Why is a star at its highest point when local sidereal time = RA (right ascension)?

At some geocentric latitude $\phi$ and longitude $\lambda$, the elevation or altitude $a$ of a star with right ascension $\alpha$ and declination $\delta$ is given by
$$\begin{aligned}
\sin a = \sin\phi \sin \delta + \cos \phi \cos\delta \cos h&&&&&&(1)
\end{aligned}$$
where
$$\begin{aligned}
h=\theta_\phi - \alpha\qquad&&\qquad&&\qquad&&\quad\,&&(2)
\end{aligned}$$
is the hour angle to the star, expressed in terms of the local sidereal time $\theta_{\phi}$ and the right ascension $\alpha$.

Differentiating (1) with respect to time yields
$$\begin{aligned}
\cos a \frac{da}{dt} = &
\phantom{+}\,(\cos\phi\sin\delta - \sin\phi \cos\delta \cos h)\,\frac{d\phi}{dt} \\
& + \,(\sin\phi \cos\delta - \cos\phi \sin\delta \cos h)\,\frac{d\delta}{dt} \\
& - \cos\phi \cos\delta \sin h\,\frac{dh}{dt} &\!\!\!(3)
\end{aligned}
$$
While not quite zero, the time derivatives of latitude $\phi$ and declination $\delta$ are negligibly small compared to the $2\pi$ radians per sidereal day time derivative of hour angle $h$. For all practical purposes, the above thus reduces to
$$\begin{aligned}
\cos a \frac{da}{dt} = - \cos\phi \cos\delta \sin h\,\frac{dh}{dt} \quad\!&&&&&&(4)
\end{aligned}
$$
The left hand side of (4) is zero at extrema of elevation angle. Thus we're looking for conditions that make the right hand side of (4) equal zero. Since $dh/dt$ is nonzero, the right hand  side is zero only if one or more of $\cos \phi$, $\cos \delta$, or $\sin h$ is zero. The first two cases ($\cos \phi = 0$ and $\cos \delta = 0$) represent conditions where elevation is constant. The only condition of interest is $\sin h = 0$, which means an hour angle of 0° or 180° (hour angle is constrained to lie between 0° (inclusive) and 360° (exclusive)).

Since latitude $\phi$ and declination $\delta$ both lie between -90° and +90°, the condition $h=180^{\circ}$ represents the minimum possible elevation, while $h=0$ represents the maximum possible elevation. From (2), $h=0$ means $\theta=\alpha$, or local sidereal time being equal to right ascension.


TASCO 49060700 60x700mm Spacestation Refractor

A telescope's magnification is its objective focal length divided by the eyepiece focal length:
$$ \frac{700~\text{mm}}{25~\text{mm}} = 28 $$
A Barlow lens multiplies the effective focal length of the objective:
$$ \frac{3 \times 700~\text{mm}}{25~\text{mm}} = 84 $$
The 10 mm eyepiece gives 70x magnification without the Barlow.
Other configurations exceed the maximum useful magnification of 2x per mm of objective diameter (60 mm → 120x).


When examining an exoplanet's atmosphere is the star's emission spectra or planet's light used?

If you look for a simple answer by easy principles and not technicalities this would be the picture...
When the exoplanet transit between our point of sight and its star, its disk block part of the light. This results in a dimming of the star light we receive.
The latter can be analyse as we do for light of any source. It is well approximated by a black body spectrum with dimmer lines corresponding to the absorption lines of the elements/species present in the star atmosphere.
If the exoplanetary disk that block the light has an atmosphere as well, the constituents of the latter will absorb their typical lines and this results on the appearance of dimmer and/or new lines in the star spectrum as received.
In principle there could be a scattering effect of the exoplanetary atmosphere, too. For instance an observer on the moon during a sun eclipse would observe the sun fainting overall, the absorption lines due to terrestrial air, and a global different spectral distribution toward the red due to scattering, also due to air.
However, in case of far exoplanetary systems, it is already astonishing that we can do the line analysis as for all the above just happen to the little portion of the star flux that is intercepted by the planet and at same time by our telescope.

My main question is, does the light that passes through the planet's atmosphere is the sun's light or the sun's emission spectra?

Emission by the star atmosphere to us or to the exoplanet is mixed with the proper photosphere emission. Practically, it is the star light whatever you like to call it. It has not really an influence on the difficult above measurement. They are difficult right because we must see tiny changes relatively to a much brighter and bigger (angular) sized source.

Considering both the stars are different but the planet are same, would the absorption bands of the planet be different in each case?

No. In principle there could be a difference in the easy of detection, but the absorption lines are typical of the absorber not of the excitation source.
Obviously we can't search for absorption happening in the UV using just visible light. But the very broad emission of stars provides quite the same spectral coverage in term of wavelengths.


Star rotational axes and Solar system orientations

No. there's in no relation as far as we know. Ecliptic plane orientations of other planetary systems with respect to the galactic plane are completely random as far as we know.

We know this also because we can find exoplanets by the transit method nearly in any direction of the sky; if we look at $N$ stars located at galactic latitude of (let's say) $20^\circ$ (thus their ecliptic plane has relative inclinations of $20^\circ$ with respect to the galactic plane) we find roughly the same number of transit events as if we look to $N$ stars at $0^\circ$ latitude (thus their ecliptic planes have relative inclinations of $0^\circ$ with respect to the galactic plane). Since we are probably not the center of a planetary conspiracy on which everything is aligned so that we see the same distribution in all the sky we can assume that inclinations are random and we just happen to detect a similar fraction of transits everywhere because we are looking at at the same fraction of the same random population in each direction.

Humanity has just started to observe other planetary systems so this might change a little but in principle if there is any "preferential orientation", for now, with our current knowledge it is completely camouflaged by the main random component of this distribution.


How did the Archimedes crater fill with lava?

Great question! The mare lava bubbled up through fractures in the floor of the crater, burying its ring system and central peak.


Getting Solar Noon from Sunset and Sunrise

The approach of using the midpoint of the sunrise and sunset times is reasonably accurate. I do not know what the accuracy is, but probably within 1 or 2 minutes. It would not be as accurate in polar regions where the sun is close to being visible all day.

The main reason it is not precise is that the sun is moving north (December through June) or south. This changes the time of sunset with respect to sunrise if the Sun were not moving north or south. Solar noon would be earlier or later, respectively, from the midpoint.


Which is the farthest distance a particular distant known- identified object from the asteroid belt can get from the sun?

tl;dr: The key part the question is the phrase

  ...from the asteroid belt...

There are lots of asteroids and only some of them are considered to be in "the asteroid belt" or to be main-belt asteroids. As @zephyr's answer to the question Do astronomers generally agree that the distinction between comets and astroids is not so clear? points out, even the definition of what is or isn't an asteroid is under debate and review. To the question of which is the farthest a main belt asteroid can get from the Sun, the answer depends on exactly how you draw your line.

There are also families and groups of asteroids, and while these have visually clear clumps and clusters when plotted in certain ways, they may not have absolute universally accepted definitions of who's in and who's out.

See this and this for some families within the main belt.

While the OP"s link says

  The Asteroid Belt is located in an area of space between the orbits of Mars and Jupiter. That places it between 2.2 and 3.2 astronomical units (AU) from the Sun. The belt is about 1 AU thick.

that wording seems awkward since "between the orbits of between the orbits of Mars and Jupiter" only really puts it between 1.4 and 5.5 AU.

The first sentence of Wikipedia's Asteroid belt says:

  The asteroid belt is the circumstellar disc in the Solar System located roughly between the orbits of the planets Mars and Jupiter

repeat, "roughly" and it goes on to say

  The asteroid belt is also termed the main asteroid belt or main belt to distinguish it from other asteroid populations in the Solar System such as near-Earth asteroids and trojan asteroids.

The article also includes the figure shown below, showing that there are plenty of other groups of asteroids besides those designated as "main belt". I am not sure if there is a 100% absolutely IAU-official definition that you can apply to any set of orbital parameters that will return Yes or No to the question "is this one main belt?" but in general it looks like asteroids with a semimajor axis beyond 3.3 AU, or a high eccentricity, or a high inclination would not be called "main belt" by most astronomers.

As @PM2Ring points out Kirkwood Gaps delineate statistical sub-groupings of the main belt.

  A Kirkwood gap is a gap or dip in the distribution of the semi-major axes (or equivalently of the orbital periods) of the orbits of main-belt asteroids. They correspond to the locations of orbital resonances with Jupiter.

You can see that while there is a minimum around 3.3 AU due to the 2:1 resonance, there are "green" stragglers extending to the right that continue on past the arbitrary 3.5 AU edge of this plot of "main belt asteroids".

Source

Source

Screen shot from Scott Manley's video What Rock Star Brian May Discovered About Interplanetary Dust

which gives a better idea that there are groups and stragglers than the image in the OP's link:

Source


Is the full moon full everywhere on the planet?

[This is intended to supplement @userLTK's answer and reaches the same conclusion.]

By definition, the Moon is full when its ecliptic longitude is exactly 180 degrees greater than the Sun's ecliptic longitude for a theoretical geocentric observer (first paragraph of https://en.wikipedia.org/wiki/Full_moon and definition of "full moon" at http://www.ncsm.city.nagoya.jp/astro/astrotool/glossary.html but couldn't find a 100% reliable resource).

By this definition, your location on Earth doesn't matter, since the full moon is defined by a theoretical observer.

Realistically, the moon's position for an observer on the surface of the Earth can differ by as much as a degree from the geocentric position. (This is the arctangent of 1/60, since the moon's distance is about 60 times the Earth's radius.)

Since the moon circles the ecliptic once every 27.321662 days or so (its "sidereal period"), its ecliptic longitude increases by about 0.55 degrees per hour.

Thus, roughly speaking, the "full moon" at a given point on Earth can differ by as much as ±1.8 hours from the true full moon. By this token, it takes the "full moon" about 3.6 hours to cross the Earth.

Of course, all of this refers to the moon's geometric position. If you allow for refraction, the results could be slightly different, especially since both the sun and moon experience over 0.5 degree refraction when rising/setting and the full moon rises right about when the sun sets and vice versa.


What is meant by "revolution of Sun's apogee"?

