Average rate of supernova x number of stars

Not that easy way. Only stars heavy enough will undergo a supernova explosion. The majority of stars is too light. The lifetime of a star is mostly determined by its mass. In some cases (supernova type Ia) a companion star provides mass to white dwarf, which originally has been too light to explode as a supernova.

Hence your technique can only work, if it is applied to stars of a certain mass and composition separately, and then summed or integrated up over all these restricted classes of stars. All together you'll get results for stars which eventually will end up as supernova, i.e. the heavy ones with more than about 8 solar masses.

... And in the meanwhile new stars will form.


Lunar Eclipses occur when a full moon occurs when the the moon crosses the ecliptic. The ecliptic is the plane in which the Earth orbits the sun, and so also the track which the sun appears to trace in the sky relative to the stars.

It appears from the diagram that there are two nodes each month (ascending and descending), but this is only approximately correct. The influence of the sun causes the nodes to move.It takes 29.5 days between two full moons, but only 27.2 days between two ascending and descending nodes). This means that a node can occur on any day of the year, and any day of the month.

A "supermoon" occurs when the moon when it is closest to the earth, the perigee. This point of perigee also moves, so the perigee can also occur at any time during the month, and the time of perigee is not related to the time of the node. Unlike terms like "node" which astro(nom|log)ers have used for many years, Supermoon is a neologism.

Thus is is possible, but rare for a full moon, a lunar node, and a perigee to occur at about the same time, and if it does, a supermoon eclipse will occur.


Extra-solar identification of binary planets

Yes indeed there could, especially in the case of a transiting planet system. Whilst the radial velocity curve would be nice and smooth, reflecting the position of the barycentre of the double planet, the transit light curve would show structure associated with the two planets obscuring the disc of the star. Essentially the transit shape would vary depending on the orbital phase of the double planet and on occasion the transit would occur in two steps as first one planet then the other
passed in front of the stellar disc. No such objects have been found.

People are already looking for this in the context of searching for "exo-moons" - for example,
the "Hunt for Exomoons with Kepler" programme. Finding a double planet with equal sized components would be comparatively easy compared with the tenuous signals expected from exomoons.


Are there heavenly bodies between galaxies?

The Big-Bang was not an explosion in empty space.
Inter-galactic space is not empty, there is an inter-galactic medium, gas clouds and material ejected from galaxies, including stars and possibly globular clusters, by various mechanisms...


How to calculate the position to be to see ISS passing in front of the moon or the sun?

The ISS Transit Finder offers such predictions up to 30 days ahead.
If a query yields no transits at the specified location,
the "Show on Map" button for a close pass shows where a transit would be visible.
As the documentation notes,
predictions beyond 10 days are rather uncertain.

Recent versions of Stellarium can show ISS's apparent position at a given place and time,
offering an alternative prediction for any given event.


Is it possible to use Hubble Telescope to observe Earth in the past?

The only way you can use Hubble to view the Earth "in the past" is to swivel it and point it at the Earth now. Assuming that this didn't break the telescope (it would) you would collect images of the Earth's surface that were approximately $500\times10^{3}/3\times10^{8} = 0.0017$ seconds old, since that is how long it takes light to travel from the Earth's surface directly below HST to its $\sim 500$ km orbital height above the Earth's surface.

The problem with what you suggest, is that although we might be able to pinpoint where the Earth was in the past with respect to its position now and the Galactic centre (actually that would be very hard in practice), the light that came from the Earth then has long since travelled beyond where the Earth is now.

For instance, let's say the Earth/Sun travel around the Galaxy at approximately 230 km every second and say we want to look at the spot where the Earth was 1000 years ago. Well, it turns out then that the Earth was at some location in space that is currently about $7\times 10^{12}$ km away (7 trillion). But the light that came from the Earth then, has already been travelling outwards at the speed of light for 1000 years, covering a distance of 1000 light-years, which is about $10^{16}$ km. So that light passed by our current location in space more than 999 years ago.


Is it possible to tell the time of day by shadows on the photo

In short: If you know length of shadow and object height (assuming vertical object casting shadow of an horizontal surface), you can tell time if you know date and you can tell date if you know time. If you know direction of shadow (e.g. relative to north) you can tell date and time without knowing any of them.

The most tricky part is to measure objects and shadows from a photograph - because you need their real length and perspective distorts object and shadow lengths -. To do this you need to make some assumptions about the geometry of the objects in the photograph. In fact, if you know the geometry of the settings, after some geometrical calculation you can find the Sun position and therefore date and time.

For example, with an image like this one and a detailed map, you can find date, time, position of camera and even focal length of lens - with a lot of measurement and geometry calculation, of course.

However, if the only objects casting shadows in your image are trees and people, measuring their exact position and size would be impossible and date and time estimations will be quite rough, specially if the image doesn't show any other identifiable and measurable features - and even worse if you can't tell the north from features in the photo. In general, finding date and time from shadows works better with buildings, specially with sharp shaped buildings with known orientation.


When exactly does a sub giant become a red giant?

The straightforward definition is in terms of where a star lies on its evolutionary track in the HR diagram (see below). The subgiant branch stars are those which have exhausted their hydrogen core and which are burning hydrogen in a shell but their He cores have not begun to contract significantly. The distinct upturn in luminosity marks the beginning of the red giant branch. This occurs when the core grows significantly more massive, cannot support itself hydrostatically and begins to contract. At the same time the envelope expands and becomes convective and the H-burning shell moves inwards and increases in temperature and luminosity.
The tip of the red giant branch is where He is ignited. This takes place "explosively" in a degenerate core if the star is below about $2M_{\odot}$ (not $5M_{\odot}$), but begins smoothly in higher mass stars. This causes the core to expand, pushes the H-burning shell out and reduces the luminosity.

HR diagram showing the location of the subgiant and red giant branches on the evolutionary tracks for stars of different mass. The tracks begin at the main sequence and evolve to the termination of the main sequence, across the subgiant branch, turn up the red giant branch, which terminates when core He burning commences and the stars become less luminous and hotter.


Extraction of error spectrum in Sloan Digital Sky Survey

The variance is the square of the standard deviation, so you can calculate 1$\sigma$ errors of the spectrum by taking the square root of one over the inverse variance of HDU1 (https://data.sdss.org/datamodel/files/SPECTRO_REDUX/RUN2D/PLATE4/spPlate.html#hdu1head).

(You will get infinity for bad pixels since their inverse variance is set to zero.)


Is the sun flipped on the other side of the world?

Yes, the orientation of the Sun will be different from the Earth's northern and southern hemisphere, just like your example of the Moon.

I would not say that a sunspot in the "northern" hemisphere would appear to be in the "southern" hemisphere just because of the change in orientation. North is fixed on the Sun and Moon, just like they are on the Earth. (If you turn a globe of the Earth "upside down", the northern and southern hemispheres do not switch, just the orientation."

The view through a telescope may change the orientation, but the directions North, South, East, and West on the Sun and Moon remain the same. The user of the telescope needs to determine which directions those are (up, down, left, right). It changes with the design of the telescope (reflector versus refractor) and the number of optical elements.


What's that limit, where we start to see infrared galaxies?

Galaxies give off a range of frequencies of light: Visible, but also infrared, microwave, and radiowaves. at longer wavelengths, ultraviolet, Xrays and gamma rays at short wavelengths. But stars give off quite a lot of their radiation in visible light, so nearby galaxies are quite bright in visible light and rather less bright in Ultraviolet.

Very distant galaxies are moving away from us due to the general expansion of the galaxy. This means that the light is stretched. Visible light gets stretched to infra-red, ultraviolet gets stretched to visible light. The distant galaxy is very very far away, and so very dim. It is a little bit brighter in infrared, although there is also a little visible light (and other wavelengths too)

Galaxies never go completely invisible in visible light, even if they are red-shifted, because there is always ultraviolet light to red-shift to visible.

The specific limit is often taken to be 700nm as the boundary between red and infrared. Although this is fuzzy because some people can see a little further than others.


From what distance can one object influence gravity of another object?

What id like to know, in what distance do they have to be from each other to create only one gravitational influence.

At whichever point you decide to call them two objects rather than one object. It's a completely arbitrary choice that depends on you rather than gravitational physics. What's going on is that gravity can be described by a mass density distribution, and which part of that distribution corresponds to "one object" and which to "another object" isn't important.

You may be confused by Newton's law of gravity that says the gravitational force force is proportional to the product of the masses and inversely proportional to the distance between them squared. But this law only applies to spherically symmetric objects. It only applies to you exactly if you're a spherical cow.

This kind of arbitrariness actually applies to the Earth as well. Even if the Earth were perfectly spherically symmetric, one could say that Earth's gravity is due to the influence of the northern hemisphere and the influence of the southern hemisphere, etc. Whether you consider the Earth to be one object, or two, or a trillion, depends on you.

The Earth isn't spherically symmetric. It's closer to an oblate spheroid, since it bulges at the equator. But it isn't exactly that either, having mountains or other topographical feature, to say nothing of mass density variations inside the Earth. In principle, one could describe its gravity including people and trees or whatnot. Gravitationally, there is no fundamental difference between you and some rock. That you consider yourself to not be "part of Earth" is a choice you make for other reasons.

If you distance yourself from the Earth, there is a difference because the mass distribution would change. But again, whether you consider this to be "only one gravitational influence" or "the Earth plus you" is up to you, and this holds true regardless of whether you distanced yourself or not.


Would a high albedo reflective substance cool down Venus?

Eventually, yes.

Interesting information about Venus: Venus is hotter than Mercury, despite being nearly twice as far from the Sun. Earth, despite being further from the Sun, receives more
energy from the Sun than Venus, due to Venus's very high albedo.

As you might guess by this information, the major factor that keeps Venus hot isn't how much energy it receives from the Sun, but rather its inability to rid itself of energy. While adding a cloud of some high albedo substance to the atmosphere may effectively prevent the Sun from heating Venus, it would also likely prevent infrared radiation from leaving Venus, thus keeping it warm.

As you know, Venus already has a very high albedo. It's thick atmosphere prevents energy from leaving the planet, which is what caused the temperature to rise far above the Earth-like temperatures we believe it once enjoyed. This is caused by IR radiation being emitted from the surface and being reflected back down towards the planet by the atmosphere more detail, which plays an essential part in the greenhouse effect.

In my opinion, a more effectual method in terms of making it habitable would be to find a way to drastically lower the albedo of Venus to allow all the excess energy to escape.


Flux vs. Velocity representation of spectra?

