB separation and its effect on the image and the image below. So if you are trying this, then it might be helpful to add a blue filter, at least to cut below say 425 nm.


The relation between the energy of star and heating effect

I'm not sure what you mean by "heating effect". The amount another body is heated by a star with luminosity $L$ is quantified by its effective temperature:
$$T_{eff}=\left(\frac{L(1-a)}{16\pi\sigma D^2}\right)^{1/4}$$
where $a$ is the albedo, $D$ is the distance to the other body, and $\sigma$ is the Stefan-Boltzmann constant. If the luminosity is given, then there is no direct dependence on the size of the star.

The only thing that matters is the luminosity. You stated yourself that the luminosities are the same, which means that the red star in question is likely a red supergiant - a member of a group containing many of the largest stars in the universe.

Why, mathematically, is this the case? Well, stellar models can tell us that this is the case, but we can also figure this out via the Stefan-Boltzmann law by assuming that the stars are black bodies. For an object of radius $R$ and temperature $T$, the luminosity is approximately
$$L=4\pi\sigma R^2T^4$$
If star 1 is a blue supergiant and star 2 is a red supergiant, then, setting the luminosities equal, we have
$$L_1=L_2\to R_1^2T_1^4=R_2^2T_2^4$$
We know that $T_2$ is much less than $T_1$ - possibly by an order of magnitude - so $R_2$ must be much larger for the luminosities to be the same. This matches what we observe and what models predict.


How to explain that Polaris is a circumpolar star in the sky of any North America city?

In Miami, Polaris will be approx 25 degrees above the horizon, not 65. Height above the horizon is equal to the latitude of the observer.  At the North Pole, Polaris is very nearly directly overhead, all the time, because the Earth's axis of rotation points very nearly towards it.

Due to precession, Polaris will not always be so close to where the Earth's axis points.


Is the structure of crystalized white dwarf related to nuclear pasta?

It's thought to be crystalline in the same sense that diamonds are a crystal. The coulomb interactions between the positively charged ions becomes much larger than their thermal energies and they settle into a crystal lattice.

This both releases latent heat and increases the thermal heat capacity of the ions (roughly doubles it). This causes a "pile up" of white dwarfs with interior temperatures at around the crystallisation phase transition. However, the pile-up is not too steep because the freezing temperature increases with density and white dwarf interiors are roughly isothermal. This means they freeze from the inside out.

The crystalline structure is expected to be bcc.

I think the only thing this has in common with nuclear pasta is that they are both phase transitions. Nuclear pasta is the aggregation of nucleons (neutrons and protons) into wacky shapes in order to minimise the total energy density of the gas. The temperature is not really an issue here and it is all to do with density.


How could a neutron star collapse into a black hole?

The scenario you describe may occur. On the other hand it may actually be that neutronisation in a white dwarf is the trigger for a thermonuclear type Ia supernova.

You may be misunderstanding the Pauli Exclusion Principle (PEP).The PEP states that no two fermions can occupy the same quantum state, not that they cannot occupy the same space or be compressed to whatever density you like. The quantum states here consist of two spin states for every possible momentum state. In a degenerate gas, all these states are filled up to the Fermi energy. All that happens when the neutron star gets smaller (or collapses), is that the Fermi energy just keeps increasing as the neutron density climbs, and the neutron degeneracy pressure just keeps increasing as a consequence.

However, in General Relativity, pressure (like mass/energy) is a source of gravitational curvature and actually increases the required pressure gradient needed to support the star. At a certain threshold radius - a small factor larger than the Schwarzschild radius, a point of instability is reached where increasing the pressure is actually counter-productive.  Beyond this, you can make the pressure as large as you like and it will not prevent the formation of a black hole.

Even inside the BH there is not necessarily a problem with the PEP. You can compress fermions to infinite density so long as they can have infinite momentum.


Viewing 2017 Solar Eclipse - How important is it to be in the center(blue line)?

