Exposure time for spectroscopy compared to photometry

Your question is a bit unclear. Compare in what sense?

The problem boils down to that you have a fixed number of photons in a wavelength band and you can choose, with the appropriate filters or dispersion elements, how many "bins" to divide these photons into.

To accumulate the same total number of photons with the same telescope will take the same amount of time, except putting more optical elements in the way will reduce the efficiency of the instrument. Filters are more efficient than spectrographs by factors of a few to an order of magnitude.

However a further factor to consider, is that although in principle we could divide the light up as finely as we want, there may be static sources of noise that are independent of the signal. In the case of a CCD detector, this means the read noise.

If you want read noise not to limit the observations, then the product of the bin size and exposure time needs to be large enough that the square root of the accumulated counts (including any sky background) in a bin exceeds the read noise.

An example might help. Suppose I have a star that I observe for 10s through a V-band filter with a width of 100 nm and it produces 1000 detected photons on a CCD, where the star image effectively illuminates 10 pixels, each of which has a read noise of 10 counts. The total read noise is $10\sqrt{10} = \sqrt{1000}$ counts (the square root of the total variance) and equals the Poisson noise from the source plus sky. Ignoring the sky, the signal to noise ratio is $1000/\sqrt{1000+1000}=22$.

Now suppose I do slitless spectroscopy, passing the starlight through a diffraction grating, splitting the light into 1 nm bins. My signal strength in each bin is divided by 100, but the read noise is approximately the same. To obtain a spectrum with a similar signal to noise ratio you would need to expose for 100 times longer, i.e.1000 s (probably longer because a diffraction grating is not as efficient as a filter).

So there is no general answer to your question, it depends on the spectrograph efficiency, the "resolution" of the spectrograph, the relative contributions of source, sky and read noise and the signal to noise ratio desired.

A simple rule of thumb to get an exposure that is just limited by read noise for an imaging filter of wavelength width $w$ and a spectrograph with a resolution element $\Delta \lambda$, would be to observe for a factor of $w/\Delta \lambda$ longer, with an extra factor of a few to account for poorer spectrograph efficiency.

Edit:
To test my ideas I go to the Isaac Newton group of telescopes' signal to noise calculator.http://catserver.ing.iac.es/signal/

I try observing a 20th magnitude star for 10s with a V filter on the WHT prime focus, assuming 1 arcsec seeing and no moon. This gives me a SNR of 46.

To get the same SNR in a low resolution spectrum withe ISIS spectrograph and 158R grating (lowest resolution available) takes 15,000 s.

Does this stack up with what I said above? Each pixel in the spectrum is 0.16 nm. If the V band filter is about 100nm wide, my crude formula above suggested an increase in exposure time of 100/0.16 = 625. But the instrument throughput is only 33% (compared with essentially 100% for the filter) so we have to add another factor 3, to get a required exposure that is 1875 times as long as for the photometry. Pretty close!


What would happen if the earth's orbit were slightly perturbed?

You're touching on a lot of related topics which isn't recommended for Stack Exchange.  It's better to keep it to one question and specific.   That said, I'll give this an answer as best I can, as I've been interested in this subject for a while and have read up on it a bit.   The powers that be might close this question for being too broad and/or lacking specifics on the specific details of the rogue planet that affects Earth's orbit.

The subject of the orbital systems around different stars is of interest to astronomers and it has been modeled, many times in fact and observed, though it's very difficult to get a good look at other solar systems.  Most of what has been observed is probably incomplete and a lot of what has been modeled involves some guess-work so it's an ongoing field of study.

  I have to think that astronomers have simulated solar systems and come
  to some conclusions about that. Is anyone familiar with that and can
  say something about it?

Yes, astronomers have studied and modeled this.   It's worth pointing out that models will only give the right answers if you ask the right questions.   The fact that astronomers didn't expect, but found many hot Jupiters is one example of the problem with running models vs learning how real systems are likely to be set up.

This is one of my favorite videos on solar-system formation which touches on your question, though this one focuses on inner-planet formation, it's entirely mathematical modeling and it suggests that 3 gas giants tends towards instability while two can be stable.  It's a little old now, but I think it's still accurate information.   The modeling begins about 20 minutes into the video and the 3 gas giant chaos is shown around 24 minutes.

Scholarpedia also has a nice article on the current solar system's stability.

When you write this:   "I have the idea that, in a complex system, orbits would tend toward roughly circular and in the same plane."

I think it's safe to say that there are both stable and unstable systems and an unstable system, after it kicks enough planets or other material out, will become a stable system.

I think it's also safe to say that the proto-planetary disk is what causes planets to form on the same plane, but planetary migration may change that.   The sum of inclination should remain fixed, but ejected can change that.

Our solar system, the 8 known planets orbit about 6 degrees off the Sun's equator and planet 9 is one possible explanation for that, but an ejected planet could explain it too.

  So I have the idea that in a complex system, where the different
  bodies can affect each other's orbits,

This is probably true and it's still true in our solar system.   Jupiter and Saturn affect Venus, Earth and Mars Milankovich cycles.    Mars, being the closest to Jupiter is especially prone to large axial tilt shifts and eccentricity shifts.     Jupiter might, in time, even pull Mercury entirely out of it's orbit where it could move significantly.  The odds may be low, but if that happens, once it starts, Mercury could pass close enough to Venus or Earth and undergo a gravity assist and basically end up anywhere, possibly crashing into another planet or crashing into the sun.

There is a circularization effect caused by tides, but it only applies in 2 body systems where the rotation is faster than the Orbit and the tidal effect pushes the orbiting object outwards.   There seems to be a circularizing effect for mutually tidally locked systems like Pluto-Charon but planetary perturbations prevent planets from having neat circular orbits in multi-planet systems.   They can still be long term stable, but there is eccentricity variation.

The only planet in our solar system with a nearly perfect circular orbit is Neptune and I suspect it's because it's the outer-most planet.   Outer planets tend to perturb the orbits of inner planets, but less so the other way around.  At least, that seems true.  I can't give a good mathematical reason for that, it's more loosely deducted on my part.

The inner 4 planets have eccentricity that changes over time (one of the Milankovich cycles, driven primarily by outer planets).   Mars' line looks more straight, but it's on a different axis, it actually changes the most.

Source of image

  they would tend to become circular and in a plane

I wouldn't make this assumption.  Though the proto-planetary disk likely begins in a plane over the star's equator, some systems have been found that don't orbit in a plane.   A plane may be most common, but I suspect that's because all systems begin that way more than they tend to become that way.

See here and here

To your main question,

  Suppose a rogue planet came zipping through the solar system and
  changed the earth's orbit a little, including the plane of the orbit.

I would think that if a rogue planet changed Earth's eccentricity, that Jupiter over time would return Earth to about where it was.   Maybe with the inclination too.  I couldn't find any good information on inclination changes over time by orbital perturbation.

If Earth's semi-major axis was changed, that seems to be more constant and that would change the length of Earth's year and might might be a much more permanent change.  Likewise, if the Semi-major axis was moved into resonance, say with Jupiter or Venus or Mars, . . . that might be bad, as it might open the door to repeating perturbations and instability.

That's my understanding anyway.  Corrections welcome.


Can a laser pointer ruin a long exposure photo?

Can laser pointers be used?
Yes and no.  If you are attending a Star Party hosted for astronomers and astrophotographers, then laser pointers are usually among the items listed that you may not use.  Some star parties hosted for astronomers (not the public), may schedule an event when a tour of the night sky (via laser pointer) is performed and all astrophotographers know this and aware that they may not want to be imaging during that particular session.  Laser pointers are not used or permitted at any other time.
If the star party is being held for the public (not for astrophotographers) then laser pointers are often used during sessions on each night and astrophotographers know that this.  They might be used at any time and without warning.
Will they ruin images?
This is a "it depends" answer.  Laser pointers used by amateur astronomers are green (typically 523nm) lasers.  This particular wavelength is used to because human eyes are very sensitive to it (it is smack in the middle of our visual spectrum) and it reflects off the particles in the air showing an obvious beam.  Other colors can be used... but unless the laser is especially powerful, the beam isn't visible.
Professional astronomers will use lasers to create an artificial star for purposes of adaptive optics.  These lasers excite gasses in the sodium layer (about 80-100km above sea level) and since the gas is sodium, they use a sodium laser to excite these gasses to create the artificial star.  Sodium lasers are yellow.  These are especially powerful lasers.
Meanwhile, the lasers amateur astronomers use are typically not very strong.  Though laws & regulations may vary by country or local government rules, these tend to be fairly low power lasers.  Here in the United States they are typically 5 milliwatts or less.  It turns out your ability to see these lasers really depends on how close you are to the source.
A 5mw laser is easily visible to people just standing a few feet away, but people a bit farther away will struggle to see it.  People on the far side of a field wont see it at all.
When I do astronomy outreach, I typically use a 5mw laser and those standing near me can see the beam.  If I were to do a tour of the night sky using a laser for a moderately larger crowd (say... 25-50 people) then I'd need a stronger laser ... maybe 50mw ... otherwise those farther away will struggle to see the beam.
If you're out with a few friends and worried you might be ruining images for someone taking images 10 blocks away... no worries.  Their cameras will not detect the laser beam.
Safety
As always, use proper care when operating a laser and avoid pointing them at people or aircraft.  This keeps the local constabulary from having to deal with the situation.


Does matter accumulate just outside the event horizon of a black hole?

What you're describing is basically the "collapsed star" (Eng) or "frozen star" (Rus) interpretation of black holes that was common prior to the late mid-1960s. It was a mistake.

Suppose you are distant and stationary relative to the black hole. You will observe infalling matter asymptotically approaching the horizon, growing ever fainter as it redshifts. Does it mean that matter "clumps" around the horizon? To find out, suppose you throw yourself towards the black hole to try to catch the matter that you see. What you will find is that it fell into the black hole long ago.

In other words, the most sensible way to answer whether or not infalling matter clumps on the horizon is to look at the situation from the frame of that infalling matter. And there, it is clear: no, it does not clump, as it crosses the horizon in finite proper time. (As an aside, for a Schwarzschild black hole, falling from rest is exactly Newtonian in Schwarzschild radial coordinate and proper time.)

