How should one rationally deal with the issue of space travelling alien civilizations?

That's basically the Fermi paradox. It seems likely that there are numerous civilizations in the galaxy, and yet we see no trace of them anywhere.

The Drake equation is often invoked to calculate the probability of existence of other civilizations, by compounding several other, more simple probabilities: the probability that a star has planets, that the planets are habitable, that life could arise on such a planet, etc. The proposed values so far indicate that there should be many civilizations in our galaxy.

Astronomy, just plain observations, might be one way to solve the paradox by detecting alien societies.

Broadcasting messages via radio has also been attempted, in the hope that someone might reply.

However, the only constant in this search has been the negative answer. There are some hypotheses trying to explain that:

the rare Earth hypothesis: maybe there's something very special about Earth that allows the evolution of intelligent life
the destruction hypotheses: maybe life tends to self-destruct, or maybe there are space marauders out there killing everyone
disasters: maybe things like gamma ray bursts tend to wipe out life periodically
the young Universe hypothesis: if there are very, very many universes in the Multiverse, then the vast majority of them are young. If so, in the vast majority of them there will be 0, or at most 1 civilizations. It's simply a matter of probability that we are in such a universe.
religious arguments such as the creation of Man was a special, unique act by a higher entity

Others argue that many civilizations might exist, but we don't see them. Possible reasons:

we're just too far from each other, in space, in time, or both
it's just too hard to travel between stars
we haven't been searching long enough, or we're looking or listening to the wrong things
they're not interested in us
they all go into a Technological Singularity pretty quickly and become god-like, undetectable and incomprehensible to us
they are too busy browsing their internets to be interested in the outer space (yes, that has been proposed seriously)
they are too alien for us to detect or comprehend
they are like the Na'vi in Avatar - a non-technological society
they deliberately avoid us (the Earth-as-a-zoo hypothesis)
they do try to talk to us, but evidence is being supressed by the government (yes, that is also a serious proposal)
they are here already, hiding amongst us (proposed by Iosif Shklovskii and Carl Sagan, so I guess it qualifies as serious - albeit highly hypothetical at best)

Truth is, we really don't know. All we can do now is speculate.


What is the planet relative to earth that shows the greatest change in apparent brightness?

$
\begin{array}{cccc}
                   \text{Body} & \text{Max} & \text{Min} & \text{Delta} \\
                   \text{Sun} & -26.78 & -26.71 & 0.07 \\
                   \text{Mercury} & -2.45 & 5.58 & 8.03 \\
                   \text{Venus} & -4.89 & -3.82 & 1.07 \\
                   \text{Moon} & -12.87 & -3.76 & 9.11 \\
                   \text{Mars} & -2.88 & 1.84 & 4.72 \\
                   \text{Jupiter} & -2.94 & -1.66 & 1.28 \\
                   \text{Saturn} & 0.42 & 1.47 & 1.05 \\
                   \text{Uranus} & 5.31 & 5.95 & 0.64 \\
                   \text{Neptune} & 7.8 & 8. & 0.2 \\
                   \text{Pluto} & 13.75 & 15.96 & 2.21 \\
                   \text{Comet Halley} & 2 & 25.66 & 23.66 \\
                   \text{Tesla Roadster} & 6.66 & 29.29 & 22.63 \\
                  \end{array}
$
The answer is Mercury (as above), subject to the proecedure/caveats below:
Procedure

I used HORIZONS to generate daily brightness data for a century for:

all the planets as viewed from Earth (except Earth itself)

the Sun, the Moon, Pluto, Comet Halley, and the Tesla Roadster

I then noted the minimum and maximum brigtness, along with the magnitude difference of these brightnesses, in the table above.

You can use HORIZONS to compute the results yourself, or view the results in the *-brightness.txt.bz2 files in https://github.com/barrycarter/bcapps/blob/master/ASTRO/

Caveats

Although Mercury's brightness changes more than Mars, the Sun's glare makes it impossible to see Mercury though Earth's atmosphere when Mercury's angular distance from the Sun is small. Therefore, Mars may be a better practical answer.

Because I used daily brightnesses, it's theoretically possible I missed absolute (intraday) minimums or maximums, especially for the Moon, whose brightness changes rapidly. However:

I took 100 years worth of data, and the moon's brightness doesn't have a period that's a multiple of one day. Since the moon's synodic period is approximately 29.5 days, its brightness is almost periodic in 59 days (2 synodic periods), but it's far enough from 29.5 days that this shouldn't be too much of a problem.