I think the velocity scale is calculated something like this:
$$v_r = c\left( \frac{\lambda - \lambda_0}{\lambda_0}\right),$$
where $\lambda$ is the observed wavelength and $\lambda_0$ is wavelength of the Mg II line, corrected for the redshift of the quasar ($z=0.868)$ it is observed towards.
The confusing thing is, for Fig.~3 in the cited paper, that the Mg II feature is actually doublet with lines at rest wavelengths of $\lambda_r =2796.35$ and $2803.53$ Angstroms. Therefore
$$\lambda_0 = \lambda_r (1+z) = 5223.58\ \ {\rm or} \ \ 5236.99\ \ {\rm Angstroms}$$
corresponding to the thick and thin solid lines on Fig.3 respectively.
In other words, zero on the velocity scale would correspond to a Mg II absorber at the redshift of the quasar (an intrinsic absorber), whereas negative velocities correspond to absorbers at lower redshift, by the velocities indicated.
For a check on this - calculate the velocity of feature "b", which appears to be at about 5221 Angstroms in the observed spectrum for the redder doublet component (the thin solid line in Fig.3). Using the above definition, this would be a velocity of -915 km/s.
Looking at Fig.3, there seems to be a major problem. Feature "b" is at -680 km/s.  However, if we look further on at Figs. 4 and 5, we see feature "b" plotted at -850 km/s in the Mg II spectrum, which is much closer, and would perhaps be in exact agreement if feature "b" was actually at 5222 Angstroms (which looks unlikely) or that a 4th, unspecified decimal place had been used in the redshift. e.g. if $z=0.8676$ then $\lambda_0 = 5235.87$ Angstroms, and the velocity of a feature measured at 5221 Angstroms would then be -852 km/s in perfect agreement with Figs. 4 and 5.
Either way, there is major disagreement between Fig. 3 (which I suspect is in error) and Figs. 4 and 5 (which could be reconciled with the above analysis if the redshift was 0.8676).
I'm sorry this is not very conclusive and I certainly wouldn't accept it without someone else verifying that the x-axis of Fig.3 appears to be problematic.


How strong would gravity need to be to prevent star formation

By way of starting a discussion, because I don't clearly see how to answer this question correctly or all the factors that may be involved.

If you increase $G$, then by the virial theorem, the central temperature of the ball of gas (that will contract quasti-statically, since radiation cannot escape unimpeded) will be proportional to $GM/R$. So for a fixed mass, the ignition temperature for nuclear fusion will be reached at a larger radius and stars would be physically larger and less dense.
$$T_c \simeq \frac{GMm_u}{10Rk_B},$$
assuming an ionised hydrogen gas and where $m_u$ is an atomic mass unit.

In order to avoid "starhood", you need to compress the gas either to within its Schwarzschild radius or to a density where it could be supported by electron degeneracy pressure without reaching a high enough temperature to ignite hydrogen fusion.

Considering the first of these. $R_s = 2GM/c^2$, so the central temperature reached at $R=R_s$ would be $\simeq m_u c^2/20k_B = 5\times 10^{11}$K, is independent of $G$ and is far higher than the $10^{7}$K required for H fusion (ignore deuterium). So direct collapse to a black hole cannot circumvent nuclear fusion.

The second possibility is likely to be more constraining.
There is a standard treatment given in Shapiro & Teukolsky (1983) thaat determines the radius $R$ at which electron degeneracy becomes important in supporting a star of a given mass $M$. The idea is that you work out the mass and radius for which the phase space occupied by an electron approaches $h^3$. The result is (for ionised hydrogen):
$$ R/R_{\odot} \simeq 0.07 (M/M_{\odot})^{-1/3}$$
However, digging into the maths a bit, this can be written as a proportionality
$$ R \propto G^{-1} M^{-1/3}$$
and thus the central temperature at which electron degeneracy pressure becomes important will be
$$T_c \propto GM/R  \propto G^2 M^{4/3}$$

Thus (for a fixed $G$) what happens is that a contracting ball of gas will reach a critical $T_c$ for fusion if the mass exceeds some critical threshold (above about $0.075M_{\odot}$ in our universe) then H fusion will begin before degeneracy sets in. If you increase $G$ then this threshold actually becomes smaller and smaller balls of gas will become stars.

But what about the Chandrasekhar mass? In our universe the threshold mass for fusion is much lower than the Chandrasekhar mass ($5.8M_{\odot}$ for pure hydrogen), so there is no problem in supporting an object just below the fusion threshold with electron degeneracy pressure. Dig a little deeper and we find that the Chandrasekhar mass is in fact $M_{Ch} \propto G^{-3/2}$. This means we can now express the critical central temperature in terms of mass as a fraction of the Chandrasekhar mass.
$$ T_c \propto G^2 \left(\frac{M}{M_{Ch}}\right)^{4/3} (G^{-3/2})^{4/3} \propto \left(\frac{M}{M_{Ch}}\right)^{4/3}$$

In other words, the threshold between a star that burns hydrogen and an object (brown dwarf) that will be supported by degeneracy pressure is a fixed fraction of the Chandrasekhar mass and does not depend on $G$.

My conclusion is therefore that although changing $G$ will do interesting things to stellar evolution (changing the rate at which it occurs for instance - see here where I show that the luminosity scales as $G^7$ and therefore lifetimes scale as $G^{-7}$ !), I do not think that changing $G$ allows direct collapse to a black hole without the central temperature becoming high enough to initiate hydrogen fusion.

Secondly, a failed star is possible in our universe - it just needs to be less than $0.075M_{\odot}$ and collapse is halted by electron degeneracy pressure. In a universe with an increased $G$ this threshold becomes smaller as $G^{-3/2}$.


What weather is present in Jupiter's great red spot?

You seem to know most of what there is to know about the Great Red Spot, except that there is not enough moisture in it for rain. It is cold enough for snow, but again there is a water deficiency. Water vapour is a comparatively rare gas in Jupiter's upper atmosphere. I think lightning would be admissible, and high winds too, and just possibly some frost. After all, there is frost on Mars despite the scarcity of water, though frost doesn't seem to be commonplace there.

The Great Red Spot would not be a good place for settlers, they'd be much better off on Mars. I presume this air base is carried around by a huge balloon, but what would they do for food and water? Normal aircraft would be no use because of the lack of oxygen, which is a pity, because there is hydrogen fuel in abundance.


'Very High Noon' - Sun directly overhead

Let me give a slightly better answer. If you visit
http://ssd.jpl.nasa.gov/horizons.cgi with these parameters:

you'll see the Sun's astrometric right ascension and declination for a
48 hour period.

Let's look at one of the results and see what it tells us:

Date__(UT)__HR:MN     R.A._(ICRF/J2000.0)_DEC
**********************************************
$$SOE
2016-Nov-13 00:00     15 13 30.08 -17 57 18.0

This says the sun's declination is 17$^{\circ}$57'18.0" which means
the sun is (almost) directly overhead at that latitude.

The right ascension is 15h13m30.08s which means the sun is (almost)
directly overhead where the local sidereal time is this value.

To compute local sidereal time we first use
http://aa.usno.navy.mil/faq/docs/GAST.php to get GMST (I'm using the
approximation):

GMST = 18.697374558 + 24.06570982441908 D

To get D, the number of days since the year 2000, I used Unix's date
command as follows, and used other commands to finish the calculation:

: per http://aa.usno.navy.mil/faq/docs/GAST.php we compute the number of
: days since 2000 January 1, 12h UT (not 0h UT)
date -d '2000-01-01 12:00 UTC' +%s;: result is 946728000
date -d '2016-11-13 00:00 UTC' +%s;: result is 1478995200
: take the difference and convert result to days for formula
perl -le 'use POSIX; printf("%.10f\n", (1478995200-946728000)/86400)'
: result above is 6160.5 (trailing digits are 0)
: computing GMST and modding to get result between 0 and 24
perl -le 'use POSIX; printf("%.10f\n", fmod(18.697374558 + 24.06570982441908*6160.5,24))'
: result above is 3.5027478917
: convert solar RA to decimal
perl -le 'use POSIX; printf("%.10f\n", 15+13/60+30.08/3600)'
: result above is 15.2250222222
: how many hours east of GMST is the Sun (in degrees)? 1h = 15 degrees
perl -le 'use POSIX; printf("%.10f\n", (15.2250222222-3.5027478917)*15)'
: result above is 175.8341149575
: converting to minutes and seconds, we first find seconds from fractional part
perl -le 'use POSIX; printf("%.10f\n", 0.8341149575*3600)'
: result above is 3003 rounded to nearest second
: and now minutes/seconds from 3003 seconds total
perl -le 'use POSIX; printf("%.10f %.10f\n", floor(3003/60), 3003%60)'
: result above is 50 and 3

So you would think the Sun is overhead at 17$^{\circ}$57'18.0'' south
latitude (south because the declination is negative) and longitude
175$^{\circ}$50'03"E (east because longitude is positive) at 00:00 UTC
on 2016-Nov-13. Let's use HORIZONS to see if this is true (it turns
out it won't be, so we'll use a minute on either side of 2016-Nov-13
00:00 UTC to see what happens):

We can confirm the Sun's RA matches local sideral time within 1 second
of 00:00UTC from these lines in the output:

Date__(UT)__HR:MN:SC.fff   R.A.        DEC         Azi      Elev    Sid_Time
2016-Nov-13 00:00:00.000 * 15 13 30.08 -17 57 18.0 104.6639 89.7692 15 13 29.2615
2016-Nov-13 00:00:01.000 * 15 13 30.08 -17 57 18.0 104.9166 89.7730 15 13 30.2643

You'll notice the local sidereal time jumps from being 0.82 seconds
behind the sun's right ascension to 0.18 seconds ahead of the sun's
right ascension in the one second period (showing the two matched
sometime within that second, probably at 00:00:00.82 or so).

However, the Sun's highest position (noted by HORIZONS as 't' for
transit) occurs a little later:

Date__(UT)__HR:MN:SC.fff    R.A.        DEC         Azi      Elev    Sid_Time
2016-Nov-13 00:00:56.000 *  15 13 30.23 -17 57 18.6 178.5844 89.9415 15 14 25.4148
2016-Nov-13 00:00:57.000 *t 15 13 30.24 -17 57 18.6 182.4651 89.9415 15 14 26.4176
2016-Nov-13 00:00:58.000 *  15 13 30.24 -17 57 18.6 186.3230 89.9412 15 14 27.4203

when the local sideral time is almost a minute ahead of the sun's
right ascension. What's going on here?