The apogee is the direction of the sun when it is at greatest distance from the Earth.  We would normally call this the aphelion because we use a heliocentric model. The sanskrit doesn't have the precise terms for apogee and aphelion, it uses a word meaning "greatest height".  The translator has converted the notion of "greatest height of the sun above the Earth" as "apogee"
This direction changes (relative to the stars) mostly as a result of perturbations of Venus and Jupiter (though the cause could not have been known to ancient authors without a model of gravity).  Currently, aphelion (or apogee from the perspective of the sun) occurs on about July 4th, but this changes slowly.
It takes about 112000 years for the aphelion to revolve by 360 degrees. The calculation you quote gives about 10000000 years for a revolution. What are we to make of these numbers?
Measurement of the rotation of the aphelion is a hard thing to observe.  It is harder than the observation of the precession of the equinoxes, which was well established by the time that the Surya Siddhanta was composed. And the value quoted here is incorrect by a factor of about  100. Apsidal precession could not have been observed at this level, and without a theory of gravity and calculus, it could not have been calculated. It is therefore reasonable to suppose that the value here is found by analogy with the apsidal revolutions of the moon (which were observed in India) although exactly how this analogy was constructed is not explicit in the text.
We should also note that the model of planetary motion was not base on ellipses, but on circles with the Earth placed off-centre and the sun moving around it. In reading ancient accounts of science, we need to put ourselves into an ancient mode of thought, even if the translation uses modern terminology.


Does this amateur observation hold a record even including robotic searches for supernovae?

No, supernovae have been observed earlier and with better cadence than this one.

The Kepler Satellite has been observing galaxies at a 30 minute cadence from months before the supernova explosion to months afterwards.  Most of these have been Type Ia which do not have detectable shock breakouts, but there was one SN with a breakout published in the Astrophysical Journal before this one: "Shock Breakout and Early Light Curves of a Type II-P Supernovae Observed with Kepler".

The Nature paper linked in the question provides 4 other SN that were also detected about as early.  I think the new thing here is that it was discovered so early by an amateur astronomer.


What does the angle between plane of earth's orbit and ecliptic vary between?

By definition, it is zero; the ecliptic is the plane of earth's orbit (or rather, the inverse: the path the Sun seems to follow along the sky).

But you're probably looking for variations in the axial tilt of the Earth. There's a lot of formulas in the article, and on the (astronomically speaking) short timescale of a few million years, it oscillates between 22 and 24.5 degrees with a period of about 41,000 years. See the graph below, the red dot is the year 2000:

(picture by Tfr000 on Wikipedia, data from Laskar (1986))


Why would discovering life on another planet be important/matter to us?

It will be very important in several different ways. Not as in Messiah having descended to the World, and maybe not during one single human lifetime. But it will profoundly change the world in our minds, and thereby change our collective long term actions. The next generation would grow up taking alien life in space for granted, their minds will be more different than we old guys can change our stiff brain flesh. Like Newton's physics dominate our everyday life now, even for those of us who don't understand how all of physics has been deduced.

If life survives in other places than Earth, then maybe we (as in life from Earth) can do so too in the future.
If there's life in other places than Earth, then maybe we came from it? All life on Earth is family, all living things are closely related with each other. The question of our origin has always had high priority in human minds, see all religions.
If alien life is biologically similar to that on Earth (as in using DNA and proteins) then we could use it as a huge laboratory for medicine and environment control and agriculture and what not in life science. Astronomy today contributes to physics in our everyday life by studying a laboratory (the universe) with scale far beyond what humans could ever construct. If biology could do the same, we would soon notice the results in the shape of useful products.
If alien life does not use DNA and proteins but is very different from us, then I can't even guess what discoveries would come out of that. Zweistein.
If alien life is not found, even when we have searched long and hard, then we can gradually make the greatest discovery of all times: That we are unique and special. Searching for ET without finding it would be the most mind boggling thing imaginable. I don't think this hypothesis has been well addressed yet, I suppose no philosopher knows even where to start.


Do we have knowledge of the geology of the moon?

We know a considerable amount about the moon, but the picture is not complete.  There are considerable resources available and so much data available that it would take too long to list every aspect of it, so instead the following are examples of resources (with annotations):

Our knowledge of lunar geology comes from the results of direct sampling (from rocks brought back) and orbiting satellite observations.  Also, with inferences and comparisons with the Earth.

NASA's Lunar Geology page, this page approaches lunar geology by focussing on the 2 main sub-branches, the geochemistry and geophysics of the moon.

The geochemistry focuses on the crustal composition, volcanic petrology and the existence of polar ice, hydrogen and helium.  The geophysics focuses on the crustal magnetism, moonquakes, outgassing, crustal features.

Both of which help scientists develop a picture of the geological history and the geological structure of the moon.  Exploration of the moon's geological features in comparisons with Earth, are described as described in the online chapter Geology of the Moon as a geological 'Rosetta Stone'.

In terms of the internal structure of the moon, observations from direct sampling, sensors on the lunar surface and from satellite data have built up a reasonable picture of what the internal geology of the moon looks like.

According to the NASA page NASA Research Team Reveals Moon Has Earth-Like Core, re-analysis of Apollo mission seismic data has revealed that there are several similarities between the lunar and Earth internal structure (as seen below):

Caption from NASA: An artist's rendering of the lunar core as identified in new findings by a NASA-led research team. (NASA/MSFC/Renee Weber)

As can be seen in the diagram (and explained in all the links), the moon's internal structure is thermochemically layered, broadly similar to Earth (there are considerable differences).  But the main point, in answer to your main question, the interior of the moon is not homogenous, but layered with a central core.


Imagine a similar asteroid as Chicxulub, the one that likely killed the dinosaurs here on earth. It is thought to have impacted at about 20 km/s.
Initial atmospheric effects: When the asteroid entered the Earth's atmosphere, it pressurized the atmosphere and caused faster than 1000 km/hr super-heated winds - https://www.lpi.usra.edu/science/kring/Chicxulub/regional-effects/ that blasted out in a 1500km radius.  Since Mars's atmosphere at the surface is less than 1% the density of earth's, the heat generated by pressurization would have been correspondingly at 1% given the ideal gas law tells us that heat and pressure have an inverse relationship. The super-heated winds, glowing heat release on atmo-entry, and asteroid ablation would be disappointing on Mars compared to Earth.
Impact and initial ejecta: On Earth, the Chixculub impactor hit in the shallow ocean near the Yucatan, causing a huge tsunami that covered most of what is now the South East United States.  The water landing softened the blow, but enough energy was released to vaporize the meteor and blow out enormous amounts of ejecta.  The big ejecta (a few centimeters to 300 meters in diameter and bigger) landed hundreds of kilometers away in just a few minutes.  Much of the smaller ejecta floated down thousands of kilometers away.  A convection reaction dust plume at the impact site pumped soot and dust into the upper atmosphere and beyond. This ejecta (which was mostly vapor) sped up as it rose. As much as 12% of this ejecta reached escape velocity and never came back down to earth! https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2001JE001532
On Mars, there wouldn't be any tsunamis of course, since there is no ocean.  Hence the impact would kick up more debris.  Since Mars's surface gravity is about 3/8 of Earth's, and since there is much less atmospheric resistance for objects moving in the atmosphere, the big ejecta would have traveled thousands of kilometers instead of hundreds.  A lot of this ejecta (especially the vapors and liquids) would have gone into orbit or reached escape velocity!  However, the convection dust plume at the impact site would have been smaller than on Earth, with less atmosphere to feed the plume.
Debris layering: Since the area of a sphere is $4\pi r^2$ and the radius of Mars isn't much larger than half that of Earth, we can estimate that the surface area of Mars isn't much more than 1/4th that of Earth.  So the debris layer would be almost 4 times thicker (really much more because there would be more debris in an impact with no water).  On Mars, the debris would be centimeters to meters thick (nearer the impact site) rather than on Earth, were it is only mm thick opposite the impact strike. Also, due to the lower atmospheric density, particles would settle out much more rapidly than the decades long global cooling Earth experienced.
Volcanism and Earthquakes: I agree with the other answer, an asteroid strike of this magnitude likely caused volcanism on Earth and would likely do so on Mars.  This would be to a lesser extent since Mars has a thicker crust, is cooler, and has less current volcanic activity. The Chicxulub strike was thought to be equivalent to an earthquake of magnitude 10 on the Richter scale. Enormous shockwaves would move across the entire planet and meet at the antipodal point, likely causing eruptions or even raising hills or mountains.
Rovers and Probes: The big effect an asteroid strike like this would have on rovers and probes is to destroy them if they were near the impact site, damage them with earthquake effects if they were somewhat distant, and envelope them with debris wherever they were on the planet's surface!


Conceptual Doubt regarding the calculation of the Solar Constant

Imagine you have a solar collector measuring one square meter, which can directly measure the power (energy per unit time) striking it. If there are layers of atmosphere above the collector, some of the incoming light is absorbed or reflected by particles in the atmosphere, so you won't measure as much power as if you place the collector in space at the distance of Earth's orbit, with no atmosphere between you and the Sun. But if you place the collector in the upper atmosphere of the Earth, the density of atmospheric particles above you is so small that they absorb/reflect only a negligible amount of power. Orientation is also an issue--you'll get the maximum power if you orient the plane of the collector so that it's perpendicular to the direction of the incoming rays. If it's not perpendicular, then if you approximate the incoming radiation as a set of parallel rays which each carry an equal amount of power (a standard approximation in optics) then fewer rays strike the collector when it's at some angle that's not perpendicular to the rays. All this is illustrated in the following diagram (from this page), with the two possible orientations of the solar collector illustrated in gray, and the light rays illustrated as black lines:

The solar constant is defined to be the power you get on such a surface of unit area if it's not being obstructed by atmospheric particles, and if it's oriented perpendicular to incoming rays so that it receives the maximum possible power at that location.