The Moon isn't a planet: I just added it for reference

Unless the Moon's actual brightness difference was higher than 22.64 magnitudes (which is probably unlikely[?]), it would remain in 3rd place in terms of brightness change, so the exact value isn't as important.

Because I used only a 100 year period, I did not include a complete orbit for either Neptune or Pluto. This shouldn't be an issue because:

The synodic period of both planets is just over a year, and much of the brightness change comes from Earth's own orbit, not Neptune's or Pluto's.

Even if the maximum magnitude change were slightly higher than in the table, it wouldn't make much of a difference.

Note that "Max" and "Min" refer to brightness, which is ordered the opposite of magnitude: lower magnitude means greater brightness.

Because I used daily data, I missed rare events such as transits and eclipses. The table above is for an "average" orbit, excluding special cases.

In some cases, HORIZONS gives "n.a." for magnitude data. I ignore these "n.a." values.

Data for the Tesla Roadster is only available from 2018-Feb-07 03:00 UTC to 2090-Jan-01 23:00 UTC, not the entire century.

Complexity
The problem is nontrivial. As you correctly note, the planet's geocentric and heliocentric distance play into the formula, but there's more to it.
Quoting Oliver Montenbruck and Thomas Pfleger's "Astronomy on the Personal Computer" (https://books.google.com/books?id=nHUqBAAAQBAJ):

Paul Schlyter's http://www.stjarnhimlen.se/comp/ppcomp.html#15 provides similar nontrivial formulas.
EDIT: https://en.wikipedia.org/wiki/Phase_curve_(astronomy) provides more information on how planetary brightness varies nontrivially with planetary phase.


Information on Planetary nebula formed from the death of our star the 'Sun'!

A Study to Show : Jupiter : Our Next Star
My 11th grade research paper 2018
Answers under -
Planetary nebula, pg 8-9
Total mass of the planetary nebula formed
= 0.49 M⊙ (99% hydrogen)
Revisiting mass loss, pg 12
Density of CSE1 on reaching Jupiter’s Orbit after Thermal Pulse 1, pg 17


What Crayford focuser to buy for a B&L Criterion 8000?

Item B is called a visual back.  The new focuser would just replace this, and screw onto item A.  The focuser you link to is designed for a refractor, but you need one for a SCT, like this
https://www.firstlightoptics.com/skywatcher-focusers/dual-speed-2-crayford-focuser-for-sct-telescopes.html
There are lots of different makes and prices.
Regarding the sizing, I'm not sure about an old scope like the Criterion: it's probably the same as the other popular current models, but you might do better with cloudynights.com to get that confirmed.


an exercise from Spherical Astronomy by Smart

Constant for every star with different declination's.

Though you would have to assume the the declination is such that it can have an azimuth of 90 each, not all stars will. Depending on your latitude some stars will never get to an azimuth of 90, they will either be always be further north or south.


How does this strategy for chasing the sunset work?

I would go to the Arctic circle, where you can experience an almost perpetual sunset, without having to drive around.

Edited to answer the question: My interpretation of the diagram, is:

The terminator at a certain time near sunset, is the one shown towards the east (darkness approaches), running NE/SW.
As time progresses, it moves into the top horizontal location, running E/W; this consistent with the more northerly latitudes having continual daytime, coinciding with a certain parallel polar circle.
As morning approaches, the terminator moves towards the west (bringing daylight with it), and runs NW/SE.

I think the arrows would be better if the tip of the one on the right, joined the base of the one on the left, showing your car journey as a continuous one.


Do supernovae produce powerful gravitational waves?

Actually the peak luminosity for the first binary black hole merger LIGO detected was $3.6 \times 10^{49} \,\text{Watts}$ (source). Whereas the brightest supernova ever recorded was ASASSN-15lh which had a peak luminosity of $2 \times 10^{38} \,\text{Watts}$ (source). In fact the energy radiated by the binary black hole merger during the $0.2 \,\text{s}$ it took to merge was over 10 times the combined luminosity of the observable universe (source).
This translates to the total energy emitted to be $5.4 \times 10^{47} \,\text{J}$ for the black hole merger and $(1.1 \pm 0.2) \times 10^{45} \,\text{J}$ for the ASASSN-15lh supernova. So as you can see, the binary black hole merger released around 100 times more energy than the brightest known supernova.
In relation to the gravitational waves emitted by supernovae, their wave-forms are not as well understood as the "chirps" you get from merging black hole binaries, however they are still expected to be strong enough to be detected in the local universe.


Accuracy of Laplace Method for determining orbital elements

In principal, Laplace's method (and Gauss's method) should be very accurate for determining orbits.