If you want to see a total solar eclipse, you just need to be between the two red lines on that map.  As long as you're within there, what we call the "path of totality", you will be able to see a total solar eclipse.

If you are outside of the red lines, you will only see a partial solar eclipse.  This map shows what percentage of the sun will be covered the further north or south you are of the line.

Graham County is entirely within the path of totality, so you should see a total solar eclipse.  If you're planning to travel, however, it might be too late.  Most campgrounds and hotels within the path are already full.  If you're already within the path of totality, my recommendation would be to stay put.  Travelling on that day, or even that week, will be difficult.

The center of the path of totality will offer the longest viewing for the eclipse.  The further from the center, the shorter it will last.  The aforementioned map shows timings at different distances from the center.  Notice that it also helps to be close longitudinally as well as latitudinally (that may not be a word).


Are there any known asteroids with average density similar to that of Earth's?

Apparently 21 Lutetia (~80x120 km) is one of the most dense asteroids known at about 3.4 g/cm$^3$. Unfortunately that is barely over 60% of Earth's density. 21 Lutetia is unique in that scientists believe that it has a primordial core, that is, it has a core from the formation of the solar system (and it hasn't been destroyed after all these years). We believe it is hidden under the surface. It would be rich in metals, explaining the density, and because it's hidden under the surface, 21 Lutetia would look no different from any other asteroid, which is the case. This asteroid is a prime target for mining because of its metal core. I suppose the answer to your question would be no, unless 3.4 is sufficient for your definition of "close".


Nature and mechanism of the short-term variability in radio strength of Sgr A*?

Nature of the variation in signal strength
From the paper Unprecedented variability of Sgr A* in NIR mentioned in Bit Chaser's comment, Do et al. record enormous variability in the strength of emissions from SGR A* in the NIR (Near InfraRed) band typically described at about 215 THz to 400 THz.  Here is a figure from their paper:

Here we see enormous variability on the order of tens of minutes.  In the radio spectrum, from Time Variations in the Flux Density of Sgr A* at 230 GHz Detected with ALMA written by Iwata et al., we see much less dramatic variability, without clear periodicity, with events again on the order of tens of minutes.  Here is a similar figure from their paper:

Iwata et al. also perform some Fourier analysis of their signal flux variability which reveals some "not-so-obvious" periodicities on the order of a minute or so.

Brinkerink et al. note that there are also periodic variabilities on longer timescales:

The emission from Sgr A* between frequencies of 20 GHz and 230 GHz
shows flux density variability of a few tens of percent on hour-long
timescales, up to 100% on month-long timescales, as well as occasional
flaring behavior (Dexter et al., 2013). In radio, Sgr A* has an
inverted spectrum (i.e., rising flux density with increasing
frequency) that peaks at the ’submm bump’, around 350 GHz, beyond
which the spectrum steeply drops in the infrared regime.

Proposed Mechanisms for the variation in signal strength
There are several proposed mechanisms for the signal strength variations.  Brinkerink et al. and Yusef-Zadah et al. propose a relativistic colliminated outflow from SGR A* that interacts with gas and dust nearby.  Yusef-Zadah et al. have this nice map from their paper: ALMA and VLA observations of emission from the environment of Sgr A*:

The authors also argue that

millimeter emission is produced by synchrotron emission from
relativistic electrons in equipartition with a ∼1.5mG magnetic field.
The origin of these is unclear but its coexistence with hot gas
supports scenarios in which the gas is produced by the interaction of
winds either from the fast moving S-stars, the photo-evaporation of
low-mass YSO disks or by a jet-driven outflow from Sgr A*.

Brinkerink et al. and Iwata et al. also talk about the synchrotron radiation, which occurs when charged particles are subject to an acceleration perpendicular to their velocity.  We would expect this for any "windy" star with a close periapsis path to the black hole, passing through the massive magnetic field associated with SGR A*.  An example of this cited in some of the literature is the S0-2 star, which has an orbital period of 16 years and the fastest known ballistic orbit, exceeding speeds of 1/60 the speed of light.