The "comoving viewpoint" was recognized by Oppenheimer and Snyder in 1939, but it was not until the 1960s, with the work of Zel'dovich, Novikov, et al., that it was generally recognized as truly significant in the community. In 1965, Penrose introduced conformal diagrams based on the Eddington-Finkelstein coordinates (1924/1958) that showed quite clearly that the stellar collapse is not slowed, but instead continues to a singularity. For an overview of the history of this change of viewpoint, cf. Kip Thorne, et al., The Memberane Paradigm (1986). These topics are commonly covered in many relativity textbooks.

Ok, but since it still takes an infinite amount time in the frame adapted to a stationary distant observer, does that mean that the horizon never forms in that frame? It does form: the underlying assumption in the argument that it does not would be either that the infalling matter needs to reach the center for the horizon to form or cross a pre-existing horizon to make it expand. But that assumption is simply not true.

An event horizon is defined in terms of future lightlike infinity, roughly speaking in terms of whether or not light rays escape if one waits an infinite amount of time. That means the location of the horizon at any time depends on not just what has happened, but also what will happen in the future. In the frame of the distant stationary observer, as matter falls towards the event horizon, it does slow down to asymptotically approach... but the horizon also expands to meet it. Similarly, the initial collapsing matter does not need to collapse all the way to the center for the event horizon to form.

  How can the finite life-time of the Black hole due to Hawking radiation be made consistant with the infinite amount of time (future) needed for the expansion of the event horizon (in the outer time-frame)?

There's no need to: [edit]that a particular time coordinate doesn't cover the full manifold is a fault of the coordinate chart, not of spacetime[/edit]. From every event, send out an omnidirectional locus of idealized light rays. The event horizon is the boundary of the spacetime region from which none of these light rays escape to infinity. This question has an objective answer--for any given light ray, either it will escape or it won't.

An external observer would need to wait infinitely long to know for sure where the event horizon is exactly, but that's a completely different issue.  With Hawking radiation, the black hole shrinks, but it doesn't change the fact that light rays from some events will fail to escape, and thus that an event horizon will exist.

Here's a Penrose diagram of a spherically collapsing star forming a black hole that subsequently evaporates:

Light rays run diagonally at ±45° on the diagram. Note that there is a region from which outgoing light rays (running diagonally lower-left to upper-right) don't escape and instead meet the $r = 0$ singularity (the bolded, undashed horizontal line). The horizon itself is the the $r = 2m$ line marked on the diagram and its extension into the star: it should actually go from the (dashed, vertical) $r = 0$ line on the left, rather than extending from the star's collapsing surface. That's because some of the (idealized, noninteracting) light rays from inside the star will also fail to escape to infinity.

Now suppose that on this diagram you draw timelike curves that stubbornly stay away from the horizon, and you insist on using a parameter along them as a time coordinate. Does the fact that you've chosen coordinates that exclude the horizon needs to be made consistent with whether or not the event horizon actually exist? The resolution is simple: if you want to talk about the horizon, stop using coordinates that exclude it.


Is there a database for identifying the falling light I saw?

This could be a satellite, such as the ISS, which appears as a bright, fast moving star. You can look for times when the ISS is visible on websites like http://www.heavens-above.com/

It might be a meteor (not a meteorite, that is the stone, not the streak in the sky). Meteors travel much faster than satellites. On any given night you will see several meteors each hour. A meteor that is "as bright as Mars" (which varies in brightness, and can reach -3 but is more likely to be between -1 and 0) is a less common event, but not exceptionally rare. Reports of Fireballs (those with magnitude -4 or brighter) are collected by various bodies, such as the uk meteor network


What causes "North-South ambiguity" when doppler radar imaging a planet surface equator on?

It's a great question!

Delay-Doppler imaging is a technique to produce a kind of image that represents the topography of a rotating body, usually when it is so far away that there's no way to resolve the object with traditional imaging, radar or optical.

A powerful radar signal is sent from Earth to an asteroid or a planet, and the reflected signal returning to Earth is recorded and analyzed. The "special sauce" that allows the technique to work is the pseudo-random sequence encoded into the modulation of the outgoing signal. Using correlation techniques, one can extract reflected components from a range of different delays due to different distances to each part of the planet, and different Doppler shifts due to different projected (radial) velocities of each part of the planet.

The raw data is pretty messy, but after you record the body rotating a substantial fraction of a full turn, using modeling, you can recover a reconstructed image of the planet's topography.

The hard part is that while Doppler shift gives you east-west discrimination since one side moves towards you while the other moves away from you (assuming the rotation axis is "up-down"), there's no easy way to tell the difference between the two hemispheres in the reflected signal.

Thus the north-south ambiguity means that there is no temporal difference between a signal reflected from upper hemisphere and the lower. Say at 50 degrees North and 50 degrees South. The signals would have BOTH the same delay (since they are the same distance away) and the same doppler shift (since they are both rotating at the same speed.

Remember that for a spherical-like object of radius R who's axis is perpendicular to us, the projected line-of-sight (radial) velocity will be:

$$v_r = \frac{2\pi R}{T} \cos(\phi) \sin(\lambda)$$

where $T$ is the period, $\phi$ is the latitude and $\lambda$ is the longitude. It's symmetrical about the equator (cosine behavior) but it varies monotonically east-west (sine).

There are solutions to this if the object is large enough though. If you have a huge dish and the object has an angular size on the same scale as the beam width of your dish, you can record data with the beam illuminating first one hemisphere more strongly, then the other. That provides some additional information, and with subtraction and sufficient modeling one can overcome the North-South ambiguity.

Of course, if the body's rotation axis is oriented obliquely towards or away from us, then the symmetry is broken and the ambiguity can also be resolved in some cases.

Related and possibly helpful:

Why do radar maps of the surface of Venus have missing slices?
Emily Lakdawalla's How radio telescopes get "images" of asteroids

Left: A very slow GIF Source.

Right: A radar image shows asteroid 3122 Florence and tiny echoes from its two moons. Here is an animation that shows them more clearly. The direction of the radar illumination (and thus the direction toward Earth) is at the top." From here. NASA / Jet Propulsion Laboratory. This is a small subset of the frames contained in the original 36 MB GIF, and the size has been decreased by a factor of 2 in order to fit in SE's 2 MB limit. From the question What is the physical geometry of this apparent "eclipse" of a tiny moon of Asteroid Florence?

Note: These two images do not necessarily have the same orientation.


Invisible Mass Affecting the Expansion of the Universe

Yes, there is enough of these things, especially dark matter which is accelerating the expansion of the universe.


Where and how to look up the coordinates and magnitude of the star in this occultation prediction?

I believe this is a misinterpretation of this tweet: https://twitter.com/Alex_Parker/status/870169900970639360

If you look at the reply chain, it starts with https://twitter.com/Alex_Parker/status/870168057032921088 where Alex Parker states "Also, last night while we were packing up I took a very quick image of the Carina Nebula with one of our #MU69occ cameras."

The tweet in question states "This is what Omega Centauri looks like after just 500 milliseconds of exposure with one of the #MU69occ cameras."

In other words, Alex Parker was taking images of various clusters with the camera he plans to use for the MU69occ event. The image targets have nothing to do with the occultation itself.

I also went to https://ssd.jpl.nasa.gov/horizons.cgi and took a geocentric view of "2014MU69" for the month of June and it confirms this object is nowhere near Centaurus.

EDIT: I emailed Amanda M. Zangari (azangari@boulder.swri.edu) and her response was:

Hi Barry,

I do not know where the star numbers come from.

Gaia IDs for each of the stars are here:

http://www.boulder.swri.edu/MU69_occ/gaiainfo.html


The azimuth of the sunrise (or sunset, or any object) is a function of the Sun's declination and observer's latitude. It can be calculated from the following forumla:
$$\cos(\theta_R)=-\frac{\sin(declination)}{\cos(latitude)}$$
where $\theta_R$ is measured from due south to the location where the object rises or sets.

For example, at 55 degrees north latitude,

declination = 0 at equinox (by definition), so $\theta_R$ = 90 degrees from due south (90 degrees azimuth or due East when rising, 270 degrees or due west when setting).
declination = 23.4 (approximately) at the summer solstice, so $\theta_R$ = 133.8 degrees from due south (or an azimuth of 46.2 when rising, azimuth 313.8 when setting). [The declination at the solstice is plus or minus the obliquity of the ecliptic which is 23° 26′ 21″ = 23.4392° in the year 2000 and changes slowly.]
Half way between those two points on the horizon is (90+133.8)/2 = 111.9 degrees from due south. Solving the equation for the declination gives a value of declination = 12.4 degrees.
Now find the date when the Sun's declination is +12.4 degrees (April 23 and August 24 according to Table of the Sun's Declination).

Of course, the Sun will not be at the exact required declination at the time that it is rising (or setting), so you will want to select the date when the sunrise or sunset is closest. It will be accurate enough for most visual observations.


Do we still use the term "astronomical unit" nowadays?

Certainly. Astronomical Unit is probably one of the most used distance units used in astronomy. It is of course only used when discussing the distances within a stellar system, such as the distances between the Sun and its planets or other bodies in the solar system. It is also used to discuss distances in other stellar systems, e.g. the distances between stars and their (exo)planets.

It is easier to get a sense of the distances between bodies in the solar system if you use astronomical units. It doesn't say much when we say that the distance between Neptune and the Sun is $5\times10^{-4}$ light year. You would have to know the distance between Earth and the Sun ($1.5\times10^{-5}$ light year), and then calculate the difference (i.e. 30 times = 30AU) to get a sense of scale.

As a side note, in professional astronomy, the light year is hardly ever used. Professional astronomers use the parsec (=3.26 light year), kilo parsec (kpc), megaparsec (Mpc), and Gigaparsec (Gpc) to specify distances between stars, galaxies, and galaxy clusters.


Hydrogen clouds don't even make up a small component of dark matter, because hydrogen is not dark. The image below depicts the emission spectrum of hydrogen in the visible regime (a.k.a., the Balmer series).

The emission spectrum of dark matter on the other hand would be completely black.