Of course "garbage in garbage out" applies, and if the observations were inaccurate the results may also be inaccurate. However, you should not get great discrepancies. I would consider careful checking of your calculation, and perhaps comparing your results with those of an independent orbital determiner, to check for a mistake in the maths.


Halley's Comet current location in sky and visibility?

In case you happen to have a very large telescope, here's where to look for Halley's Comet in 2019. I got these from JPL's Horizons ephemeris.

date (UTC)    h  m  s        d  m  s        T-mag    N-mag
2019-Jan-01   08 26 12.95   +01 38 11.0     25.48    28.99
2019-Feb-01   08 22 37.55   +01 48 26.3     25.48    28.97
2019-Mar-01   08 19 35.31   +02 04 26.5     25.49    28.99
2019-Apr-01   08 17 21.46   +02 24 37.7     25.52    29.04
2019-May-01   08 16 49.80   +02 41 21.1     25.55    29.07
2019-Jun-01   08 17 59.21   +02 51 29.8     25.58    29.10
2019-Jul-01   08 20 21.84   +02 52 27.8     25.60    29.10
2019-Aug-01   08 23 27.41   +02 44 30.4     25.61    29.10
2019-Sep-01   08 26 24.57   +02 29 35.3     25.60    29.10
2019-Oct-01   08 28 23.88   +02 12 04.1     25.58    29.10
2019-Nov-01   08 28 56.98   +01 55 29.1     25.56    29.08
2019-Dec-01   08 27 45.99   +01 45 32.8     25.53    29.04
2020-Jan-01   08 24 59.69   +01 45 05.9     25.51    29.00

T-mag N-mag =
   Comet's approximate apparent visual total magnitude ("T-mag") and nuclear
   magnitude ("N-mag") by following standard IAU definitions:
    T-mag =  M1 + 5*log10(delta) + k1*log10(r)
    N-mag =  M2 + 5*log10(delta) + k2*log10(r) + phcof*beta
   Units: MAGNITUDES

From Wikipedia's Halley's Comet; History:

  After 1986

  Grainy, white-on-black image showing Halley as a barely distinguishable black dot
  Halley's Comet observed in 2003 at 28 AU from the Sun

  On 12 February 1991, at a distance of 14.4 AU (2.15×109 km) from the Sun, Halley displayed an outburst that lasted for several months, releasing a cloud of dust 300,000 km across.49 The outburst likely started in December 1990, and then the comet brightened from magnitude 24.3 to magnitude 18.9.112 Halley was most recently observed in 2003 by three of the Very Large Telescopes at Paranal, Chile, when Halley's magnitude was 28.2. The telescopes observed Halley, at the faintest and farthest any comet has ever been imaged, in order to verify a method for finding very faint trans-Neptunian objects.9 Astronomers are now able to observe the comet at any point in its orbit.9

  9New Image of Comet Halley in the Cold". European Southern Observatory. 1 September 2003. Retrieved 26 February 2018.

  49J. C. Brandt. "McGraw−Hill AccessScience: Halley's Comet" (PDF). McGraw-Hill. Retrieved 27 November 2009.)

  112D. Prialnik; A. Bar-Nun (1992). "Crystallization of amorphous ice as the cause of Comet P/Halley's outburst at 14 AU". Astronomy and Astrophysics. 258 (2): L9–L12. Bibcode:1992A&A...258L...9P.

You can view Halley's Comet's orbit in 3D here: https://ssd.jpl.nasa.gov/sbdb.cgi?sstr=1P;orb=1

  caption: Seventeen years after the last passage of Comet Halley, the ESO Very Large Telescope at Paranal (Chile) has captured a unique image of this famous object as it cruises through the outer solar system. It is completely inactive in this cold environment. No other comet has ever been observed this far - 4200 million km from the Sun - or that faint - nearly 1000 million times fainter than what can be perceived with the unaided eye. This observation is a byproduct of a dedicated search 1 for small Trans-Neptunian Objects, a population of icy bodies of which more than 600 have been found during the past decade.

Source

  A record observation
  At 28.06 AU heliocentric distance (1 AU = 149,600,000 km - the mean distance between the Earth and the Sun), this is by far the most distant observation ever made of a comet 2. It is also the faintest comet ever detected (by a factor of about 5); the previous record, magnitude 26.5, was co-held by comet Halley at 18.8 AU (with the ESO New Technology Telescope in 1994) and Comet Sanguin at 8.5 AU (with the Keck II telescope in 1997).

  Interestingly, when Comet Halley reaches its largest distance from the Sun in December 2023, about 35 AU, it will only be 2.5 times fainter than it is now. The comet would still have been detected within the present exposure time. This means that with the VLT, for the first time in the long history of this comet, the astronomers now possess the means to observe it at any point in its 76-year orbit!