Yes.  Moons around Jupiter (Io, Europa and Ganymede), Saturn (Titan and Enceladus) and Neptune (Triton) all have some form of geological activity.  Charon also may have geological activity, being in a binary system with Pluto.  However, while Earth's geological activity is caused by internal heating and tectonic plates, the geological activity of the moons around Jovian planets comes in the form of tidal forces.  Io is the most iconic instance of tidal stress, because Io's plumes are frequent, volatile and make the world look extremely chaotic, with its surface frequently being altered and renewed by its non stop volcanic activity.  (Because it is chaotic)
As for tectonic plates, Europa is the closest you get to tectonic plates with moons in our star system.  Water replaces lava when it comes to ice worlds.  Ice worlds being worlds that have ice instead of rock for their crust.  This means that water mantles are a frequent occurrence, with the core of ice worlds being mineral rich stone.  This is the case for Triton as well, which has cyro-volcanism from the sheer tidal stress Neptune exerts on the captured dwarf planet.
Enceladus and Titan have water mantles, Enceladus being the world notable for its massive plumes, high albedo and tiger stripe surface fractures.  Titan may also have tectonic activity for similar reasons to Europa.


Are any bow shocks seen in the visible spectrum?

I'm far from an expert in this topic, but I was curious about this question and looked into it myself. Sadly, I was only able to come up with a pretty weak answer.

The very Wikipedia page you got the images from contains this line,

  Bow shocks are also a common feature in Herbig Haro objects, in which a much stronger collimated outflow of gas and dust from the star interacts with the interstellar medium, producing bright bow shocks that are visible at optical wavelengths.

But it is not cited. Perhaps it should have a [citation needed]. That sentence does link to this page about the Herbig Haro Object, which does contain a cited line,

  HH objects associated with very young stars or very massive protostars are often hidden from view at optical wavelengths by the cloud of gas and dust from which they form. The intervening material can diminish the visual magnitude by factors of tens or even hundreds at optical wavelengths. Such deeply embedded objects can only be observed at infrared or radio wavelengths.

So it seems to me the answer to your question is that no, images of bow shocks in visible light are not produced because the visible light from the bow shock cannot reach us due to there being stuff in the way.

I understand that an answer based on 2 Wikipedia pages as sources is pretty weak by Stack Exchange standards, and that these Herbig Haro Objects are not the only thing in the universe that create bow shocks, and therefore the possibility exists that visible bow shocks around other objects may exist. I hope this answer inspires an expert to give an answer deserving of that green check mark.


Earth's precession and the position of the Sun in the sky

Yes, the Gregorian calendar is based on a good approximation of the mean tropical year, so the dates of the equinoxes are stabilised.
Geometrically, the equinoxes occur when the Sun's apparent path on the ecliptic crosses the celestial equator. The March equinox point (aka the First Point of Aries) is the 0° point for ecliptic longitude and also the 0 hour point for Right Ascension.
Due to the precession of the equinoxes, the First Point of Aries moves relative to the stars by about 50 arc-seconds per year, which gives rise to the 72 year value you mentioned.
The upshot of this is that a table of RA and declinations of stars must specify the epoch that it uses, so that the given positions can be adjusted to give the actual RA on a given date. A very popular epoch for the last few decades is J2000 which is 12:00:00 noon (Terrestrial Time) on the 1st of January, 2000, which is Julian date 2451545.0 TT. In UTC, that's 11:58:55.816.
There's a brief explanation on Wikipedia of the effects of precession on Right Ascension.


First, I feel like the article is confusing, so it would be better if you read peer-reviewed articles instead.

(If I recall correctly, Kaspi & Beloborodov 2017 per your reference discusses more towards observations. It might be easier to understand if you check theoretical papers instead).

  How does the field decay of a magnetar power the emission of high-energy electromagnetic radiation?