Dark matter is gets its name from the fact that it doesn't absorb or emit light, i.e., electromagnetic radiation. From an elementary particle standpoint, this means dark matter particles can't interact with photons via the electromagnetic force at all, so dark matter particles must be electrically neutral, which electrons and protons certainly are not. No electrons are protons means no hydrogen atoms.


Dish antenna as parabolic mirror for OPTICAL telescope?

It's not possible I'm afraid.

Optical wavelengths (light) are typically of a wavelength under a micron, and an optical surface needs to be accurate to this level or better to be useful.

Radio wavelengths are typically 10-20cm or longer, and an adequate reflector can be made with surfaces accurate to a few centimetres (at a guess as I don't know the exact tolerance).

Also - I once experimented using a frequently mentioned substitute, a convex shaving mirror(!) as a reflecting telescope.  Even that was so poor it was unusable at the lowest possible magnification.

(But it is possible to make your own reflector out of reasonably stable glass - it just won't be as big as a satellite dish! Search around for "Amateur telescope making".)


What are some dark features on (or qualities to) the moon?

not sure about the dark side, but there are numerous other lunar features:

Wikipedia Lunar Features

Maybe Catena (a chain of craters) or Rima (lunar rilles)?


How can it be known that Venus does not have plate tectonics?

There's much less data available from Venus.

Some data exists. As mentioned in HDE 226868's answer, maps of Venus's surface exist. Like Earth's atmosphere, Venus's atmosphere is transparent to some low frequency electromagnetic radiation such as those used by radar. These observations are consistent with a planet that has stagnant lid tectonics and inconsistent with a planet that has active plate tectonics; more on this below.

In addition to these remote observations, the Soviet Union successfully sent several spacecraft into Venus's atmosphere, some of which landed and briefly operated on the surface of Venus. The initial attempts failed because nobody thought the surface conditions on Venus would be as brutal as they are. To make landing successful, the Soviet Union had to significantly downsize the parachutes and they had to use materials and avionics that could withstand very high temperatures.

Every piece of evidence gathered to date regarding Venus is inconsistent with a planet with active plate tectonics:

Venus surface temperature and pressure are well above water's critical point. Venus cannot have any liquid water on its surface. Water is widely (but not universally) thought to be critical as a lubricant that enables plate tectonics to occur.
The radar observations show a planet with a nearly universal surface age, about half a billion years old. This is inconsistent with a planet with active plate tectonics but consistent with a planet with stagnant lid tectonics.
Venus's atmosphere is very thick, much thicker than the Earth's atmosphere, and is dominated by carbon dioxide. Plate tectonics recycles carbon dioxide at subduction zones. Stagnant lid tectonics does not. A planet with plate tectonics will see a gradual reduction in the amount of carbon dioxide in its atmosphere over geologically long periods of time. A planet with intermittently active stagnant lid tectonics will instead see a gradual increase in the amount of carbon dioxide in its atmosphere over geologically long periods of time.
Multiple physics-based models of a planet with very high surface temperatures suggest that such planets will have rather thin and rather ductile crusts that can readily repair themselves against damage caused by subsurface tensions.

Note very well: The stagnant lid tectonics of hot terrestrial planets such as Venus  and possibly Titan (Titan is "hot" because its geology is ice-based rather than rock-based) is rather different from the stagnant lid tectonics of cold terrestrial planets such as the Moon and Mars. The surfaces of the Moon and Mars are very old. The surface of Venus is much younger in comparison. Venus has undergone at least one somewhat recent nearly global resurfacing event. This has not happened on the Moon or Mars. Plate tectonics appears to require a planet whose surface is neither too cold nor too hot, and that has a good amount of liquid water on the surface.

Some references:

David Bercovici and Yanick Ricard, "Plate tectonics, damage and inheritance," Nature 508.7497 (2014): 513.
DOI: 10.1038/nature13072.
A. Davaille, S. E. Smrekar, and S. Tomlinson, "Experimental and observational evidence for plume-induced subduction on Venus," Nature Geoscience 10.5 (2017): 349.
DOI: 10.1038/NGEO2928.
James F. Kasting and David Catling, "Evolution of a habitable planet," Annual Review of Astronomy and Astrophysics 41.1 (2003): 429-463.
DOI: 10.1146/annurev.astro.41.071601.170049.
Mikhail A. Kreslavsky, Mikhail A. Ivanov, and James W. Head, "The resurfacing history of Venus: Constraints from buffered crater densities," Icarus 250 (2015): 438-450.
DOI: 10.1016/j.icarus.2014.12.024
Ignasi Ribas et al., "Evolution of the solar activity over time and effects on planetary atmospheres. I. High-energy irradiances (1-1700 Å)," The Astrophysical Journal 622.1 (2005): 680.
DOI: 10.1017/S0074180900182427.
Accessible pdf: https://iopscience.iop.org/article/10.1086/427977/pdf.


Is C/2012 S1 (aka ISON) the fastest comet on record to-date?

A comet that travels to the Sun from the Oort cloud will be travelling roughly at its escape velocity at perihelion.

The fastest comets will be those that travel from the Oort cloud and plunge directly into the Sun.

If you want the comet to emerge from around the Sun, then its perihelion distance must be no lower than $\sim 7 \times 10^{8}$ m. At that point it will have approximately the escape velocity at that radius $v_{esc} = (2GM_{\odot}/r)^{1/2} = 617$ km/s or 1.39 million miles per hour.

Therefore I expect that lots of comets have attained something like or exceeding this speed before. Comet Lovejoy for example achieved a perihelion distance of $8.36\times10^{8}$ m and was travelling at 536 km/s (1.2 million miles per hour, and in fairly good agreement with the formula I quoted above - the aphelion was not infinity, but $\sim 157$ au, which largely accounts for the difference).


How long for a complete solar system cycle

It depends how precisely you mean the same position.

Take 2 since my previous attempt was ... optimistic.

In the late 1970's, the outer planets were moving into a configuration that made a Planetary Grand Tour much cheaper and quicker. They will be in a similar configuration in about 175 years.

But when will the planets be in the exact configuration? Never, well, not before the Sun turns into a red giant and swallows the inner planets.

But from time to time the planets will all be on the same side of the Sun. If the planets were all in a line, then after 495 years they would all be on the same side of the Sun, then again at 500 years, 522 years, 523 years (these last two are, of course 88 days apart, the period of Mercury), 995 years, 997 years, 1006, 1183 years, etc.


That is not correct.

The area is the total area between the two radius lines, so there is a curved side.

Imagine you have two points almost 180 degrees from each other. Using just a triangle, the area is close to zero. Now, two points placed closely together can have the same area in between them. Then you have an equal are, but not equal time, so your hypothesis can not be correct.


Variable and Multiple Stars in Hipparcos

Yes. If you have the Hipparcos data from ftp (http://vizier.u-strasbg.fr/viz-bin/VizieR?-source=I/239&-to=3), you will have several data files.

In hip_main.dat you have a field MultFlag at position 347 that indicates whether the star is a double or multiple star:

Note on MultFlag: indicates that further details are given in the Double
     and Multiple Systems Annex:
     C : solutions for the components
     G : acceleration or higher order terms
     O : orbital solutions
     V : variability-induced movers (apparent motion arises from variability)
     X : stochastic solution (probably astrometric binaries with short period)

To determine whether the star is a double or multiple star system, look in one of the files for double star solutions. For instance in h_dm_com.dat there is a field at bytes 24-25 that gives the number of components (2 for double and >2 for multiple stars).

24- 25  I2     ---     Ncomp    Number of components in this solution   (DCM2)

The Hipparcos number identifying the star can be found in bytes 43-48.

Of course this only works for multiple stars with orbital solutions.

For variability, three fields are provided in the main Hipparcos file:

314-320  F7.2  d       Period    ? Variability period (days)              (H51)
     322  A1    ---     HvarType *[CDMPRU]? variability type               (H52)
     324  A1    ---     moreVar  *[12] Additional data about variability   (H53)

with two notes (NB. if HvarType is C then the star is constant, i.e. not variable!):

Note on HvarType: Hipparcos-defined type of variability (a blank entry
     signifies that the entry could not be classified as variable or constant):
     C : no variability detected ("constant")
     D : duplicity-induced variability
     M : possibly micro-variable (amplitude < 0.03mag)
     P : periodic variable
     R : V-I colour index was revised due to variability analysis
     U : unsolved variable which does not fall in the other categories
Note on moreVar: more data about periodic variability are provided

There are two additional files on variability: hip_va_1.dat and hip_va_2.dat, which provide information such as variable type (GCVS) and max and min magnitude and also the variable star name.
You can find a description of the variable type in the General Catalogue of Variable Stars (GCVS) in the file vartype.txt if you download the GCVS at http://cdsarc.u-strasbg.fr/viz-bin/Cat?B/gcvs


Dark matter emitting EM radiations beyond our currently known spectrum?

The namesake of dark matter is because it does not emit electromagnetic radiation. Humans have built detectors on all ranges on the electromagnetic spectrum and a single particle will likely not emit wavelengths that are very large.

However, dark matter may produce light in collisions with other dark matter particles or normal matter. That is the premise of large scintillator detectors.


It is purely a matter of convention.  Longitude and latitude are normally expressed in degrees, longitude from 0 to 360⁰ and Latitude from -90 to 90⁰.  If you express these in radians, longitude would be from 0 to $2\pi$ and latitude from $-\frac\pi2$ to $\frac\pi2$.
If you are writing software, your functions should be robust enough to accept any value out of this range, for example when calculating the position of a planet, it may be convenient to have a continuous mapping from time to longitude, this would necessitate longitudes that are greater than $2\pi$.


How does the geometry for constructing a declining vertical sundial work?

You might find the following useful; scroll down to "Vertical Declining Sundial".

http://www.walkingshadow.info/Publications/Scratch.pdf


Any programme that shows actual planet information?

This seems like something that would best be provided by a simple data table which you can open as needed rather than as a feature of the language.  There's one at https://nssdc.gsfc.nasa.gov/planetary/factsheet/ that has the basic statistics for the planets (plus Pluto); is that the sort of thing you're after?


Do astronomers, in principle, have more information about older phenomena than younger?

There are more stars and galaxies but not necessarily more information.

