How does making a refracting telescope very long reduce the chromatic aberration of an uncorrected lens?

The actual math is a bit complicated, but there's a simple intuitive explanation.

Longitudinal chromatic aberration happens because, when you cut a convergent lens in two, and you look at the cross-section, the edge of the lens looks a bit like a prism, doesn't it? (look at the diagram that you've posted above, the top of the lens) And it does exactly what a prism does - it makes a "pretty" little rainbow. Not so pretty when you're trying to do astronomy, however.

Short, fat, triangular prisms make widespread rainbows. Thin, elongated, thin-wedge shaped prisms make tight, narrow rainbows. The spread of the colors increases as the two sides of the prism are separated by a wider angle. It decreases as the two sides get closer together at a narrow angle. At the extreme, if the two sides of a "prism" were parallel, there would be no rainbow - but then it's not a prism anymore, it's just a glass plate, completely flat.

All else being equal, a single-lens-objective refractor has a "fatter" lens when focal length gets shorter (the two sides of the lens are more strongly curved, and/or less parallel); it has a "thinner" lens when the focal length gets bigger (the two sides of the lens are less strongly curved, and/or closer to parallel). A "fat" lens at the edge is like a fat, blunt prism. A "thin" lens at the edge is like a sharp, slender prism. The former makes a big spread of color. The latter makes a much more tight rainbow.

So that's what the old telescope makers did, back in the day (Hevelius, late 1600s). They just pumped up the focal ratio, made these super-long refractors with weakly curved lenses to combat the rainbow edge images which are so annoying and ultimately can falsify your observations.

Now, if you study the properties of lenses, you notice that divergent lenses (the kind that make the image smaller) have a chromatic aberration opposite to convergent lenses (lenses that magnify, and that were used as primary, or objective lenses in telescopes back then). It could be argued that by combining a convergent and a divergent lens, chromatic aberration might be greatly reduced; as long as the convergent lens is stronger than the divergent companion, the combination should remain convergent.

Also, different glasses need to be used, so that the convergent lens is strongly convergent, but weakly dispersing, while the divergent lens is weakly divergent, but strongly dispersing. That's how you get near zero dispersion, but still pretty decent convergence.

In the early 1700s this was accomplished by British opticians. Chester Moore Hall is credited with the making of the first achromatic objective lens using the scheme shown above. But the actual work of grinding/polishing the glass was contracted to George Bass, who noticed that the two lenses he was making, when put together, had near-zero chroma. Hall failed to recognize the practical applications of the achromats, neglected his own invention, and years later Bass mentioned the achromat scheme to John Dollond, who understood its value and started to popularize the new design.

https://en.wikipedia.org/wiki/History_of_the_telescope#Achromatic_refracting_telescopes

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

The achromatic combo guarantees zero chromatic aberration at two wavelengths. So you basically correct the aberration in the red and blue, or something like that, and the rest is "good enough". Things can be improved further by adding another lens - the apochromatic objective, or the "apo", or the "triplet". This was invented by Peter Dollond, John's son, in the mid-1700s.

The apochromat guarantees zero chromatic aberration at three wavelengths - usually in the middle and then closer to the two extremes of the visual spectrum. So that's overall better than the previous scheme.

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

The best apo refractors have extremely low chromatic aberration, basically invisible to the eye in most situations. But they are expensive.

Further improvements can be made. The most expensive amateur refractors are "quad" schemes, whereby using 4 lenses zeroes out chroma at 4 wavelengths. Takahashi is a brand nowadays known for excellent quality quad refractors, used in top-level astrophotography rigs.

BTW, not just chroma, but basically all aberrations can be reduced by increasing the focal ratios of single lens objectives. Pretty much every single aberration gets better if you put a piece of paper in front of it, with a hole smaller than the diameter of the lens. But the aperture gets smaller, resolving power gets worse, there's less light captured by the instrument, etc. Tradeoffs...

W.r.t detailed math of single lens chromatic aberrations, see chapter 5.3.1. Chromatic Aberration of a Singlet in 'Telescopes, Eyepieces, Astrographs' by G.H. Smith, R. Ceragioli, R. Berry. I believe that chapter is exactly the answer to your question, complete with working out the math for the aberration as a function of lens parameters.

http://www.willbell.com/tm/TelescopesEyepiecesAstrographs.html

Given your interest in the theory of optical systems, you should read that book.

Also worth mentioning:

Introduction to Lens Design, by J.M. Geary

http://www.willbell.com/tm/lens_design/index.htm

Practical Computer-Aided Lens Design, by G.H. Smith:

http://www.willbell.com/tm/lensdesn.htm

Some good theoretical info, as always, can be found on telescope-optics.net

https://www.telescope-optics.net/chromatic.htm


Requirements for a satellite/planet to be tidally locked to a planet/star

Wikipedia gives the formula

$$t_{\text{lock}} \approx \frac{\omega a^6 I Q}{3 G m_p^2 k_2 R^5}$$

where

$\omega$ is the initial spin rate expressed in radians per second,
$a$ is the semi-major axis of the motion of the satellite around the central body (given by the average of the periapsis and apoapsis distances),
$I\approx 0.4 m_s R^2$ is the moment of inertia of the satellite, where $m_s$ is the mass of the satellite and $R$ is the mean radius of the satellite,
$Q$ is the dissipation function of the satellite,
$G$ is the gravitational constant,
$m_p$ is the mass of the central body, and
$k_2$ is the tidal Love number of the satellite." source

Notice that the semi-major-axis is the power of 6. A small change in the orbital distance can have a very large effect on whether the body will become locked.

Also note the terms $Q$ (a measure of how elastic the body is, and hence how much energy is lost in distortion source) and $k_2$ (a measure of the rigidity of the body: how much it is distorted by tides source), which are both hard to measure. The 7 detected planets in the Trappist-1 system are all much closer to their star than any of the sun's planets and for reasonable values of $Q$ and $k_2$, the planets will be tidally locked within a few million years, much as the moons of Jupiter are tidally locked.

In our solar system, mercury is asynchronously locked with the sun (on a 3:2 resonance) The Earth is not locked: it is too far from the sun, and the moon has a big effect. Mars is much too far out, and Venus is odd.


How should I interpret this result obtained using lomb scargle?

Not being familiar with Lomb-Scargle (never having used it myself) the first thing I would say is that your graph is not really suitable for this kind of analysis.  If I graph it in my spreadsheet I get what looks more like a "bumpy line" not a periodic pattern :

The purpose of this algorithm is to take data that has a periodic pattern and find try and determine the frequency spectrum of that periodic data.  If you use data that is not periodic or does not properly exhibit a range of data values that allow discovery of the period then you unsurprisingly do not get sensible results.  So what I'm saying is that I suspect your data is not going to give you useful results with this algorithm.  I think you need more data points over a longer timescale.
Note my limited understanding is based on how similar algorithms are used for analyzing light-curve data.  This is well explained in some detail in this paper on Arxiv.  THe first section has no mathematics at all and illustrates a typical use case of the algorithm.  Make particular note that you need data covering more than than the period of the light curve to get a sensible result.  Note also the way the first "maximum peak" in the power spectrum is the period.  Figure (1) is that paper shows the raw data (a jumbled mess covering multiple periods) and figure (2) shows the way the period is identified and how the raw data can be graphed using this discovered period to properly align out of phase values.  I am dubious that your data is suitable, possibly because the underlying data is not periodic or is only a partial window of a period.
Section 8 of that paper has several pieces of advice about effectively using this algorithm, so I would recommend you read that even if you have to skip the mathematics in the middle of the paper.


Is Earth slowing down in order to stay on its orbital track?

As the Sun's mass is reduced and it's gravitational attraction is thus reduced the Earth will slowly spiral out from the Sun and yes slow down (to preserve angular momentum). But the effect is ridiculously small (do the math yourself) it will not be measurable directly.


How can we be sure that we have identified very distant stars correctly?

We determine the spectral type (i.e. temperature) of a star using multicolour photometry, or (ideally) spectroscopy. By guesstimating the temperature, mass and radius of a star, we can say that two stars that have pretty similar observational properties probably are closely related to each other.

Cepheid variables, for example, display periodic pulsations that depend quite strongly on their intrinsic luminosity -- this is why they're reasonably good standard candles. Their characteristic variability makes them clearly identifiable as a Cepheid, and observations of their pulsation are backed up by data that place them in the same region of the HR diagram.

If it looks like a duck, swims like a duck, and quacks like a duck, then chances are it's a duck. If it's as hot as a Cepheid and pulsates like a Cepheid, then it probably is a Cepheid.

So to answer your final paragraph: yep. Though with no spatial resolution and often (in the absence of parallax measurements) only tenuous distance data, classifying stars is a rather messy business.


If we accelerate a neutron star to move at speeds comparable to light would it become a blackhole because of the added energy?

In special relativity, changes of reference frame are symmetric between observer and observee.  If moving a neutron star gave it enough kinetic energy to become a black hole, then you could effect this change by jumping into a moving reference frame yourself.

What matters is the neutron star's mass in its rest frame.


Why did it take five years to "figure out" how to use astrometric calibration sources to deblur LOFAR images?

The point of the paper is not that no calibration has been used previously. It points out that they now use a more refined method and set of parameters to calibrate the data.
That in general is usual procedure: the more data you gather, and the longer a mission runs, the better you understand your instrumentation and all external factors which influence the data you gather - and especially the sensitivity towards different systematic and instrument errors. Thus with data becoming available you only have the people to work on the problems, often Bachelor, Master or PhD students. Thus prior to instrument availability and detailed data analysis you can only do a coarse calibration based on general knowledge of the problem. Five years is a reasonable and good time interval two have one to two generation of students work on it and get an good overall picture which considers all aspects you mention (and probably more) in the necessary detail and assess their contribution in a way which satisfies scientific scrutiny including the review process.


Is the LSST (Vera C. Rubin) telescope going to change position every 20 or 35 seconds?