A young magnetar (i.e., strongly magnetized neutron star) has two main energy components: rotational (aka. spin) and magnetic. Rotational energy has shorter timescale as shown in Eq. 2, i.e. ~ one year (https://ui.adsabs.harvard.edu/abs/2010ApJ...717..245K/abstract). Magnetic energy would dominate later (see https://ui.adsabs.harvard.edu/abs/2012Ap%26SS.342...55G/abstract; note, I have never read this paper but I notice a good number at its abstract). Therefore, this gives you some sense that timescale matters. You might want to also check how a magnetar/pulsar links to the pulsar wind nebula, accelerated particles, and radiation (e.g., https://ui.adsabs.harvard.edu/abs/2014MNRAS.437..703M/abstract).

  Is it as simple as ∂/∂ produces an electric field which accelerates charged particles that happen to be there, and then those accelerating particles radiate photons?

Check Eq. 14 of the second paper. Simply yes. Note: not all particles radiate.

  Does the energy radiated by these photons cary any significant fraction of the total energy in the field?

I don't understand this question. If you mean the total radiation field, significant fraction of energy in the radiation depends on factors such as timescale.

  Is the word done producing these photons the reason that the magnetic field decays?

Not sure what you are referring to, but it sounds like a "not quite so" to me.

noted:

Pulsars are objects when observed we see pulse of light. Some pulsars are white dwarfs, while most of them are neutron stars.
The misalignment of rotation and magnetic-dipole axes + beam direction relative to observers are important.
While a young magnetar might produce strong jets, which can create the pulse, we might not observe the pulse because its environment does not support. You might want to check "choked jets" (https://arxiv.org/abs/1906.07399).


Where can I find Helium 3 in our solar system?

You wouldn't need to travel too far, the moon has been bombarded with large amounts of Helium3.

https://www.esa.int/Enabling_Support/Preparing_for_the_Future/Space_for_Earth/Energy/Helium-3_mining_on_the_lunar_surface


Maritime telescopes: Stabilisation requirements for optical vs. radio telescopes?

You asked for arguments against this, but I don't think there are any other than money.  Consider the AN/SPQ-11 passive radar that was aboard the US Navy ship Observation Island.  The actual receiver area is not as large (7M according to https://fas.org/irp/program/collect/cobra_judy.htm), but we could imagine a larger antenna being placed on a larger ship such as the USS Enterprise.  The beam (maximum width) of the Enterprise is 41 meters, which could certainly accommodate a 10-20m dish.
I think the only reason this would not be built would be lack of funds.  From an engineering perspective, it seems doable.


What would be the average distance between each 1-degree latitude on Jupiter?

Yes, they have coordinate systems as well. For example, the Wikipedia page on the Great Red Spot mentions it lies 22 degrees south of the planet's equator. Jupiter has a polar radius of 66,854 km; a latitude degree is $2 \pi / 360$ times that, so almost 1,167 km.

Note that because Jupiter lacks a solid surface, longitudes are less well-defined. In fact, there are three different systems for this:

  Because Jupiter's visible features do not rotate uniformly at all latitudes, astronomers have defined three different systems for defining the longitude.

Rocky planets like Mars do have a well-defined coordinate system, as shown in this map and on this page.

These coordinates are used to mark and find interesting places on either the surface (like on Earth) or in the atmosphere of a planet. There are also tables available showing which latitude/longitude is directed towards us at a certain moment, so that you can compute whether a particular region of the planet is visible from Earth or not.


Estimating a star's radius, temperature, and luminosity based on its mass

You could plot $L$ vs $T$ and compare your model result to the measured/estimated $L$ and $T$ of main sequence stars?

You have not explained or labelled what your diagram is, but it isn't $L$ versus $T$; I suspect that it absolute $V$ magnitude versus $B-V$ colour.