You might choose to measure the information content in terms of photons received, in which case, for a given star/galaxy, this decreases as the inverse square of the distance, so the two effects cancel out.

Astronomy is a constant struggle with the tension between these two competing dependencies. As we look further, not only do we see more of particular types of object, but we also get to see examples of rare objects - i.e. the tails of the distributions. On the other hand as we expand our horizons, so our grasp of what is going on becomes increasingly blurred in terms of spatial, spectral and temporal resolution. It is for that reason that bigger and bigger telescopes are built!

The issue you raise concerning redshift is interesting. The co-moving volume versus look back time in any direction will not be a simple function (i.e. doesn't just go as the square of the lookback time) and depends critically on the adopted cosmological parameters.

There is of course a limit in any case. Once we get back to the microwave background at $z=1100$, then further probing backwards with electromagnetic radiation is stymied by the optically thick nature of the universe at early times.


Is there a flaw with the newer purposes and correlations attributed to the HR diagram? (And would a third axis of mass correct the enclosed flaw?)

Yours is a very, very interesting work. If you publish it it may happen that 3D diagrams will be used in the future. Please publish it.

But this does not mean that there is a flaw in the H-R diagram. It is correct as it is: it displays Luminosity Vs. Color and is perfect at that. And then it happens that some extra relations appear: stars are not randomly scattered on the HR diagram, but they group in branches, and these branches are related to stages of their life. The position on each branch is related to each star's mass. This is not a general relation holding for all the diagram. Almost the same happens with size, but this is more tightly related to Luminosity, per definition.


Finding the SN explosion time by t-square fitting

I think you misunderstood.

$(y-k) = a (x-h)^2$ is a parabola with its apex at $(h,k)$. Therefore, the formula you mentioned above has $t_0$ as the time at peak, not at the explosion.

For the case of SN, we normally do not know when the explosion happened. Mostly, we have an image before SN detection that can set a constraint for the explosion happening in between the last non-detection to the first detection.

However, most analytical models require the explosion time as the reference, so the parabola fit is applied as an approximation.


Is there any telescope on Earth that can see the lunar rovers on the moon?

The largest optical wavelength telescope that we have now is the Keck
  Telscope in Hawaii which is 10 meters in diameter. The Hubble Space
  Telescope is only 2.4 meters in diameter.

  Resolving the larger lunar rover (which has a length of 3.1 meters)
  would require a telescope 75 meters in diameter.

Information extracted from The Curious Team answers: Are there telescopes that can see the flag and lunar rover on the Moon?

Update: In a comment to my answer, @Envite mentions Astronomical Interferometers which are:

  ...an array of telescopes or mirror segments acting together to probe
  structures with higher resolution by means of interferometry. The
  benefit of the interferometer is that the angular resolution of the
  instrument is nearly that of a telescope with the same aperture as a
  single large instrument encompassing all of the individual
  photon-collecting sub-components.

It also must be noted that probably this wouldn't be the best idea for watching the moon rovers since:

  The drawback is that it does not
  collect as many photons as a large instrument of that size. Thus it is
  mainly useful for fine resolution of the more luminous astronomical
  objects, such as close binary stars.


Black holes distort the geometry of spacetime. So you need to take care about that.

Changes of gravity propagate with the speed of light. That's thought to occur for accelerated mass, e.g. a binary of black holes.

A second fundamental point: Speeds don't add up in an additive way, but in a subadditive way: speed of light + speed of light = speed of light.

If you or the black hole travel with the speed of light, you resp. the black hole needs to be massless at rest, otherwise your relativistic mass, or that of the black hole would be infinite.

Photons are massless at rest, so you're actually talking of light, or something which behaves rather similar to light in many aspects.

So at the end you're talking of light directed to a black hole and reflected by a mirror inside the black hole.

The light can get inside the black hole, but it cannot leave the hole after being reflected from the mirror due to the high curvature of spacetime below the event horizon.

If you are travelling towards a binary of black holes, you get blue-shifted changes of gravity.


This is a part answer to your question, based on some recent research of the photochemical behaviour modelled and observed for Titan's tholin haze and modelling of Titan's stratosphere.

The process appears to begin, according to the paper Ice condensation layers in Titan’s Stratosphere (Barth, 2012) (Abstract only - paywalled), with

  Photochemical destruction of methane along with the destruction of nitrogen molecules from energetic electrons in Titan’s upper atmosphere result in the production of a number of hydrocarbon and nitrile compounds which may be capable of condensing at the colder temperatures of Titan’s lower stratosphere.

then, according to the paper Laboratory experiments of Titan tholin formed in cold plasma at various
pressures: implications for nitrogen-containing polycyclic aromatic
compounds in Titan haze (Imanaka et al. 2004), in particular in reference to Titan's stratosphere,

  . In the stratosphere (100–
  300 km), further chemical reactions are induced by the catalytic
  $CH_4$ dissociation by such molecules as $C_2H_2$ and
  $C_4H_2$ absorbing the long UV (> 155 nm) irradiation

The significance of these UV absorbing molecules is explained in the article Photochemical activity of Titan’s low-altitude condensed haze (Gudipati et al. 2013) (Abstract only - paywalled), they state that tholin haze could form on condensed aerosols in Titan's atmosphere, demonstrating that, at least part of Titan's atmosphere is photochemically active. Through modelling, they found that

  Detected in Titan’s atmosphere, dicyanoacetylene ($C_4N_2$) is used in our laboratory simulations as a model system for other larger unsaturated condensing compounds. We show that $C_4N_2$ ices undergo condensed-phase photopolymerization (tholin formation) at wavelengths as long as 355 nm pertinent to solar radiation reaching a large portion of Titan’s atmosphere, almost close to the surface.

and evidence of these ices is suggested in the article Titan’s aerosol and stratospheric ice opacities between 18 and 500 μm: Vertical and spectral characteristics from Cassini CIRS (Anderson and Samuelson, 2011) stating that the ices and aerosols

  appear to be located over a narrow altitude range in the stratosphere centered at ∼90 km. Although most abundant at high northern latitudes, these nitrile ice clouds extend down through low latitudes and into mid southern latitudes, at least as far as 58°S.


Not really, for the same reason that you cannot travel west by jumping up in the air and let Earth rotate underneath you, such that you land a little farther to the west.

The reason is that standing on Earth's surface, you already have a velocity toward the east which matches exactly the speed of the surface. Thus, in the reference frame of Earth, you simply jump up and down.

In the same manner, if you use your mad hypothetical spacecraft to fly "up", i.e. away from the Galactic plane, you already have a velocity of $\sim$250 km s$^{-1}$ in the direction of the rotation of the plane, so that in the reference frame of the plane, you simply fly straight up.

Apart from this, in order to leave the Galactic plane, you need to fly roughly 500 lightyears. This will take a long time. You should stay home.


How to calculate when Mars was in retrograde around 129 BCE?

Mars' retrograde motion happens in the summer of 129 BC.

The JPL Ephemerides are an excellent source of solar system predictions, and are widely used. DE422 is about 600 MB and covers your period of interest.

Since your profile shows that you can program a computer, I've used the Python package Skyfield to download it, do the interpolation and calculate the apparent positions of Mars and the Sun from Alexandria in 129 BC. If I understand correctly, this would be year -128 in UTC, since there is no year zero.

You can also get similar data online using the JPL Horizons web interface.

I'll leave the problem of checking when the Sun was below the horizon but Mars was above the horizon in Alexandria on each day to you, but I've included the altitude methods in the script.

Mars is red, Sun is yellow:

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Loader, Topos

load = Loader('~/Documents/SkyData')  # avoids multiple copies of large files
ts   = load.timescale()

data    = load('de422.bsp')
earth   = data['earth']
mars    = data['mars']
sun     = data['sun']

ts      = load.timescale()

Alexandria  = earth + Topos(latitude_degrees  = 31.2,
                            longitude_degrees = 29.9,
                            elevation_m       = 5.0)

# NOTE! The year 129 BC is year -128 in UTC I believe
# SEE   https://en.wikipedia.org/wiki/Year_zero

# Check when Mars is "up" and Sun is "down" on a given day

hours    = np.arange(0, 24, 0.1)
one_day  = ts.utc(-128, 7, 1, hours)  # July 1st

mars_obs = Alexandria.at(one_day).observe(mars)
sun_obs  = Alexandria.at(one_day).observe(sun)

mars_alt, mars_az, mars_dist = mars_obs.apparent().altaz()
sun_alt,  sun_az,  sun_dist  = sun_obs.apparent().altaz()

# Check RA and DEC in 129 BC

days     = np.arange(1, 366)
one_year = ts.utc(-128, 1, days)

mars_obs = Alexandria.at(one_year).observe(mars)
sun_obs  = Alexandria.at(one_year).observe(sun)

mars_RA, mars_Dec, mars_dist = mars_obs.apparent().radec()
sun_RA,  sun_Dec,  sun_dist  = sun_obs.apparent().radec()

if True:
    plt.figure()

    plt.subplot(3, 1, 1)

    marsalt, sunalt = mars_alt.degrees, sun_alt.degrees

    plt.plot(hours, marsalt, '-r')
    plt.plot(hours, sunalt,  '-y')
    plt.plot(hours, np.zeros_like(hours), '-k')
    plt.title('altitude of Mars and Sun from Alexandria, July 1, 129 BC')
    plt.xlabel('hours (UTC)')
    plt.ylabel('altitude (degs)')
    plt.ylim(-10, None)
    plt.xlim(0, 24)

    marsra, marsdec = mars_RA._degrees, mars_Dec._degrees
    sunra,  sundec  = sun_RA._degrees,  sun_Dec._degrees

    # get rid of ugly lines
    dsunra, dmarsra = sunra[1:] - sunra[:-1], marsra[1:] - marsra[:-1]
    sunra, marsra = sunra[:-1], marsra[:-1]
    sunra[dsunra<-100]   = np.nan
    marsra[dmarsra<-100] = np.nan

    plt.subplot(3, 1, 2)
    plt.plot(days, mars_RA._degrees, '-r')
    plt.plot(days, sun_RA._degrees,  '-y')
    plt.title('RA of Mars and Sun from Alexandria, year of 129 BC')
    plt.xlabel('days')
    plt.ylabel('RA (degs)')
    plt.xlim(0, 365)

    plt.subplot(3, 1, 3)
    plt.plot(mars_RA._degrees, mars_Dec.degrees, '-r')
    plt.plot(sun_RA._degrees,  sun_Dec.degrees,  '-y')
    plt.plot([0, 360],  [0, 0], '-k')
    plt.title('RA vs Dec of Mars and Sun from Alexandria, year of 129 BC')
    plt.xlabel('RA (degs)')
    plt.ylabel('Dec (degs)')
    plt.xlim(0, 360)

    plt.show()


Brown dwarfs are heavier than Jupiter (jupiter has mass 0.000949 solar masses), although about the same size. Their compositions are fairly similar, with brown dwarfs depleted in deuterium if they are heavy enough (it is burned over a few million years after formation).