How the Vera C. Rubin telescope will physically carry out the actual survey of the sky (the Legacy Survey of Space and Time; LSST) is still subject to evolution and refinement. The broad survey strategy is set out in the LSST System Science Requirements Document which is (intentionally) relaxed on how it is actually carried out. The main requirements are:

The main survey, of at least 18,000 square degrees must be uniformly covered do that each field of the camera's 9.6 square degree field of view gets an average of 825 30s visits over the 10 years of the survey. (The 825 visits are totalled over all 6 filters)
Rapid revisits (anywhere from 40 seconds to 30 minutes) must be acquired over at least 2000 square degrees to discover fast transients. (This can mostly be done by overlapping the fields of view of adjacent fields slightly

Other science requirements such as needing to detect moving objects (one of the four main science goals) and being able to reach the stated parallax and proper motion goals also impose some constraints on how often and when you need to revisit a field.
These constraints apply to the main "wide-fast-deep" (WFD) survey which is what will cover the 18,000 $\deg^2$ area above. Within the main WFD survey area, the baseline plan is make 2 visits per camera field (9.6 $\deg^2$) in either the same or different filters each night (This allows the moving objects and rapid transients to be found and alerts sent out). This pair of visits is repeated every 3-4 nights during the time the field is visible. Each visit consists of 2 back-to-back 15s exposures in the same filter, separated by the readout time of 2 seconds. These two 15s exposures are then combined in the pipeline to make the final 30s image.
Exactly how the WFD field visits are strung together, how far the telescope moves between fields and in what filters it observes and when, and what the observatory spends the other ~10% of time on, is a very complicated problem and depends a huge amount on what objects you want to be sensitive to and what type of science you want to do. These various tradeoffs are still ongoing as part of the Survey Cadence and Optimization Committee. The report of the large number of survey cadence simulations has recently come out but a final survey strategy will not be decided on until April 2023 (timeline summary) after more meetings, reports and simulations. Also, while 1x30s exposure at each visit (rather than 2x15s which has higher overhead) and a shorter time spent on each field is preferred (it increases the survey throughput, letting you cover more fields/area per night), this will depend on the final LSST Camera performance. This is because the susceptibility of the cameras' CCDs to cosmic rays, and whether you need the 2x15s exposures to remove them, is still partly unknown until it arrives at the telescope and has been operating for some time.
The simulations are based on a telescope model, whose code is available on github. The time taken to go from field to field depends on a quite a few things such as how far the telescope has to move (as there is both acceleration and deceleration limits as well as optics settling time depending on how far it moved), how much the dome has to move and whether you are changing filters (a 120s operation).
In general, the total time spent on a field is going to be:
$$
t_{slew}+t_{settle}+t_{exp}\times num. exp. + t_{shutter}\times num. exp. + t_{readout }\times (num. exp.-1)
$$
where $t_{slew}$ is the slew time (depends how far you move), $t_{settle}$ is settling time (also depends how far you move but minimum of 3s), $t_{exp}$ is exposure time, $t_{shutter}$ is shutter open/close time (1s; it's really big...) and $t_{readout}$ is the readout time (2s; the $(num. exp.-1)$ comes about because you can slew to the next field while reading out the last exposure).
So for 2x15s exposures, we have minimum of $3s+(15s\times2)+(1s\times2)+(2s\times1)=37s$ between fields. For the 1x30s, we have minimum of $3s+(30s\times1)+(1s\times1)+(2s\times0)=34s$ between fields. Exactly how often the telescope will be moving about to a new field will depend on the survey cadence decided on as outlined above, but there will be a strong preference to keep the slew times short and the number of filter changes small to maximize the amount of fields and sky covered per night.


What methods can high school students use to measure the radius of the earth?

It is hard.  You should consider what you are willing to allow them to research and what you will ask them to calculate.

First consider the classic "Shadow of a stick" method: On a certain day you measure the length of a shadow cast by a stick at two locations. The locations should be at the same longitude. By some trigonometry you can work out the number of degrees of latitude between the locations. By combining this with the distance between the locations you can calculate the radius of the Earth.

The biggest problem is how will your students know the distance between your chosen locations?  Traditionally this would be done by surveying. The process of surveying is a lot harder than measuring the length of a shadow. If your students are allowed to google for the distance, they might (reasonably) argue that they could have just googled for the Earth's radius.

Moreover if students are actually going to use this method, they need to travel thousands of km North or South, this may not be practical, so your students can't actually measure the length of the two shadows.

A second method uses time. The time of sunrise (or sunset, or noon) is measured at two locations on the same latitude. This lets you find the angular distance between the locations and that can be used with the actual distance to find the radius of the Earth.  Again this requires you to know the distance between the two points. And there are the same difficulties in travelling.

Looking at the shadow of the Earth as it crosses the moon can be used to calculate the radius. The is described in a document "Measuring the size of the shadow of the Earth" The difficulty with this method is that it depends on knowledge of the size of the moon (and the distance of the sun, though this can be approximated as infinity) This knowledge is difficult to get directly. However, this can demonstrate that the radius is finite and fixed, ie "the Earth is round".

In other words, to calculate the radius of the Earth, one needs additional information that can't be easily measured: Either the distance of two distant locations, or knowledge of the size of other astronomical bodies.


Would Europa be an ocean planet if it were in the habitable zone?

According to Arnscheidt et al. (2019) "Atmospheric Evolution on Low-Gravity Waterworlds", the transition between "planet-like" and "comet-like" (escaping) atmospheres occurs at a surface gravity of around 1.48 m/s2. As a caveat, the waterworlds being considered in the paper are objects which have water reservoirs of 40% of the total mass, which is a much higher fraction than Mars or Europa, such worlds would be more like scaled versions of Ganymede.
The surface gravity on Europa is 1.315 m/s2, putting it on the comet-like side of the transition. According to this, it seems unlikely that Europa would be able to maintain long-term habitable conditions in the face of atmospheric escape.
Martian gravity is 3.71 m/s2, putting it on the planet-like side of the transition, so a "super-Ganymede" with Mars-like gravity would likely be able to maintain long-lived habitable conditions.
They also note that the high albedo of ice means that deglaciating ice worlds tend to skip the long-lived habitable state entirely, going straight from frozen to runaway greenhouse/atmospheric escape:

The ice-albedo feedback can hamper transitions from snowball states to temperate states: this has
already been demonstrated for terrestrial-mass worlds
(Yang et al. 2017). We can incorporate the ice-albedo
feedback into our model using the simple albedo step
function
$$
A(T_s) = \begin{cases}
\alpha_I & T_s < 273\,\mathrm{K} \\
\alpha_L & T_s \ge 273\,\mathrm{K}
\end{cases}
$$
where $\alpha_I$ is the albedo of the icy (snowball) state, and
$\alpha_L$ is the albedo when there is surface liquid water. Hysteresis plots for different choices of $\alpha_I$ are shown in Figure 5. We observe that a snowball state experiencing
a stellar flux-driven deglaciation generally bypasses the
long-lived state entirely, except for very low $\alpha_I$ values.
Although the mechanism setting the inner edge of the
habitable zone is different, the conclusion of the habitable state likely being bypassed upon stellar flux-driven
deglaciation is the same as that of Yang et al. (2017).

(emphasis mine)


What lies at the center of a neutron star if any?

Certainly not a black hole! That would not be stable situation at all.

The content of the core of a neutron star is the subject of much speculation. The possibilities fall into a number of categories. (i) An increasingly hard neutron equation of state, such that neutrons retain their identities as they are squeezed closer, but an increasingly repulsive many-body strong nuclear force provides support. (ii) Additional hadronic degrees of freedom, such that the neutrons (and protons) transform into other heavy hadrons such as lambda or sigma particles. (iii) Some sort of quark plasma. (iv) Boson condensations involving the neutrons decaying into pions or kaons with zero momentum.

There are a number of diagnostics of these possibilities: primarily, the maximum possible mass of a neutron star should decrease from about 3 solar masses for (i) down to about 1.5 solar masses for (iv). Secure measurements of a 2 solar mass neutron star would seem to rule out (iv), but even that seems not completely agreed. Another diagnostic is how quickly neutron stars can cool. The presence of quark matter or boson condensations should lead to much more rapid cooling by neutrino emission. Again, nothing conclusive has emerged yet.


Your guess was correct. It is the Andromeda Galaxy, M31.

Here is a map of the part of the sky near zenith at the place and time you provided: Sky map for Taganrog, Russia on 11/23/2013 5:00:00 PM UTC. Even the rotation is small. The sky map is rotated approximately 30° counter clockwise relative to the photo. You were approximately facing south when taking it.

On your picture there is a part of the Andromeda constellation. I will use only a small bottom part of your picture (below the imaginary horizontal line going through M31) to identify the constellation.

The bright star to the left of M31, very close to the edge is μ Andromedae (marked 37 μ And on the sky map). It is the star between Mirach (out of photo) and M31.
On the bottom of the right part of the photograph you see three stars on a imaginary vertical line: ψ, κ and ι Andromedae.
Between ψ and κ, to the right there is λ Andromedae (marked 16 λ And and connected with μ on the sky map).

The bottom of the three stars in the upper-left corner of your photo is φ Andromedae (connected by the vertical line with 37 μ And on the sky map).

The 1 ο And on the sky map is not on your photo.


In sky surveys, what does inverse variation really mean? How do we use inverse variation to remove noise?

It is inverse variance. A larger inverse variance means the value in that pixel is less uncertain (in a relative sense).

The inverse variance can be used as a weighting factor - it directly gives you level of significance for the measurement. It can also be set to zero for measurements that should not be used (for whatever reason).

An example would be finding the mean value of a set of pixels. A standard mean would apply equal weight to all pixels. A weighted mean would be calculated as
$$ \bar{x} = \frac{\sum_i x_i w_i}{\sum_i w_i},$$
where $w_i$ here would be the inverse variance of each pixel.


How long would it take for a black hole the mass of our sun to swallow our solar system?

Is a common missconception to think black holes suck everything around them by definition as if the fact they are black holes gave them some kind of different gravitational influence than the one excerted by any other object with a similar mass.