Since the Earth is an ellipsoid, and not a sphere, it turns out that
"up", the direction perpendicular to a location on Earth, is NOT
directly away from the center of the Earth (it turns out gravity acts
in a direction that is neither towards the center of the Earth nor
perpendicular to the surface, but that's not relevant for this
question): see http://www.oc.nps.edu/oc2902w/c_mtutor/shape/shape4.htm
for details.

If you want really accurate calculations you'll have to use
http://naif.jpl.nasa.gov/naif/tutorials.html 's
http://naif.jpl.nasa.gov/pub/naif/toolkit_docs/C/cspice/surfnm_c.html
function.

Although this doesn't cause the discrepancy above, another
consideration is a location's elevation above the surface of the
Earth, which can also have slight effect (since "up" isn't always
towards the zenith). Happily, one thing you can ignore is refraction,
since refraction at the zenith is effctively zero.

My http://test.barrycarter.info/sunstuff.html is a very approximate
representation of where the Sun is overhead and is generated by
https://github.com/barrycarter/bcapps/blob/master/bc-sun-always-shines.pl


Why there are less massive stars than low mass stars?

One popular argument for the Initial Mass Function is that the mass spectrum -- i.e., the fact that there are more low-mass than high-mass stars -- is inherited from the density spectrum of molecular clouds. In other words, stars form from the collapsing cores of molecular clouds, and the distribution of cores is already biased towards there being more low-mass cores than high-mass cores (this is in fact observed in molecular clouds). So you naturally end up with more low-mass stars than high-mass stars.

Of course, the natural next question is: why is the distribution of clouds cores biased towards low-mass cores? The usual suggestion is that it is the result of supersonic turbulence, and simulations of supersonic turbulence for conditions similar to those of a molecular cloud tend to produce density distributions similar the observed distributions.


$$v_r = c\left( \frac{\lambda - \lambda_0}{\lambda_0}\right),$$
where $\lambda$ is the observed wavelength and $\lambda_0$ is the wavelength at rest. This gives an equivalent velocity with respect to the central wavelength of the line.

In your plot, what is shown is a line profile where the wavelength separation from $\lambda_0$ for that line (or from the measured centre of the line, I can't tell which) has been converted into a velocity.

Note that it makes little sense to plot more than a single line profile in this way.


Meteor traveling through atmosphere without hitting Earth

yes it is possible,they are called Earth grazers or Earth-grazing fireball.
they are not rare but only few incidents are recorded.

for more details read this Wikipedia page https://en.wikipedia.org/wiki/Earth-grazing_fireball


From Mean Moon to True Moon in an old procedural calendar

How would you interpret this? "(1780 + 80) * 3 on base 808, and add 2"

As before, there is a confusing avoidance of specifying the unit under consideration, but the pattern is still that the last number is in minutes of arc, the middle in degrees, and the first in $30^\circ$ for the Zodiac divisions, so:
$$\begin{eqnarray*}
\frac{(1780+80)\times 3}{808} &=& 6+\frac{27}{30} + \frac{10}{30\times 60} + \frac{41}{30\times 60^2}+\ldots \\
&=& (6;\;27\!:\,10)\text{.}\end{eqnarray*}$$
The $+2$ makes this $(6;\;27\!:\,12)$, although this makes the second addend in minutes of arc while the first one is not. As before, the calculation is not rounded.

  Eade has (8; 11 : 7) - (6; 27 : 12) = (1; 3 : 55), but that doesn't work.

  That specific substr. gives me (1; 13 : 55), which could be a typo in the paper, but that value doesn't carry things forward either.

I think that you are correct that this is a typographical error, but the corrected value probably is carried forward--the error is just repeated again in Step 16. Note that $(1;\;13\!:\,55) = 43^\circ 55'$, which makes the corrected Step 16
$$296\sin(1;\;13\!:\,55) = 205.309 = (0;\;3\!:25)\text{,}$$
following the convention that the numerical value is re-interpreted as minutes of arc. On the other hand, if we follow what Step 16 actually says, we get a very different value:
$$296\sin(1;\;3\!:\,55) = 165.164 = (0;\;2\!:45)\text{.}$$

Granted, the value actually carried forward to Step 17 is $(0;\;3\!:24)$ instead, but the former is definitely much closer. This further discrepancy could also be an undocumented geographical correction (the procedure does, after all, have things like $+3$ and $+2$ at times), but I'm more inclined to guess that this is due to imprecision in the calculation of sine, as the paper indicates that a table was used.


How did Meeus calculate equinox and solstice dates?

The mean instant of an equinox JDE_{0} (or solstice) is the statistical mean calculated over a sample of equinox's instants (since they do not happen exactly in the same instant/day every year).

The first eight equations are derived by interpolation and the corrections (the 24 terms) are derived as a truncated iteration, I suppose from the JDE{N}-JDE{0,N}.

The truth is that Meeus does not explain anything at all (though he mentions interpolating the apparent longitude of the sun in 3 dates for higher accuracy). Without an explanation in the book, it is pretty difficult to guess how he did it.

You can find some more details about the iterative method here:

http://www.spacebanter.com/showthread.php?t=171014

I hope this helps :)


How are the photos of the Milky Way taken?

We are situated inside the galactic disk. The first image below, showing our approximate location, is a fabrication rather than a photograph. Such a photo would require a space probe to be tens of thousands of light years above the plane of the disk, which is patently infeasible with current technology. We don't even know exactly how many spiral arms our galaxy has!

But from where we are, we can still see a great density of gas and dust stretching 360 degrees across our sky, as you look along the galactic plane. These "inside views" are indeed real photographs - much the same as if you go outside on a dark night, well away from the city lights, you yourself will see its faint, dusty outline much like in the second image (which I can assure you is real).


Do ASKAP and ALMA have "fast dump" interferometric modes? Can they see and perhaps report Fast Radio Bursts in real time?

Short answer: No and yes.

Do ASKAP and/or ALMA have "fast dump" interferometric modes? Or do they have at least some way to extract dispersion of transient events at the millisecond level?

ASKAP has FREDDA which searches in total-intensity filterbanks (aka dynamic spectra) and triggers the voltages for imaging or localizing the FRB. general disclaimer that I am not 100% sure but do know about their searchpipeline
ALMA operates in a whole different frequency regime ($\sim 230$ GHz).
Dispersion delay scales $\sim \frac{1}{\nu^3}$ where $\nu$ is the observing frequency. So you'd need ridiculously high DMs to produce a sweep there. You could say there is almost no dispersion at these frequencies. There is also the question if FRBs are produced in that regime. The highest frequency any FRB (repeating or one-offs) has been detected is around $8$ GHz (FRB121102, Gajjar+2018 etal).
Having said all of that, the essence of what you are suggesting is captured in realfast which runs commensally on the VLA and does fast de-dispersed imaging at around $1-4$ GHz. Meaning, it produces an image on the milli-second timescale for many dispersion sweeps and then searches for bursts. There is also an AI running which vets many of the candidates it gets to test. All of this is done in realtime.
Check out this realfast detection of a periodic FRB detected last April.

Can either see and perhaps report Fast Radio Bursts in real or near-real-time?

These pipelines have the motivation to save the raw data and so employ realtime searches which issue triggers and then slices of the raw data are saved. There is AI running which decides if the trigger is to be issued or not. And even with it, there's a lot of False positives.
Reporting FRBs is a different story. Usually, confirmed FRBs are reported in Astronomer's Telegram (ATel), or Transient Naming Server (TNS). There is no one unified network (like LIGOs-VIRGO-FERMI) which shares triggers because of various reasons.
Or, you could choose to do follow up observations and then report.


What causes the aurora of other planets to have different colors?

Ultimately, you are correct. However, there may also be some clarification in order: what blue aurora of Saturn are you referring to? The most well-known auroral images of Saturn like this one, are not true color images. That is, what we see in the photo isn't what you'd see with the naked eye; instead, the photo has been enhanced to emphasize something interesting, in this case the aurora. In reality, that beautiful blue you see in the picture isn't even of light that's visible with our eyes. It's UV light which would be completely invisible to us. People working with data taken by a specialized camera have to give the UV light some visible color or we wouldn't see it at all. The Saturn aurora images I am aware of all have been modified in this manner.

However, aurora themselves are the result of charged particles funneling down a planet's magnetic field and crashing into the planet's atmosphere. On Earth, the aurora occurs because of the atoms in the atmosphere getting energized by these collisions, and then emitting light when they de-energize. Here, it's the oxygen and nitrogen in the atmosphere that are responsible for aurora light (primarily greens from oxygen, but also reds at higher altitudes from oxygen and occasionally blues from nitrogen). In other planets' atmospheres dominated by different atoms and molecules, the reactions will be different, and the wavelengths of light emitted will be different as well. For example, Saturn's aurora are due to energized H2 molecules. On Jupiter, it can be the infalling particles themselves (sulfur and oxygen) that are responsible for auroral emission.


We could not have formed "outside" the expanding universe, as there would not be any substance to form our galaxy, and solar system -including our Sun and planet.


Would a direct hit by a small comet be catastrophic?

Here, have some fun with this...

http://impact.ese.ic.ac.uk/ImpactEffects/

It doesn't go into long term effects, but the short term is pretty devastating for a significant portion of the planet.  'Nuclear winter' effects seem inevitable for those not atomised, scorched, crushed, or covered in ejecta.

Not good, nope.


Where to find spectra from Las Campanas Redshift Survey?

LCRS catalog data should be accessible at VizieR:VII/203.
The associated paper is Schectman et al. 1996.


How long until the Earth and Moon become a binary planet?

I'm not sure I agree with the double planet POV, but the calculation is pretty simple.   The earth weighs 81 moons, so for the Barycenter to be outside the earth, the distance (center of Moon to surface of earth), = 81 earth radii.

or about 515,000 KM.   It's current farthest distance is 405,000 KM, average distance 384,000 KM and closest 363,000 KM  Source, so, it depends on whether you mean, temporarily outside, outside more than 50% of the time or always outside - each would provide different answers.

But moving away 4 CM per year (same source) or 1 KM every 25,000 years, it would need 2.75 billion years to move away the necessary 110,000 KM necessary to have the barycenter move outside of earth at the moon's furthest point.  (less if the orbit gains eccentricity which is possible).   But probably more than that, cause as the moon moves away from the earth, the tidal forces that continue to push it away grow weaker and the earth slows down a little.    If the earth ever gets tidally locked with the moon, the reverse will happen and the sun's tidal forces will draw the Earth and Moon slowly towards each other, so a precise answer is too mathematically difficult for me, but longer than 2.75 billion years seems a pretty good guestimate.


Are these parallel lines I'm seeing in this image a real pattern on the Moon?