Adams & Laughlin modelled the collisions of brown dwarfs, mainly to estimate whether two dwarfs could merge to make a star heavier than 0.075 solar masses that would begin hydrogen fusion properly. They found that this could happen, and not too much mass appears lost as they splatter into each other if it is a somewhat glancing collision. Obviously, a near glancing hit might just scatter some of their atmospheres but not stop them.

Jupiter would likely behave the same. The result would be a heavier brown dwarf. It would be heated significantly by the energy of the merger, which in turn would make it expand to a larger size until it cooled down. Jupiter would also contribute deuterium so if the dwarf was heavy enough it would further heat up for a while. Given Jupiter's minimal mass compared to the 0.075 limit it is not likely it would be enough to turn the new object into a red dwarf.


Why could I so easily see and photograph such a bright totally eclipsed Moon from a bright city road? (May 26, 2021 total lunar eclipse)

Based on your description of your experiences, my guess is that it’s not that this eclipse was especially bright, but rather that the previous one you saw was unusually dark.  My experience (late 1990s to present) has always been that a totally eclipsed moon is easy to see even under bright urban conditions, but this page says that there were some eclipses in the 1980s and early 1990s that were hard to see due to recent volcanic eruptions.  Do you know the date of the previous eclipse you saw?


Can dark energy or dark matter affect black holes

Dark matter can, and probably does, fall into black holes and affects them just as any other form of matter or energy falling into them does. It doesn't fall in a huge amount because, like anything else approaching a black hole it usually isn't aimed straight in, but rather misses a bit and swings by and escapes again. Unlike normal matter, dark matter experiences almost no friction or collisions from other stuff near the black hole (it goes straight through everything, including itself, we think) so it doesn't get caught in the accretion disk or anything. Dark matter falling into the central SMBH doesn't do much to galaxy, because its mass is still there pulling on the stars and so on just as it did before it fell in.

Dark energy is another story. In most of the various competing theories in which it exists, it is an energy inherent in space. Every cubic meter of space contains a certain (small) amount of dark energy, which can't move around or anything. It shows up as something like a pressure, causing space to tend to expand whenever there isn't a larger force (like the gravity of nearby masses) causing it to contract.


Answering your first question:

  If I understand it correctly then the black hole is created by
  collapse of a super massive star at the end of its life. This star
  will have a finite mass. Then what leads to the black hole to have
  such massive gravity so that nothing can escape it (because if it was
  a star before with a finite mass then it would also have some finite
  gravity, right)?

You are right that a black hole will have a finite mass that is smaller than that of the massive star that collapsed (some matter is carried off by the supernova). At large distances the gravitational field of this black hole will be no different than that of a star with the same mass.

The difference arises because in a black hole this mass is compressed into a much smaller volume. Consequently, you can get much closer to the concentration of mass, than you would be able to for a star (where the mass is scattered over a relatively large volume). It is here where the gravitational fields can become really "strong" and prevent light from escaping.


In what ways does the mass and energy of the milky way curve space time?

In short, not by much.

Out by us the Earth the gravitational acceleration towards the centre of the galaxy is $v^2/r = \frac{(2\times2\times 10^{20}\times \pi\  \mathrm{metres} / 230000000\  \mathrm{years})^2 }{ 2\times 10^{20} \mathrm{metres}}$, which is of the order of $10^{-10} ms^{-2}$ : in other words, it is very small. This is because spacetime is not highly curved except close to massive bodies.

This spacetime is approximately flat. It can be well approximated by Minkowski space. If you need to include in your model the distribution of mass in the galaxy, then Newtonian mechanics is sufficient, for example https://physics.stackexchange.com/questions/62637/the-potential-and-the-intensity-of-the-gravitational-field-in-the-axis-of-a-circ

In the centre of the galaxy is a black hole, and it is in the neighbourhood of the black hole that GR really comes into play. Near the black hole, the Kerr solution of GR is a good model (the Kerr solution describes a symmetric uncharged spinning mass)

Now on the massive scale, a galaxy, or a cluster of galaxies can have a significant effect on light, given enough time, and enough distance. Light, travelling for many millions of years, past a galaxy, can have its path significantly bent by gravity. (The Schwartzchild solution is a first approximation to the shape of spacetime around a galaxy, though for a better approximation, numerical relativity can be used) This causes gravitational lensing. For light traveling inside a galaxy this is not a significant factor, There isn't time in a few years for light to be bent significantly by the general gravitational field of the galaxy. However light from a star passes close to another star, then there can be measurable bending.

The measurement of the bending of light of stars behind the sun during a solar eclipse was one of the first experimental confirmations of General Relativity (1919) https://www.wired.com/2009/05/dayintech_0529/ Note that even light that passes right over the Sun's surface would only be deflected by less that 2 arcseconds.


Simple rule to find some of the zodiac constellations after sunset?

This is far too complicated. You can find any asterism in the sky far more simply by remembering its shape and its position relative to other (perhaps brighter) constellations.

For the Zodiac constellations, they are all in a band that (in the Northern hemisphere) goes from East through South, to West. In the winter, that band will be high is the sky. In summer it will be low.

And then you learn the sky.

In winter you first learn Orion and from there you can find your way to Gemini, the V of the Hyades in Taurus.  Cancer is dim (but has that nice cluster), but next in line, Leo, is nice and bright.

If you want to find out what is actually visible right now, go outside! If you want to check on the location or visibility of a particular constellation right now, use planetarium software.

You don't need to calculate anything.


Measuring the size and the geometry of Universe

The relative recent WMAP result shows, the Universe seems infinite.

However, the "is the Universe infinite" question has a hidden problem: it doesn't matter, how far away can we see, we can't be sure, if anything out of it is finite or not. :-) Thus, it is about a definitely unknowable thing.

The WMAP measured essentially the curvature of the Universe and it found a zero curvature (i.e. it seems to be planar). If there is a minimal curvature, below a measurement precision, and we have a cyclic, spherical geometry, then its radius must be at least 300billion light years.


How long for the sun to traverse its diameter at the pole?

The Sun's daily motion does not affect its altitude on the poles, because it revolves around the zenith (and the zenith coincides with the celestial pole). This means the Sun's altitude is only altered by its movement on the ecliptic. The ecliptic is tilted at a roughly 23.5 degree angle to the equator (which is the horizon if the observer is on the North pole), and only the vertical component of its yearly motion contributes to its increase in altitude. Therefore, in order to estimate how long the Sun takes to rise/set on the North pole, let us assume the Sun is moving across the ecliptic at a constant velocity and that Euclidean geometry holds for the small area on the celestial sphere around the Sun. If the Sun takes one tropical year to traverse 360 degrees, its "speed" will be $\rho=0.98565 \frac{\circ}{d}$, and the vertical component $\rho\sin(\epsilon)=0.39\frac{\circ}{d}$. The time it takes to traverse a solar diameter is (roughly $0.5^\circ$) is $0.5/0.39 \sim 1.25$ days. The same is for the South pole, everything is rotated around the equator. You can find more details about the topic here.
As for the ratio of sunrise duration to daytime duration, these are affected by different phenomena, and the ratio of sunrise duration to daytime duration is not constant throughout the year and depends on the latitude of the observer so comparing them is quite ambiguous.


There is no tilting cycle.

The Earth continues to tilt the same way all year round.  The axis points towards the pole star. But as the Earth goes round the sun, in December the northern hemisphere points away from the sun. In June the Northern hemisphere points towards the sun, as the Ascii art below shows (not to scale)

N                    N
 \                    \
  O         *          O
   \                    \
    S                    S
Earth                 Earth
in Dec     Sun       in Jun

So no special explanation is required. The Earth points the same way all year, it just moves relative to the sun.

The tilt of the Earth relative to the stars doesn't change. So the tilt of the Earth relative to the sun is entirely due to the Earth's orbit.  This proves that the cycle of tilting relative to the sun will take exactly one orbit to complete. It is no coincidence; the cycle of seasons will take exactly one year.

(There is variation in the direction of tilt, called precession, but this is a much slower cycle, lasting 25000 years)


What is the last nuclear reaction in a binary system before supernova?

I guess it depends on the type of Supernovae.

Type 1a supernovae are produced by a white dwarf star orbiting a companion close enough that matter transfers from the companion to the surface of the white dwarf.

Assuming only your star system of a 3M star and 6M star, the 6M star would become a white dwarf star first.  At the end of it's life, it would have been furiously converting hydrogen to helium, and helium to carbon via the C-N-O cycle and the triple alpha process (C-N-O being dominant in stars above 1.3M sol).  The highest reaction in this stars main sequence lifetime would have been the creation of oxygen, but oxygen is a catalyst in the C-N-O cycle, and therefore the last probable reaction in the star would be that of Helium in some form of shell burning in the outer core (more on that next).

The white dwarf, while orbting the companion, may siphon off hydrogen from the surface of the donor star.  This hydrogen will accumulate evenly over the surface of the dwarf star, slowly growing the mass of the dwarf.  Also, the hydrogen will be shell-burned into helium during this time increasing the density.

If sufficient mass can be accumulated, then the carbon within the white dwarf can ignite in a runaway reaction - literally blowing the entire star apart.  These supernovae produce every element up to Nickel (Ni-56).


Can a photon have a stable orbit around a black hole?