One aspect of the orientation issue that's important to understand is that if you orient your solar collector so that it's parallel to the ground (so that it would be lying flat if it was actually at ground level), then it will only receive the maximum power if the surface of the Earth is itself perpendicular to incoming rays from the Sun at that point, which is not true in the above diagram. It would only be true at the point on Earth that's the dead "center" of the Earth as seen by an observer looking at it from the Sun's position, or the point where the gray line connecting the centers of Earth and Sun intersects with surface of the Earth on this diagram:

That is also the only point on Earth where an observer on the surface would see the Sun "directly overhead", at the very top of the apparent dome of the sky. So even if you stripped away the Earth's atmosphere, only at this point on the Earth would the power reaching a square meter of the surface be equal to the solar constant, the average power per square meter over the entire illuminated half of the Earth's surface would be less than the solar constant since most of the surface is at an angle relative to incoming rays. Incidentally, the angle of the surface relative to incoming rays is also the main reason why at Northern latitudes, it's warmer in the summer (when the surface at a Northern latitude is closer to being perpendicular to oncoming rays) than in the winter (when the surface is at a more inclined angle relative to incoming rays), as illustrated in a diagram from this page:

It's true that since the Earth is not a point, an observer in the upper atmosphere directly above that point on the surface would not be at exactly the same radius from the Sun as the radius of Earth's orbit (defined in terms of the position of the center of the Earth), but if you actually do the calculation of the solar constant at the radius of Earth's center vs. the radius of the point on the surface that's closest to the Sun, the difference is very small so it can be ignored if you're only stating the solar constant to nearest Watt/meter^2 or even the nearest tenth of a Watt/meter^2. The nominal solar luminosity is defined to be $3.828 \times 10^{26}$ and 1 AU is defined as 149597870700 meters, and if you use the formula you mentioned to calculate the solar flux at 1 AU it works out to 1361.1664654 watts per square meter. The radius of the Earth at the equator is about 6378 meters, so at an equinox when the Sun is directly overhead at the equator, an observer seeing the Sun directly overhead should be at a distance of 149597870700 - 6378 = 149597864322 meters, if you use the same formula to calculate the solar flux at this radius you get 1361.1665815, so it only makes a different if you want precision beyond the second decimal place.


How can a galaxy in-between our view and the galaxy behind it have a 'lensing effect'?

It's called gravitational lensing. Here's a link to the wikipedia article on the subject: http://en.wikipedia.org/wiki/Gravitational_lens.
Gravity affects everything, including light. A massive object such as a star, a galaxy, or in this case, a cluster of galaxies, bends the path of photons that pass very close to the massive object. Bending light is the basis of the lenses in an optical telescope.
The behavior of a gravitational lens is different from that of an optical lens., but the end result is the same: Gravitational lenses enable astronomers to see more distant objects than they would without them. Gravitational lenses create multiple images of the remote objects behind the lens. Oftentimes these multiple images are in the form of streaks. See the below image of gravitational cluster Abell 1689, which is the cluster that let those astronomers see A1689-zD1.

(source: hubblesite.org)
The cluster magnified the remote galaxy A1689-zD1 by a factor of 9.3. Without that magnification, that remote galaxy would be invisible to current technology.


How was the hypothetical ninth planet kicked so far out of the Solar System?

It's hard to say much about this planet, given that most of its properties are unknown. It hasn't been directly observed; instead, its effects on Trans-Neptunian Objects (TNOs) have been simulated and match observations. That said, its mass can be estimated, which is why it is conjectured to be the core of a giant planet.

One of the papers that Batygin & Brown cite is Morbidelli et al. (2012). Morbidelli has done prior work on the evolution of the Solar System if a 5th gas giant formed early in its history, which led to the Jumping Jupiter scenario. This is all described in a modification of the Nice Model.

Jumping Jupiter models with the four gas giants lead to one getting ejected from its stable orbit via interactions with the others (see
 Nesvorný (2011)). This is one of the ice giants - Uranus or Neptune. However, if we postulate the existence of a 5th planet - a third ice giant - then it can be ejected from its orbit, leaving the other four in a stable arrangement.

Batygin & Brown are implying that the object perturbing the TNOs may be the core of this 5th gas giant.

Since writing the first draft of this answer, I've found out that Batygin has made some estimates that this planet would have been ejected long before the Late Heavy Bombardment - thought to be the result of a Jumping Jupiter scenario - occurred, meaning that it could not be that ejected ice giant, unless we accept a different cause of the LHB. This makes things more interesting, as different initial setups are possible.


Hipparcos measured the parallax (distance) of stars, along with their colors in two wavelength regions (B_Tycho, "blue" and V_Tycho, "green"). Using the color information you can make a reasonable estimate for the stellar temperature, and coupled with the parallactic distance, this would allow you to get an estimate for the stellar radius.

But Hipparcos does not provide us with any good information about stellar metallicities, ages*, radial velocity, rotation, or surface gravity.

* - aside from stars that you can group in position and velocity space to identify as belonging to stellar moving groups with defined ages.


Could binary gas giants have ring and moon systems?

It would depend how close the gas giants were too each other, but in theory it would be possible.   Similar to the binary star system (image below), provided you ignore the AUs and scale this down from stars to planets, the same principal applies to orbits around two gas giants orbiting each other.

Source article.

They would have relatively small hill spheres depending on how close they were and corresponding, even smaller true regions of stability. But planetary ring systems tend to be quite close to the planet anyway and larger ring systems need to be almost exclusively inside the Roche Limit.

The binary gas giant orbits would have to be outside each other's respective Roche Limits anyway and provided there was sufficient distance between then, like maybe 4-6 times the larger of the two Roche limits (ballpark guess), then it's possible to have both gas giant planets with their own ring systems.

What's not possible is to have two orbiting disks around two bodies where the orbiting disks intersect each other, like they had in the Startrek Discovery premier episode.   And that was forming star systems, not gas giants, but it's still impossible for any orbiting ring or debris disk to intersect the way they did it.   It was a cool looking image, but clearly impossible.

I also want to point out that a ring system around both planets like the top of the first image with the writing "disk around both stars", that's not possible with a typical ring system.  Ring systems need to be in tight orbits around their planet, though faint/disburse ring systems like Saturn's E ring can exist further out.   The further out a ring system, the more likely it would be disturbed by the binary planet.


What class of elements are providing the many electrons needed to make the H- ion?

Easily ionised ones like sodium and potassium. Not much from lithium because that is rare.

You don't need many free electrons. The fraction of H$^{-}$/H is very small, something like $10^{-7}$ in the solar photosphere. And indeed the number densities of sodium and potassium to hydrogen are of that order.


Could a black hole pass quiescently through the Oort cloud?

An approach would be to use the Bondi-Hoyle (B-H) accretion rate and then assume that some large fraction of the gravitational potential energy was radiated from the innermost stable circular orbit (ISCO) as the material spiralled into the black hole via an accretion disc.

To calculate the B-H accretion rate we need how fast the black hole traverses the Oort cloud and the density of matter there.
$$\dot{M} \simeq \lambda\frac{4\pi G^2 M^2 \rho}{v^3},$$
where $M$ is the black hole mass, and $\lambda$ is an accretion efficiency parameter that is judged to be $\sim 0.01$ on the basis of observed compact objects (see Fender et al. 2013).
The luminosity will be
$$ L \simeq \eta \frac{GM \dot{M}}{R_{\rm ISCO}},$$
where $R_{\rm ISCO}= 6GM/c^2$ for a Schwarzschild black hole. Thus
$$ L \sim \eta \lambda \frac{4\pi G^2 M^2 \rho c^2 }{6v^3}.$$

Getting all my information from wikipedia, I assume there are 5 Earth masses in the Oort cloud in a spherical shell from 20,000 to 50,000 au. This gives an average density $\rho \sim 2\times 10^{-23}$ kg/m$^3$.

This is actually lower by a factor of a thousand or so than the density of the interstellar medium. Thus accretion from the ISM is far more important on average. Putting the numbers in for a 1 solar mass black hole and $\lambda=0.01$, $\eta =1$:
$$L = 3\times 10^{23}\left(\frac{\rho}{10^{-20}\ {\rm kg m}^{-3}}\right)\left(\frac{v}{10\ {\rm km s}^{-1}}\right)^{-3}\ {\rm W.}$$

This is quite an appreciable luminosity and much would emerge at UV and X-ray wavelengths. It would be readily detectable. The flux at a distance of 50,000 au would be $5\times 10^{-10}$ W/m$^2$, which is 4 orders of magnitude above the faint flux limit of the ROSAT all sky X-ray survey.


These are known as
two-line elements
(see also nasa.gov,
satobs.org).
On both lines, the first field after the line number is a NORAD ID,
which you can use to query a database such as
N2YO.
The second line gives geocentric
orbital elements
for the epoch given in the first line.
The last line looks like an error estimate by observation analysis software.

The first convenient visualization service I found was
Orbital Predictor,
where I naively pasted the above TLE data with a name less appropriate than "Mentor 7" or "2016-036A."


Why does the solar wind switch to flowing straight by the time it reaches Earth?

When plasma moves in a magnetic field then the charged particles follow helical paths around the field lines, due to the $q\vec{v}\times \vec{B}$ Lorentz force. If the radius of gyration is small, then the plasma is effectively tied (or frozen) to the field lines.

However to decide whether it is the motion of the magnetic field that enforces motion of the plasma or vice versa, we must compare the magnetic pressure with the gas pressure (or equivalently, compare the magnetic and kinetic energy densities).

The magnetic energy density scales with magnetic field squared and the field drops further from the Sun (e.g. a dipole field falls as $r^{-3}$), whereas the plasma kinetic energy density (which depends on the plasma density and temperature) could remain relatively high (the solar wind density drops only as $r^{-2}$ and the temperature is roughly constant).  There is a transition between the magnetic field dominating the plasma close to the Sun, to where the plasma dominates and effectively carries the field with it. This transition point is known as the Alfven radius and is usually at a few tenths of an au from the Sun.

Below this radius, the plasma (sort-of) co-rotates with the field, which is anchored to the photosphere. Beyond that the plasma is (sort-of) free to move radially outwards. This how the Sun loses angular momentum.


Why do Auroras only occur in polar regions?

The reason is because of Earth's global magnetic field. Take a look at the image below which shows a representation of what the magnetic field lines around Earth look like.

What you see above is what's known as a Magnetic Dipole field. The reason being that there are two (Di means two) magnetic poles, namely the north pole and the south pole. This is completely analogous to a giant bar magnet you may have played with in science class. You can see here that the field lines leave the magnetic south pole and travel around, going up into the magnetic north pole. The field lines don't really leave from or enter the Earth's equator.

But why does this mean aurora's happen at the poles? Because, according to the theory of electromagnetism, charged particles are forced to travel only along magnetic field lines. They cannot travel across them or break free from them. Once they get stuck, they're stuck (this isn't 100% true but for the purposes of this answer, it is). Note that the particles can travel both forward and backward on the field lines. They're not constrained to going only in the direction of the field line. This means that the primary location in which these charged particles can enter the Earth's atmosphere is where the field lines tend to hit the atmosphere and we already noted that that location was the magnetic north and south pole (which coincidentally is near the physical north and south pole).