The transformations between $L,T$ and $V$, $B-V$ are highly non-linear, especially at low temperatures (which is why your plot fails to reproduce the M-dwarfs) and involve folding a model atmosphere through some standard filter responses or using relationships between $T$ and $B-V$ and bolometric corrections and $T$. (e.g. Mamajek 2019)

EDIT: For the clarified question.

Your relationship for $T$ vs $B-V$ is not applicable at lower temperatures.

Your relationship between absolute magnitude and luminosity takes no account of the spectral energy distribution of the light from the star (i.e. it assumes the same bolometric correction for all temperatures, which isn't correct). You must calculate the bolometric magnitude from the luminosity and then apply a temperature-dependent bolometric correction to calculate a V-band magnitude.

The solution is to construct relationships that are valid over the full range of temperatures using the resource I referenced above.


Is it thought that the majority of Earth size or larger terrrestrial planets have retained most of their original atmospheres?

Too long for a comment, so . . .

It's important to remember that the field of primordial atmospheres has a good bit of uncertainty to it.  I've tried to read up on it as a layman and there's a good deal of contradicting theories and not much certainty.  Your first article addresses this uncertainty several times.

  The reason I mention the ratio of CO2 to N2 is because I read
  here, for example, that out gassing produces the molecules in
  about a 4.6:1 ratio.

Could you pull the quote to that 4.6:1 ratio.  I didn't see it in that article and that sounds much too accurate for an estimate in this field.   4.6 to 1 suggests about a 2% range of uncertainty.   I find that hard to believe.   Certainly a ratio with that kind of accuracy should cite a respected research study.

  And noticing that the ratio of CO2 to N2 is far higher on Venus than
  that just expected from out gassing

What ratio is expected from outgassing?  Your 2nd article mentions a neat and tidy percentages:

  The composition of outgassing is similar for Venus, Earth and Mars and
  is composed of 58% H2O, 23% CO2, 13% SO2, 5% N2 and traces of noble
  gases (Ne, Ar, Kr). The latter evolution of this outgassing is driven
  primarily by the surface temperature and chemistry of the planet.

But I call BS, cause one, this doesn't site a source and some basic google-research on Earth volcanic activity suggests the numbers are far from that neat and tidy - see charts below.

Source

and

Source

The 2nd source, above, doesn't even mention Nitrogen (N2), Wikipedia does, though they don't give percentages beyond 60% for water, 10-40% for CO2.

Earth's volcanic gas today is probably not a good measuring stick because of 3 plus billion years of life (some nitrogen fixing, lots of long chain carbon molecules) and some 4 billion years of plate tectonics.

And to my knowledge, a sample of volcanic gas from Venus has never been studied.  In fact, there are no known active volcanoes on Venus at all.   Source.   Outgassing can perhaps be estimated by atmospheric content, but I'm skeptical on the accuracy of those numbers they gave.  I think your first article from NASA far more reliable.   They quote a ton of sources and several times in the article, indicate legitimate uncertainty.

What I took away from the NASA article:

Yes, when the solar system was very young, the t-tauri phase of the sun blew most of the atmosphere off the 4 inner planets (or 3 Earth-like if you want to ignore Mercury, which the first article does).  This is backed up in part by the lack of Xe-129 which should form as a gas based on the 15 million year half life of I-129.   If the atmospheres weren't blown off, we should see a certain percentage Xe-129 in the atmospheres of Earth/Mars/Venus.   The percentage is much too low, suggesting the blow-off did happen over the first, perhaps 100 million years, continuing well after the end of the T-tauri phase, when the sun was still very active with solar flares.

The giant impact that is thought to have created the Moon (about 30 million years after planet formation) may have blown off a chunk of Earth's atmosphere as well, but the Solar wind probably did too.

Where the primordial atmosphere came from? Whether out gassing or comets (or a combination of both), I don't think that's a resolved question.  What seems true, based on the first article, is that Venus's water evaporated and it lost it's water by a combination of photo-chemistry and it's higher temperature and perhaps some jeans escape.   No water on it's surface may have slowed down any hope it had for plate tectonics and that reduced the rate that it outgassed.    (a general summary of the NASA article).  To get a better idea of Venus history, they'd like some rocky samples, but they don't have any.