For any massive object the gravitational potential energy is given by Newton's law:

$$ V(r) = -\frac{GMm}{r} $$

The gravitational potential energy is due to the attractive gravitational force, but for an orbiting object there is also a (fictitious) centrifugal force pushing it outwards. If we calculate the potential energy due to the centrifugal force and add it to the gravitational potential energy we get an effective potential energy:

$$ V_\mathrm{eff}(r) = -\frac{GMm}{r} + \frac{L^2}{2mr^2} \tag{1} $$

where $L$ is the angular momentum, which is a constant for an orbiting object (because angular momentum is conserved in a central field). If we plot this graph for the Earth-Moon system we get a graph like this:

(this comes from my answer to the question Could we send a man safely to the Moon in a rocket without knowledge of general relativity? on the Physics SE)

Note that there is a minimum in the potential energy curve, and this minimum gives the radius of the stable orbit. Note also that because it's a minimum if we displace the object away from the minimum it will fall back towards the minimum again i.e. this is a stable orbit.

Now, for light we cannot simply use Newtonian mechanics because light is massless, but we can do the calculation using general relativity. The details are a bit intimidating, but we end up with an effective potential just as described above. For light the effective potential turns out to be:

$$ V_\mathrm{eff}(r) = \sqrt{1 - \frac{2GM}{c^2r}}\frac{L}{r} \tag{2} $$

And if we graph this it looks like this:

This looks very different from our first graph, and it's different because light is massless and only ever travels at the same speed of $c$. The graph of $V_\mathrm{eff}$ for light has a maximum not a minimum. The maximum corresponds to the position of a circular orbit, just like the minimum in the first graph, and we find the radius of this circular orbit is given by:

$$ \frac{r}{r_\mathrm s} = 1.5 $$

where $r_\mathrm s$ is the Schwarzschild radius. This radius gives the position of the notorious photon sphere.

But for light $V_\mathrm{eff}$ has a maximum not a minimum. That means if we displace the light by even the tiniest distance from this maximum it will lower its potential energy by moving either inwards or outwards. The orbit at $1.5r_\mathrm s$ is unstable and the tiniest perturbation will cause the light to spiral into the black hole or away from it. This means we cannot accumulate light in the photon sphere as the question asks. Any attempt to put light into this orbit is doomed to failure as even the tiniest perturbation (e.g. from other objects orbiting the black hole, or even from passing gravitational waves) will destabilise the orbit and the light will be lost.


Why are Shack-Hartmann sensors so expensive (4k+ USD)?

I have been deeply involved in both Shack-Hartmann and lateral-shear polarization interferometers. Now I want something simple and slow for hobby projects and had the same question. I don’t think such a project is as daunting as might be imagined.  One would like 12 bits in the camera, the Sony CMOS cameras are so good, the last bit is nearly noise free.
The camera need not be expensive: Any amateur telescope web store will have dozens for less than $400.  You want monochrome, preferably USB 3, with the largest diagonal focal plane size you can afford.  A Global shutter is nice if you have a dynamic situation. Rolling shutters are generally lower noise.
A lens array will be expensive.  Thorlabs, Seuss, and RPC Photonics, and others sell them for \$400-\$600.  The Thorlabs arrays are mounted.  In the end you want the focal spot to be at least 2 pixels across. Choose your micro lens F/# accordingly, usually quite slow, F/20 or more. Likely you want the closest pitch you can find as this sets the spatial resolution. A 4 x 5 mm focal plane will view 40 x 50 micro lens spots. Good enough.
You will need to craft a mount that mates the camera and the lenslet array at the micro lens focus distance.  They are slow, so the depth of field should be forgiving.
The real crux is the software.  There is public domain software that is close (git has AOtools) and there may be more. You will need to take a dark exposure for the shutter time used (I typically take 100 frames and average them).  Then the reaL GOTCHA issue: What to use as a reference?  You can use a shear plate to focus a collimated light source (a single mode VCSEL works well), but you are unlikely to have a shear plate about.  Place the VCSEL or other point source bright enough to see (an LED?) as far away in a dark room as you can manage.  Don’t use a mirror or lenses, turn off heating and cooling, etc. This gives you a known curvature. for 100 μm lenslets at F/20 the focal length is 2mm so a tilt of 1/2 a subaperture (to keep things easy) or 50 μm is only 25 milliradians (mr), something over a degree.  The edge of the camera must see the source as no larger than this angle.  So if you have 50 subapertures across your camera the edge is 2.5 mm from the middle.  Thus 2.5e-3 meters / d = 2.5e-3 radians so the source needs to be more than a meter away.  Easy!  With the distance from the source to the focal plane known you can now easily calculate the tilt distribution (should look spherical).
Use openCV or your favorite utility to get images into your computer and a) subtract background b) make sure the camera does not saturate (adjust exposure time, retake the dark background and try again) c) calculate the peak intensity of each spot and which pixel coordinates that is in d) Draw a box around each focal spot which is several spot diameters across each spot e) calculate the centroid of each spot using only pixels in the box (the rule is add no pixels that do not contribute to your signal, the centroid). You can play with the box size.  An F/20 blue-green 500 nm wavelength spot from a 100 μm lenslet will be about lambda F/# across (just 20 wavelengths, about 10 μm) which should be 3 to 4 pixels so a box 16 x 16 pixels should be a good place to start.  The centroids should be good to a small fraction of a pixel.  I have done better than 1/100 pixel with a really good focal plane array and 1/20 pixel for standard cameras such as the above. Now you can calculate the centroid location within the box, and reference that to the whole array.  A plot across a slice will give you a linear function which is somewhat less than the lenslet pitch.  Do this in both X and Y.  You may have a rotation in the spot coordinates. You can physically correct this, or use math.  The slope in X and Y should be the same.  If not, the lens array is tilted.  If you move the light further away, the spots should move to be more directly behind the lenslets.
In passing I would note that I found the accuracy of the lithography on the imager, and used to make the lenslet array was better than my best tilt calibration! Thus for a 3.5 μm pixel and 100 μm lenslet the spots should be 100/3.5 pixels apart.  This will be independent of focus etc.  I also found trying to uniformly illuminate the array to correct for the variation in responsivity pixel by pixel generated more noise, not less.  Silicon is awfully good.
The last calibration needed is to define where zero tilt is.  This should be close to normal to the focal plane array, so pick the middle lenslet and the pixel you declare is on boresight. Now you have the subaperture tilts calibrated.
The last messy task is to convert from wavefront subaperture tilt to wavefront.  There are books written on the subject. Notionally if you start at your zero tilt and add up the subaperture tilts as you move across the array, subtracting the zero tilt position of each subaperture, you have wavefront.  Note you can take any path from the center to the subaperture you want to know and add up the X and Y tilts and you should get the same answer.  Ideally you want to take the average of all possible paths (that are illuminated) as that would be the best average value you could get.  This is referred to as the minimum RMS error.  There are many ways to calculate this.  If you think of the tilts as the gradient in phase, and calculate the divergence of the gradients, you have a measure of the wavefront curvature.  It is also simply Laplace’s equation which can be solved by successive over-relaxation, or if you like fancy algorithms, use a multigrid method (it’s way faster). Greg Allen’s thesis has another approach, https://dspace.mit.edu/bitstream/handle/1721.1/120381/1084482108-MIT.pdf?sequence=1&isAllowed=y .   There are many more.
In summary, you can build a Shack-Hartmann wavefront sensor for about $1000 and a lot of software development.  The Thorlabs sensor seems cheap given the effort, at least until someone comes up with an open source sensor and software.
Jon


Since when do astronomers have the notion that space is void?

(This is a mash up of the answers in Who was the first to postulate that space was a vacuum? and What was the first astronomical measurement which demonstrated that “the Earth is surrounded by vacuum”? )
The notion that earth is surrounded by vaccum has been around since at least the atomist philosophers like Democritus (c. 450 BC), albeit without demonstration outside of speculation it was open to debate.
Galileo himself had the notion that air is not weightless, which implies that the amount of air weighting on our bodies is limited, suggesting there is vacuum above it.
His disciple Torricelli invented the barometer, allowing him to estimate the amount of air above our head and demonstrate that space around the Earth is a vacuum.
Conclusion, Galileo most certainly had a clear notion that the planets are moving in a vacuum (which does not mean it is what inspired his laws of motion), and it was firmly established shortly after his death.


Can an amateur astronomer bounce a laser off the moon?

This experiment is still done frequently, but you need a powerful pulsed laser, precise pointing (error below about 20 arc seconds) and a sensitive detector.

Amateurs will fail most likely, because they won't get access to an appropriate equipment.

Here is a youtube video, which may give an impression of how difficult it is to perform the experiment.


What is the 'scale factor' equation for a dark-matter dominated universe?

You answered your question actually. You have said

Note that this solution is not valid for domination of the cosmological constant, which corresponds to an w=−1

then you are saying

But if 'w' is negative-one then 'a' is proportional to tt

The $w=-1$ does not apply to $a(t) \propto t^{2/3(1+w)}$


Could we estimate the age of the universe based on the planar property of the Solar System?

First, the Big Bang did not happen at a point. I cannot emphasize this enough. It happened everywhere in space at the same time. You can't think of it as an explosion from one location pushing everything else away.

Furthermore, planets and stars didn't exist at the time of the Big Bang. The Solar System, for example, formed about 4.5 billion years ago, while the oldest stars - Population III stars - formed within a few million years after the Big Bang. At the beginning of the universe, everything was in an exotic soup-like state: a quark-gluon plasma.

Therefore, your hypothesis is flawed in two places.


Who is considered as the father of modern astronomy?

I would say Copernicus. He did his work maybe 100 years earlier and formulated the heliocentric view of the universe.


As the Universe expands, do things like planets expand also, or just the space in between?

Things that are bound by other forces will not be seperated by the expansion of the Universe. For instance, planets do not expand due to cosmic expansion. The solar system does not either because it is held together by gravity. However, the space between galaxies or between galaxy clusters does expand.


Why does the face of the moon 'sync' with the earth?

It is called "Tidal locking", or "gravitational locking" or "captured rotation". According the the Wikipedia page on "tidal locking" (check the references for sources), it is due to Earth's gravity causing a small tidal bulge on the Moon, which affects its rotation. Over time, the Moon's rotation, affect by Earth's gravity, makes it's orbital rotation synchronise with its lunar rotation.