Auroras themselves are just emissions from our atmosphere in response to being bombarded by high energy particles. To put this all together, here's what happens: The Sun spews out huge amounts of energy and tons of charged particles. Some of those charged particles hit the Earth's magnetic field and get trapped in it, being forced to travel along the magnetic field lines. They are then constrained to primarily hit the Earth's atmosphere at the magnetic poles where their extreme energy excites the atmosphere, causing it to emit light in shimmering red and green sheets. The auroras can't occur at the equator simply because not enough particles can hit the atmosphere there to cause auroras.

One day though, we may see auroras at the equator. The Earth's magnetic field is not fixed and moves with time. About every 450,000 years the field flips around completely. Its been about 750,000 years since the last reversal so we're due for one potentially soon (and by soon I mean in the next few thousand or tens of thousands of years). During the flip, while the magnetic poles are over the equator, you'll see auroras there.


I don't think there is a "recommended" FL for DSO observation. If you only observe visually, the eyepieces you use are just as important und the magnification depends on them. If you are concerned about the field-of-view you will get from these telescopes, I recommend checking out FOV calculator tools such as astronomy.tools. There you can enter your telescope and eyepiece information and show the FOV compared to a deep sky object of your choice.
I personally have been very happy with Skywatcher telescopes (I own an Esprit 100ED Triplet and a 200/1000 Newtonian), however I never really tested them visually as I only use them for imaging.


The size of Copernicus’ and Ptolemy’s corrections

I gather the worse problem for Copernicus was Mercury.  Not only is Mercury's orbit quite elliptical, it has relatively high precession due to the gravitational influence of other planets.

I found a quote attributes to Copernicus in this document (sorry, only direct link available) where he says :

  It is interesting to note that Copernicus had difficulty explaining Mercury’s motion and once commented that ”this planet has .  ..  influenced many perplexities and labours on us in our investigation of its wanderings".

This is apparently originally from this source :

Baum, Richard and Sheehan, William.In Search of Planet Vulcan:  The Ghost in Newton’s Clockwork Universe. Plenum Trade, New York. 1997.

The gravitational perturbation of the planets, particularly by Jupiter, and the concept of barycenter for even ideal elliptical orbits required the work of Newton (following Kepler's concept of elliptical orbits) to explain.  Up until Newton there was no solid theory for the motion at all and people were "just" (poor choice of words) trying to find the simplest model.  Copernicus provided a simple model, but also a less accurate one.


Is the sphere of influence of our milky way's center super massive black hole same as the Milky way's radius?

No. The black hole at the centre of the Milky way is only a few million solar masses. The same amount of mass, made up mostly of stars, would be enclosed within $\sim 10$ parsecs of the Milky Way centre. The bulge of the Milky Way, which is a few thousand parsecs across contains more than 20 billion solar masses.

Thus the dynamics of our Galaxy, outside of the central few parsecs are hardly influenced by the black hole at all.

What holds the Milky Way together? The gravity of everything else - stars, gas, dust, dark matter (the latter probably being the most important once you get to galactocentric radii beyond the Sun).


10 Jupiter masses at around 2 Earth radii?
That for sure doesn't exist / would be quite the sensation to discover.

When you look at data of any kind, one should pay attention to the measurment errors at least as much as to the actual value.
A regular physics result (for example for a measurment of the gravitational acceleration $g$ of where you stand) looks like
$$g=(9.81 \pm 0.02) \frac{m}{s^2} $$
or if for any reason you have asymmetric errors
$$g=(9.81^{+0.02}_{-0.01}) \frac{m}{s^2} $$
and errors always give an idea about how uncertain the method is with which the value was derived.
Now if you do take a look at the errors reported for the Mass quoted on the website you see that they are
$$M_{planet} = (10.41^{+0.0}_{-10.41}) $$
or so to say highly asymmetric, which should make one suspicious.
A look then into the original publication makes it clear that this cited mass is in fact only an absolute upper limit.
The authors of the paper were using two methods to estimate the masses of planets.

Looking for transit timing variations of known, and seen transiting systems. That means they had the system Kepler 52, with transiting planets K52b,c . K52b because it transits way more often than c has a well-determined period (Period with small errors!) and because of that any deviation in expected future transit time could be attributed to the \textbf{maximum masses} of K52c.
The more massive and the more compact a system is, the quicker it will destabilize. This fact is often used in reverse, to take the system age and at given distances derive maximum masses below which the system must lie, or else it would have already flown apart.

Both methods can only give maximum masses and I'll just leave here fig. 5 from the original paper with the planet you're interested in:

Now remembering, that $1 M_J \approx 320 M_{\oplus}$, you see where your 10 Jupiter masses for K52c come from: That's the planets possible maximum mass for system stability. The TTV method gives already a constraint that is 100 times lower ($37.4 M_{\oplus} \approx 0.11 M_J$).
Thus $37.4 M_{\oplus}$ is the planets true maximum mass.

This is clearly an error on the side of exoplanet.eu, but then probably there are too many planets and papers to read for whoever puts those data in there.

Summarizing
What we have here is only a maximum mass. Also the wrong one. To say what is now more probable, if $M_{K52c} = 37.4 M_{\oplus}$ or $M_{K52c} = 3.74 M_{\oplus}$ i'm not certain enough if I understand their anticorrelation method for the TTV signals.


In which direction is the sun travelling?

The direction of solar motion is referred to as the solar apex. It is at around RA=18h28m and Dec=+30d in the constellation of Hercules (you can see a map on the page linked to). Note that this is the direction in which you would see the "star wars" effect of onrushing stars diverging from the apex point. i.e. It demonstrates the sun's motion with respect to the local stellar population.

On your 2-dimensional cartoon of the Earth's, the circle is the ecliptic plane and the Earth's northern hemisphere (at winter solstice) is tilted away from the Sun at an angle of  23 degrees. The celestial coordinate system defined above is geocentric and the Sun is at 18h -23d at the winter solstice. The apex point is therefore towards  the Sun but 57 degrees above it.

Look at the map below (produced by Christian Ready), you can judge where the solar apex is with respect to the Sun and ecliptic plane (green line), which is roughly the path (drawn in your diagram) that the Earth takes around the Sun.

The very bright star Vega is reasonably close to the solar apex direction.

The Milky way centre is at RA=17h45m, Dec=-29d in the constellation of Sagittarius - i.e. almost where the Sun is in December. In Galactic coordinates (see Best approximation for Sun's trajectory around galactic center? ) the Sun moves at about 10 km/s towards the Galactic centre, about 5 km/s faster than the average circular speed (which is itself about 220 km/s) of local stars in the tangential direction and 7 km/s upwards out of the Galactic plane.

Thus the motion of the Sun with respect to the Galactic centre, as opposed to the local stellar population, is basically at $\sim 230$ km/s towards a point that is slightly less than 90 degrees around the Galactic plane and slightly above it. This would be at 9h30m -50d in the constellation of Vela. I think that is into the page and down about 30 degrees in your cartoon.


There is a preferred standard of rest, that of the Cosmic Microwave Background (CMB). You can detect motion with respect to the microwave background. So move fast with respect to the background and the CMB in the direction of motion will be blue-shifted and in the opposite direction will be red-shifted compared to a direction orthogonal to the direction of motion.

The Earth motion has been measured with respect to the CMB, see here

You could/can also detect your motion from the anisotopy of galactic red-shift in the same way.


How is the polar angle of a pulsar beam determined?

The simple answer is that yes, we can determine the pulse width if we know the angular radius of the emission cone and a couple other geometric quantities about the pulsar and its orientation relative to us.
Pulse profiles are complicated - with some exceptions, they're more than just quick spikes or smooth Gaussian curves. Most involve multiple components added together into complicated features; the beams have structure, and that structure leads to a fingerprint of sorts. Some switch between different emission modes over various cycles. There are plenty of fascinating nuances to study.
Nonetheless, even with these complicating factors, we can indeed define a pulse width. It depends on three things: the angular radius of the emission cone, $\rho$, the angle between the magnetic and rotational axes, $\alpha$, and the impact parameter, $\beta$, which defines the angle between the magnetic axis and the line of sight of the observer.
After a mess of spherical trigonometry, we can arrive at an expression for the pulse width, $W$ (see Gil & Han 1981, or any other work on emission geometry):
$$\sin^2\left(\frac{W}{2}\right)=\frac{\sin^2(\rho/2)-\sin^2(\beta/2)}{\sin\alpha\sin(\alpha+\beta)}$$
Therefore, the beam size, the location of the magnetic axis, and the orientation of the observer do uniquely determine the pulse width, as you guessed.
Obviously, it's unlikely that an observer can know these additional parameters entirely a priori, but that doesn't mean we don't know anything about them. You might expect that $\alpha$ would be randomly distributed, and it's possible that pulsars are born as such, but it is expected that $\alpha$ decreases as the pulsar ages due to the same torques that increase the spin period (Young et al. 2010). The two axes align on timescales of $\sim10^{6\mathrm{-}7}$ years. If you could determine the age of the pulsar, and you assumed that it was born with $\alpha\approx90^{\circ}$, you could make some estimates about the likely distributed of $\alpha$ in the present day.


Composition and Proton Flux from the Solar Wind

That's by number-- you get a lot more helium than that by mass.  Also, when they talk about proton flux, they just count the bare protons, not protons embedded in larger ions like helium.  Those are the kinds of things that should be made clear, but end up being common assumed conventions.


How can the author of this article conclude that earth-shine on the moon is 100 times brighter than moon-shine on earth?

The author says that the difference between the factor of 100 and the geometric increase due to area and albedo is because of "atmospheric absorption and directional effects". The former is probably only about 20% or so (i.e. about 20% of moonlight is absorbed/scattered in the Earth's atmosphere). The latter effect could be quite big.
The reflectance of solar system objects shows an opposition surge at small phase angles.
It is difficult to see the full Moon at really small phase angles ($<2$ degrees) from Earth because you would get a lunar eclipse. However, it has been measured by spacecraft and according to Buratti et al. 1996, the brightness of the Moon increases by "more than 40%" between phase angles of just 4 degrees and zero.
When the Sun illuminates the Earth as seen from the Moon then it would be much easier to see at very small phase angles because the Moon's shadow is that much smaller.
I suspect this effect plus the ~20% absorption in the Earth's atmosphere is what the author of your reference is talking about.