The main difference between the Sun and a solar-mass black hole its the density. The Sun would contain a solar mass in a sphere with 1.3 million km in diameter while the Schwarzschild diameter of a solar-mass black hole would be below 6 km. But the mass is still the same. Take a look to Newton's Universal Law of Gravity:

$F = -G\frac{m_1m_2}{r^2}$

As you can see the gravitational force depends on the masses of the two bodies interacting, $m_1$ and $m_2$ (for example, the Sun and the Earth), but does not depend on the density of the bodies in any way, thus the interaction has the exact same behaviour in both the Sun and the solar-mass black hole scenarios. The planets would continue to circle around the barycenter of the Solar System as if the Sun was there.

Obviously, besides the gravitational influence, a lot of horrible consequences would take place. The absence of the solar light flux would freeze the entire system pretty fast. The first few days of darkness would be fine for Earth (even if air currents, migration patterns and circadian rythms would be totally disrupted). In a matter of months the Earth's surface would start to be sterilized for complex organisms unless moved to geothermal facilities. By the first year life on Earth would have been eradicated as the world oceans become a thick ice crust. After 10 years or so even Earth's atmosphere would start to collapse as air turns liquid and freeze upon arrival to the ground. With some luck the ice shell would still insulate the water below and microbial life could still be there, thriving with the geothermal energy of Earth's interior. The best option for us would be to transfer to Venus since atmospheric greenhouse gases would probably allow for larger thermal inertia and thus the surface would freeze slower than on Earth. In the end, after a few decades, even Venus would be sterilized. The entire Solar System would be a dark cold planetary ghost.

There is a mechanism to allow for the black hole to suck the planets eventually and that is gravitational radiation. We know that orbiting bodies emmit gravitational waves. These waves carry angular momentum away from the system shrinking the sizes of the orbits. We in fact have detected strong gravitational waves coming from inspiraling black holes. Therefore, with enought time, gravitational waves radiated by the planets would put them so close to the black hole as to be "swallowed", but this makes no difference with the Sun scenario we live in, orbits are decaying right now by the same process. But we need to have some orders of magnitude in mind, this process is crazily insignificant, more for small masses like the ones of planets, and even more for the huge orbital periods our planets have.

How long you would have to wait for this to happen? In the case of Earth, it would be swallowed in $10^{13}$ times the history of the universe. Yep, those are not 13 zeros of years but 13 zeros of times the universe age. Mercury, is not swallowed much earlier, in fact only half that time (which is still trillions of times the age of the universe). And for the entire Solar System to be swallowed (considering Neptune the last object of it), you would have to wait 145.000 times that. Not to mention the exponentially increasing waits for Kuiper belt objects and Oort cloud comets.

With these crazy spans of time we also have to think about other strange effects. Hawking radiation would evaporate the solar-mass black hole slowly, meaning that the central mass of the system would start to decrease. This means that planetary orbits would start to get larger (since a planet orbiting a less massive body with the same orbital speed would climb the gravitational dwell by some amount). The solar black hole would evaporate in $10^{58}$ times the age of the universe. Until the planets get free from the influence of an ancient mass that no longer exists in the Solar System. This is a way larger characteristic time period than the one required for orbital decay by gravitational radiation. So the gravitational decaying of the orbits would be the predominant effect even if the wait would be a bit larger (since the mass of the black hole decreases slowly).

So yeah, you can wait $10^{29}$ years until all the planets get swallowed by your black hole. The Sun would swallow them in the same way only that they would touch the Sun's surface even before than the event horizon of your black hole.

Not only that, the Earth is probably going to be swallowed by the Sun way before than in the solar-mass black hole scenario. We know because of stellar evolution that the Sun is going to get larger to the point of reaching our current position in just 6 billion years at max (even if there is some discussion about the solar mass loss in this era enlarging Earth's orbit and making it "safe").

If you want to keep your solar-mass black hole scenario think about those crazy trillions of times the age of the universe wait. Many things are going to happen way before that would disrupt the entire Solar System. It is expected that the orbital perturbations between the planets will eject several planets in a matter of a few billion years. That is a blink of an eye in the wait for them to decay. At these extreme timescales even close flybys of other stars will certainly dirsupt the system if planets don't do it by themselves in the first place.

So my answer is that in real life, turning the Sun into a black hole, would never end in the Solar System been swallowed by it, more like everything disspersed in interstellar space in some hundred of billion years at most.


The short answer is: probably nothing much, because galaxies are very fuzzy objects without "edges".

If you look at the stellar disk, it just fades out in density, with no evidence for a sharp cutoff (this is true of disk galaxies in general). There may or may not be a "break" at some radius, but this just marks the beginning of a steeper falloff in density, not an "edge" in any real sense. The gas disk extends even further out (that is, further out than we can easily trace the stars in the disk), and doesn't show much evidence for a sharp cutoff either.

There's also a round halo of stars, which is low density but extends to at least 400,000 light years; two stars were recently discovered  which are plausibly part of the halo and located at distances of about 800,000 light years. (Note that there are other galaxies -- such as the Large and Small Magellanic Clouds -- which are closer to us and to the center of the Milky Way, so you have to accept that individual galaxies can be found "inside" other galaxies.)

And then there's the halo of dark matter...

I can think of two ways in which you might talk about the outer boundary of an isolated galaxy (which the Milky Way sort of is) from a theoretical perspective. The first is the idea of a halo "accretion shock", in which relatively cold gas falling onto the galaxy from intergalactic space runs into the galaxy's halo of hot gas ("hot" = millions of degrees), creating a shock where the infalling gas slows down and the local density of gas goes up. (This is a little bit analogous to the heliopause, though the direction of gas flow is the other way around: the heliopause is where the outflowing solar wind shocks into the more or less stationary interstellar gas.) The problem is that extensive, hot halos of gas are thought to form only around massive galaxies, and it's not clear the Milky Way is massive enough to qualify. It also won't work for galaxies in massive groups and clusters, where the individual galaxies aren't accreting gas any more (the cluster is what has the hot halo and the accretion shock).

The other possibility is a region where the infalling dark matter particles merge into the dark-matter halo (whose particles are orbiting within their common gravitational potential, and so are not as a whole falling towards the center any more). Simulations suggest that as one moves inward the local density of dark matter starts to increase rather rapidly at this point (and the average velocity of the dark matter particles changes), but it's still a fairly smooth process (unlike gas, the dark matter particles aren't "shocking" into each other). There's no way of determining where this is for the Milky Way, other than something really vague like "maybe 1 million light years"?


What causes the abundance of dust to 'float' in the thin lunar atmosphere?

A recent study in Geophysical Research Letters — "Dust charging and transport on airless planetary bodies" (Wang et al. 2016) and reported in the ScienceAlert article "NASA just explained why Moon dust is 'levitating' above the lunar surface" — seems to explain the phenomenon through laboratory simulation experiments which observe what occurs when micron-sized dust is exposed to ultraviolet (UV) radiation or electrically-charged gases known as plasmas.

From these experiments, the researchers observed that:

  In both cases, the dust particles would leap several centimetres above the surface.

The experiment observations can be seen below:

Image source: Science Alert article linked above

Thus, one can conclude that on the Moon light would scatter through these levitating particles to create the horizon glow.

The mechanism behind the 'levitation' is stated as being:

  This is due to the reaction [with the plasma and UV light] causing an emission and re-absorption of electrons inside 'micro-cavities' formed between neighbouring particles, which can generate unexpectedly large electrical charges and intense repulsive forces.

  These forces cause the dust particles or clusters to lift off the
  surface, or 'levitate'.

Although the effects are observed with the lunar horizon glow, the researchers expect that effect would cause a suspension of dust across the entire lunar surface to some degree.

A significant implication of this research is stated in the NASA article "Dust “Floats” Above Lunar Surface: Electrostatic Dust Transport Reshapes Surfaces of Airless Planetary Bodies" as:

  The laboratory observations also showed dusty surfaces becoming smooth as a consequence of dust mobilization. These electrostatic dust processes could help to explain the formation of the “dust ponds” on asteroid Eros and comet 67P, and the unexpectedly smooth surface on Saturn’s icy satellite Atlas.


Are there only a set number of atoms in the whole universe?

There are several processes at work:

Fission (of unstable elements) increases the number of atoms.
Fusion (occurs in stars and supernovae) decreases the number of atoms.
you could claim the formation of neutron stars decreases the number of atoms: it reduces atoms to just neutrons
you could claim the formation of black holes decreases the number of atoms: it moves atoms out of reach of the 'normal' universe.

The next question is which of these processes produce/annihilate the largest number of atoms. Haven't found data for that yet. As Peter Erwin pointed out, any atom undergoing fission must have been built up from hydrogen and helium to start with.


Is dark energy evenly spaced throughout Universe?

If dark energy varied by location, then plots of 1a supernova brightness vs redshift should vary depending on which direction you look in the sky. AFAIK, that's not the case. For example, although coords are not accounted for, there's not a lot of scatter in this plot:

  In the relationship between the distance and redshift of Type 1a supernovae, the data (points) agree with the equation in which light propagates through the expanding universe on the least-time path (solid line). Image credit: Annila. ©2011 Royal Astronomical Society

Read more at: http://phys.org/news/2011-10-supernovae-universe-expansion-understood-dark.html#jCp


Is this the best non-radio image of whatever's at the center of M87? How was it taken?

Your top image is this from Dr Jean Lorre stock science images.  Just a photo showing stars and the optical jet with a 22 MB CCD camera through a small telescope.  It is unfortunate that he chose a false color brightness scale with black at the top intensity.  But, that black region does not correspond to the true blackness in the interferometric image which has resolution 100,000 times higher.

For comparison, here's a False-col. optical image of spiral galaxy NGC 1097 by the same "artist", also with the dark center where the visible light intensity is actually a maximum.


I see a non flickering star. How is this possible?

Stars are, to all intents, infinitely far away and hence present themselves as points - thus the small changes in the atmosphere's refractive index caused by different patches of density and temperature of air make them appear to twinkle.

Planets, on the other hand, are relatively close by and (with the exception of the Moon and the Sun itself) whilst we cannot resolve any solar system body into a disk with the naked eye, they are still larger than (most) variations in the atmosphere, hence they don't appear to twinkle or the twinkling is much reduced.

The evidence suggests you saw a planet, quite likely Jupiter or Saturn.


What is the hottest thing in the universe?