How to show that the Jeans Criterion for Mass, Radius and Density are equivalent?

What you want to do is use the relation between Mass, Radius, and Density. The proper expression should be

$$\rho_c = \frac{M_c}{(4/3) \pi R_c^3}$$

Your three Jeans conditions are $M_c > M_J$, $R_c > R_J$, and $\rho_c > \rho_j$. By using the relation above, you can transform any of the three conditions into one of the others.

For example, going from $M_c > M_J \rightarrow R_c > R_J$ is done as follows.

$$M_c = \frac{4}{3}\pi R_c^3 \rho_c > (\frac{5kT}{G \mu m_H})^{3/2} (\frac{3}{4 \pi \rho_c}) ^{1/2}$$

Now, solving for $R_c$ gives

$$R_c^3 > (\frac{5kT}{G \mu m_H})^{3/2} (\frac{3}{4 \pi \rho_c}) ^{3/2}$$

$$R_c > (\frac{15kT}{4 \pi G \mu m_H \rho_c})^{1/2} \equiv R_J$$

The other transformations follow in a similar manner.


The Sun and Regulus together at the summer solstice

Using sky simulator software (in my case Guide 9.1), the sun and regulus rose at the same time in about 2500 bc.  I think you would need to go back another 1500 years to be able to see regulus in the pre dawn glow. At that time, the sun would be about 7 degrees below the horizon as regulus was rising.

However the surrounding woodland wasn't cleared until about 5500 years ago (per Wikipedia), so it may not have been possible to make this earlier observation.


If the earth lost 1/81th of its mass how would it orbit change?

I will answer the specific case of the Moon disappearing since the more general question, is a bit open-ended.

Of course the Moon cannot just "disappear." However, there are very real circumstances where a planet or other stellar object might be disrupted. The best example I can think of is if a sufficiently large asteroid or moon from another planet somehow make their way into Earth's orbit and collide with the Moon. Such collisions between Solar System objects are rare, but occasionally occur in models of the future evolution of the Solar System (see work by Jacques Laskar).

However, one must be careful to first consider whether that mass is actually lost from the system. For instance, in the Giant Impact Scenario a Mars-sized object is theorized to have slammed into the Earth, but the material never left the system. If a similar collision were to disintegrate the Moon and the mass remained as rocks and dust in orbit around the Earth, the effect on Earth's orbit would be minor. This is because it is the Earth-Moon system that orbits around the Sun. Note that there is a small change since the Earth-Moon barycenter has shifted position, but the effect is tiny so let's ignore it for now.

Instead, let's consider a case where the collision disrupts the Moon, and its material leaves Earth's orbit. If this occurs sufficiently quickly, (faster than the orbital period, i.e., 1 year) then from an orbital dynamics standpoint, the mass is effectively lost instantaneously. The Dutch astrophysicist Adriaan Blaauw, first calculated the relevant physics. Earth's orbit will expand, but by an almost miniscule amount. The reason is that the dynamics of the Earth's orbit around the Sun is dictated by the total mass of the system (i.e. Earth's mass plus the Sun's mass). To lowest order, the new orbit's orbital separation ($a$) will be:

$$ a \approx 1 + \frac{\Delta M}{M} {\rm AU} \approx 1 + \frac{1/81 M_{\oplus}}{M_{\odot}} {\rm AU} \approx 1 + 4\times10^{-8} {\rm AU} $$

So, the orbit will expand by roughly 40 parts in a billion. A quick calculation using Kepler's third law ($a^3\sim P^2$) shows that the length of a year would increase by about 2 seconds. A similar change in the Earth's orbital eccentricity will occur ($\Delta e \sim \frac{\Delta M}{M}$), but since this change is smaller than the precision to which we have measured the Earth's orbital eccentricity, the effect is negligible.


Why is a meteorite impact on the moon visible as white light?

The velocity of the impactor of tens of kilometers per second provides enough energy to heat the impactor and parts of the target to several thousands of Kelvins, so that parts are converted to plasma or to vapor, at least.

According to Planck's law the color at these temperatures is white or bluish.

According to the Stefan Boltzman law the total emitted energy is proportional to the forth power of the absolute temperature. Hence material which isn't heated to thousands of degrees, and may glow yellowish or reddish is too dim to be noticible in comparison.

Since most of the heated ejecta are finely distributed, they cool down rapidly by the emitted radiation, and also after contact with the surface.