They are not real.

The images used in the animation use a mosaic of photographs taken by an orbiting satellite, the Clementine orbiter. This satellite has a polar orbit, and so returns images in strips that run roughly north-south. Due to variations in lighting between orbits, when the strips are joined together, there are artefacts at the joins.

The whole moon image from a ground-based telescope doesn't have such artefacts. The lunar orbiter view was used as it covers the whole moon, including the far-side. As the moon orbits Earth, it appears to wobble (an effect known as libration.) By starting with a whole moon map, and then projecting it to a sphere, this libration can be simulated. You couldn't do that with a single ground-based image.


Is Planet Nine shepherding the Kuiper Belt?

Estimates of Planet 9's orbit currently suggest that it doesn't have any major effect on the Kuiper Belt, because its orbit is too far away and on a different plane than the Kuiper belt, and from the inner Oort cloud, at 200 and 1200 aphelion and perihelion, However it does have an effect on Trans-Neptunian objects that transit through the Kuiper belt. It doesn't get closer than 4 times the distance of Pluto and it transits at a different plane to the Kuiper belt.
Shepherding is the act of actually creating and directing the asteroid belt by providing many objects with a semi stable orbit with too violent collisions to make a planet (biggest is Ceres at ${\rm 500 km}$ diameter), and having a massive effect on the unstable collection of asteroids, like making large stable groups around Jupiter's Lagrangian points at 60 degrees to Jupiter, and making fast resonant orbits that are not round like stable asteroids but periodically go close to Jupiter and then go straight through the inner asteroid belt and back to towards Jupiter actually in a triangle.

Neptune's orbit averages 30AU, and the Kuiper belt is generally held to be between Neptune and Pluto's orbits, Pluto orbits at 31-48AU.
In that sense, Pluto and Neptune are the two objects which have the biggest effect on the Kuiper belt.
The Trans-Neptunian objects all traverse the Kuiper belt, with their Perihelions ranging from 17 to 50 AU, and if Trans-Neptunian objects swing out towards Planet 9, it's a less stable orbit for them, than if they swing out away from planet 9, so all the large Kuiper belt objects swing away from the alledged direction of planet 9, where they can have a stable course without a transiting planet to interfere with them. Trans Neptunians tend to not align themselves with Planet 9's orbit and tend to find a balance away from Planet 9 and away from Jupiter and the majority of them don't get closer than Neptune, about 7/8ths of them are further than Neptune. https://en.wikipedia.org/wiki/List_of_trans-Neptunian_objects
The Trans-Neptunian objects are often referred to as Kuiper belt objects, and they didn't start their course to 100ds of AU's distance because of Planet 9, but now they are affected by planet 9 because they are in its vicinity over millions of years. In a million years, they have about 500 chances to approach planet 9, and 500,000 chances in a billion years.
The shepherding effect is complex, Jupiter accelerates objects very fast in stable orbits that resonate with it, it prevents planets from forming from the asteroid belt, and it disrupts the otherwise stable asteroid belt, sending objects fast through the asteroid belt and wherever, towards and away from the Sun. Jupiter created and maintains the asteroid belt which is unstable space debris, and it protects Earth from comets that want orbits that through its path.
Planet 9's inclination is about 30 degrees, so it can't interact with the inner Oort cloud, and the outer Oort cloud is too far away.


trying to understand stellar nucleosynthesis

The elements you are talking about are created by the r-process - the rapid neutron capture onto heavy elements that can occur in neutron-rich environments. The r-process is probably responsible to some extent for all the most neutron rich (i.e. more neutrons than protons) nuclei with atomic masses above that of iron and is probably responsible for all elements heavier than lead and bismuth. However, there are three broad peaks in the solar-system abundances, centred at Germanium, Xenon and Platinum (the first, second and third r-process peaks), where the r-process is thought to dominate the production over competing processes such as the s- and p-processes (plot taken from here).

If one subtracts the expected contributions from other sources one can estimate the fraction of a given element that comes from the r-process (plot taken from here).

It is a contemporary astrophysical problem (i.e. there is no definitive answer) for the source of the r-process. In other words, it is an unsolved mystery as to what fraction of each r-process element comes from which astrophysical location and it may well vary as a function of cosmic time or location. The two main candidates for the sites of creation for r-process elements are core-collapse supernovae, the neutrino driven winds from supernovae and neutron star mergers.

The main problems to be solved are the detailed physics of supernova explosions and neutron star mergers, the relative populations and numbers of these events as a function of cosmic time and location and even the nuclear physics ingredients that go into calculating the yields of such elements.

A recent paper by van de Voort (2015) performs simulations that suggests most of the r-process elements in the universe could be created by neutron star mergers, and they point out that it is especially difficult for core-collapse supernovae to produce what are known as the "third-peak" r-process nuclei. These claims are all contentious, although the latter point about the heavier r-process elements is echoed in a number of previous works. (e.g. Wanajo 2013). Nevertheless, even this is disputed; for example Nishimura et al. (2015) perform simulations of rapidly rotating supernovae progenitors with strong magnetic fields and find that they can reproduce the abundance pattern of all the r-process elements seen in the Sun.

So if I wanted a vague stab in the dark at directly answering your question, I would choose the group of elements in the third r-process peak, with mass numbers $191 < A< 197$ as being most likely to have been (mostly) produced by neutron star mergers. This includes Osmium (192), Iridium (191, 193), Platinum (194, 195, 196) and possibly gold (197).


The answer to the headline question is: No. Most of Saturn's rings are below the Roche limit of about 2.5 Saturn radii. Hence tidal forces will prevent that part of the rings to form a (large) moon.

Actually, part of the rings may be caused by loss of material from some of Saturn's moons, as suspected from observations of Enceladus.

Accretion of Earth is still going on. So any number between millions of years and billions of years of accretion time for a planet can be justified.
Half of Earth's mass should have accreted within 10 million years, see this paper.


What is the name of that which exists beyond the Universe?

By definition, the capitalized word Universe denotes everything there is, so even if we one day discovered we're just a part of a Multiverse, all the parallel universes of it would still be parts of the Universe as a whole, where Multiverse would just describe its nature. Or, if some yet undiscovered regions of it would defy our current understanding of its physical laws and constants as we know them, all of it would still be a part of the whole Universe. So it really doesn't matter, if beyond the known universe, there are regions of honey, milk and chocolate biscuits and all of it is carried by a giant tortoise. All of it would be the Universe, the physical universe as we can observe, the honey, milk, chocolate biscuits and the tortoise. The lot. All of it. The whole shebang.

Notice that I use capitalisation here, i.e. you can have more than one universe, but they're all a part of the Universe. Without capitalisation, it's just any domain, a particular sphere in physical or metaphysical sense, and only a part of the whole Universe. Sadly, this capitalisation is often neglected or used inconsistently, as is often the case with earth vs the Earth (the top soil vs the planet), sun vs the Sun (any star with planets vs our Sol), moon vs the Moon (any natural satellite vs our Luna), even galaxy vs the Galaxy (any galaxy vs our Milky Way). For example, observable universe is a sphere, a region of space, within the Universe.

The beauty of this naming convention is, that we already know the name for (but not necessarily of) everything there is, even if we don't know or can agree on what all that encompasses, or what laws govern some regions of space, time, or some other, yet unknown existence of it. It is universally true regardless of anyone's beliefs, even if they choose to call all of it by other names or attribute this existence to a sentient being, super being, or God. We're all a part of everything there is - the Universe.


Would there be any difference for our solarsystem to be not in the Milky Way?

No it wouldn't make a difference unless you moved it near another star or body. If it was moved such that something was close enough to exert significant gravitational attraction it would have an effect on our solar system.


Where is the intervening light in the M87 black hole?

Based on the comment from @SteveLinton

The radius of M87 is given to be 60,000 light years, which is a volume of 900 trillion cubic light years.  Given that it contains roughly one trillion stars, there is an average of 900 cubic light years for each star.

Several resources referred to the diameter of the black hole in the range of 1.5 light days (38 billion km).  That's 0.004 light years.

If you consider the cylinder1 extending from the black hole to the edge of M87, the cylinder is 0.004 light years in diameter with a height of 60,000 light years.  The volume of that cylinder is just over 3 cubic light years.  Since there is one star for every 900 cubic light years, it follows that the cylinder encompassing our line of sight between the black hole and the edge of M87 is empty of any stars.

Whether or not any star would even be bright enough to be visible is another matter.

1 Technically a conic frustum but I don't think the difference in volume is significant.


Does the shape of a supernova remnant depend on the progenitor star's magnetic field?

The factors that affect the shape of supernova remnants include:

Interstellar medium (the gas and particles between stars)
Planets that surrounded the star
Magnetic field (in the same way magnetic fields cause auroras)
Mass distribution of the star before supernova

There are probably more, but that's all I can think of at the moment


What is meant by Electronic Beam Steering and how is it achieved?

This can actually mean a couple of different things, but in the case of steering an MWA tile, Jeremy is quite correct. Each tile has a 'beamformer' box that all of the individual antenna cables plug into. This box physically delays each signal appropriately (by having a longer path to go through for more delay) based on the signals sent to it for each pointing. Because of the quantised path lengths there are what are called 'sweet spots' that are used, in other words arbitrary pointings don't normally happen you just dial in the closest 'sweet spot' to where you'd like to point.

There are also version that do this in software (and in fact in a special phased mode the MWA does this between tiles sometimes), where 'turns of phase' are used instead but it is essentially the same idea.


What would Aurora look like if no magnetic field was present around the Earth?

I was looking through some sites and came across this - https://www.livescience.com/earth-magnetic-field.html
This clearly mentions how, if the magnetosphere surrounding the Earth weakens, the Auroras might penetrate into the Atmosphere and light up the skies closer to the equator. So, I also looked it up but, to my astonishment, I found no sources that answered this question to my satisfaction.
Though, intuition says that If there is no Magnetosphere, there should be no Aurora either.


With all the stars in the universe, why isn't the sky saturated with light?

Logic1: If the universe is infinite AND static, the sky will be saturated with light.

Note "AND."

Logic2: If the sky is not saturated with light, the universe is not infinite or not static.

Note "OR."

Logic1 = Logic2

Observation: the sky is not saturated with light.
Therefore: the universe is not infinite or not static.

We know that the universe is dynamic, at least.


Can protoplanetary disks form main-sequence stars?