Energetic neutrinos have been observed from the core of a supernova (SN 1987A). The inferred temperature at the "neutrinosphere" is about 4 MeV (equivalent to 50 billion K - ($5\times 10^{10}$ K, Valentim et al. 2017). Hence it is observable and has been observed.

The very centre of the proto-neutron star that is responsible for the neutrino emission is likely to be a factor of two or so hotter, but cannot be observed, even with neutrinos, because the "neutrinosphere" is opaque to neutrinos. By the time this "clears", the proto-neutron star is much cooler - its surface would be orders of magnitude cooler.

Arguably we could study the very core of a supernova through gravitational waves if one were to explode in our own Galaxy. Whether this counts as "observing" a hot object, I'm not sure.

In a similar vein, we have observed "kilonova" that appear to be due to the merger of two neutron stars. The temperatures generated in these events are also likely to be of order 100 billion K ($10^{11}$ K), but again these temperatures are not observed directly - the gravitational waves and gamma rays produced in these events are caused by "non-thermal" mechanisms.


Why are solar systems stable and not chaotic?

The solar system is chaotic, but it is also stable!
The fixed and linkages between the bars of a double pendulum allow for very rapid energy transfer between the arms.  This makes the chaotic motion develop rapidly.
The interactions between planets are gravitational and much much weaker, moreover, the planets are heavier and it takes a lot more energy to deflect them. This means that in the medium to long-term it is possible to predict exactly where a planet will be.
In the very long-term the system is chaotic, but the gravitational linkage between planets is not strong enough to actually cause one of them to deviate very far from its orbit, unless a resonance develops.  So it is chaotic in the sense that we won't know which side of the sun Mars will be on.  But it is stable in the sense that we know that Mars will continue to be the planet between Earth and Jupiter (but perhaps slightly closer to Earth or slighty closer to Jupiter)
There is only one potential resonance: it is possible that in the very distant future, Mercury could develop a resonance with Jupiter, which could (in 3 to 4 billions years) eject it from the solar system.
So it is the very rigidity of the double pendulum that makes chaos develop so quickly.  Whereas the looseness of gravity means the motion of the planets can be predicted far into the future.


Estimate prevalence of ʻOumuamua-style interstellar obects inside Jupiter orbit now that C/2019 Q4 (Borisov) is found?

The enhancement of a cross-section due to gravitational focusing is given by
$$ \sigma_{\rm eff} = \pi a_J^2 \left(1 + \frac{2GM_{\odot}}{a_J\ v^2}\right),$$
where  $a_J$ is the semi-major axis of Jupiter's orbit (assumed circular), $v$ is the relative velocity (at infinity) and I have ignored the mass of Jupiter.

Thus, using $v=30$ km/s (as specified in the question) the term in brackets is the gravitational focusing enhancement of the cross-section and equals 1.38.

We now assume an isotropic velocity distribution and a homogeneous density of such objects of $n$ per unit volume. The flux of such particles is $f =n v \sigma_{\rm eff}$.

If $f = 0.5$ yr$^{-1}$, then $n \sim 5 \times 10^{12}$ per cubic parsec.


How Did Early Astronomers Measure Distances?

They were not very successful, but not for a want of trying.

The ancient greeks tried to measure the angle between the sun and the moon when the moon is half full. They got a value of 87 degrees, (when really it is 89.85). From this they could estimate the relative distances of the Moon and the sun.

Further observations of the moon from different locations on Earth could, in principle find the relative distance to the moon in terms of the radius of the Earth. And the radius of the Earth could be determined by noting the different lengths of shadows at the same time in different locations.

Several Greeks attempted to put these together to find the distance of the sun, and while their methods were geometrically correct, their observations were not accurate enough to achieve an accurate value for the distance of the sun. Typically they estimates were about 5% of the true distance. Attempts based on similar methods were made in other centres of astronomy, in China, India and the Middle East. They all tended to dramatically underestimate the distance.

Even after the invention of the telescope allowed the observations to be done more accurately the distance was usually under-estimated. only when careful measurements of the transit of Venus were made, was Jerome Lalande able to calculate the distance of Venus, and use Kepler's laws to derive the distance of the sun and other planets.

Kepler had created a model of the solar system that predicted the relative distances of each planet very successfully, but not the absolute distance

Nearly all other distances are based on the Earth-sun distance. The distances of the planets can be found using Kepler's laws. The actual distance of stars was unknown we were able to measure the slight change in position of a star as the Earth orbits the sun. If you know the distance to the sun you can calculate the distance to a star. Prior to the observation of this "stellar parallax" estimates of the distanct to stars was essentially guesses, and usually significant underestimates.

So prior to radar ranging, the methods were geometric, and based on a ladder of measurements. The size of the Earth was used to determine the distance to the moon and sun. Then the distance to the sun was used to find the distance to the stars.

For more distant stars there can be considerable uncertainty about the distance. If a star is too far to have parallax, then it's true distance will not be known to accuracy.


It depends on the definition of Earth-like planet.
In terms of size, density and gravity Venus is very Earth-like, but in terms of atmospheric and  surface conditions and its axis of rotation, and period of rotation Venus in not Earth-like.
I will assume you mean a terrestrial or rocky planet similar in size to Earth.
Earth's crust has two subdivision: oceanic and continental. Oceanic crust is between 5 km and 10 km thick whereas continental crust is between 30 km and 50 km thick:

Oceanic: 5 km (3 mi) to 10 km (6 mi) thick and composed primarily of denser, more mafic rocks, such as basalt, diabase, and gabbro.
Continental: 30 km (20 mi) to 50 km (30 mi) thick and mostly composed of less dense, more felsic rocks, such as granite.

Earth's crust is fractured into plates as described by plate tectonics. The boundaries of some plate collide causing uplift in the collision zone. The Himalayas is an example of such a zone. At other boundaries one plate subducts underneath the other plate. It's at these boundaries that most of Earth's volcanoes form.
For plate tectonics to occurs large amounts of water must be present to lubricate the subducting plate. The implication of this is that for plate tectonics to occur large oceans need to be located near subduction zones.
The other type of volcanoes that can occur are intraplate volcanoes which are located far from tectonic plate boundaries in the interior of tectonic plates. Such volcanoes are thought to arise from mantle plumes.
Earth has a diameter of 12 740 km, and Mars has diameter of 6780 km.
The crust thickness of both planets is 5-50 km for Earth and 10-50 km for Mars.
The ration of maximum crust thickness to diameter for both planets is 0.003925 for Earth and 0.007375 for Mars. The ratio for Mars is nearly double that of Earth, yet the Olympus Mons volcano on Mars is larger than any volcano on Earth.
For volcanoes to form, the temperature of the magma must be high enough to melt a hole through the crust and the pressure within the magma chamber must be high enough to overcome the resistance to flow posed by the friction of walls the volcanic vent and the gravity of the planet.
All terrestrial planets will experience volcanism at some period during their formation, it is a matter of when volcanism stops and why. Volcanism on Mercury ceased early due to the planet contracting early in it formation. Even the Moon experienced volcanism.


Calculating RA/dec from JPL ephemeris data

Read the fine print under the Horizons output when you select an OBSERVER table. It says (emphasis mine)

  R.A._(ICRF/J2000.0)_DEC =
     J2000.0 astrometric right ascension and declination of target center.
  Adjusted for light-time. Units: HMS (HH MM SS.ff) and DMS (DD MM SS.f)

You could do the same, but it will take some extra work. The Mars you see at some time $t$ was emitted from Mars several minutes earlier than that. You need to find the $\Delta t$ such that the distance between Mars at time $t-\Delta t$ and Earth at time $t$ is equal to the speed of light times $\Delta t$. One correction will get you very close, a second will get you even closer.

Another issue is time. JPL uses its own time scale, JPL ephemeris time (Teph). This is a relativistic time scale. It differs from Terrestrial Time (TT) by at most a few milliseconds, so for most purposes you can use TT in lieu of Teph in your homebrew ephemeris calculator. TT currently differs from UTC by 67.184 seconds, and beginning in July (there's a leap second at the end of June), it will differ from UTC by 68.184 seconds.

Or you could just live with it. If you are using your calculations for your backyard telescope, the handful of arc seconds of error that results from not accounting for the finite speed of light is small.


Transformation from Geocentric Coordinates into Equatorial Coordinates?

If I understand your question correctly than your problem is that the primary direction of vCR[3] is the intersection between the prime meridian and the Earth's equator, instead of the vernal equinox. The geographical latitude is the same as the declination, but to get the right ascension you will need to rotate your coordinates around the z-axis by an amount that is equal to the sidereal time at Greenwich ($\theta_G$). This is done, simply by replacing the the longitude (lon) with the sidereal time at your location $\theta = \theta_G-lon$.

x =  R* cos(lat) * cos(theta);
y=   R* cos(lat) * sin(theta);
z =  R* sin(lat);

vCR[3]={x,y,z};

where theta is the sidereal time $\theta$.

The sidereal time is not a time but actually an angle that gives the right ascension of objects that are at that moment crossing the local meridian (i.e. the line between South and the Celestial north pole, for observers on the northern hemisphere). It is in a sense the longitude of your position on Earth if the geographical coordinate system were to rotate along with the celestial sphere.

EDIT: To calculate the Sidereal time, you can use the method explained in Astronomical Algorithms by Jean Meeus:

$$T = \frac{JD-2451545.0}{3625}$$
$$\theta_0 = 100.46061837 + 36000.770053608 T + 0.000387933 T^2 - \frac{T^3}{38710000}$$

The sidereal time is expressed in degrees.

Then express the UTC time as an angle:

$$t = 15h+0.25m+\frac{0.25}{60}s$$

where $h$ is the number of hours, $m$, the number of minutes, and $s$ the number of seconds.

The sidereal time at Greenwich at the specified time is then:

$$\theta_G = \theta + 1.00273790935*t $$

and the local sidereal time is:

$$\theta = \theta_G - L$$

Where $L$ is the geographical longitude (positive west from Greenwich).


Mars vs Venus: the retention of atmospheres in relationship to Earth

One thing we have to keep in mind:
All atmospheres escape, always. Only the degree to which this happens is different on different planets.