When do Mercury/Venus reach greatest elevation at sunset/twilight for a given location?

You can take a look at this table - it's in Dutch, but the dates and numbers in the table shouldn't be too hard to understand.  The first date is that of the greatest elongation (GE), the second (halfway the columns) that of the best visibility (BV).  You should ignore the lines with red dates - they're unfavourable appearances, and the code may find a date far from that of the GE.  The table shows that the difference between the moment of GE and that of the BV is typically a few days, although it can be about a week. (You can click on the first date for a page which includes a sky map of the appearance at constant Sun altitude (-6° I believe)).

Venus is a different story, because it is usually visible for months.  As you may know, in the northern hemisphere morning appearences of the inner planets are unfavourable in spring and favourable in autumn, and for evening appearances this is the other way around.  The table shows that Venus had its GE (west/morning) on 8 June 2001, but was best visible on 22 August.  In the mean time, the planet had moved somewhat closer to the Sun, but the effect of moving from spring towards autumn was clearly stronger (click the date link for the weird map that goes with this).  Hence, for Venus the difference can be several months.

Of course, these are just examples for one location (52°N).  I think the process may be too complex to be described by a single, simple equation.


What is the tightest orbit binary planets can orbit each other?

One can certainly imagine double planet "Rocheworlds" (like the one in Robert Forward's eponymous novel) where two terrestrial worlds fill their Roche lobes until they touch.
On its own this is perfectly fine: there are indeed a fair number of asteroids and trans-Neptunian objects forming tight pairs or a merged dumbbell shaped contact binary. The planets would be tidally locked to each other, not dissipating energy through tides.
The problem is that when the rest of the solar system is taken into account tidal dissipation begins to matter. The tides will cause the planets to move a bit relative to each other, and strong tidal dissipation will dampen that motion - dragging them inwards. Hence, relatively quickly since the tidal dissipation would be strong in big and soft terrestrials, the double planet would merge.
The orbital speed calculation looks a bit messy. Basically, the mass of the touching planets (that we need for the speed) is their density times the volume of their Roche lobe, which lacks analytical expression. However, one can approximate them fairly well as spheres of radius (see also Eggleton) $r=0.38 R$ where $R$ is their orbital distance. So we get $$M\approx 0.2298 \rho R^3.$$ Kepler's law gives us the period $$T^2 = \left(\frac{4\pi^2}{2GM}\right)R^3$$ (note the $2GM$ term rather than $GM$ - this is for a double pair), or $$T = \sqrt{\frac{2\pi^2}{ 0.2298 G \rho}}.$$ For Earth-density worlds I get $T=4.2512$ hours. Note that this is independent of the size of the system!
Can systems like this form? Probably. Can they remain long enough that life emerges on the planets and they become Earth-like? Hard to tell, but I suspect the answer is something like "yes, if they start sufficiently far away from each other then they will spiral in slowly enough that there is a biosphere when things get really tight." I suspect such coincidences in time and formation are rare. But the universe is large.


A supermoon does not occur every three or four months. There may be 2 or 3 consecutive supermoons (that is, separated by 1 month) that occur at about the same time each year.

If you ignore the precession of the Moon's orbit, then there is one time of the year when the Full Moon and perigee occur (point 1 in my diagram below). Two weeks later, the New Moon occurs near apogee (point 2). At any other time of the year, the Full Moon occurs at a different point in its orbit around the Earth, so it is farther from the Earth than at perigee. Six months later, the Full Moon is occurring at apogee (point 4), and no one cares about that!
 (not to scale!)

Full Moon at perigee (closest to the Earth)
New Moon at apogee (farthest from the Earth)
New Moon at perigee
Full Moon at apogee

The Wikipedia article on the Supermoon has a nice graphic showing the Full Moon and distance from the Earth. Depending on how close to the Earth the Moon needs to be to be "super", you can see that there is a "season" when the supermoon occurs. (I have copied the image here, and added a dashed line at 360,000 km to show which moons might be "super" and which ones are not.)

Now, if only people would care about the Super First Quarter Moon. Then we would be celebrating on May 12, 2019! (Not really. That is farther than the 360,000 km criteria.)


Determining Horizontal Coordinate System data (azimuth, altitude) for geographic locations

This can be done using the Python package Skyfield fairly easily.

Here is a shell of a script; I used some dictionaries to hold data, you may want to do something else. I've stored a lot of goodies but only printed azimuth, altitude and distance in kilometers.

By default Skyfield's .altaz() method calculates atmospheric refraction for anything higher than -1 degrees (1 degree below the horizon geometrically). You can turn that off by replacing it with .altaz(pressure_mbar=0).

OUTPUT:

 place     altitude  azimuth     distance (km)
Caracas    -33.67   258.26         7075.28
Edinburgh  -2.39    338.98          533.65
Canberra  -76.25     64.93        12377.25

Here's the Python script:

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

load        = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files
data        = load('de421.bsp')
ts          = load.timescale()

Earth       = data['earth']

London      = Earth + Topos(latitude_degrees  = 51.51,
                            longitude_degrees = -0.1275,
                            elevation_m = 11.)

places      = {'Caracas':(10.48, -66.90, 900), 'Edinburgh':(55.95, -3.19, 47.),
               'Canberra':(-35.29, 149.13, 578.)}

now         = ts.now()

answer_dict = dict()

for name, (lat, lon, elev) in places.items():
    place     = Earth + Topos(latitude_degrees  = lat,
                              longitude_degrees = lon,
                              elevation_m       = elev)
    vector_km            = place.at(now).position.km - London.at(now).position.km
    distance_km          = np.sqrt((vector_km**2).sum())
    dic                  = dict()
    answer_dict[name]  = dic
    dic['lat']  = lat
    dic['lon']  = lon
    dic['elev'] = elev
    dic['vector_km'] = vector_km
    dic['vector_distance_km'] = distance_km
    alt, az, d    = London.at(now).observe(place).apparent().altaz()
    dic['altaz']  = alt._degrees, az._degrees
    dic['dist']   = d.km

for name, dic in answer_dict.items():
    print(name, dic['altaz'], dic['dist'])


Tycho Brahe Danish Astronomer Religious?

Like nearly everybody at the time, he "believed in God" and was a member of the Danish Lutheran church.  He from time to time used arguments based on bible verses to back up his solar system model (in which the sun orbits the Earth, and the planets orbit the Sun).  According to Tycho, the idea of a rotating and revolving Earth would be "in violation not only of all physical truth but also of the authority of Holy Scripture, which ought to be paramount."

However he was not notably religious. He was suspected of heresy by some in the Lutheran church. He was not personally on good terms with the local bishops in Denmark. His interest in medicine and astrology were not sanctioned by the church and while he was a follower of Philip Melanchthon, he was not viciously sectarian, at a time when many others were.

He believed in God, but was not especially religious.


Biosignatures for anaerobic or other early microbiology?

Anaerobic life has much less impact on the atmosphere than aerobic life. Anaerobic life could exist in the deep sea and have no gas exchange with the atmosphere at all.

Methane would get everyone excited. Biotic methane might be expected to have a different isotope mix compared with abiotic methane. But it is hard enough to distinguish isotope mixes when you have a rover on a planet.

More complex compounds would be very exciting, but a very sensitive detector would be needed to find the very low concentrations required.

It may be that the marker of abiotic life is the detection of anomalously low CO2 levels. Most abiotic atmospheres have very high levels of Carbon dioxide, but life will progressively reduce this level. A low level of carbon dioxide, in an atmosphere containing water, especially if coupled with methane would look bery interesting.


I've seen some stars tonight. What could I possibly have been looking at with these pictures?

in-the-sky.org planetarium mode is your friend. It will pick up your location and time zone from your ip address, but you can manually set it to anywhere else.

I've added screenshots at both 20:52 where the Moon is about 26 degrees above the horizon, and at the stated time of22:52 where the Moon is more like 53 degrees.

20:52:

22:52:

Here's a plot of altitude and azimuth of the Moon on that night. Python script as well.

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

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads = 180/pi, pi/180

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

data    = load('de421.bsp')
earth   = data['earth']
moon    = data['moon']
ts      = load.timescale()

print earth.at(ts.now()).position.km   # for no reason at all

local_hours = np.linspace(18, 24, 13)
UTC_hours   = [h-8 for h in local_hours]
times       = ts.utc(2018, 12, 22, UTC_hours, 0, 0)

KL     = earth + Topos(3.1333, 101.6833, elevation_m=66)
Ampang = earth + Topos(3.1255, 101.7588, elevation_m=66)

alt, az, dist = KL.at(times).observe(moon).apparent().altaz()
KLalt, KLaz   = alt.degrees, az.degrees

alt, az, dist        = Ampang.at(times).observe(moon).apparent().altaz()
Ampangalt, Ampangaz  = alt.degrees, az.degrees

if True:
    plt.figure()

    pts = [6, 8, 10]

    plt.suptitle('Moon  22-Dec-2018 KL & Ampang', fontsize=16)

    plt.subplot(4, 1, 1)
    plt.plot(local_hours, KLalt)
    plt.plot(local_hours, Ampangalt)
    plt.plot(local_hours[pts], Ampangalt[pts], 'ok')
    plt.ylabel('Altitude (degs)', fontsize=16)
    plt.xlabel('hours (local)', fontsize=16)
    plt.ylim(0, 90)

    plt.subplot(4, 1, 2)
    plt.plot(local_hours, KLaz)
    plt.plot(local_hours[pts], Ampangaz[pts], 'ok')
    plt.ylabel('Azimuth (degs)', fontsize=16)
    plt.xlabel('hours (local)', fontsize=16)
    plt.ylim(0, 90)

    plt.subplot(2, 1, 2)
    r = (1. - Ampangalt/90.)
    r[r>1] = np.nan
    theta = rads*Ampangaz
    x, y = r*np.sin(theta), r*np.cos(theta)
    plt.plot(x, y)
    plt.plot(x[pts], y[pts], 'ok')
    th = np.linspace(0, twopi, 361)
    xc, yc = np.sin(th), np.cos(th)
    plt.plot(xc, yc, '-k', linewidth=2)
    d = 0.05
    plt.plot([-d, d], [ 0, 0], '-k')
    plt.plot([ 0, 0], [-d, d], '-k')
    plt.xlim(-1.1, 1.1)
    plt.ylim(-1.1, 1.1)

    plt.show()


How could a supernova seed solar nebula?