Earth maintained it's oceans, perhaps because it had more water to begin with, or a lower temperature or combination of the two.  The lunar impact may have sped up Earth's outgassing as it liquified the entire crust for thousands, perhaps as long as a couple million years.  It's unclear if Venus ever had it's crust liquified.

But after Earth cooled after the giant impact, it had enough water for oceans and oceans made it possible to form carbonate rock and remove much of the CO2 from it's atmosphere.  Life assisted with the removal of CO2 and fixed some nitrogen too.  Ammonia also dissolves in water, so much of the nitrogen that came to Earth by comets could have been stored in the oceans, while ammonia that came to Venus by comet would have been vaporized as gas and light enough where some of it would inevitably be lost to jeans escape and subject to photo-chemical dissociation, and, a weaker magnetic field.

Earth and Venus have similarish amounts of CO2, it's just that Earth has most of it's CO2 stored in carbonate rock.    Earth and Venus may have similar amounts of Nitrogen too, though Venus atmosphere has about 4 times as much as Earth's atmosphere, currently, that doesn't take into account Nitrogen in the oceans and fixed or bonded within the Earth.

I don't think Venus "extra" CO2 is a mystery because it didn't have oceans long enough to pull the CO2 out of it's atmosphere.  We also don't know enough about the materials that formed Venus might have been different than those that formed Earth, and we don't know that much about the process that followed, so I don't think it's fair way to say it has too high a CO2 to N2 ratio.

It's entirely possible that the giant impact did blow away a portion of Earth's atmosphere, leaving it somewhat atmosphere depleted compared to Venus, but your NASA article points out that there's some uncertainty on how much atmosphere the Earth had after the giant impact leading to uncertainty on how quickly the Earth cooled.

It could, I think, work in the other direction, where the enormous heat of impact and liquification of the crust may have enabled more atmosphere to outgas, giving Earth a thicker atmosphere early on but post impact, where, the Venus equivalent of that gas might still be trapped inside Venus.

There's a lot about this subject that remains unresolved.   This is just my takeaway from the articles.   Take this answer with a grain of salt.


Will this affect image quality?  It depends on "where" the dust is located.  If on the telescope, then not very much.  If on the sensor, you'll get the shadows of the dust on your images (use of dark-frames can be used to "subtract" the dust shadows when calibrating the images via software).
Based on the sharpness of dust, I suspect your dust is probably very close to the image sensor.
When light travels through the telescope, you can think of it as taking all possible paths to reach the sensor.   This means that while some photons are blocked by the dust, other photons are taking paths that bypass the dust.  The dust wont show up on the image because the farther away it is much too far away from any focal plane -- so it's completely out of focus to the point of being invisible.  But as some photons are technically blocked... it may fractionally reduce some contrast if the dust gets pretty bad.
The closer the dust is to the image sensor, the easier it is to spot.  Since your dust shows up fairly clearly, my guess is that it's fairly close to your camera sensor.  You mentioned the telescope is a C14, but you did not mention the camera and/or any filters being used.
Cleaning Telescope Optics
After about 4 years without any cleaning, I removed to the dew shield and found this:

I spent a few hours very careful cleaning the corrector plate (which is where I suspect you'll find the bulk of dust) using Dr. Clay's cleaning solution recipe (you can find that here:  http://arksky.org/asoclean.htm ).
The technique has you gently daubing a lint-free cleaning pad with cleaning solution applied to it.  This pulls off the dust but leaves the optics rather messy.  Avoid rubbing action while the optics are still caked in dust as this will just drag the dust around on the surface.  Again, the corrector plate is much more durable than mirror surfaces and I generally never need to clean the mirror surfaces since Schmidt Cassegrain Telescopes aren't "open" to the air.
Once the dust is mostly eliminated, I'll give it a second cleaning pass and this time I use gentle cleaning action (almost no pressure).
Lastly, I'll use a mist of distilled water as a rinse and daub dry with a clean microfiber cleaning cloth.
The result was this:

While this "looks" clean, my advice is ... never get too close to your telescope optics while holding a bright light.  You'll see all kinds of imperfections.  My advice is to leave them be -- getting too aggressive can result in damage.
With that aside, a corrector plate is not the same as a first-surface mirror.  Mirrors have a soft coating and are easily damaged.  Corrector plates have anti-reflective coatings, but they are hard and not so easily scratched as long as you only use very mild cleaning solutions and are gentle.
Cleaning Camera Sensors
I use a solution called "Eclipse" (the manufacturer is "Photographic Solutions, Inc." and it is available from most stores that sell supplies to professional photographers.  Eclipse is nearly pure methanol.  This means it will act as a solvent to remove not only dust, but oil splatter (some DSLRs will occasionally splatter oil if the reflex mirror was generously lubricated).  Since the substance is nearly pure methanol it will evaporate almost completely and leave virtually no trace.
I tend to clean camera sensors in phases of aggressiveness.

Some cameras have a self-cleaning cycle ... a piezoelectric effect which vibrates a filter in an effort to shake the dust loose (helps to point the camera down while doing this.)

Use a soft blower with ordinary air.  Avoid canned compressed air because those usually aren't "air" and the propellant in the can can leave a haze on the camera ... and that has to be wet-cleaned.

Use a very soft and pristinely clean paint brush (not the type you use to paint a house... the type an artist uses to paint a picture.)  There are versions of this that have a grounding wire which can help release static cling.

Wet cleaning.   This is where I use the Eclipse solution.  The same vendor makes lint-free cleaning pads (PEC Pads).  It only requires a few drops and often just one or two wipes to clean the sensor.

After each of these phases, I inspect the sensor with a magnifying loupe and/or take a sample image.  Cover the front of the telescope with a piece of plain white fabric (e.g. white t-shirt is a popular choice) and stretch it so that there are no wrinkles.  Set the camera to a middle exposure so that the image comes out as a middle gray.  The telescope need not be focused.  Inspect the images carefully to for signs of dust shadows.  Keep in mind that the images on the camera is rotated 180°.  E.g. a dust speck appearing in the upper-right corner is really in the lower-left corner on the camera sensor.


My first thought is that you saw a flare from a satellite, meaning a favorable reflection off of a solar panel or something. I use the software "Heavensat", and it shows the satellite Nextsat (Norad number 30774) passed very close to Arcturus at 23:40 pm from your location (45.7288° N, 24.3784° E). Note that I am assuming that your time is 3 hours ahead of UT (Universal Time).

I do not know if Nextsat typically flares, although reflections from satellites are common. Nextsat is in a low altitude orbit, so its motion would normally be detectible. Perhaps the duration of the brightening was shorter than you estimate (so that the motion was not detectible), or it was so unexpected that you did not notice the motion.

My second thought would be a head on meteor like you suggested.


This kind of location is called a peak of eternal light.

Apparently there is no certain identification of such a peak on the Moon, but there are definitely places that get more than 86% sunlight during the lunar year (ignoring eclipses). Kaguya did not find any permanent peaks, but there is apeak near Peary at the North Pole that gets 89%, and several places near the South Pole that have multi-month illumination.

Calculating the requirements for such a peak is complicated since the moon is definitely not spherical on the hundreds of meter scale; at the very least one needs an ellipsoid model. The south pole situation is complicated because there is a large basin around the actual pole.


Can lunar occultation of Venus occur during solar eclipse?

Yes, but such an event would be very rare. Venusian transits last only about 7 hours, and there are only two in 113 years, So Venus is in transit for about 0.0015% of the time.

So as a back-of-the-envelope calculation, you would need to see about 70000 eclipses before you had a combined transit/eclipse.  There are about 2 solar eclipses a year, so you might hope to get this unusual syzygy once every 35000 years or so.