I do appreciate your question a lot, as somehow people always forget about the existence of Venus when asking "what keeps an atmosphere in place?". As we will see, the classic story of "a magnetic field protects, that's all" is more of an urban myth.

Also all factors you have mentioned play a certain role. But let's pick them apart one-by-one, starting from the back in order to tell the story of escaping particles.

1. Distance
As you've correctly noted, distance plays a role indirectly by determining solar wind and irradiation at a given distance, so this factor is out as a direct cause.

After all, when modeling escape from planetary atmospheres one would rather take solar wind strength and UV-irradiation as free parameters rather than distance. This would make certain that one can compare planets around different stellar types, or stellar ages with each other.  Thus, distance is usually not considered a free parameter in the literature.

2. Planetary Temperature
Temperature is not equal to temperature. The surface of a planet has a variation in temperature from equator to poles $\Delta T_{\rm EP}$.
Then again, the temperature varies with altitude above the planet. Similar to the sun, the temperature difference from surface to high up in the atmosphere/corona is $\Delta T_{SC}$ and one finds $\Delta T_{SC} \gg \Delta T_{\rm EP}$.
On Earth temperature variations go (very roughly) from $+40^{\circ}C$ to $-60^{\circ}C$, so $\Delta T_{\rm EP} \approx 100 K$, while the temperatures in the exosphere at 500km height go up to $T_{exo} = 1000K$.

Modern measurments and calculations for Mars and Venus give low exospheric temperatures of around $300K$, due to efficient cooling to space of $\rm CO_2$ (see this presentation by Coates et al. for Venus and a recent review by Lammer et al. (2008))

For the escape of particles into space, it is the lower boundary of the exopshere, the exobase, that determines escape rates. Particularly, it is the temperature and the density at the exobase that determines escape rates, as from here the fast tail of the Maxwell-Boltzmann distribution can directly head off into space. This is a particular escape process called Jeans-escape. Below the exobase particles that travel upwards at escape speeds will encounter on average more than one particle and thus be scattered without escape. This is btw. also the definition of where to find the exobase.

This is to illustrate that neither the surface temperature of planets, nor the variations thereof play any role for atmospheric escape. Very strictly this is not true, as the atmosphere of a star and possible young rocky planet can get hot enough to launch a hydrodynamic wind into space, greatly increasing escape rates. For the sun we call this phenomenon the Parker solar wind.

Now let's take a look how particles can escape from the exobase.

3. Particle mass
High up in the atmosphere, at around 100km height, there lies another atmospheric boundary, the homopause. Here mixing of particles due to large-scale motions of the air becomes so inefficient, that particles start forming layers according to their masses.

Heavier atomic and molecular species will thus, once they reach the exosphere $\sim$500km be there in much smaller numbers than, for example Hydrogen. Thus, hydrogen escape rates will always be higher than other species.

This is the 'classical' picture, which cannot explain some mysteries involving the noble gases. Noble gases are chemically inert, meaning they don't form molecules and once in the atmosphere, they should just hang around there for billions of years. Except for those who escape. This is why noble gases are giving important datapoints when constructing theories of what happened to Earth's atmosphere.

An important problem is that Xenon is missing, relative to its lighter counterpart Krypton and the other noble gases. This shouldn't be if the classical picture is correct. I recommend reading the introduction in the article of Zahnle et al. (2018) for a very detailed picture of this "missing Xenon problem".

In order to address this problem and this article, let me introduce another particle parameter:

4. Ionization energies
So particle mass only doesn't explain escape rates. the article of Zahnle et al. (2018) (and references therein, this is not a new idea) however proposes that if escape rates are mostly given by ions, then a different picture emerges, that could reconcile escape via mass fractionation and missing Xenon.

The general idea is that because the ionization energies differ as we move through the periodic system of elements, it is much easier to ionize Krypton and Xenon. A mass of ionized hydrogen, or protons will be pre-existent in the high atmosphere. Ion-ion collisions are much more efficient in coupling species together than ion-neutrals.
So if we now assume that instead of the neutral hydrogen, the protons mostly escape, they'd have dragged Krypton and Xenon ions along.
Except that under Earth atmospheric conditions the Krypton ions quickly recombine to neutral Krypton, while Xenon doesn't. So the Zahnle article concludes that the missing Xenon is a signpost of past, dominant ion escape.

So if ions are this important for atmospheric escape, as opposed to the neutral species, we probably have to think more about the magnetic field lines that they follow. Finally we get to discuss the planetary magnetic field.

5. Magnetic field and solar wind

Some magnetic field lines connect to the field in interplanetary space. Ions travelling along those will be inevitabely lost to space and picked up by the solar wind. The strength of this effect is dominated by the geometry of field lines intersecting with the solar wind, and can lead to net protection or net erosion of the atmosphere.
This, among with other effects like pick-up and sputtering have been summarized in a comparative article on Earth, Mars and Venus in Gunell et al. (2018). Their key finding is

  While a planetary magnetic field protects the atmosphere
  from sputtering and ion pickup, it enables polar cap and cusp escape,
  which increases the escape rate. Furthermore, the induced
  magnetospheres of the unmagnetised planets also provide protection
  from sputtering and ion pickup in the same way as the
  magnetospheres of the magnetised planets. Therefore, contrary
  to what has been believed and reported in the press (Achenbach
  2017), the presence of a strong planetary magnetic field does not
  necessarily protect a planet from losing its atmosphere.

They find, that with all those complications of different escape processes and different ionized and neutral species, still for any single planet the ion escape rates can outcompete the neutral ones by a factor of $\sim 4$. Is this enough to explain the retention of the Venusian atmosphere and the escape of the Martian one? I think not. There is another factor that we've ignored so far.

6. Planetary mass

To close off this escape-story, I want to come back to the beginning of my answer, where I was happy that you've mentioned Venus. This is because Mars and Venus form a small set of a nearly identical case-study on atmospheric escape. Only two factors are different, while it is much harder to compare any other two planets in the solar system.

For the sake of comparing Mars and Venus to zeroth order, one could say that their atmospheres have identical composition, which is mostly $\rm CO_2$, as you have already stated. Both have no intrinsic magnetic field. Then, Venus has a hotter exosphere, stronger solar wind conditions, but still is somehow able to retain an atmosphere that has several thousand times more mass than the martian one.

If we now add to our perspective that a magnetic field doesn't play such a huge role for the retention of atmospheres, then the only remaining parameter that is different between Mars and Venus is the mass, which differs by a factor of about 10.
A factor of 10 is significant here, because if we go to planets now with 10 Earth-masses we already get into the regime of the Ice-giants Uranus and Neptune, which are able to hold on to their neutral hydrogen.
Understanding this is comparatively simple, as the totel escaping flux $\Phi_0$ is the integral of the Maxwell-Boltzmann-tail which goes very roughly as
$$ \Phi_0 \sim n(z_{\rm exo}) \cdot v_{\rm rms} \cdot \left( \frac{v_{\rm esc}^2}{v_{\rm rms}^2} + 1  \right) \cdot \exp(-\frac{v_{\rm esc}^2}{v_{\rm rms}^2}) $$
(source: Coates, or homework problems...)
where $n(z_{\rm exo})$ is the number density of a species at the height of the exosphere, $v_{\rm esc} = \sqrt{2GM/R}$ is the escape velocity of a planet with $M$ the mass and $R$ the radius and $v_{\rm rms} = \sqrt{2 kT_{\rm exo}/m}$ is the root-mean-square velocity of the MB-distribution at a given exospheric temperature $T_{\rm exo}$ for a given mean molecular mass $m$. So as there is an exponential factor involved, the Jeans-escape rate must go up quickly as planetary mass goes down.
So as ionic escape rates are a factor of a few times the neutrals, even if they dominate the scape, they're still bound to the exponential in the function.

More information can also be found in this, a bit older article by the same authors as the Zahnle paper, where they speculate also a little bit about escape from exoplanets, and the role of atmospheric chemistry for this effect.

Summary

The escape rate of ionic species can dominate over the escape rates of neutral species.
Ionic escape rates however don't react as strongly as one could think to the presence of a magnetic field. This leads to the interesting picture that the presence of a magnetic field is only a second order effect for determining escape rates.

The dominant parameter is then simply how deep the gravitational well of a planet is, that ions need to escape from. A escape rates go exponentially with planetary mass vs. molecular weight and available thermal energy at the exobase this is what determines escape rates most strongly.

Noble gases and their depletion can possibly resolve some mysteries of Earth's past, all the while telling us how much hydrogen has escaped from Earth.
As technology progresses, we might be able to address the same questions for Mars and Venus one day, but we're not there yet.

This was a long rant, please tell me if something is unclear.


What is considered "contact" for contact binary white dwarfs, and roughly how long can they stay that way before merging?

The issue I see here is that as a white dwarf gets more massive, it gets smaller. So in white dwarf–white dwarf mass transfer, the accretor will shrink, i.e. it evolves away from contact. Meanwhile the donor will tend to expand and lose mass faster through the L1 point.

So it might not even be possible for white dwarfs to form a stable contact binary. Proving a negative is somewhat tricky though.

There has been research into simulating white dwarf mergers. Dan et al. (2011) "How the merger of two white dwarfs depends on their mass ratio: orbital stability and detonations at contact" show several scenarios in their figure 7. The 0.6+0.55 M⊙ collision looks a bit like the Mark Garlick artwork from the press release, but it's on the line where disruption occurs on a timescale of ~20 orbits so it wouldn't look like that for very long until it merges and (judging by figure 10) undergoes contact-induced helium detonation.

From what I can tell, the usual use of "contact" in regards to white dwarf binaries refers to when matter from the secondary impacts (contacts) the primary, either by direct collision or mass transfer.

In terms of visible contact binaries, the brightest of the W UMa variables is Epsilon Coronae Australis which varies between magnitudes 4.74 and 5 over the course of an orbit. The next one is the secondary of 44 (i) Boötis, which is somewhere close to magnitude 6 (the magnitudes listed for its entry in the General Catalogue of Variable Stars include the contribution from the primary star).


What caused gravitational waves detected by LIGO?