The Uranium on Earth did not come from a single supernova. The Earth formed from material that had been enriched by perhaps a billion stars that lived and died before the Sun and solar system were born. That material, enriched by heavier elements, is turbulently mixed within the galactic interstellar medium and is then gathered together to form new generations of stars. The mixing timescale is much shorter than the age of the Galaxy (see for example How can there be 1,000 stellar ancestors before our Sun?)

However, it is possible that one or two local supernovae did imprint a pattern of odd abundances for particularly rare or short-lived (radioactive) isotopes. It is widely believed that an explanation for the inferred high ratio of $^{26}$Al/$^{27}$Al (the former being radioactive with a half-life of less than a million years) in the early solar system, indicates that it was enriched by the ejecta from a nearby supernova. This may suggest that the Sun was born in a cluster of stars that was numerous enough to have contained very high mass stars capable of exploding as supernovae on short timescales.

Finding the solar systems birth place or its siblings is proving to be a difficult (and perhaps impossible) task, given the time that has elapsed and the multitude of effects that act to disperse stars (and their natal gas) on much shorter timescales (see Parent stars of our Sun - Where are its remains?).


In what manner is the spectrum of a black body really a continuum?

A blackbody spectrum is continuous.

As Dean comments, there are no true blackbodies. But this is not really the issue here:

It's true that most emitted photons from, say, a star, which is quite close to a blackbody, is the result of some discrete process. E.g. an electron is brought to an excited state by a collision, the atom de-excites, and a photon with the wavelength corresponding to the given transition is emitted.

But two mechanisms make the total spectrum of the star truly continuous:

Firstly, ionized atoms recombine with free electrons. The electron will go to any state, ground or excited, and with various probabilities. The energy of the photon emitted is equal to the energy it would take to remove the electron from the atom (i.e. ionize it again), plus the kinetic energy it had before recombination. And this kinetic energy can have any value (although if it's too large, the probability of recombination decrease).
Secondly, even without recombinations, you will not observe only the discrete lines. The reason is that the atoms of the blackbody are not stationary, but move with a ("thermal") velocity $v=\sqrt{k_\mathrm{B}T/m}$, where $k_\mathrm{B}$ is Boltzmann's constant, $T$ is the temperature, and $m$ is the mass of the atom. This motion causes a Doppler broadening of the emitted lines, resulting in a continuous, and Gaussian$^\dagger$, probability distribution.

$^\dagger$Actually the probability distribution is a convolution of the Gaussian random motion, and the Lorentzian natural line profile, resulting in a so-called Voigt profile. This profile is dominated by the Gaussian in the line center, and only becomes Lorentzian in the wings.


Could the E.H.T. produce an image of the human artifacts on the moon?

This is a really interesting question!

tl;dr: A definite maybe, but you would have to engineer a clever way to focus the transmitted power to a much smaller spot first; possibly several orders of magnitude smaller than what any one dish can do. Phasing several widely-spaced dishes alone would not be enough.

Let's see what can be checked easily.

Resolution:

Using round numbers of $f$ = 300 GHz, $c$ = 3E+08 m/s gives a wavelength $\lambda$ of 1 millimeter. With a baseline $D$ = 10,000 km, the angular resolution $\lambda / D$ is 1E-10. At a lunar distance $L$ of 360,000 km that's a resolution of 36 millimeters. Not only could you spot the donut, but the donut's hole would be resolved (barely)!

Radar at lunar distance:

The second image in your linked article New Radar Images Uncover Remarkable Features below the Surface of the Moon include images of Luther and Aristillus craters which are one and five million centimeters in size, respectively. The images already look noisy at this scale, so going a factor of 1E+05 smaller in resolution is likely to require a stronger transmitted radar signal, as measured in focused power per unit area on the lunar surface.

You might think that that can be fixed by making many of the EHT dishes (distributed across the globe) into transceivers, using very long baseline to focus the radar spot's power, and that could work to a limited extent but the problem is that it would not work the same way a filled aperture works. A sparsely populated arrayed aperture will still put most of the power into a plethora of ugly sidelobes.

There is a name for exactly this problem. I can't remember it now but I'll look for it.

Something similar, but different:

In the question Why was the 100m Green Bank dish needed together with DSN's 70m Goldstone dish to detect Chandrayaan-1 in lunar orbit? I talk about the use of two very large dish antennas to image detect the existence of a small object at the lunar distance. The experiment is described here and here as well.

They used two "tricks" to make this feat possible with only two dishes.

The DSN's Goldstone 70 meter dish transmitted the signal, and the large diameter of the dish made it possible to put most of the transmitted power into a circle which intercepted the spacecraft at its maximum elongation from the Moon (shown below) so that very little of the power would intercept the much larger Moon.
They took advantage of the Doppler shift of the reflected signal caused by the spacecraft's orbital velocity to further separate reflected signals from the Moon from the reflected signals from the spacecraft.

above: "This computer-generated image depicts the Chandrayaan-1's location at time it was detected by the Goldstone Solar System radar on July 2, 2016. The 120-mile (200-kilometer) wide purple circle represents the width of the Goldstone radar beam at lunar distance. The white box in the upper-right corner of the animation depicts the strength of echo. Inside the radar beam (purple circle), the echo from the spacecraft alternated between being very strong and very weak, as the radar beam scattered from the flat metal surfaces." Credit: NASA/JPL-Caltech.  From here

above: "Radar imagery acquired of the Chandrayaan-1 spacecraft as it flew over the moon's south pole on July 3, 2016. The imagery was acquired using NASA's 70-meter (230-foot) antenna at the Goldstone Deep Space Communications Complex in California. This is one of four detections of Chandrayaan-1 from that day." Credit: NASA/JPL-Caltech.  From here


How were the distances to stars measure before parallax?

As far as I know direct parallax measurements are the only way to directly measure the distances to stars.
Once parallaxs of hundreds of stars were known and diagrams of the relationship between stellar luminosity and spectral types, such as the Hertzsrpung-Russel diagram, were made, it became possible to estimate a star's absolute magnitude more or less accurately and thus calculate its distance from its apparent magnitude.
It may be noted that stars in some spectral types can have any one of up to nine luminosity classes.  A star's luminosity class is included as Roman numerals in its spectral classification if it is known.  Calculating the distance to a star without knowing its luminosity class can be very inaccurate.
https://en.wikipedia.org/wiki/Stellar_classification#Yerkes_spectral_classification[1]
The heliocentric theory, that the Earth revolves around the Sun, was first suggested in classical Greece.  Aristotle (384-322 BC) rejected that theory, because of the lack of detectable stellar parallax caused by Earth's motion around the Sun.
Exploration of the Universe Brief Edition, George Abell, 1964, 1969, page 18.
That same argument was used against the heliocentric theory in the early modern age.

Stellar parallax is so small that it was unobservable until the 19th century, and its apparent absence was used as a scientific argument against heliocentrism during the early modern age. It is clear from Euclid's geometry that the effect would be undetectable if the stars were far enough away, but for various reasons, such gigantic distances involved seemed entirely implausible: it was one of Tycho Brahe's principal objections to Copernican heliocentrism that for it to be compatible with the lack of observable stellar parallax, there would have to be an enormous and unlikely void between the orbit of Saturn and the eighth sphere (the fixed stars).2

https://en.wikipedia.org/wiki/Stellar_parallax[2]
As astronomers gradually accepted the heiocentric theory after about 1600, there were attempts to meaure stellar parallaxes which ended in failure after failure.  And each astronomer who failed to measure stellar parallax could have calculated the smallest stellar parallax he could have measured, and thus the distance that the star(s) he studied must have exceeded.  So astronomers should have gradually increased their estimates of the minimum possible distances to various stars whose parallaxes had failed to be measured, while perhaps hoping that some stars might possibly be a lot closer.
And as the distance between the Sun and the Earth was repeatedly measured more and more accurately, astronomers could have estimated the Sun's luminosity more and more accurately.  By assuming that any particular star was as luminous as the Sun, and by measuring its apparant magnitude, astronomers could have estimated how far away it was.
But of course the actual luminosities of stars vary greatly, and in the 17th and 18th centuries measurements of the apparent brightness of the Sun and of distant stars would probably have been very inaccurate.
I doubt if any estimates of the distances to stars were very accurate before the first stellar parallaxes were measured in the 1830s, because the first measured parallaxes seemed too small, and thus the distances too great, for astronomers to believe.
For example, the first actual measurement of a stellar parallax was made by Thomas Henderson, but his is usually listed as the second measurement.

The large proper motion of Alpha Centauri AB was discovered by Manuel John Johnson, observing from Saint Helena, who informed Thomas Henderson at the Royal Observatory, Cape of Good Hope of it. The parallax of Alpha Centauri was subsequently determined by Henderson from many exacting positional observations of the AB system between April 1832 and May 1833. He withheld his results, however, because he suspected they were too large to be true, but eventually published them in 1839 after Friedrich Wilhelm Bessel released his own accurately determined parallax for 61 Cygni in 1838.[62] For this reason, Alpha Centauri is sometimes considered as the second star to have its distance measured because Henderson's work was not fully acknowledged at first.[62)

https://en.wikipedia.org/wiki/Alpha_Centauri#Observational_history[3]

The distance to Vega can be determined by measuring its parallax shift against the background stars as the Earth orbits the Sun. The first person to publish a star's parallax was Friedrich G. W. von Struve, when he announced a value of 0.125 arcseconds (0.125″) for Vega.[37] Friedrich Bessel was skeptical about Struve's data, and, when Bessel published a parallax of 0.314″ for the star system 61 Cygni, Struve revised his value for Vega's parallax to nearly double the original estimate. This change cast further doubt on Struve's data. Thus most astronomers at the time, including Struve, credited Bessel with the first published parallax result. However, Struve's initial result was actually close to the currently accepted value of 0.129″,[38][39] as determined by the Hipparcos astrometry satellite.[4][40][41]

https://en.wikipedia.org/wiki/Vega#Observational_history[4]
So Henderson and Struve suffered failures of nerve, being unable to accept how distanct Alpha Centauri and Vega were according to their measurements, and so lost the honor of making the first measurement of stellar parallaxes to Bessel.
And in my opinion that would not have happened if astronomers of the time commonly accepted that even the nearest stars must be several hundred thosuand Astronomical Units, and multiples of light years and parsecs (units which had not yet been invented) from the Sun.
So any estimates of the distances to stars made before the first successful parallax measurements probably greatly underestimated the distances.


iteration to cover the whole sky with right ascension, declination, angle

Imagine you are patching a plane with circles, instead of patching the sky sphere with circles (which is the same as filling the space with cones). Now, for every circle of diameter D define a square of diagonal D (D=2d). It is easier to patch the plane with squares, isn't it? So you have squares of side D/2*sqrt(2), and that is exactly your DEC_step, for each vertical sweep, and your RA_step, from sweep to sweep.