Yes, unless you want to get really particular with the "protoplanetary" part. For example, there are stars forming in the circumstellar disks around Wolf-Rayet stars [reference]. If we were picky, we might not call the circumstellar disk around this Wolf-Rayet star a protoplanetary disk (and instead refer to the planet-forming circumstellar disks of the newly formed stars as such), but there exists no clear line between the disks of gas within and the larger disk of gas around the Wolf-Rayet star of which they are a part.

Source:

Triggered star formation surrounding Wolf-Rayet star HD 211853 - Liu, Wu, Zhang, and Qin


The simple answer is yes, an asteroid can have rings. The current known example is 10199 Chariklo, whose rings were discovered in 2014 (see Braga-Ribas et al. (2014)). It actually has two rings, at distances of 391 km and 405 km, 7 km and 3km in width. As the discovery paper notes, they may be the (young) remnants of a debris disk around Chariklo, and have some moderately-sized bodies embedded in them. The observations indicated that water ice is likely present. Another small body, 2060 Chiron, may also have rings (see Ortiz et al. (2015)). Again, water ice is thought to be a major component of the putative rings.

Both ring systems were hypothesized in part due to evidence from stellar occultations, i.e. cases where the minor body passed in front of a star. Stellar occultations have proved useful in many cases involving small bodies, such as Kuiper Belt Objects.

As Ortiz et al. noted, Chariklo and Chiron are the two most massive members of a class of minor planets called centaurs. They orbit (unstable) in between the giant planets of the Solar System. Insights into ring formation around centaurs could perhaps provide information about centaur formation and evolution. It seems likely that the rings form from either

The impact of a smaller body on the surface of the centaur
The breakup of a satellite orbiting the centaur (as it crosses the Roche limit, as you speculated)

Now, rings have not been observed around Trans-Neptunian Objects in similar stellar occultation data, leading the authors to hypothesize that centaurs are a separate population. More observations could shed some light on this hypothesis.

A final interesting point I'd like to make is that many moons around minor planets have been found, but only one or two ring systems are known.


What are the biggest problems about the numerical, finite-element GR models?

If you can provide examples of numerical methods in GR you've seen/heard of that would help focus the question.

From the article you linked to: "The technique keeps track of a vast number of quarks and gluons by describing the space and time inside a proton with a set of points that make up a 4D lattice". This almost gets to the main issue with Numerical Relativity. There is no natural computational grid on which to simulate space-time. The whole game with GR is that gravity is space-time so first you have to simulate the space-time and then you have to simulate the objects (neutron stars, black holes, gravitational waves) on top.

As the links below go into, its very difficult to create a consistent computational grid since the physical space-time your trying to simulate for a black hole has "funny" things in it like singularities, or an event horizon pas which we can't really know what's going on.

I think this article: http://astronomy.com/magazine/2016/02/putting-einstein-to-the-test?page=1

does a good job of summing up the field, and its quite accessible.

For something more rigorous please see: http://arxiv.org/pdf/1010.5260v2.pdf
That paper gets into some of the math behind the article linked to above.


Periodic error correction in automatic telescopes

Quoting the most relevant parts from the Richard McDonald's Wiki on Astrophotography Mounts: Periodic Error Correction:

  Periodic Error can be reduced to an acceptable level using a variety
  of techniques, only some of which are in the range of a beginner or
  mid-level astrophotographer.

  Throw Money at the ProblemVery high-end mounts for astrophotography have very small periodic error because of the time
  and money spent on manufacturing high-precision gears. You can also
  buy higher-precision gear upgrades for some mounts. For the beginner,
  let's call this impractical. Buying a \$1000 replacement worm gear for
  your \$1000 mount is probably not a good balance.
  Don't Use GearsThis is really an extreme case of the "throw money" solution, mentioned just for fun. There are some
  very-high-end experimental drive systems showing up on the market that
  don't use worm gears, and that don't exhibit periodic error. Examples
  include direct drive systems and harmonic drives. 'Way out of our
  price range.
  AutoguidingA second guide camera and computer can be used to make frequent small corrections to the mount's pointing,
  and this can reduce or eliminate Periodic Error if the onset of the
  error is not too sudden. This is the subject of a separate article.
  Periodic Error Correction FeatureTraining PEC with hand-control Most mounts intended for astrophotography include a
  feature called Periodic Error Correction (PEC), which can be used
  alone, or in conjunction with Autoguiding. PEC is used in two phases:

  Training. During this phase you use the control panel or menus to say "Hey! Pay attention to this!" to the mount, then you manually
  keep a star perfectly centred for several worm periods. You do this by
  centering a star with a high-magnification eyepiece that includes a
  cross-hair reticle. You then stare at the star for 10 to 15 minutes
  and use the mount's hand controller to manually make the small
  adjustments necessary to keep it perfectly centred on the cross hair.
  The mount records the error corrections you supply, remembering where
  in the worm position each one was needed. By training through more
  than one worm cycle, the mount can record an average correction, in
  case you reacted slowly or over-reacted.
  Playback. Once you have trained the mount, you can turn on PEC. The mount will "play back" the recorded error correction information
  by slightly changing the speed of the Right Ascension drive to move
  ahead or back each time your manual error corrections did the same.
  This will cancel the periodic error, resulting in a smoother track.

  Training PEC with
  hand-control: Periodic Error Correction (PEC) button on the control
  panel of an equatorial mount.

  The manual training phase is quite tedious and a more modern
  alternative is to use a camera and computer - usually the same one you
  will use for auto-guiding - to track a star while the mount records
  the training information.

  There is even specialized software available to help collect Periodic
  Error data, analyze and smooth it, and upload it to the mount. You do
  not need such software, as the manual techniques mentioned above will
  work just fine. However, it makes a tedious job simple and pleasant,
  and you may find it a worthwhile investment. I use PEMPRO and am
  very impressed with it. It's not free, but it is inexpensive and works
  very well (and includes another feature to help achieve perfect polar
  alignment).

Quoted text and photograph source, copyright and courtesy of Richard McDonald, no copyright infringement intended. I would also highly recommend reading same author's extensive article on Autoguiding.


What is the effect of the axial precession on the orientation of the planet's orbits?

When talking about the longitude of ascending node you must be very careful to define the reference plane that you are using.  As you state, due to the Earth's axial precession, the First Point of Aries moves along the ecliptic over ~26,000 years.  This is because the celestial equator is slowly precessing about the ecliptic.

Now, in the case of the orbital elements of other planets, the longitude of ascending node is again with respect to the First Point of Aries.  As your intuition suggests, the orbits of the outer planets won't be influenced by the precession of the Earth's equinox.  This means that the longitudes of ascending node will not be constant over long time periods.  If we suppose that the physical orbits are fixed (i.e., the planets are not perturbing each other), the change in the longitudes of ascending node will be entirely due to the Earth's precession.  When you take both these changes into account, you will find that they cancel each other out and the position of these orbits remains fixed with respect to the distant stars.

You can see in the orbital elements provided by NASA that they are given with respect to the J2000.0 epoch.  In other words, the orbital elements are provided for a single point in time, and to get the orbital elements today you must take into account the Earth's precession.

http://ssd.jpl.nasa.gov/txt/p_elem_t1.txt


How to convert apparent visual magnitude to SI units?

I found this from Wikipedia:

$$\mathrm{illuminance} = 10^{(-14.18 - M_\mathrm{V})/2.5}~\mathrm{lx}$$

https://en.wikipedia.org/wiki/Illuminance#Astronomy

Looks to be right.  I checked the formula with the luminous flux of Sol given by Wikipedia as 3.75×1028 lumens, with an (extraterrestrial) luminance at 1 AU of
$$\frac{3.75 \times 10^{28}~\mathrm{lm}}{4\pi~\mathrm{au}^2} = 133000~\mathrm{lm/m^2}$$
The formula above gives instead a (terrestrial) illuminance of 105700 lm/m2 for an apparent visual magnitude of -26.74 (which makes sense since there would be less illuminance on the surface beneath an atmosphere).

https://en.wikipedia.org/wiki/Sun


What is the current angle of motion of the sun with respect to the Galactic Plane?

In an answer to this question, you can see that I describe the trajectory of the Sun around the Galaxy. Its current motion can be defined in a number of ways, according to your preferred frame of reference. From your question I judge that you want it in a "Galactocentric" frame, so we take the velocity of the Sun with respect to the "Local Standard of Rest" (LSR), which is the Sun's motion with respect to an average of local stars, and add something to it to account for the fact that local stars are orbiting the Galactic centre every 220 million years or so.

The velocity of the LSR is 10 km/s towards the Galactic centre, 5 km/s faster than an average star at the Sun's location tangentially to a line towards the Galactic centre, and 7 km/s upwards out of the Galactic plane (see for example here).

Thus, if we take this frame of reference, the component of the velocity of the Sun in the Galactic plane is $\sqrt{10^2 + 5^2} = 11.2$ km/s amd the component out of the Galactic plane is just 7 km/s. The angle this trajectory makes with the Galactic plane is $\arctan(7/11.2) = 32$ degrees. Note that there are uncertainties/disagreements about the exact value of the Sun's motion with respect to the LSR at the level of 1-2 km/s, so this final angle is also somewhat uncertain.

However, this rather steep angle does not take into account that the LSR is moving around the Galaxy at rather high speed. This speed is around 250 km/s in the Galactic plane and at right angles to the Galactic centre (e.g. Reid et al. 2014). There may be a more modern figure, but I cannot immediately locate it, but it is probably safe to assume an uncertainty of at least 5-10 km/s in this. With this we now have velocity components of 11.2 km/s out of the plane and 255 km/s in the plane. The angle you are seeking is then $\arctan(11.2/255) = 2.5$ degrees, with an uncertainty of only $\sim 0.3$ degrees. So if you were trying to make a drawing of the solar system motion with respect to the Galaxy, that is the angle you would use. Note that the Sun is currently above the Galactic plane by about 20 pc.


How bright would Sagittarius A* be from Earth if it became an Active Galactic Nucleus?

If Sagittarius A* were to ignite and feed on a massive accretion disk, blasting the massive jets we call quasars, it wouldnt be noticed by anyone here on Earth unless they were looking for it.  The relativistic jets accelerate out of the poles of the Supermassive Black Hole (SMBH) shooting out of the plane of the Milky Way, not through the disk structure which would cause massive damage to the galaxy. And with the naked eye, relativistic jets wouldnt be visible to us through all the gas and dust of the Milky Way, combined with the 26000 LY away from the galactic center we are, we really wouldnt even know the quasar came to life, and the Moon would still be the brightest heavenly body at night.

For more information, check out the Wikipedia article about quasars.


How can Earth-Sun Lagrange points L1 & L2 even be semi stable considering the moon?