It couldn't be Jupiter or the moon because

The frequency of the detected waves is too fast. The detector found something that rotated several hundred times a second, and then stopped.
The amplitude is too big. Even though Jupiter is close, the gravitational waves it produces in its orbit are extraordinarily weak. Far too weak for us to detect them (good thing too, because if it was producing powerful gravitational radiation, it would start to spiral into the sun)

We know one thing that can produce a chirp of this frequency and amplitude, and that is a black hole merger. In the last few moments before merging, these two black holes converted about three solar masses into gravitational radiation: That is a colossal amount of energy. About $$500,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000$$Joules.  That is why for one moment it was the "brightest"  point in the gravitational sky

(Note. standard form is for wimps)


Statistics of elements abundance in exoplanets

The observational determination of the chemical abundances in exoplanets is in its infancy. In terms of terrestrial type planets, i.e. those of size less than a few Earth radii, the constraints are confined to comparing the measured densities (obtained from the masses and radii of transiting planets found by Kepler and CoRoT) with models of what planets with certain assumed composition would look like. An excellent recent example of this can be found in Dressing et al. (2015). In this paper they make the claim that all of the low-mass
planets are consistent with a single, simple 2-component models (a mixture of 83% MgSiO$_3$ and 17% iron, but that this changes at higher masses, where more volatile elements or significant water are needed to explain their lower densities. The plot below, taken from that paper, illustrates the available data and should be quite up-to-date. Note how all the low-mass planets (and the Earth and Venus) can lie on the same family of models.

I don't think that the authors are claiming that this is exactly what all the planets are made of, but simply illustrating that at present there do not seem to be any large deviations from such a composition (for example, planets that are made solely of iron).

There are relatively few planets on this diagram, because it is difficult to obtain the masses of small transiting planets (it requires detection of the doppler shift caused by the pull of the planet on its host star).

Of course different models yield somewhat different results. For instance, Wagner et al. (2012) used the same data for Kepler-10b and CoRoT-7b and their own detailed models to argue that these planets have an iron core that makes up about 60% of the planet - i.e. a lot more than makes up the Earth.

At the moment the data for the lowest mass planets currently indicate that there could only be a limited amount of diversity. But the information that we are working with, the sample size, and the fact that only masses and radii are
determined, is too sparse to be sure.

From a theoretical point of view there are many ideas. The basic concept about the formation of the terrestrial-type planets is that they form (relatively) close in to the parent star and have compositions that are reflective of what elements and minerals can condense out of the protoplanetary disk at high temperatures. This in turn depends on the balance of elements that are present in the protoplanetary disk, where in the disk the planet forms, the detailed structure of the protoplanetary disk, how it cools and how planets migrate in the disk.  Unsurprisingly, by varying some of these conditions it is possible to create planets with a wide variety of compositions, which as I said above, appears to be mildly contradicted by the available evidence. At the most basic level we understand why the Earth and other terrestrial planets are comparatively lacking in "volatile" elements like hydrogen and helium (and to some extent C, N, O and noble gases), because they (or minerals containing them) could not condense at the high temperatures close to the proto-Sun.

Examples of these theoretical approaches can be found in Moriarty et al. (2014) (of which you are familiar), but also see Carter-Bond et al. (2012) for examples of how chemical diversity might arise. It seems that the Mg/Si and the C/O ratios have the biggest influence on the final compositions of the formed planets. A low C/O ratio favours the formation of silicates and fewer carbon-carrying compounds; but if there is more carbon than oxygen then it becomes more favourable to form carbon and silicon-carbide (I guess this is what you mean by "carbon planets"), but this also depends on the temperature in the region where the planet forms. For reference, the solar C/O ratio is 0.54 and the relative abundance of carbon in the Earth is much lower (than in the Sun) but the C/O ratio measured in other stars can be higher.


How to find the radius of a satellite not knowing the mass:
If I know the

semi-major axis,
density of the planet, and
orbital period..

how can I find the radius of the satellite? I can't think of any formula to use.

I have a hunch that you've misunderstood the question, and that it asks for the radius of the planet not the artificial satellite, and that's certainly doable. Otherwise, why would the planet's density be a given?

The equation that links semi-major axis, mass of the planet, and orbital period is:
$$T=2 \pi \sqrt{\frac{a^3}{GM}} $$
where $G$ is the gravitational constant 6.674E-11 m^3 s^-2  kg^-1.
Moving things around we get
$$\frac{4 \pi^2 a^3}{T^2G} = M $$
Density is mass per volume, or:
$$\rho = \frac{M}{\frac{4}{3} \pi R^3} = \frac{3M}{4 \pi R^3}$$
Moving that around we get
$$R = \left(\frac{3M}{4 \rho \pi}\right)^{1/3}$$
Put in the earlier result for $M$, we get
$$R = a \left(\frac{3 \pi}{\rho T^2G}\right)^{1/3}$$

For Earth's average density, Wikipedia says 5514 kg/^3, for its radius, try 6,371,000. meters. For the ISS at 400 km (400,000 meters) above the Earth, use 92.5 minutes (5550 sec). It should all work out nicely!
The reason this works with average density, even though Earth's density is higher in the core than the surface is the application of Newton's shell theorem.


What area of Proxima Centauri would need to be occluded for the effect to be observable on Earth?

The photometric precision achieved by satellites like TESS is around 100 parts per million. The original Kepler mission achieved a bit better on bright stars - maybe 20 ppm, but the question of precision may be moot since stars themselves are quite variable;  McQuillan et al. (2012) https://arxiv.org/abs/1111.5580 suggest that the floor level of intrinsic variability in M-dwarfs may be as high as 1000 ppm.

Anyway, let's assume that a rapidly encoded signal at 100 ppm could easily be distinguished from stellar noise and maybe even be generous and say 10 ppm might be possible. That would mean that this fraction of the stellar surface would need to be obscured.

The radius of the star is 0.15 that of the Sun. So the physical area to be obscured is $\sim 10^{-5} \pi R^2 = 3.4\times 10^{11}$ m$^2$.

The web page referred to in the question suggests that Starshot consists of a thousand space craft with light sail areas of 16 m$^2$. It thus seems to fall short of providing detectable modulation by 7 orders of magnitude (unless I have misunderstood the proposal, which isn't mentioned on that web page).

There is also going to be the question of the low bit rate and what limited information could be transmitted using such a crude technique. Recall that the spacecraft will be travelling at an appreciable fraction of the speed of light, so won't have long to send their message.

A further point is that I don't think it can be assumed that the trajectory of the spacecraft will be such that they are travelling in a direct straight line
between the Earth and Proxima Centauri, which are of course in relative motion.
Thus the spacecraft would approach Proxima Cen "at an angle" as viewed from the position of the Earth some 30 years after the spacecraft were launched, unless the trajectories can be tuned quite carefully en-route.
Even if they were, the parallax motion of the Earth around the Sun would mean that the spacecraft would not necessarily be aligned be when they reached Proxima Cen, although I suppose the launch time and speed could be arranged so that was the case. However, from what I have read, the idea is just to pepper the inner au of the Proxima Cen system with these 1000 spacecraft and so only a few of them would actually be lined up so that they could obscure the star as seen from Earth.


Sun and moon's orbit in ecliptic coordinates using skyfield

I cant help you with skyfield, but i usually use JPL Horizons Web interface. No installation required, you can also print it in a text file if you want:

https://ssd.jpl.nasa.gov/horizons.cgi

Otherwise i found the documentation for skyfield:
https://rhodesmill.org/skyfield/toc.html

And if nothing of that works, i made a little astropy script for you

from astropy import units as u
from astropy.coordinates import SkyCoord, EarthLocation, AltAz, get_body
from astropy.time import Time
import numpy as np

# Create 1000 Timepoints between Time 1 and Time 2 (one year later)
t = np.linspace(2451545, 2451545+365, 1000)

pointlist = []

#Loop through this times
for tn in t:
    # For every timepoint, create an astropy_time object
    astropy_time = Time(tn, format="jd")
    # Get Planet (as string, "earth", "moon", "mercury" etc. in aequatorial coordinates
    planet_aequatorial = get_body("moon", time = astropy_time)
    #Transform to Barycentric True Ecliptic (relative to the center of mass of the solar system).
    planet_ecliptic =  planet_aequatorial.transform_to("barycentrictrueecliptic")

    # Add a point to the orbit. Every point is described as (longitude [deg], latitude (ecliptic coords), distance (km))
    pointlist.append([planet_ecliptic.lon.deg, planet_ecliptic.lat.deg, planet_ecliptic.distance.km])
    print(planet_ecliptic.distance.km)
    # So pointslist is a 2D array. The rows are all the 1000 points of the orbit

# In every point there is 3 columns for [Long, Lat, Distance]
print(pointlist)

# You can also save the result with
pointlist = np.array(pointlist)

np.save("results.npy", pointlist)


What type of camera instruments are used to record optical images in space telescopes like HUBBLE & JWST.?

The HST and JWST are both mirror telescopes. There are many instruments on each, but the best known is the wide field and planetary camera (wfpc1), and its successors (wfpc2 and wfc3). These are all CCD cameras, and as such are based on the same basic technology that is found in many digital cameras, but there the similarity ends.

The camera is placed at the primary focus of the telescope. The telescope, therefore, acts as the main "lens". There are secondary optics, particularly in wfpc2 and 3 there are optics to correct the miscurvature of the main mirror. These optics are relay mirrors, not lenses.

Each camera is a one-off. Designed and built in-house by NASA. Compared to modern cameras, the resolution of the CCD is not so high. The original WFPC had an 800x800 pixel CCD, the WFPC2 had 4 800x800 pixel CCD sensors (at different pixel densities) If these seem low, remember that Hubble was designed and built in the '80s.

The cameras have to deal with a lot of things that the cameras we use don't. There are high-energy cosmic rays that can damage CCDs, and servicing required a shuttle launch. The CCDs are sensitive in a wide range of frequencies, from infrared to ultraviolet.

The latest camera, installed in 2009, has two light paths as shown, separating the visible from the IR imaging. As noted by Sean Lake in a comment, CCDs are read out by moving charge across the detector, infrared arrays read out individual pixels. IR arrays have more problems with stuck pixels or variable sensitivity across the array.