This will lead to a lot of duplicated examination (in fact you will examinate exactly the same place at the Poles once for each sweep) but you can be sure you'll not leave anything out of your squares.


Unbelievably high masses/ densities in NASA website data

Starting from the index you mentioned, I clicked through the links for some individual planets, which in turn link to discovery papers or other relevant observations. For planets around Kepler-23, -24, -25, -26, -27, and -28, the relevant papers are Ford et al. (2012) and Steffen et al. (2012), two out of a series of papers. Both papers used transit timing variations (TTV) and something called dynamical stability, which gives an upper limit to the mass of the planets such that the relevant n-body system is stable over long timescales. This leads to plots like the following (Fig. 3, Ford et al.):

The arrows indicate simulations where the system was stable over timescales greater than $\sim10^7$ years.

Additionally, as Ford et al. note,

  We caution that the uncertainty in the masses and sizes of the host stars directly translates into uncertainties in the planet masses and sizes.

Furthermore, shorter observations by Kepler also lead to larger uncertainties. Thus, the maximum masses of the planets as given in these simulations may very well be much greater than their actual masses. Steffen et al. have similar results, and they use a very similar method.

After more reading, it's clear that the same is true for planets around Kepler-32, Kepler-48, Kepler-52, Kepler-54, and Kepler -57 to -60, inclusive.

Kepler-39b is interesting. The paper NASA cites, Bouchy et al. (2011), states that the object is larger than one would expect, even taking into account that there is a large uncertainty in radius. It may be a fast rotator (see Zhu et al. (2014)), which would explain its oblateness, but that still isn't enough to explain the discrepancy.

Now, I referred to Kepler-39b as an "object", not a planet. NASA lists its mass as over 20 Jupiter masses; Bouchy et al. placed it at about 18 Jupiter masses, with a very low uncertainty. This could mean that it's more like a brown dwarf than a massive gas giant, and remember, the distinction is fuzzy when it comes to sub-brown dwarfs. This, then, may be the most likely explanation.


The role of gravity during planetesimal accretion

The very first stage of planet formation involves purely inelastic (i.e., sticky) collisions between pairs of small particles of dust to make slightly larger particles of dust, pairwise collisions between these slightly larger particles to make even larger particles, and so on. So far, the only gravitation that comes into play is that of the central protostar and the disk as a whole. The gravitation of these little objects is very small.

Then some magic happens that makes the little rocks that result from this very first stage grow into planetesimals that do gravitate. How do little rocks (a few centimeters in diameter) grow into little planetesimals (a few hundred meters in diameter)?  This is the key open problem in the theory of planet formation.

The core of the problem is that objects the size of a spec of dust to an object the size of a small pebble move with the gas that forms the bulk of the disk. The outward pressure from the gas in the disk makes the gas and small objects move at a slightly less than orbital speed. Once an object reaches the centimeter scale or so, the pressure from the gas no longer provides the support needed to keep those objects moving at that slightly suborbital speed. The objects start to become separated from the gas. By the time an object reaches a meter or so in diameter, the gas is equivalent to a hurricane force headwind.

One consequence of this is that collisions become ever more violent as objects grow beyond a centimeter in size, and hence are more likely to break colliding objects apart than to make them coalesce. An even bigger problem is that that headwind makes centimeter- to meter-sized objects spiral into the protostar, and very quickly (a few hundred years from 1 AU). A number of proposals have been put forth to overcome this meter barrier problem. So far, none appears to have stuck. (Sorry for the pun.)

Once objects reach some critical size, the drag from the gas reduces in importance thanks to the square-cube law and the gravitational attraction increases. That critical size varies from hundreds of meters to tens of kilometers, depending on which paper one reads (experts vary). These objects are "planetesimals," and there are lots of them. These once again combine readily to make planetary embryos. From then on, its just a mopping up process as gravitation either makes these planetesimals / planetary embryos combine or ejects them from the star system.


If black holes swallow stuff, will a big black hole eventually swallow the universe?

As far as I know, no. Because the universe is expanding, and we have good evidence that the expansion is accelerating. Yes,  some people talk about the Big Crunch, but there's no evidence for it. I don't think there ever will be. I say that because it's space that's expanding, and a gravitational field isn't collapsing space. The universe didn't collapse when it was smaller and denser, so I see no way it can collapse in future.

  The universe will eventually go dark, because there will not be enough matter to make stars, and all that will be left is a bunch of black holes and they can merge. So will the universe get swallowed by a big black hole someday?

As above. Even if star formation ceased and a whole lot of matter ended up in black holes, the universe would still be expanding. As far as we know. The expansion would have to reverse for the black holes to coalesce. Mind you, it might be a good thing if it did. As Stephen Hawking said, the universe is like a black hole in reverse. If the universe ended up as one big black hole, it would be ending up the way (we think) it started out. Then you'd have a cyclical universe.


How close to Earth were the asteroids moving through this Hubble composite image?

I visited the linked article in Astronomy Now and found the URL of the image contained the string STSCI-H-p1733a. STScI stands for Space Telescope Science Institute.

Searching that string leads to

http://hubblesite.org/images/news/release/2017-33 (four images)
http://hubblesite.org/news_release/news/2017-33
http://hubblesite.org/image/4080/gallery

the last two of which identify the location as Abell 370, at about RA: 2h 40m, Dec: -1.6°. At about RA = 0h the ecliptic and equator coincide, but by 2h 40m the ecliptic has a declination of about +16°, so this photo is looking about 18° below the ecliptic.

The first paragraph of the caption contained in both of those links says:

  Like rude relatives who jump in front your vacation snapshots of landscapes, some of our solar system's asteroids have photobombed deep images of the universe taken by NASA's Hubble Space Telescope. These asteroids reside, on average, only about 160 million miles from Earth — right around the corner in astronomical terms. Yet they've horned their way into this picture of thousands of galaxies scattered across space and time at inconceivably farther distances.

If the distance is about 160 million km, then they are suggesting these asteroids are in the inner part of the main asteroid belt.

The Earth is moving at about 30 km/sec around the Sun, and the Hubble Telescope is "only" moving 7.6 km/sec with respect to the Earth, in roughly the same direction. So much and probably most of the motion is due to the Earth's velocity. I'm suggesting this because if the distance to the asteroids is really roughly 160 million kilometers, then they would be near opposition to the Sun and the Earth's motion would be roughly perpendicular to the viewing direction.

There are a lot of asteroids! According to Wikipedia:

  One hundred asteroids had been located by mid-1868, and in 1891 the introduction of astrophotography by Max Wolf accelerated the rate of discovery still further. A total of 1,000 asteroids had been found by 1921, 10,000 by 1981, and 100,000 by 2000. Modern asteroid survey systems now use automated means to locate new minor planets in ever-increasing quantities.

Asteroids are discovered constantly, and sometimes this happens in images that were not intended to be searches for asteroids, as in this case. While most most asteroids orbit close to the plane of the ecliptic, some of their orbits do stray substantially above and below it, as in this case.

While the date of the News Release is 02-Nov-2017, I don't know the time frame of all of the exposures that contributed to this final composite image.

Here is a screen shot of the Planetarium Mode of Dominic Ford's in-the-sky.org. The marker shows the position of Abell 370 just below the equator, and the Ecliptic is the thicker line above it.


How wide is the Moon's umbra and penumbra at EM-L2?

The angle between the inner and outer edges of the penumbra is the same as the Sun's apparent angular diameter.
At 1.0 au from the Sun, that's 32 arcmin or 0.0093 radian.
At the distance from the Moon to EM-L2, this angle spans 0.0093 * 64700 km = 602 km.
The Moon's diameter is 3474 km, so the umbra diameter would be 2872 km and the penumbra diameter would be 4076 km.


How did Fermi manage to see neutron star merger?

The Fermi instruments have a very wide field of view; the gamma ray burst detector covers the whole sky not occulted by the earth with low angular precision, and the Large Area Telescope covers about 1/5 of the entire sky with arc-minute precision.  I don't know if the LAT happened to be pointing in the right direction (I guess ~20% chance?) but the burst detector would at least be able to identify the timing of the gamma ray signal from the collision from nearly anywhere in the sky.


Does the Sun have permanent geographical features?

As you suspect the Sun does not have any permanent surface features. Up till 1951 the longest actually observed sunspot group lasted 134 days see http://adsabs.harvard.edu/full/1951ASPL....6..146P. That duration gives you an idea how long sunspot groups last. There are associated with coronal holes which would last about as long. However, the structure (shape size) of the group and any coronal hole would be constantly changing.

Temporary maps are made and there is a heliographic co-ordinate system, with a latitude and longitude.

The sunspots show the rotation of the Sun, and so a north and south solar pole can be found. North is in the same side of the ecliptic as the Earth's north pole).

Solar longitude is much more arbitary. The zero helographic meridian is defined as the plane (solar pole to solar pole) passing through where the solar equator crossed the plane of the ecliptic (earths orbital plane) at Greenwich mean noon on January 1, 1854. see http://wso.stanford.edu/words/Coordinates.html
Since then astronomers have just used a constant rotation rate based upon the suns equatorial rotation rate. These are called Carrington Solar Coordinates

A similar system is used for the longitude of the gas giant planets like Jupiter and Saturn.


Using your scenario, and lets ignore dark energy expansion and red-shifting at greater distances,.  There would be no distance where the sun will disappear.  That might seem counter intuitive but it's true.