The L1, L2, L3 points are not stable. (full stop) Small deviations grow exponentially, even in the perfectly circular restricted three-body problem. In reality, we have a non-circular many-body problem, when (1) these points are not strictly well-defined and (2) deviations from the simple case have to be accounted for to obtain the exact trajectories (and to keep any artificial satellites near these points).

The L1 & L2 points are attractive for space missions exactly because they are unstable, since this prevents the accumulation of debris (natural and artificial in origin) and hence reduces the probability of collisions.


How often are new astronomical objects (variable stars, supernovae, comets, etc) discovered by amateurs?

The answer is: frequently. There are many amateur astronomers that make it their ambition to discover new supernovae or to observe and report on new variable stars.

As an example, let me cite amateurs Robert Evans, who apparently holds the record for most supernovae found by visual observation, or Tom Boles, who holds the record for supernova discoveries by an individual.

Observations of variable stars can be reported to the Information Bulletin of Variable Stars (IBVS). Supernovae discovery or new comets would normally be reported to the International Astronomical Union Circulars.


That's a timelaps video, most likely created from a sequence of photos. It's easy to choose sensitivity and ISO settings on normal cameras such that the night sky at the darkest night is over-exposed. The moon light looks like sunlight - as it simply is reflected sunlight.
E.g. see https://cloud.planetmaker.de/index.php/s/yJ7AXWG4dEC6csW - it's a an exposure of several seconds taken several hours after sunset in November in Kiruna, Sweden... the over-exposed part in the center are headlights of a car, the rest is illuminated by the nearly-full moon at that time.
As children of modern civilisation living in urban areas, we are simply not accustomed anymore to how bright even nights can be when you had sufficient time to adopt - and cameras can do that instantly. In arctic (or generally snowy) regions, the snow adds a lot to the overall brightness as it reflects the little available light at night extremely well.


Kinetic energy and galaxy arm rotation rate

It is miniscule. General Relativity is not required to understand the dynamics of galaxies. Motions are non-relativistic, a few 100 km/s at most, so the kinetic energy of objects is always much smaller than their rest mass energy.
Another way of seeing this is to calculate $GM/Rc^2$, a ratio which tells you the relative size of errors you will get in ignoring GR in any calculation.
For our Galaxy, we might say there is about $M\sim 10^{11} M_{\odot}$ within $R\sim 15$ kpc, and the ratio is then $3\times 10^{-7}$, which indicates Newtonian gravity is fine for most purposes.


Can red dwarf stars have a spectral type of L?

Yes. Stars (those objects that are supported by hydrogen fusion) can be as cool as spectral type L2.
Brown dwarfs can be as warm as M4/5 when they are young.
i.e. there isn't a clear spectral type Vs mass relationship. It depends on age too.


Do all galaxies have a black hole in the center?

The answer here is certainly no. Many dwarf galaxies (e.g. the Magellanic Clouds) don't contain supermassive black holes, although some may contain less massive intermediate-mass black holes (Mezcua et al. 2018). As they are the most numerous type of galaxy in the universe, it's quite possible that most galaxies, in fact, do not contain supermassive black holes. This shouldn't be too surprising; if the bottom-up theories of supermassive  black hole formation are correct, many intermediate-mass black holes must coalesce to form a massive one. Dwarf galaxies simply wouldn't have (or be able to form) the requisite population, in most cases.

That said, a number of dwarf galaxies have been observed to host supermassive black holes (notably including M60-UCD1 and Hen 2-10, among others). These have provided, and continue to provide, insights into the formation and evolution of low-mass galaxies.

Some users here have also identified some massive galaxies without supermassive black holes, including M33 and A2261-BCG. The former has a bulge, but the central stellar velocity dispersion is low, implying there is no black hole at its core (Merritt et al. 2001), assuming the required parameters are known well enough.  In A2261-BCG, it's possible that one or more central supermassive black holes were ejected, leaving a "puffy" core without a black hole. Keep in mind, of course, that this is only one possible explanation for the observed distribution of stars in A2261-BCG.

In general, a central supermassive black hole does not influence the large-scale kinematics of stars and gas within a galaxy. Most supermassive black holes are much less massive than their host galaxy; I believe Sag A*, at the center of the Milky Way, is on the order of about 0.001% of the total mass of the galaxy. Even massive ellipticals like NGC 4889 have black holes about 0.1% of their total mass. As far as I know, ultra-compact dwarfs have the highest black-hole-to-galaxy mass ratios, with an extreme example being the previously-mentioned M60-UCD1.


Is there any legitimacy to this report of planet nine being discovered by an amateur?

This has not been confirmed so whether it is legit is unknown at this time.  A recent report (https://arxiv.org/abs/1706.05348) calls into question the original analysis that leads to the initial report of planet nine. The report that fueled initial interest is given below.

(EVIDENCE FOR A DISTANT GIANT PLANET IN THE SOLAR SYSTEM
Konstantin Batygin and Michael E. Brown
Division of Geological and Planetary Sciences, California Institute of Technology, Pasadena, CA 91125, USA; kbatygin@gps.caltech.edu
Received 2015 November 13; accepted 2016 January 10; published 2016 January 20)


How do radio telescopes gather information about visible light?

They don't; the picture you saw all over the news was a false color image, where frequencies are shifted from the radio spectrum to the visible spectrum (blue meaning shorter radio wavelengths and red meaning longer radio wavelengths) and/or the intensities mapped to different colors (instead of just affecting the brightness as with true color images).

Or, to quote an old StarChild question:

  The term "false color" is used to describe what astronomers (and others) often do to images to make them more comprehensible. Long ago, when radio astronomers first started generating images of sources, they wound up with essentially images that were just shades of gray - ranging from pure black to pure white. Each shade represented the intensity of the radio emission from a particular part of the object. Radio astronomers took their shades-of-gray images and converted them to color ones by assigning red to the most intense radio emission and blue to the least intense emission recorded on the image. Intermediate colors (orange, yellow, and green) were assigned to the intermediate levels of radio intensity. Black was assigned to places in the image in which there appeared to be no radio emission.


What is the angular diameter of Earth as seen from the Moon?

You can calculate the angular diameter of the Earth using the equation:
$$a = \arctan \frac{D}{d}$$
where $a$ is the angular diameter, $D$ is the physical diameter of the Earth, and $d$ is the distance from the Moon to the Earth.

The equatorial radius of the Earth is $r_E = 6378.1 \textrm{km}$, the diameter is therefore $D= 2 \times r_E = 12756.2$.

The mean distance to the Moon is $d=384399\textrm{km}$. This gives a (mean) angular diameter of $a=1.90065$°.
At Perigee the distance is $d=362600\textrm{km}$. This gives an angular diameter of $a=2.01482$°
At Apogee the distance is $d=405400\textrm{km}$. This gives an angular diameter of $a=1.80226$°

These Moon-Earth distances are as seen from the centre of the Moon. To calculate the diameter from the surface of the Moon, you'll have to subtract the position of the observer along the Moon-Earth axis.

If the observer is on the Moon's equator and the Earth is at zero hour angle (i.e. on the local meridian), the distance to the Earth needs to be subtracted by $r_M=1738.14\textrm{km}$. This gives the following values:

The mean distance to the Moon is $d=384399-1738 = 382661\textrm{km}$. This gives a (mean) angular diameter of $a=1.90928$°.
At Perigee the distance is $d=362600-1738 = 360862\textrm{km}$. This gives an angular diameter of $a=2.02452$°
At Apogee the distance is $d=405400-1738 = 403662\textrm{km}$. This gives an angular diameter of $a=1.81001$°

The angular diameter of the Earth from the surface of the Moon is, therefore, between $a=1.80226$° (at apogee and the Earth is near the horizon) and $a=2.02452$° (at perigee and for an observer at the equator and when the Earth is at maximum altitude on the meridian).

Or about 2 degrees.


How well is Earth’s motion through the universe quantified?

Surprisingly, we will soon be able to do this well enough to go back several hundred million years and place ourselves, if not in the solar system, inside the Galaxy and roughly in the right part of the Galaxy.   It turns out that we know, with high accuracy, the Earth's motion with respect to the microwave background radiation, which provides a well defined frame of reference for motions.  Therefore, we will be able to work out which direction and about how far the MW has moved with respect to this universal frame and separately the Earth with respect to the center of the galaxy. It requires solving for the growth of what is called the 'microwave dipole velocity' over time.  Some of this velocity develops from nearby objects like the MW and the nearby groups of galaxies.  So, we need to know their masses and relative motions a bit better than we presently do.  But, by far, most of the velocity is from objects quite far away, and this allows us to use the simpler and well understood 'perturbation' techniques. The biggest uncertainty for the Earth's position might be the interactions with local stars back in time and what effects they had on perturbing the Suns orbital motion about the center of the galaxy.  Again we need to know better the masses of local stars and their relative velocities to calculate the Earth's trajectory back in time.  As it happens, the space telescope GAIA, which is about to make its second release of data, will greatly improve our knowledge of the motions and, indirectly, the masses of all these stars and nearby galaxies.

So, if you have got that time machine working, we are nearly ready to go!


Why does the Hubble flow does not carry photons with it?

Expansion isn't some mysterious force that "carries" galaxies. There's a very common misconception that it is, but it's not true. The expansion is just the recessional motion of galaxies and other matter.
Light does participate in the expansion. Even though it moves at a constant speed (and therefore can't move with the galaxies as you correctly observed), it still thins out with time. This is counterintuitive, but it happens even in a special relativistic cosmological model (the Milne model, which is a special case of FLRW cosmology). If you think of the big bang as an explosion of light bulbs, the bulbs that you see from a particular era in the universe's history have higher and higher relative speeds as time goes on (since the light from nearer, slower bulbs has already reached and passed you), and therefore are more redshifted, and therefore emit photons at a lower rate (since redshift affects the perceived rate of every process). So the relic photons from a particular era of the universe's history get more redshifted and less dense as time goes on.


Is there enough hydrogen left after a star dies so another star will have enough to light up?

There are several misconceptions in your question.

First, a star does not vacuum everything in its vicinity. Rather it forms from a condensation in a gas cloud, which in turn collapses to a proto-star surrounded by a gas disc, which can contribute further material. Once formed in this way, a star typically does not acquire more gas (exceptions are symbiotic binary stars etc).

Second, a star with mass in excess of $\sim8$M$_\odot$ will (typically after a long time) suffer from a supernova, when most of its envelope is flung back into space. That gas is still mostly hydrogen, though enriched by 'metals' (non-primordial elements). However, the gas is hot and fast moving and hence not in a condition to form another star.