Schematic from the space telescope science institute.

So the cameras on Hubble, and the cameras that will be placed on JWST are distant relations of commercial digital cameras. They have no lenses, can see in a wider range of frequencies. The camera and its housing are also about 2m wide. However, the basic technology of converting an image into a digital file is shared between commercial digital cameras and the HST


The corona varies dramatically in brightness from the inner portion near the sun to the outer portion a few solar radii away. The innermost portion is the brightest; at ISO 100 and f5.6, an exposure of ~1/125 second will capture it. So it's roughly the same brightness as the full moon. The outer corona will require much longer exposures of perhaps a half second, but note that during this time the sun will move noticeably with a long lens if you're not using a tracking mount, so some blurring will likely occur. Overall, practicing on the crescent moon will give you a pretty good idea of the dynamic range of the corona.

The diamond ring is dangerous to look at directly, as is any unobscured part of the sun's disk; you should look away at first glimpse, or earlier.

There's lots more detail scattered around the web, which you've probably found by now. A good overview is available at http://www.mreclipse.com/Totality2/TotalityCh12-3.html#SEExpoTab


How thick or deep does solid carbon dioxide accumulate on Mars?

1-2 m in the north and 8m in the south. https://en.m.wikipedia.org/wiki/Mars#Physical_characteristics https://en.wikipedia.org/wiki/Martian_polar_ice_caps#North_polar_cap

It is the northern cap that disappears in summer.


Would a viewer see Earth from the rim of Shackleton Crater?

The moon has a monthly motion called libration Of interest to you is libration in latitude. The orbit isn't exactly in the plane of the ecliptic, and the rotation axis is slightly different from each. This means that about a week after passing its ascending node, you can see up to about 6.5 degrees past the moon's South pole. Two weeks later, the south pole is invisible, but then north pole can be seen.

From the point of view of an observer on Shackelton it would mean that over the course of a month the Earth would slowly rise and then set again.


Does this CHEOPS first light image imply bad astronomy?

As the article you reference makes clear, the defocusing is deliberate. It spreads the light of bright stars (the main targets for CHEOPS) over more pixels and hence mitigates saturation and non-linearity problems in the detectors.
The first light images look very similar to simulated pre-flight images (e.g. Hoyer et al. 2020; Futyan et al. 2020). The first light images are in line with expectations. There is therefore no unexpected problem with the telescope or optics, as there was with HST, and therefore no unexpected compromise of the mission sensitivity or goals.
From Hoyer et al.

The focal plane is defocussed to deliver a large PSF with a 12 pixels radius encircling 90% of the
flux. As a result of the combination of the Ritchey-Chrétien design and other specific features of the building of the telescope,
the PSF exhibits sharp and peaky features at sub-pixel level.

The reasons for the weird defocused PSF shape are discussed in section 11.1 of Benz et al. (2020). The triangular pattern is caused by strain on the primary mirror from a three-point fixation mechanism. This fixing mechanism was added to ensure that it wasn't shaken out of alignment during launch. In fact, Benz et al. say that the measured PSF (in zero gravity) is a little bit more symmetric, wider and smoother than found in the laboratory calibration tests. As a result there are some positives for observing bright stars - reducing the possibility of image saturation, but the situation is slightly worse for fainter stars because the wider PSF introduces more background signal into the photometry aperture.


What were the odds that ʻOumuamua passed so close to the Sun?

How could it be found otherwise? It is currently (November 2018) only at about the orbital distance of Mars and cannot be detected by the world's largest telescopes.

Such objects are small and only seen in reflected light - they have to get close to the Sun (and Earth) to be seen. There may well be a substantial population of interstellar rocks passing through the solar system unseen at greater distances.


For small values of practical an interstellar Orion craft could achieve ~3.3% c (~133 years to Alpha Centauri, one way with no deceleration at the end of the trip)). Which is in principle within reach of present technology if we chose to pursue it. But cost and aversion to nuclear technology on such an extravagant scale make it unlikely that we will see Orion space craft of any ilk built in the foreseeable future.


What is the most extreme weather found on another planet?

I'd say that HD 189733b is a good candidate for the most extreme known weather on another planet (outside our Solar System).

According to some recent news accounts, the atmospheric temperature is believed to be over 1000° C, with 7000 kph winds. (For comparison to the data in Rory Alsop's answer, that's about 1900 meters per second.)

And it rains molten glass. Sideways.

UPDATE : As Guillochon points out in a comment, HD 80606 b likely has even higher winds, though they're not continuous. It's a Jovian with an extremely eccentric orbit. Quoting the Wikipedia article:

  Computer models predict the planet heats up 555 °C (1,000 °F) in just
  a matter of hours triggering "shock wave storms" with winds that move
  faster than the speed of sound, at 3 miles per second.

which, in civilized units, is about 4800 meters/second. Probably no molten glass rain, though, so it's not clear that it's more "extreme".


Yes of course, there are other forms apart from the four stated above. Basically you have just two kinds:

the radio-quiet, and
the radio-loud.

But then again, there are two Seyferts

Seyfert I, and
Seyfert II

Apart from Quasar, Blazar and Radio galaxy, we also have presently,

BL Lac (Named after its prototype BL Lacarte (original member of Blazar type)), and
OVV (Optically Variable Violent Quasar or OVV Quasar; subtype of Blazar)

Best source to begin with is wikipedia.org of course.


Why was Neptune rather than Uranus chosen as an archetype?

NASA distinguishes four types of exoplanets: Gas Giant, Super-Earth, Neptune-like, and Terrestrial. The question asks why ice planets termed "Neptune-like" and not "Uranus-like".
Let's start with some statistics: What is curious, that (at least according to scholar.google.com), intially more icy exoplanets were termed Uranus like planet rather than Neptune like planet before eventually, the latter term started growing:
1980 to 1990

2 citations with Uranus like planet
0 citations with Neptune like planet

1990 to 1995

3 citations with Uranus like planet
5 citations with Neptune like planet

1995 to 2000

2 citations with Uranus like planet
4 citations with Neptune like planet

2000 to 2005

4 citations with Uranus like planet
12 citations with Neptune like planet

2005 to 2010

4 citations with Uranus like planet
43 citations with Neptune like planet

This supports the hypothesis that it was a "vote with the feet", rather than something coordinated, much like initial density fluctuations in a protoplanetary disk which might serve as the seed for a planet, or not.
@uhoh already suspected this in a comment, plus also mentions that the pronouncaction of Uranus might lead to unintended puns when talking about "planets like ...". It is sometimes these linguistic pecularities which make a brand successful and another one not.

I'm 100% sure that

it's because of the way Uranus sounds when said out loud, and
the decision was made in the collective unconscious, we'll never find a smoking gun, or a...

After all, there quite some similarities between Uranus and Neptune, see e.g. the concise comparison between the two planets at WolframAlpha, although there is also reasoning Why Neptune and Uranus are different.


Can we detect pulsars that aren't "pointed" towards us?

They don't need to be pointed exactly at us, since the angular beam radius is a few degrees (Pulsar Astronomy, P. 212) but there is a portion that we can't detect, described by a beaming factor (P. 211):

The larger beamwidth at higher energies can be seen in these graphs:

R. N. Manchester

We mostly can't see "ordinary neutron stars". "neutron stars can only be easily detected in certain instances, such as if they are a pulsar or part of a binary system" - Wikipedia

If the pulsar is radiating gamma rays more isotropically (but still varying with phase as in the Manchester charts), then these could be detected.


How to calculate right ascension of Greenwich?

I think you mean to ask, "What is the local sidereal time of Greenwhich?".

To calculate the GMST (Greenwich Mean Sidereal Time):

  To an accuracy within 0.1 second per century, Greenwich (Mean) Sidereal Time (in hours and decimal parts of an hour) can be calculated as GMST = 18.697 374 558 + 24.065 709 824 419 08 * D , where D is the interval, in UT1 days including any fraction of a day, since 2000 January 1, at 12h UT (interval counted positive if forwards to a later time than the 2000 reference instant), and the result is freed from any integer multiples of 24 hours to reduce it to a value in the range 0–24.

[source]


What is it about the layout of the three gratings in a Bahtinov mask that makes its indicate focus in this way?

The movement of the diffraction spikes is similar to the phase detection autofocus that's been used in SLR cameras for decades. Before the days of autofocus, many SLR cameras had a split-image viewfinder that used the same principle as well1.
To keep things simple, let's consider a mask with just two slits out at the very edge of the aperture.

Now let's consider what that does as we're focusing. I'm going to draw things in terms of a refractor, but using a reflector doesn't really change things. So, here's a simplified (and drastically not to scale) diagram of light coming from a star, going through the two slits above, and coming into focus on a plane behind the lens:

So, we have a star on the left, two rays of light leaving the star and traveling to the two slits, going through the slits, being refracted by the lens, and coming into focus on the plane. The image is in focus when the two rays of light come back together precisely at the surface of that plane. So what we see when we look at this should be basically a single line. We really have two diffraction spikes, one from each slit, but since we've brought it into perfect focus, they'll line up with each other perfectly. So, in simplified (and hand drawn) form, what we see looking at this, is a picture like this:

Two diffraction spikes, but they align perfectly with each other. I've drawn a boundary between the two, but in real viewing, there won't be an obvious boundary at the edge of each diffraction spike--we'll just see basically a single line.
So let's consider what happens when we move that plane, so the image is out of focus. The rays of light continue to be refracted to exactly the same degree, so we end up with something like this:

In this case, we can see that the two aren't hitting the focus plane at the same point, so what we'll see will be two mis-aligned diffraction spikes:

That explains how (some of) the spikes appear to move as we focus. That leaves another obvious question: how do we get the other spikes that don't move as we focus? That's actually pretty simple. If you look back at the first picture above, there's a crucial fact to note. The slits are (at least approximately) tangent to the aperture.
To form the X-shaped diffraction spikes, the Bahtinov mask uses slits that are radial instead. Technically, these do still move (a little) but it involves the diffraction spike moving along its length instead of sideways, so it's much less visible even at best. There's also little or nothing at the end of each diffraction spike, so you don't have much to compare against to see the movement.
Beyond that, it's all "user interface" enhancements to make it easier to use: More slits, carefully aligned with each other, project diffraction spikes on top of each other, so you get what looks like a single diffraction spike that's a lot brighter. The X-shape guides you eye toward the place you should be looking to see whether the spikes are properly aligned, and so on.
Now don't get me wrong: those are important, and definitely part of the genius of the design--but the basic principle of how/why the spikes appear to move relative to each other remains the same (and why some move but others don't appear to).