On Earth, we receive about 10^17 photons per square CM.  (source).  Using the inverse square rule, to receive 10 photons per second on a pair of human eyes, that distance would be 100 million AU - give or take, or about 1,580 light years.    We probably can't see just 10 photons, but in pitch black, with eyes adjusted to the dark, maybe we could.   That's about the distance where the sun would give of steady but very very dim light.

Further than that, what happens has more to do with the quantum nature of light than the brightness and distance.   As you double the distance you see 1/4th the frequency of photons.   At 5,000 light years you see 1 photon per second, basically a very dim blinking light, but just as bright as it is at 1,580 light years, just appearing less often.     Double the distance again and it flashes once every 4 seconds.    There's no distance where you stop seeing the sun, you just see it less and less often, but it never appears dimmer than 1 photon.


Are Gamma Ray Bursts of galactic or extragalactic origin?

Based on our understanding of what might be possible causes for GRB, there's nothing to suggest that this phenomena is isolated to other galaxies alone, but we have so far only detected GRB originating from outside of the Milky Way. Good that we did too, because if any of them happened in our cosmic neighborhood and was pointing towards us, it would be the last thing we'd ever see and would effectively sterilize our planet. More information is available on Wikipedia on Gamma-ray burst, it's actually a pretty good article presenting extensive information on GRB (I guess that's why it's featured there).


Why does earth's north pole roughly point in the same direction throughout the year?

It is just conservation of angular momentum. In order to change the direction of the spin axis of the Earth (which would change the direction of its angular momentum) you would have to subject it to an external torque. The curvature of space has no significant bearing on the question.

In fact the direction in which the Earth's spin axis points does change, but on much longer timescales. The "precession of the equinoxes" (as it is called) is cyclical with a 26,000 year period.


Definition question: continuum-subtracted spectrum

A spectrum of, say, a star or a galaxy, consists of wavelengths emitted from various physical processes. In the case of the star, there is the blackbody spectrum that reflects the temperature of all of its gas, whether it is hydrogen, helium, or metals. On top of this, there are lines, i.e. emission from certain atomic transitions that are excited and then de-excite emitting photons in a very narrow wavelength interval. The blackbody spectrum is said to comprise the continuum of the total spectrum.

In the case of galaxies, the continuum consists of the combined spectrum of many different stars' different blackbody spectra, infrared light from dust, etc. On top of this, there are lines, e.g. from nebular processes such as my favorite line Lyman $\alpha$, which is emitted when hard UV radiation from hot O and B stars ionize their surrounding neutral hydrogen, which subsequently recombines emitting mostly Lyman $\alpha$.

If we want to measure the total flux in a given line, we are usually only interested in the flux that comes from a particular physical process, and not the part of the flux that comes from the "underlying" sources which contribute to the continuum. Hence, we need to subtract the continuum. This is usually done by fitting some functional form to the continuum (e.g. a straight line or a second order polynomial), substract this fit from the spectrum, and finally integrate from one side of the line to the other in the resulting "continuum-subtracted spectrum".

In the top figure below, the total spectrum's flux density$^\dagger$ is in black, and the fit is the red lines. Underneath is drawn the continuum-subtracted spectrum, and a few dashed green lines show some lines that rise above the continuum. The integrated line flux is indicated by light red.

$^\dagger$Note that many authors use the term "flux" where it should actually be "flux density". Integrating flux density in $\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Å}^{-1}$ over wavelength in Å$^{-1}$ gives flux in $\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}$.


Precautions when retrieving material originated from space

Meteorites are composed of rock and iron. They are not radioactive, at least no more so than any other stone.  A banana is probably more radioactive than a meteorite.

You do not need protection for foreign viral agents either. Meteorites are stones that have been orbiting in space, usually for billions of years. They do not contain any living things. The have also passed through the atmosphere so fast that their surface can begin to melt from the friction.

You can pick up a meteorite with you unprotected hands

For pedants: A meteorid is "space stone". Meteor is the flash of light in the sky as a meteoroid passes through the atmosphere. Meteorite is the stone from space that you find on the ground. It is the meteorite that you retrieve for study.

There is a suggestion that ancient meteorites bought the ingredients for life to Earth, or even seeded life on the planet, but it is speculative as we don't have the evidence to prove it.


Are brown dwarfs which don't sustain any fusion considered stars?

Brown dwarfs are technically stars, but not in your typical sense, as they cannot start or sustain nuclear fusion. Think about it this way: stars are formed in clumps inside giant gas clouds called nebulae, along with planets, but what if one of these clumps is too big to become a planet and too small to become red dwarfs. These are brown dwarfs. Some consider them to be stars, others consider them to be planets, and some neither. So, opinions vary.


Will the Sagittarius A* Black Hole eventually swallow the entire Galaxy?

Black holes do not suck in matter any more than stars and planets do: an object in orbit would remain in stable orbit if nothing perturbed it.
However, in the long run ($10^{19}$ years and more) interactions between stars will perturb their orbits, making many of them end up in the central black hole. An easy way of seeing this is to consider a random close encounter between two stars. They will change velocities, and potentially one will get a sufficiently high velocity to escape the galaxy - but the other one, by energy conservation, would be more tightly bound and orbit in a closer orbit. Eventually the galaxy "evaporates" through this kind of interaction, with heavier stars congregating near and eventually into the black hole and the lighter stars mostly expelled. (For all the mathematical details of this and related processes, see Binney & Tremaine's Galactic Dynamics).
Even if this did not happen, gravitational radiation makes a star orbiting something slowly lose angular momentum and spiral in, but this takes a far longer timescale.
Hence stars either fall in or are expelled over very long timescales.


Time dilation on an object circling earth

At 99% the speed of light the behaviour would be almost completely determined by special relativity. The scenario is well-investigated for synchrotrons. In principle a synchrotron or a storage ring, e.g. around the equator of Earth, could be built.

At 99% the speed of light the frequency $f_s$ of the circling object should occur red-shifted by a factor of a little more than 7 to an observer in the center of the circle due to the transverse relativistic Doppler effect:
$$f_o = f_s/\gamma=f_s\cdot\sqrt{1-v^2/c^2}=f_s\cdot\sqrt{1-0.99^2}=f_s\cdot\sqrt{0.0199}=0.141067 f_s.$$
For an observer immediately near the ring, the frequency $f_o$ is the same as for the observer at the center for the signal emitted, when the particle was diametral on the other side of the ring. When approaching the observer along the lign of sight, the signal of the particle is blue-shifted to
$$f_o=f_s\cdot\sqrt{(1+v/c)/(1-v/c)}=f_s\cdot\sqrt{1.99/0.01}=f_s\cdot\sqrt{199}=14.1067\cdot f_s.$$
When leaving the observer along the lign of sight, the signal of the particle is red-shifted to
$$f_o=f_s\cdot\sqrt{(1+v/c)/(1-v/c)}=f_s\cdot\sqrt{0.01/1.99}=0.070888\cdot f_s.$$

By this we now have calculated the observed frequencies for three positions of the circling particle/radio to give some idea about the oscillation of the observed frequency.

More details about the relativistic transverse Doppler shift, see e.g. Ives-Stilwell experiment. Close to the experiment with the observer in the center of the circling particles are Mössbauer rotor experiments. In this cases ions or atomic nuclei emitting or absorbing at known wavelengths are used as "radios".

This paper describes a slower version of the transverse Doppler shift, as observed by using the GPS satellites moving with just 4 km/s. In this slow case the gravitational frequency shift, as predicted by general relativity to be induced by the gravitational field of Earth, plays a relevant role, relative to the (in this case) small transverse Doppler shift. Here the GPS satellites are the moving radios.


Well, because the axis of the rotation of the Earth is not the same as the axis of rotation of the disk of the Milky Way (and also because we're transforming a 2-dimensional spherical map into a 2-dimensional cartesian map), the path of the disk of the Milky Way galaxy looks something like this:

So, there is actually a wide range in declination that the Milky Way can be seen at. The range of declination you can see depends on your latitude (for a review of RA and declination, coordinates used in the celestial coordinate system, see this post). For example, here in Philadelphia (just about $+40^{\circ}$ declination), I'd be able to see from from $-50^{\circ}$ to $+90^{\circ}$ in declination. For Nebraska, find the latitude of your location. To calculate the lower limit, add your latitude (which will be positive since you're in the northern hemisphere) to $-90^{\circ}$ (mine was $+40^{\circ}$, so: $-90^{\circ} + 40^{\circ} = -50^{\circ}$). To find the upper limit in declination, it's even easier. Since you're in the northern hemisphere, you can actually see the north celestial pole. This means that the upper limit is simply the maximum it can possibly be, which is $+90^{\circ}$. The larger the latitude the more circumpolar your night sky gets.

The good news is that you should definitely be able to see it in Nebraska. The only question during what season will you see it at its best? Take a look at the constellations which contain parts of the Milky Way and find the season you can see it in - now you know where to look.

The last thing I want to add is that you need nice dark conditions. Local weather conditions or sources of light pollution may very easily hide the Milky Way. From Philadelphia, even on perfectly clear nights we have no chance of seeing it. If you've got a city on one of your horizons, try to plan around that - either go to a darker location or try looking opposite in the sky from any sources of light pollution.


What was the entity/particle/state at the time of the Big bang?

The notions of today's universe don't make much sense for the (hypothetical) pre-Planck epoch of the universe. All four known fundamental forces should have been united. Matter didn't exist. It can form only in cooler conditions. Even cause and effect don't work well below Planck scales.

The structure you eventually get can be treated in terms of a groupoid, as described e.g. in this paper. Groupoids are related to groups (algebra). Groups actually are - in some sense - very simple groupoids. You may construct some groupoids from disjoint groups by connecting them via isomorphisms. So you get kind of a network of abstract entities.

Now try to adjust quantum mechanics and General Relativity, such that it fits with these structures, meaning without the smooth space and time, as usually assumed for standard physics. Let known or extrapolated physics of the Planck- and GUT (grand unified theory) epoch emerge from the pre-Planck physics to start the actual Big Bang.

It's possible to construct physical theories for a pre-Planck era. But it's very hard to find observational or experimental evidence to test these theories with respect to actual physics.

Those pre-Planck groupoid theories can be defined in a way that they don't need a predecessor epoch, e.g. by eliminating the macroscopic cause and effect paradigm, which is based on a notion of an arrow of time.
Causality and a concept of before/after on isomorphic entities doesn't make much sense.

Although, there should be some mechanism which eventually disrupted a strictly non-local pre-Planck configuration.