Third, the gas from the supernova will eventually mix with other gas and dissolve into the general pool of interstellar medium (ISM). Some of that may cool to form a molecular cloud (as gas cloud where molecular $H_2$ dominates), which in turn may become the site of new star formation.

We know that the Sun has been formed from enriched material, which is a mixture of primordial gas with the ejections of several supernovae.


Did the Late Heavy Bombardment Period Happen Because of a Stellar System Collision?

There are several potential explanation for the cause of the late heavy bombardment. One of them includes a passing star that disturbed the Oort cloud. A "stellar system collision" would have destroyed the solar system, though. While galaxies are mostly empty and a galactic collision won't involve that many actual collisions (if any), the gravitational interactions between the galaxies actually changes star orbits around the galactic centers. Similarly a "stellar system collision" would heavily change planetary orbits, even if there were no collisions at first.

Another hypothesis is a totally "interior" one: A 2:1 orbital resonance between Jupiter and Saturn changed the orbits of Neptune and Uranus, potentially even causing them to swap positions. Neptune's new orbit then affected many Kuiper belt objects, either sending them way out to form or join the Oort cloud, or sending them in as part of the late heavy bombardment.


Difference in redshift between 2 interacting galaxies

From eq. 10 in Hogg's classic paper, assuming that the peculiar velocity $v_\mathrm{pec} \ll c$:
$$v_\mathrm{pec} = c \frac{z_\mathrm{obs} - z_\mathrm{cos}}{1 + z_\mathrm{cos}},$$
where $z_\mathrm{obs}$ is the observed redshift, and $z_\mathrm{cos}$ is the redshift from cosmological expansion only.

Let me invert that and wrap it up in Python for ya:

def zobs(vpec,zcos):
  c = 3e5
  return (1+zcos) * vpec/c + zcos


calculating rising time from culmination time

For a star, the time from rising to culmination is the same length as the time from culmination to setting. From knowing the time of culmination and setting, you can calculate that difference and subtract it from the time of culmination to obtain the time of rising.

This method is generally accurate for the planets and Sun, but it is not always accurate for the Moon. Culmination refers to when an object is highest in the sky. For a star, this is when the object is on the meridian. For the Moon, there can be a significant difference between the time the Moon crosses the meridian (transits) and when it culminates. See Transit and Culmination (although the images on that site were not working today).

P.S. I have never seen tables that give the time of culmination for the Sun and Moon. Times are generally given for transits (when the object is on the meridian).


Heavy bombardment - Why has things cooled?

According to the Nice model -- that's Nice, France (rhymes with Reese or geese), not nice (rhymes with mice) -- the solar system formed with the giant planets (the outer four) in the following order: Jupiter, Saturn, Neptune, Uranus. Outside of these planets existed a large field of asteroids (also called planetesimals), the remnants of which form our Kuiper belt.

The Nice model hypothesizes that Neptune and Uranus actually swapped places due to an orbital resonance between Jupiter and Saturn. This had the effect of moving Jupiter inward (towards the Sun) and kicking Neptune farther out. This orbital disruption also caused a thinning of the Kuiper belt, where some of the planetesimals gained kinetic energy and were knocked into the Oort cloud, and some of them slowed down and fell into the inner solar system, causing a period of heavy bombardment.

It is thought that the period of heavy bombardment ended once the orbits of the outer four planets stabilized and an equilibrium was reached. The result was the current order of the outer planets we see today (Jupiter, Saturn, Uranus, Neptune), a thinner Kuiper belt, and an outer Oort cloud. Of course, this is just a model; the Nice model itself has already undergone some revisions, even though it is relatively recent. A nice video illustrating the Nice model can be found here.

I hope this answers your question.


Redshift observation at different wave length

It's not impossible, but does require a bit if a special set-up.

Normally, the light from a star is redshifted mainly due to its velocity toward/away from us (a small, but important, contribution comes from gravitational redshift, when the light climbs up the potential of the star). Because stars tend to rotate, light coming from the side of the star rotating away from us is slightly more redshifted than the light coming from the side rotating toward us. If a large prominence is present on one side, then because the prominence is hotter than the average surface temperature of the star and thus emits at shorter wavelengths, the combined spectrum will exhibit a larger redshift at short wavelengths.
Since the flux from the prominence alone is, after all, small compared to the total flux, the effect would be rather small, however.

A more common mechanism that may cause a differential redshift is stellar outflows. If large amounts of gas are being expelled (e.g. in protostars or AGB stars), then atoms and molecules in the outflowing material will absorb some of the light from the star. These absorption lines are then blue shifted wrt. the "systemic" velocity (i.e. the velocity of the star wrt. us), again skewing the redshift as a function of wavelength. A spectral line may even be seen both in emission and absorption, known as a so-called P-Cygni profile. The figure below (from the Wiki article linked to) shows the same intrinsic wavelength (here H$\alpha$) both redshifted and blueshifted (by rotation and thermal motion), but the blueshifted light is absorbed by the outflowing atmosphere:


Is stacking welder's glasses a safe way to watch at the eclipse?

YES

You have summarized pretty much everything available on the subject. There are loads of people saying you cannot, but fortunately for us, the Royal Astronomical Society of Canada, that you quote, includes evidence to back up their claims. They are a respected institution and the author of this article appears immensely qualified. The article itself does not include evidence that welding goggles can be stacked, but the inclusion of such for all the major points adds credibility to the entire article.

It does appear like the glasses are not exactly additive, this Reddit post goes through some calculations, but they are pretty close such that aiming for 1 tint higher will more than offset it and if anything is overkill.


How does the amount of dark matter in measured galaxies vary?

tl;dr: In general, smaller galaxies have relatively more dark matter than larger galaxies. The atypical galaxies that you refer to are discussed in the last paragraph.

Mass fractions in the Universe

Matter in the Universe is dominated by two components; baryons (i.e. atoms in various forms) and dark matter (DM). The baryons predominantly exist in two forms: gas (including plasma) and stars (some gas condenses to form dust, planets, etc., but that's a minor part).

The fraction of the Universe's total energy in the form of matter is $\Omega_\mathrm{M}\simeq0.31$ (most of the rest is dark energy). The baryons and DM have fractions $\Omega_\mathrm{b}=0.05$ and $\Omega_\mathrm{DM}=0.26$, respectively (numbers from Planck Collaboration et al. 2016).

Hence, the baryons and DM comprise fractions
$f_\mathrm{b}  = \Omega_\mathrm{b}  / \Omega_\mathrm{M} = 16$% and
$f_\mathrm{DM} = \Omega_\mathrm{DM} / \Omega_\mathrm{M} = 84$% of the total mass, respectively.

Galaxies have more dark matter than the average Universe

Galaxies formed when DM+gas overdensities decoupled from the Hubble flow and started collapsing, and so were "born" with the "cosmic" fractions. However, observationally it is found that the baryon fraction in galaxies is much smaller; for instance, a "typical" Milky Way-sized galaxy roughly has $f_\mathrm{b}$ only half of the cosmic average, and the DM fraction is correspondingly larger, >$90$% (e.g. Werk et al. 2014).

Gas and dark matter dynamics

The reason is the different mechanisms that affect the dynamics of the particles.
Whereas the DM is collisionless, the gas particles collide and dissipate their energy, making it easier to cool and contract. Thus, the part of the galaxies that we can observe is much smaller than "whole" galaxy; the  typical "half-light radius" is but a few percent of the emcompassing DM halo (Kravtsov 2013, Somerville et al. 2017). The half-light radius is the radius within which half of the light is emitted; when you see an image of a galaxy, it typically looks perhaps 4-5 times larger than this. So, something like this:

A significant fraction of the baryons reside in the halo as well. This is the hot gas of millions of Kelvin, and the high temperature and low density makes it difficult to detect, as it glows only in faint X-rays.

Feedback blows out gas, but dark matter stays

But various physical processes act so as to try to blow the baryons — but not the DM — out of the galaxies. These processes are known as feedback, and are due to energy being "injected" in the baryons.
For massive galaxies with halo masses $M_\mathrm{h}\gtrsim10^{12}\,M_\odot$ this feedback is dominated by their central black hole accreting gas, resulting in extreme luminosities manifesting itself as a quasar, or active galactic nucleus (Silk & Rees 1998; Croton et al. 2006). For smaller galaxies with $M_\mathrm{h}\lesssim10^{12}\,M_\odot$, the feedback is mostly due to stellar winds and exploding stars depositing thermal and kinetic energy in the surrounding gas (Dekel & Silk 1986; Hopkins et al. 2012). At even lower masses, the galaxies are so small that a significant fraction of their gas may be ejected altogether during early epochs of star formation (e.g. Bullock et al. 2000), resulting in galaxies that consist virtually only of DM and a few stars.

The smaller a galaxy is, the shallower its gravitational potential, and thus the easier it is for the gas to escape the galaxy.

Hence, in general the smaller a galaxy is, the larger its amount of dark matter is.

Conversely, the larger a halo is, the more its mass fractions converge toward the cosmic fractions. These are only reached for halos of masses $M_\mathrm{h} \gtrsim 10^{15}\,M_\odot$, which are no longer individual galaxies, but rather groups and clusters. This is seen in the figure below (from a paper from last week by Henden et al. 2018), which shows the stellar mass fraction (left) and gas mass fraction (right) as a function of halo mass:

The baryon fraction is $f_\mathrm{b} = f_\mathrm{stars}+f_\mathrm{gas}$, and the DM fraction — which is what you're asking for — is then given by $f_\mathrm{DM} = 1 - f_\mathrm{stars} - f_\mathrm{gas}$. The stellar fraction is seen to decrease with halo mass, due to the AGN feedback discussed above, but the total star+gas fraction increases.

Note though that although the DM — being collisionless — is not exactly blown out of the galaxy like the baryons, the gravitational attraction between the two components still affects the DM and alters the density profile of the halo (Duffy et al. 2010).

Galaxies with more gas than dark matter (?)

Thus, a galaxy without baryons is not a huge mystery, as long as it's small. Even larger galaxies are apparently able to rid themselves of most the baryons; van Dokkum et al. (2016) reported a MW-sized halo with 98% DM.

A galaxy with baryons, but without DM is more spectacular, but was reported recently by van Dokkum et al. (2018) (yes, the same guy).
In another answer about DF2 I discuss various processes that might lead to such a galaxy, including misinterpretation of the data.


The key for stellar classification is the Hertzsprung-Russell diagram (or HRD for brevity), also know as the color-magnitude diagram.