When the universe expands does it create new space, matter, or something else?

Yes, space is constantly being created. The new space does not hold any matter (like atoms) or dark matter. This means that the density of normal and dark matter decreases at the same rate as the volume increases. However, dark energy, which is something completely different and thought to be a property of vacuum itself, is being created with the new space, so the density of dark energy stays constant.

This in turns mean that while the early Universe (i.e. from it was 70,000 years old and until it was almost 10 billion years) was dominated by matter, the Universe is now dominated by dark energy.

And it will only get worse.


What does the amino acid found in the atmosphere of Venus mean?

An amino acid is any acid that has an amine in it. They are not necessarily found in proteins. They can be found in conditions similar to deep space, for example:

In a laboratory at NASA Ames Research Center in California’s Silicon Valley, the team of astrobiologists shone ultraviolet light on deep-space-like "ices," simulating conditions that are commonplace in interstellar space. Deep-space ice is common water ice laced with simple molecules. The team subsequently discovered amino acids, molecules present in, and essential for, life on Earth.

Amino acids are really simple molecules, that could arise from a number of abiotic reactions. Do not be impressed if they are found on a barren planet, for they have been found in meteorites before:

Murchison contains common amino acids such as glycine, alanine, and glutamic acid as well as unusual ones such as isovaline and pseudoleucine. A complex mixture of alkanes was isolated as well, similar to that found in the Miller–Urey experiment. Serine and threonine, usually considered to be earthly contaminants, were conspicuously absent in the samples. A specific family of amino acids called diamino acids was identified in the Murchison meteorite as well.

I'll personally only be impressed when at least a protein or sequence of DNA/RNA is found outside Earth. Then we can speculate about life out there. I do believe it exists - I just think we are getting too excited now with stuff that may not necessarily be biotic.


Why is the Innermost Stable Circular Orbit (ISCO) of a non-rotating black-hole 3 times the Schwarzschild radius?

There's a fairly full treatment in these lecture notes but it's not really amenable to a summary in a stackexchange answer. However they end up with the following equation for the radius of a circular orbit of a particle with angular momentum $L$ and mass (ie not a photon).

$$GMr^2 - L^2r + 3GML^2 = 0$$

For suitable values of $L$ this has two solutions for $r$ and, one inside the ISCO radius (which is $6GM$ in these units) and one outside. Further analysis shows that the inner orbit is unstable and the outer one stable. See around equations 7.55 for more details.

Photons have an unstable circular orbit at $3GM$ and the event horizon is at $2GM$.


Intuition about why gravity is inversely proportional to exactly square of distance between objects

Imagine "gravity" spreading out in a sphere, like light from a bulb.
For each doubling of the distance, the sphere has four times the area.  The surface area of the sphere is proportional to the square of the radius.  If the same gravity is stretched over that sphere, the force of gravity would be inversely proportional to the square of the radius.
This gives sufficient intuition for most physicists of the time to believe that gravity was probably inverse square.  Newton's real genius was that he was able to prove using mathematics that a planet moving in an inverse square law would obey Kepler's laws, something that his contemporaries had failed to do.
A well known story, related by de Moivre:

In 1684 Dr Halley came to visit [Newton] at Cambridge. After they had been some time together, the Doctor asked him what he thought the curve would be that would be described by the planets supposing the force of attraction towards the sun to be reciprocal to the square of their distance from it. Sir Isaac replied immediately that it would be an ellipse. The Doctor, struck with joy and amazement, asked him how he knew it. Why, saith he, I have calculated it.


What happened to the 2017 proposal on redefining planethood? Is this information available?

In the linked article by A. Bouchard and another by J. Daley,
the words "in the journal Lunar and Planetary Science" link not to a journal article
but to a poster in a K-12 education session
at the 2017 Lunar and Planetary Science Conference.
First author Kirby Runyon told Universe Today in 2017 that he would not submit this geophysical definition to the IAU process:

We in the planetary science field don’t need the IAU definition...
If [the geophysical definition] is the definition that people use and what teachers teach, it will become the de facto definition, regardless of how the IAU votes in Prague.

It was not presented at the 2018 IAU General Assembly in Vienna.
I don't know where Bouchard got "has been submitted to the IAU for review."


When was visible light for the first time created?

The best model we have is that of electro-weak unification.

The photon is an quantum of the electromagnetic field, and as long as there has been an electromagnetic field, excitation of the field have existed.

The "weak" nuclear interaction is another field, (it has the effect of changing up quarks to down quarks, and so is responsible for beta decay of radioactive material). Quanta of the weak field are rather different from photons. They are charged and have mass, but at very high energies (a million billion degrees or so) the electromagnetic and weak fields are unified into a single "electroweak" interaction.

But after the universe had cooled (it took sometime between $10^{-36}$ and $10^{-32}$ seconds) the weak and electromagnetic fields split, and the first true photons formed.

These photons didn't get very far before colliding with other particles that were being formed. As you note, it took nearly 400000 years until the universe cooled enough to become transparent.

Now you mention visible light. The photons formed after the split would have had very high energies, we would call most of them gamma rays if they existed today. But as photon of all energies could be formed, a small fraction would have been in the visible range.

At recombination, when the universe became transparent, the temperature was about 3000K, at which temperature most photons are in the infra red range, but enough are visible to have given the universe a very bright orange-red colour, similar to the colour of a lightbulb.


Assuming a sufficient amount of mass above the density threshold, does the actual concentration of the mass matter in creating a black hole?

... does that mean that any sufficiently large volume of mass over that density is also a black hole? Or does the actual concentration of mass within the event horizon matter?

I'm not completion sure what distinction you're drawing between concentration and density, but I will assume that what you mean by the former is the details of the matter distribution, e.g., whether it's concentrated at the center, spread throughout, or whatnot.

For a spherically symmetric isolated body of mass $M$, it is completely irrelevant. The reason is Birkhoff's theorem: outside the gravitating body, spacetime geometry is necessarily Schwarzschild. This is the general-relativistic analogue of Newton's shell theorem. Therefore, it doesn't matter whether the (radial) distribution is uniform, concentrated at the center, or some kind of shell, or anything else: once it is compact enough that its outer surface gets to the Schwarzschild radius $2GM/c^2$ or below, it is fully enclosed by an event horizon, and is therefore a black hole.

So under those assumptions, the answers to your two questions are 'yes' and 'no', respectively, although you might want to be careful about how you define 'volume' when comparing overall densities.

What happens if we get rid of the assumption of spherical symmetry is a bit more complicated. If we're in an asymptotically flat universe, then we can think of a black hole as all the event from which an ideal light ray fails to escape to infinity, and the boundary would be the event horizon; more generally, we might have to be more careful about how we define 'inside' and 'outside'. Note that this makes the event horizon depend on the future, i.e. it depends on what light rays escape or don't escape even if you wait an arbitrarily long time for them. Hence, in a dynamic situation (such as a collapse to a black hole), where the location of the event horizon depends on not only on the past and present, but also on what will fall into the black hole in the future.

This makes general statements about density quite difficult in situations that don't have some simplifying assumptions. Density is too simplistic; the general notions of a black hole and event horizon are highly non-local.

Nevertheless, there is a general result that is morally similar to the above that is very relevant to your second question: the no-hair theorem. In general relativity, any isolated black hole is fully characterized by conserved quantities at infinity (mass, angular momentum, electric charge...). That means that the details of the matter distribution inside the event horizon do not matter at all. Of course, the singularity theorems guarantee that at least under some general assumptions about the behavior of the matter, it will collapse to a singularity, but that's a separate issue.


Clarification of radio spectrum terminology

Terminology
The terminology of thermal and non-thermal emission is somewhat unfortunate. Formally, non-thermal emission is continuum radiation from particles with non-Maxwellian energy spectra. Practically, we generally consider it to be emission which is not described by Bremsstrahlung or black body radiation (source).1 Note that this is true for the entire electromagnetic spectrum, not just the radio region.
Synchrotron emission is a form of non-thermal emission which occurs when particles are accelerated perpendicular to their translational motion, specifically when the particles are relativistic (otherwise it is called cyclotron emission).
Self-absorbed synchrotron emission is described well in the introduction of this paper, and occurs when the synchrotron emission becomes a non-negligible source of heating. It occurs only when the self-absorption frequency is larger than the cooling frequency, i.e.
$$
\nu_{c} < \nu_{a}
$$
The spectral index, $\alpha$, describes the dependence of the radiative flux density on the frequency of the emission. With frequency $\nu$ and radiative flux density $S_{\nu}$,
$$
S_{\nu} \propto \nu^{\alpha}
$$
An inverted spectrum results from $0 < \alpha < 2.5$,2 and this is generally indicative of thermal emission. However, the observed emission can be depressed by absorption processes, so a positive spectral index (i.e. an inverted spectrum) is not robustly indicative of thermal emission on its own.

 Radio Astronomy
Most radio astronomy uses radio interferometry instead of direct observation, there are a number of reasons for this: most emitters are very weak in the radio portion of the electromagnetic spectrum, radio telescopes for direct imaging have to be much larger than their shorter-wavelength counterparts, and radio interferometry allows the highest angular resolution of any technique (it is possible to create a radio interferometery array the size of the Earth, which we have done).
That being said, there are a number of phenomena and objects whose activity in the radio region lends them to direct imaging:

Masers - stimulated spectral line emission, usually in the range of $3$mm to $0.3$m. This phenomena can occur in the atmospheres of gas giants, the expanded fuel sheath of late-type stars, protosolar disks and nebulae, supernovae remnants, and black holes. The produced spectrum is characterized by high brightness, extraordinarily high black-body temperature equivalent typically on the magnitude of $10^9$ K, but as high as $10^{14}$ K, and very high polarization that is predominantly circular. A spectra which contains a sharp peak that would correspond to a black-body temperature above $10^9$ K would likely be the product of a maser.

Quasars - The supermassive black holes in the centers of galaxies sometimes have accretion disks or clouds which produce immensely luminous emissions as the material falls into the black hole. The sheer scale of the energy involved (often orders of magnitude larger than the total output of the rest of the galaxy) leads to a relatively uniform emission across much of the electromagnetic spectrum, from X-rays to far infrared.