1. Note that many rangefinder cameras also used split-image focusing, but it worked rather differently--what I'm talking about here is strictly the version used with SLRs. An SLR with a split-image viewfinder used the same principle as shown here. The difference is that instead of using an aperture with slits, it used a pair of prisms at the center of the viewfinder to collect light from near the edges of the lens. But it still used the basic idea of collecting light from two opposite edges of the aperture, and showing them next to each other, and your focusing by aligning the pictures showing from each. It didn't use diffraction spikes, because in normal photography you typically have objects in the picture with obvious lines in them. Diffraction spikes just let us take subject matter that's mostly just dots, and creating lines from them.


Pulsars with accreting disk in binary system

As far as I know all accretion disks are prograde. When the pulsar starts accreting mass from it's companion this mass is pulled to follow the pulsar's rotation.


Why don't the stars in a binary star system of equal masses always orbit their center of mass in a circular orbit?

There's no reason why 2 bodies of equal mass couldn't have elliptical orbits around each other.   There's an example of that here

.

The simple way to think about this is, if two bodies of similar mass approach each other, one of two things can happen, they either have sufficient velocity to pass each other with some hyperbolic curving of both objects or there's enough gravitational attraction that they capture each other in orbit, but the shape or eccentricity of the orbit depends on the ratio between the tangential velocity and the orbital speed at the closest pass.   For a circular capture the ratio needs to be exactly 1, and that kind of exactness is rare, so we nearly always get an elliptical orbit with an orbital capture.

Solar-system planet or moon formation (not moon capture) tends to be much more circular, because elliptical orbits of objects tends to cancel out as large bodies coalesce, but you're never going to get 2 similar mass objects orbiting each other in a condensing gas cloud.   Most of the mass inevitably collects in the gravitational center.  There can be, however, a tidal influence that circularizes orbits over time, never reaching a full circle, cause that's impossible, but slowly becoming more circular.


when are you able to see Halley's comet, and for how long?

Halley's comet orbits the sun and its orbit lasts 75 years. The orbit is a long elliptical orbit. For most of those 75 years it is a cold black dot, and frozen solid in the outer solar system. But a short period during that orbit it gets close to the sun. The ice and gas begin to boil of its surface and gets blown back by the solar wind of the sun. It then starts to look like a comet.

The last time Halley's comet was close to the sun was in 1986, and so the next time will be in 2061. You won't be able to see Halley's comet until then.

(image from University of Wisconsin)

I remember 1986, and the comet was visible but hard to see. It was a very dim fuzzy dot, that was just about visible in my binoculars. This is because, unfortunately, the Earth was a long way from the comet. In 2061 Earth will be even further from the comet, however Earth will be better positioned when Halley is closest to the sun, so it should make a good show, and be clearly visible to the naked eye. Comet predictions are hard as much depends on exactly how much water vapour boils off, and this is hard to predict. Comets can have sudden changes in activity.

Moreover, Halley's comet isn't the only comet. There are others that can be as visible as Halley will be in 2061. There are also long period comets that are undiscovered until they come close to the sun. These can be very bright. Comet Hale-Bopp was a "great" comet in 1997 and was visible for many months. Another great comet could come along any time but we can't predict when.


What percentage of Helium-3 is primordial vs produced in stars

My question is 2-fold:

    1. Is the primordial amounts of $^3$He significant or insignificant compared to what stars produce and eject by coronal mass ejections or blowing up into nebulas, and ...

The $^3$He composition of CMEs can vary significantly, see: "Unusual composition of the solar wind in the 2-3 May 1998 CME observed with SWICS on ACE" (Jan 1999), by G. Gloeckler, L. A. Fisk, S. Hefti, N. A. Schwadron, T. H. Zurbuchen, F. M. Ipavich, J. Geiss, P. Bochsler, and R. F. Wimmer-Schweingruber, DOI: 10.1029/1998GL900166

  "Early work [Bame, et al., 1979; Schwenn et al., 1980; and Zwickl et al., 1982] showed that He and heavier elements are overabundant in CMEs and there is enhanced He$^{+}$. Observations from the SWICS instrument on Ulysses revealed some compositional differences in CMEs, such as a high O$^{7+}/\,$O$^{6+}$ ratio, indicating significant heating in the corona [Galvin,
  1997].

  ...

  SWICS is especially well suited to measure solar wind $^4$He$^+$
  and the isotopic helium ratio, $^3$He$^{++}/\,^4$He$^{++}$, as described by Gloeckler' and Geiss [1998a].".

    2. are there specific stars, due to the internal heat and rate of reaction that product and eject more $^3$He into their mass ejections.

Most of $^3$He was originally produced by stellar processes, but see also my answer above. Each star produces varying amounts at various times, see: "The Origin of Helium and the Other Light Elements" (Nov 4 1998), by G. Burbidge and F. Hoyle:

  4. D and $^3$He
  The light isotope $^3$He is produced in large quantities in dwarf stars where the masses are not large enough for it to be destroyed by $^3$He ($^3$He, 2$p$) $^4$He. It is also the case that there is a class of stars in which it has been shown from measurements of the isotope shift that most of the helium in their atmospheres is $^3$He. These stars include 21 Aquilae, three Centaurus A, and several others (Burbidge & Burbidge 1956; Sargent & Jugaku 1961; Hartoog & Cowley 1979; Stateva, Ryabchikov, & Iliev 1998). The stars are peculiar A, F, and B stars having He/H abundances that are $\sim \frac{1}{10}$ of the normal helium abundance. The $^3$He$/\,^4$He ratio can range from 2.7 to 0.5. These stars occupy narrow strip in the (log $g$, T$_{eff}$)-plane between the B stars with strong helium lines and those with weak helium lines that show no evidence for the presence of $^3$He. However, the detection of $^3$He from the isotope shift will fail if the $^3$He$/\,^4$He ratio is $\le$ 0.1. Thus, many of the weak helium-line stars may well have $^3$He$/\,^4$He abundance ratios far higher than the abundance ratio that is normally assumed to be present, namely, $^3$He$/\,^4$He $\approx$ 2 x 10$^{-4}$.

  The high abundance of $^3$He in these stars has been attributed by G. Michaud and his colleagues to diffusion (Michaud et al. 1979 and earlier references). Whether or not this is the correct explanation, what these results do tell us is that stellar winds from such stars will enrich the interstellar gas with He in large amounts. This $^3$He is in addition to the $^3$He that will be injected from dwarf stars. The final abundance required is $^3$He$/\,$H $\; \approx \;$ 2 x 10$^{-5}$. It has been argued by those who believe that $^3$He is a product of big bang nucleosynthesis that there has not been time to build up the required abundance by astrophysical processes.

  However, not only do we not know what the rate of injection from stars is, but in the QSSC, the timescale for all of this stellar processing is $\sim 10^{11}$ rather than H$_0^{-1}$ $\approx$ 10$^{-10}$ yr. Thus, we believe that He may very well have been produced by stellar processes.

Another Wikipedia link not mentioned in your question is: "Helium-3 - Natural abundance - Solar nebula (primordial) abundance":

  "Solar nebula (primordial) abundance

  One early estimate of the primordial ratio of $^3$He to $^4$He in the solar nebula has been the measurement of their ratio in the atmosphere of Jupiter, measured by the mass spectrometer of the Galileo atmospheric entry probe. This ratio is about 1:10,000,$^{[43]}$ or 100 parts of $^3$He per million parts of $^4$He. This is roughly the same ratio of the isotopes as in lunar regolith, which contains 28 ppm helium-4 and 2.8 ppb helium-3 (which is at the lower end of actual sample measurements, which vary from about 1.4 to 15 ppb). However, terrestrial ratios of the isotopes are lower by a factor of 100, mainly due to enrichment of helium-4 stocks in the mantle by billions of years of alpha decay from uranium and thorium.

  Terrestrial abundance
  Main article: Isotope geochemistry

  $^3$He is a primordial substance in the Earth's mantle, considered to have become entrapped within the Earth during planetary formation. The ratio of $^3$He to $^4$He within the Earth's crust and mantle is less than that for assumptions of solar disk composition as obtained from meteorite and lunar samples, with terrestrial materials generally containing lower $^3$He$/\,^4$He ratios due to ingrowth of $^4$He from radioactive decay.

  $^3$He has a cosmological ratio of 300 atoms per million atoms of $^4$He (at. ppm),$^{[44]}$ leading to the assumption that the original ratio of these primordial gases in the mantle was around 200-300 ppm when Earth was formed. A lot of $^4$He was generated by alpha-particle decay of uranium and thorium, and now the mantle has only around 7% primordial helium,$^{[44]}$ lowering the total 3He/4He ratio to around 20 ppm. Ratios of $^3$He$/\,^4$He in excess of atmospheric are indicative of a contribution of $^3$He from the mantle. ...".

  [43] "The Galileo Probe Mass Spectrometer: Composition of Jupiter's Atmosphere" (Science 10 May 1996: Vol. 272, Issue 5263, pp. 846-849) by Hasso B. Niemann, Sushil K. Atreya, George R. Carignan, Thomas M. Donahue, John A. Haberman, Dan N. Harpold, Richard E. Hartle, Donald M. Hunten, Wayne T. Kasprzak, Paul R. Mahaffy, Tobias C. Owen, Nelson W. Spencer, and Stanley H. Way, DOI: 10.1126/science.272.5263.846

  [44] "Non-Lunar $\underline{^3}$He Resources" (Presented at the Second Wisconsin Symposium on Helium-3 and Fusion Power, 19–21 July 1993, Madison WI), by LJ Wittenberg - fti.neep.wisc.edu