This shows the relation between the surface temperature (on the x-axis) of a star and its luminosity (on the y-axis). The temperature is an index of the Spectral type. For distant stars also spectroscopy is taken into account, to classify spectral types according to their absorption lines here some examples. The usual classification is called MKK (from the astronomers that had developed it - Morgan, Keenan and Kellman in 1943), and divides the stars in 7 stellar types O, B, A, F, G, K, M. Please note that from the 1943 other stellar types have added, but these are the most used. This sequence goes from the hottest (O) to coolest (M) stars. Each letter class is then subdivided using a numeric digit with 0 being hottest and 9 being coolest. Our Sun is a G2 type star. This classification also uses an additional notation to avoid the "degeneracy" you mentioned: a roman numeral between "I" (one) and "V" (five). The higher the number, the wider the lines in the observed spectrum. However, despite of the quantity these roman numbers indicate, they are an index of luminosity class. This is based on the width of certain absorption lines in the star's spectrum which vary with the density of the atmosphere and so distinguish giant stars from dwarfs. Luminosity class I stars are supergiants, class III regular giants, and class V dwarfs or main-sequence stars, with II for bright giants, IV for sub-giants, and VI for sub-dwarfs (from here). Then, our Sun is "completely" identified by the type G2V. You can take a look at the figure below, and check out by yourself where it belongs:
.
Also note that other indexes are used to characterize the stars, as for the width of the lines in the luminosity class, or for some peculiarities.

Finally, if you are interested in the study of stars, you should always keep in mind the Stefan-Boltzmann law:

$L=4\pi\sigma R^2 T^4$

By this you can identify isoradii lines in the HR diagram:

Also, there is a proportionality between luminosity and mass:

$L\sim M^{3.5}$

Then you can identify "isomasses" lines in the HR diagram, but I can't find such a figure to show.


Can a magnetic field of an object be stronger than its gravity?

It depends on what object it's acting on.  There are many objects, including stars, that have magnetic fields where Lorentz forces on charged particles like electrons and protons are stronger than the gravitational force on them.

Also remember that the strength of the Lorentz force depends on the speed of the particle moving through it, so a fast enough moving electron even here on Earth will receive a larger magnetic force than a force of gravity. This is how the Earth's magnetic field is able to contain charged particles in the Van Allen belts that its gravity could not contain.


How are we identifying chemicals in space?

The science that studies the problem you described, is called "spectroscopy". Every element absorbs some of the wavelengths. Same is with molecules. The spectrum is like the footprint of the molecule.
When you observe the light from emitted from some object, you see the continuous spectrum, emited from the light sources (stars, etc.). But it has some black lines present. These are the wavelengths that are absorbed by the molecules and elements. Then you compare it with different spectrums on the Earth (where we can identify the substance) and when you find the same spectrum, you know with a extremely high probability that this is the substance you are searching for.
Example for methanol: Astronomers observed the object with spectroscopy and found the spectrum with some black lines present. They compared it with various other spectra on the Earth and found out, that methanol has same spectral lines. Both measurments can be showed with a graph:
https://webbook.nist.gov/cgi/cbook.cgi?Spec=C67561&Index=1&Type=IR
Spectral lines are normally showed with a spectrum (with colors) and some black lines, but this graph shows the dark lines more physically (not all is absorbed) on a graph as deep valleys.
Be aware! There may be redshift present in distant objects. That means, that the lines would shift to the red part of the spectrum. (Similar to Doppler effect.)


What are the key coordinate systems used in Astronomy, and where are they used?

Navigating the solar system, you'd probably track your probe using Earth-based radio telescopes in the equatorial ICRF, though you might describe the overall trajectory in terms of the ecliptic.

Navigating at sea, you might measure the Sun's position in horizontal coordinates at a time in GMT to determine your own geographic coordinates.

Practical uses in backyard astronomy:

horizontal: telling someone where to look relative to an object on the horizon; pointing a telescope on an altitude/azimuth mount
equatorial: finding or plotting an object in a star atlas; choosing atlas pages to use on a given night; pointing a telescope on an equatorial mount
ecliptic: importing comet or asteroid orbital elements into planetarium software; planets are easiest to observe when their longitude differs most from the Sun's
galactic: low latitudes have more Milky Way star clusters and nebulae; high latitudes have more external galaxies
supergalactic: pondering the distribution of galaxy clusters?


Why celestial objects appear brighter in the night sky than in photographs from closer proximity?

The Moon, Mars, Venus are seen against a dark sky using an eye whose aperture is adjusted to the average brightness of the visual field. So bright small sources are over-exposed and so appear bright.

Photographs of planetary surfaces are taken with aperture and exposure set for proper exposure. So appear closer to how they would look if they filled your eye's field of view and so the aperture would be set to properly expose the surface.


Solution for counting atmospheric layers of planets according to relational data

While the question is rather hard to follow, the formula simply doesn't work.

For Mercury the formula using a naive application of the numbers predicts:
four layers

Orbital Velocity (km/s)        47.4
Density (kg/m3)              5427
Orbital Inclination (degrees)   5.1
n                            -328.65368
Aphelion (10**6 km)            69.8
Layers                          4

That is clearly not true, since Mercury has no atmosphere. You can vary the formula, by choosing different units, or taking the inclination relative to a different plane, but no matter what I try, can't get the formula to give a sensible value for all the planets.

The formula is inconsistent, since you generally can't add velocity (km/s) to sqrt(density)*inclination (kg^0.5 degrees/m^(1.5)) as these quantities have different units. It rarely makes sense to add quantities with different units.

There is, moreover, no reason to suppose that the orbital characteristics of a planet, and its density would have any effect on the number of layers in the atmosphere, which is not a very well defined quantity anyway.

Sorry but this formula is entirely wrong.


How will Starlink affect observational astronomy?

tldr; some modelling has been done with a lot more to do, but generally the impact of these constellations is fairly negative, but potentially manageable.

Ok, there's a lot to unpack here. First things first, while people have been aware of Starlink and have thought of it, modelling the impacts to observatories hasn't happened as of yet in quite a few places.

The main observatory which has had modelling done, LSST, released this statement;

  LSST is particularly sensitive to bright trails of these satellites, due to LSST’s unprecedented product of field-of-view and light collecting area.  Most LSST images will contain such a trail if plans for multiple LEOsat constellations materialize. The first group of Starlink satellites are sufficiently bright during dawn and dusk (when LSST would be surveying) that the trail would exceed sensor saturation, generating uncorrectable artifacts in the data
  LSST Statement

AURA (the owners of LSST) provided some additional information;

  The LSST Project Science Team has been simulating the potential impacts to LSST observations. Their latest update of preliminary results from November 2019 indicates that (assuming the full deployment of planned satellites) nearly every exposure within two hours of sunset or sunrise would have a satellite streak. During summer months there could be a 40% impact on twilight observing time (less in winter) and saturation of sensors by the satellites can continue well past astronomical twilight. Because of scattered light in the optics by the bright satellites, the scientific usefulness of an entire exposure can sometimes be negated. Detection of near-Earth asteroids, normally surveyed for during twilight, would be particularly impacted. Dark energy surveys are also sensitive to the satellites because of streaks caused in the images. Avoiding saturation of streaks is vital. The LSST team notes that “For the full Starlink constellation, at any one time during the night over 200 Low Earth Orbit (LEO) satellites will be visible. This will increase by the mid-2020s with additional planned constellations of LEO satellites.”
  AURA Statement

I've found some other astronomy authorities that have released statements relating to the problem at hand, but they don't have that much information as to potential impacts of these constellations.

As of June 2019, the IAU stated that

  We do not yet understand the impact of thousands of these visible satellites scattered across the night sky
  IAU Statement

In regards to this lack of information the AAS created this survey is to gather as much information as possible about the impacts to various observatories so that they can be better informed:
AAS Survey
From what I can tell, the results of the survey aren't out yet and of course when released it will provide a lot of insight into the potential impacts and how to manage them.

These are the properties which the AAS expect of a constellation of 1,584 Starlink satellites.

Consider satellites to be visible only when altitude ≥ 30 degrees and in direct
Sunlight or Earth's penumbra
magnitude 5
satellites cross sky in 4 minutes
6-9 satellites visible at any time during 1-hour starting (ending) at 12-degree -evening (morning) twilight
spatial density on sky ~ 7E-4 deg-2
angular speed across sky ~ 0.5 - 1 degree / sec

In regards to radio astronomy, the NRAO were happy enough with SpaceX Starlink, though there might be concerns with other constellations such as Oneweb.
NRAO Statement
Oneweb concerns

In reference to the previous point, all of this comes with the caveat that Starlink is potentially only one of perhaps 4 or 5 LEO constellations. Even if SpaceX acts responsibly you still have to convince all of the others to do so as well, which might be difficult.

SpaceX and AAS are working together on ways of reducing the impact of Starlink to astronomy. AAS on mitigating Satellites Constellations impact


Extract anomalies from two-line element set (TLE) on ISS (International Space Station)

First you should get an idea of what "Anomaly" means.  An anomaly is an angle which can describe the position of the satellite in its orbit. The true anomaly is the actual angle made between the position of the satellite and the perigee. The true anomaly changes quickly when the satellite is close to the Earth and slowly when it is far from the Earth.

The mean anomaly is much simpler: it just increases linearly with from 0 to 360. It is zero when the satellite is at perigee. So if a satellite takes 100 minutes to orbit and it is perigee at time t=0, then it will be at 90degrees at t=25, 180degrees at t=50.  The mean anomaly is just proportional to time.  "Mean" here means "average", not "meaning".

The orbital elements will tell you the mean anomaly at a particular time (given as "days in the year" In your data, at day 320.81254693 of 2018 (the "epoch") the mean anomaly was 308.7017.

You also know that the mean motion is 15.53998777. That is the number of orbits the satellite makes each day. One orbit is 360 degrees, so in 'n' days the mean anomaly will increase by 15.53998777 × 360 × n

You can now find the mean anomaly at any future time. For example 2.5 days after the epoch, at day 323.31254693, the anomaly will be 308.7017 + (15.53998777 × 360 × 2.5). You may need to reduce this angle to the range 0 to 360.

To add accuracy you can also use the fact that the orbital period is slowly changing (It is getting shorter as the ISS falls towards Earth due to drag). But this change is not quick and can be neglected over short periods of time.

Now you can work out how many days pass between the epoch and Dec 1st and you can work out the mean anomaly.

The question is worded strangely. It is an odd use of "which" and is unclear about the time. It should ask "What is the mean anomaly of the  ISS at the epoch" and "What is the mean anomaly of the ISS at 00:00 on Dec 18th 2018"