None are forecast.


I just went through this by myself.  The principal thing you have to do is choose the vector format for your ephemeris, and then download the ephemeris as a plain text (ascii) table. This is way easier than using the Spice tools.


One should distinguish the "Roche limit" from the "Roche lobe," as the latter is what you need when you have two objects of similar mass, like a binary star.  The Roche lobe is an effective equipotential in a frame that cororates with the orbit of the binary (assumed circular).  The equipotential is "effective" because it includes both stellar gravities, and also the centrifugal force from the rotating frame.  The Roche lobe is not only an effective equipotential, it is the last one that is closed-- any equipotential further out connects to infinity and cannot correspond to material in equilibrium that is corotating with the orbit.  From the side, this last equipotential looks like a kind of infinity sign.  Material outside that "infinity sign" has to start moving in the rotating frame, bringing in coriolis forces and in generally leading to complicated motion.  However, if close to the Roche lobe, the gas tends to flow through the nodal point at the center of the "infinity", so tends to be transferred from the domain of gravitational attraction of one star to the other.  That's "mass transfer."

Setting up the calculation of the Roche lobe is fairly easy because you only need the two gravities and the centrifugal force at the orbital period.  But it is a difficult set of equations to solve, and even at the research frontier, is usually approximated rather than calculated exactly.  No doubt this is the source of your difficulty.

Why one star fills its Roche lobe is a consequence of stellar evolution, often as one star attempts to evolve into a red giant or asymptotic giant.  Before it can reach its giant radius, its outer layers encounter the Roche lobe, and pass along it through the central node (the "Roche point") and may ultimately end up attracted to the other star.  This is what is thought to have happened in Algol, for example, which is a binary where the star with the greater mass now started out as the star with the lesser mass (because it gained from its companion).  The rate of mass transfer is also difficult to compute, because it depends on how fast the mass-loser is evolving.  It is only its evolutionary rate that causes the "Roche overflow," so the mass transfer rate is pegged to the evolutionary rate, which is largely due to rate of fusion that adds mass to its core.


Are the bright dots in this image individual stars?

A query to the Gaia Archive within 6 arc min of target name M51 finds 20 sources with phot_g_mean_mag <= 17.
Aladin Lite helps to locate them visually.
Five are circled in the lower half of the image in question:

Gaia source_id
ra °
dec °
parallax (mas)
magg

1551894661845524736
202.4034
47.1522
8.854 ± 0.017
14.3

1551988532650838016
202.4205
47.1790
2.387 ± 0.041
16.4

1551988670089792256
202.4236
47.1979
0.632 ± 0.025
15.1

1551988425277200640
202.4323
47.1537
0.813 ± 0.025
15.2

1551988567010595840
202.4417
47.1800
1.536 ± 0.012
13.3

These parallaxes indicate distances between 113 and 1600 parsecs, well within the Milky Way.
Of the three sources circled in the upper half, two seem to be listed in Gaia EDR3 but do not have good parallaxes there.
Aladin Lite shows SIMBAD matches near all three in
Hwang and Lee 2008, a catalog of 2200 star clusters in M51.

HL2008 id
ra °
dec °
magV
Gaia source_id

89770
202.494
47.216
20.0
1551988944968489984

13162
202.505
47.154
21.1
--

28066
202.531
47.167
19.8
1551987909882986368


Position of the Sun in the constellation Pisces around the year 1500

The position of the sun, relative to the stars, at the time of the equinox changes slowly and predictably.  This means that we know to a high degree of precision exactly where the sun was at the time of the equinox in 1500.

It is pretty easy to calculate  The position of the sun at the time of the equinox changes by about 1 degree every 71.3 years. The sun is currently in pisces at the time of the Equinox.  520 years ago it would have been 7.3 degrees closer to Aries.

Dbachmann / CC BY-SA (http://creativecommons.org/licenses/by-sa/3.0/)