Pulsars - The mechanisms for emission here are poorly understood, but in the formation of a neutron star (or even a white dwarf occasionally), the conservation of angular momentum results in a very rapid rotation. A strong magnetic field causes protons and electrons to accelerate on the star's surface which produces cyclotron and synchrotron emission. The emission exits the pulsar in a beam.

Radio galaxies - Active galaxy nuclei which are exceptionally luminous in the $30$ m - $300$ km region. The radio emission in these nuclei is generated by synchrotron action characterized by a smooth, broad-band radio spectra with strong polarization.

1 Not a published source, but does a pretty good job of explaining this in the introduction.
2 $2.5$ is the rough upper limit for $\alpha$ in the power law description.


Why has there been no moon for 2 nights straight

The Moon is out of sight for a few nights every four weeks, more or less. It moves around the sky slowly. Sometimes it's far from the Sun (as seen from our planet), and that's when you see it at night, up in the sky. Other times it seems close to the Sun, and then of course you won't see it at night. Then the cycle repeats, over and over again. Takes a bit over 4 weeks to repeat.

The star-like things moving in the sky can be many things. Usually they are satellites in orbit around Earth, or airplanes flying at high altitude.


How Soon Could a Waxing Crescent Moon Be Seen?

From the US Naval Observatory: Crescent Moon Visibility:

  Naked-eye sightings as early as 15.5 hours after New Moon have been reliably reported while observers with telescopes have made reliable reports as early as 12.1 hours after New Moon.

USNO is usually a very reliable site for info like this.


What causes the raise of temperature in the outer layer of a red giant when the core gets contracted?

The temperature increase of the core and shell is governed by the virial theorem. As contraction occurs, gravitational potential energy is released. According to the virial theorem, half of this is radiated (or conducted or convected) away and half goes into the internal energy of the contracting material, thus raising its temperature. You can find a simple proof here.

Radiation pressure is unimportant, but continues to increase in any case. Only the surface of the star becomes cooler, not the interior.


Amateur astronomers regularly broadcast their viewing sessions over the Internet on the LiveSkies network. Pluto is within the viewing capabilities of most of the equipment used to broadcast but it just looks like a dim star that can be seen to move over long periods of time.
https://www.liveskies.org/
Scroll down to the  View As A Guest Link then click on a live broadcast. There are no schedules. People can broadcast from anywhere but the majority broadcast from North America. If it is dark and clear somewhere in North America someone may be on line.
Many amateur astronomy clubs have their own observatories that are open to the public on occasions. Google amateur astronomy clubs. There is probably a group near you.
Professional observatory time is a valuable commodity. Observatories with "high powered" equipment and skies dark enough for good viewing will have limited if any public access.


Can a supernova or a close passing star create a transient sun gravity focus "habitable zone"?

Supernovas have a light curve going from normal to peak and back on the order of a month. Hence the spot would go from too cold to habitable and back very quickly.
The same is true for stars passing the focal point: their motion (typically tens of km per second) makes the heated point move equally fast.
Habitable zone denotes a region where a planet could have liquid water (and maybe other benign environmental properties). This implicitly assumes that this is a stable state, since the context is planetary orbits and the emergence of life, a process that takes long time. A ultra-brief habitable zone is missing the spirit of the concept, even if it may be correct to state that a certain location is habitable over a short span of time.
As the deadly focal point post discusses, the brightening is also not enormous: to get a habitable spot, you need a fairly nearby supernova or star, so the gain is not enormous.


Doesn't gravity attract objects in space until they collide?

When an object is in orbit, there are two factors at play, not just one. The first, as you mention, is the force of gravity pulling the objects together. However, each object also has a momentum component which is generally (in the case of circular orbits) perpendicular to the direction of the gravity.

If we look at the common situation of a small-mass object orbiting a large (massive) object, then we can ignore the perpendicular velocity (momentum) component of the larger object and reach a simplification: The smaller object is continually pulled towards the primary but perpetually 'misses' due to its own perpendicular momentum.


How are the element abundances calculated for a meteorite in the Hydrogen log10 scale?

Meteorite abundances are referenced to their silicon abundance and then that is bootstrapped onto the hydrogen scale by assuming that the silicon abundance of meteorites is the same as that in the Sun. See for example Asplund et al. (2009).
This means there is an additional uncertainty of a couple of hundredths of a dex when comparing meteoritic and solar photospheric abundances.

So the answer to your second question is work out the abundances relative to Si and then use the solar Si/H ratio to put them on the hydrogen scale.


Why didn't the Event Horizon Telescope team mention Sagittarius A*?

There was a mention of Sagittarius A* during the Q+A portion of the press conference; the team indicated that they hope to produce an image sometime in the future (although they were careful to make no promises, and they're not assuming they'll be successful).

That said, I'm not wholly surprised that we ended up seeing M87, rather than Sgr A*, for a couple reasons which the team mentions in their first paper:

As Glorfindel said, Sgr A*'s event horizon is much smaller, meaning matter orbiting the black hole has a shorter orbital period. This contributes to variability on the timescale of minutes. The observations of M87 took place over the course of a week - roughly the timescale over which that target varies, meaning the source should not change significantly over that time.
Second - and this is the reason I've seen cited more often - Sgr A* lies in the center of our galaxy, and so thick clouds of gas and dust lie between it and us. That results in scattering, which is a problem. There are ways to mitigate this, of course, and the team has spent a long time on this, but it's simpler to just look at the black hole that doesn't have that problem in the first place. That's why M87's black hole is an attractive target.

Neither of these are impossible hurdles to overcome, but they're certainly very real difficulties that can't be ignored.


Do sungrazing comets leave a field of meteoroids near the Sun?

The SOHO image you attached, provides some evidence, that most of the debris continues the trajectory of the original body.
Most of the debris hence won't replenish the Vulcanoid population.
But due to the disruptive event you get an additional delta-v, which may change the orbit a bit, however insufficient to slow down to a Vulcanoid orbit; compare the velocity of the comet near perihelion with the speed of sound as an order of magnitude estimate of the upper limit of the delta-v for debris.
Dust may either be slowed down in this or in future orbits in the solar corona to end up in the Sun, or blown away by solar radiation pressure, again no Vulcanoids.


The redshift of the CMB is not measured, it is calculated.

The brief explanation is that as the universe expands and cools, it becomes energetically favourable to form bound atoms. The reduction in free electrons allows the universe to become transparent and photons escape as a blackbody radiation field and eventually form the CMB.

The temperature at recombination, $T$, can be calculated using well understood physics (see https://en.m.wikipedia.org/wiki/Recombination_(cosmology)) and if the current temperature $T_0$ is known, then the redshift is simply given by $ z = T/T_0 -1$.


Can a single location on earth witness solar eclipse totality 2 times during a single eclipse event?

No.

The video has been made so that the sun remains at the centre, but so that "up" is still "up relative to the horizon". This means that the shape of the crescent of the moon is how it will appear in the sky.  The turning of the Earth makes the direction of the moon relative to the sun appear to change.


How do we know Milky Way is a 'barred' spiral galaxy?

There are several different lines of evidence which together form a coherent picture: that of a barred galaxy. Moreover, as most disc galaxies are barred, we should expect the same from the Milky Way. The various evidences are:

The observed light distribution (2MASS) shows a left-right asymmetry in brightness and the vertical height. This is explained by the near end of the bar being located on that side.

The observed gas velocities show velocities which are "forbidden" in an axisymmetric or near-axisymmetric (spiral arms only) galaxy. These velocities occur naturally from the orbits of gas in a barred potential

The velocity distribution of stars in the Solar neighbourhood shows some asymmetries and clumping which is most naturally explained by orbital resonance with the bar rotation.

The extent, pattern speed, and orientation of the bar is consistent between all three of these.


Was the use of silver on 'optical' telescope mirrors more common in the past? Why?

Aluminium coating is a relatively recent process - it became available around the 1920s or 1930s. The Hale telescope arrived just in time to take advantage of this new technology.  (It requires a reasonably good vacuum to work, which probably explains why it took a while to come along.)

Before that - around the mid-1800s - various chemical "silvering" processes, usually involving real silver, were used. These weren't ideal as silver tarnished over time - the silver literally oxidised on exposure to air.  (Aluminium also reacts with air, but forms a thin oxide layer which doesn't harm the image and prevents further oxidation.)  Also some of these silvering processes were either toxic or explosive in application...

Before that - the first telescope mirrors were made of metal that was cast and ground to the required curve then highly polished. The material was "Speculum" a copper-tin alloy. They were poor in quality; far less reflective than later mirrors.  This account gives a nice description of casting mirrors for the Parsonstown telescope using over a ton of molten metal. Two mirrors were necessary for the larger telescopes because the mirror in use had to be taken out and re-polished so often...


The IRAS Point Source Catalog, Version 2.0, is a catalog of some 250,000 well-confirmed infrared point sources observed by the Infrared Astronomical Satellite (IRAS), i.e., sources with angular extents less than approximately 0.5, 0.5, 1.0, and 2.0 arcminutes in the in-scan direction at 12, 25, 60, and 100 microns (um), respectively.

This includes some comets, planetary disks, and even minor planets. Some sources are likely to be star forming, but an IR source meeting the spec above could be many things.


I come from a similar latitude and enjoyed the night skies and two things that can add to "the amount of light in the sky" that I can think of. The first is the near-full moon as mentioned above, and the second is unusually low clouds. Being near Toronto and outlying communities, there is a great deal of light produced, and some goes up. On nights when the clouds or even semi-transparent haze are present and unusually low, they can reflect light pollution more often than normal. Add to that the light from the Moon and it should be enough.
The reason I mention latitude is that when the ground is covered in white, reflective snow, the light pollution is even worse; lights projected preferentially downward are more strongly reflected back upwards. I don't know if there's any snow on the ground still there; I haven't seen snow in way too many years.
Since you didn't mention that your brother noted the full Moon, I'm guessing it was cloudy and so they didn't see it. So I'm guessing that instead they saw clouds or haze lit by it and/or by local light pollution.