After cooling there will certainly be some fine dust, which still has to fall back to Moon's surface, but you won't see this on a video from a distance, since it's cooled down and distributed over a wider area.


Which telescope to choose and some general questions

Macs also need to be collimated, but much more rarely than Newtonians.
 They hold their alignment better (for more time).

For astrophotography you also have to think about the field of view each scope will give with the camera a that you will use and the fact that their mounts are not equatorial (less important if you are into planetary photography).

I see that the Mac has a Vixen dovetail, which means you can later use it on other compatible head mounts or put some other scope with the same dovetail type to the SLT mount.

It might be cheaper to wait more time and buy from the beginning the "right" equipment for your task, than buy something cheaper and then replace its parts.

Cheers!


How much magnification is needed to see the planets of solar system?

You're probably asking the wrong question - which I am going to answer anyway, and after that I am going to answer the question you should have asked instead.

As a general rule, there isn't much point in pushing the magnification above 2x the diameter of the instrument, measured in mm. 3 inch, that's 75mm, that's 150x max. Beyond that limit, even under ideal skies the image is large but blurry.

After that, seeing (or air turbulence) pushes that limit further down. Your aperture is small enough that it almost never suffers from seeing, but larger instruments are often affected. It varies greatly with time, place and season. There are times when a 12" dobsonian, that in theory could do 600x, is clamped down by seeing to 150 ... 180x. There are times when you could take a 20" dobsonian all the way up to 1000x - but that's very, VERY rare, it's the stuff of legends.

Assuming average seeing conditions and instruments of usual size (refractors of 3...4" aperture, reflectors 6" or larger), here are some rules of thumb:

Jupiter is seen best under mid-high magnification. It's rare that more than 200x is beneficial. This is because it's a very low contrast object, and additional magnification comes at the cost of less contrast, which makes things worse.

Saturn works best at high-ish magnification, bit more than Jupiter but maybe not much more. Around 200 ... 250x usually works. It depends on what you do - if you're trying to see the ring divisions, push it a bit higher.

Mars can use the highest magnification that you could generate, given the instrument and the conditions. It's a very small object, contrast is not bad, so crank it all the way up. Most instruments are limited by seeing when observing Mars.

Moon is the same as Mars.

As you can see, magnification is never an issue for you. More magnification will not make it better. In fact, more magnification always means the image is more blurry, not more crisp - it's always a compromise between size and blurriness that decides the optimal magnification.

Don't worry, everyone begins thinking that more is always better. Soon enough, experience shows them what's really going on.

That being said, I believe it's not magnification that's giving you trouble, but the general condition of the optical stack that you're using. These are things that are extremely important, and yet are ignored by many, many amateurs - and the results are not optimal. Here are a few things that you should investigate:

Collimation

Is your scope collimated? In other words, are all optical elements aligned on the same axis? The likely answer is no. It makes a huge difference in the scope's performance, especially for planets. Here's a collimated scope, compared to the same scope out of collimation:

Further information on Thierry Legault's site, which is extremely informative.

A series of articles and documents regarding collimation:

http://www.cloudynights.com/documents/primer.pdf

Gary Seronik: A Beginner’s Guide to Collimation

Gary Seronik: Collimation Tools: What You Need and What You Don’t

Gary Seronik: No-Tools Telescope Collimation

Note: Some telescopes (e.g. pretty much all refractors) do not require collimation; they are collimated from factory and hold collimation pretty well. But most reflectors (SCTs, all newtonians including dobsonians, etc) do require this periodic maintenance.

Thermal equilibrium

At 3" aperture, this is probably not a big issue, but there's no reason why you should add another problem to the existing ones. Your scope should be at the same temperature as the air around it, otherwise its performance decreases. Take it outside 1 hour before you start observing, and that should be enough for you.

Larger telescopes (around 10" ... 12" and larger) should use active ventilation for better cooling (a fan on the back of the mirror). More details here:

Gary Seronik: Beat the Heat: Conquering Newtonian Reflector Thermals — Part 1

Gary Seronik: Beat the Heat: Conquering Newtonian Reflector Thermals — Part 2

In your case, simple passive cooling for 1 hour should be enough, but it's worth reading those articles.

Focal ratio

A 3" scope, at 300mm focal length, that's an f/4 instrument. That's a pretty steep f/ ratio. Most eyepieces will not do well with such a blunt cone of light, and will start to exhibit aberrations that blur the image. Only very expensive eyepieces work well at such low focal ratios - things like TeleVue Ethos, or Explore Scientific 82 degree eyepieces.

Try and keep the planet in the center - most aberrations are lower there. Even very simple eyepieces do better in the middle of the image.

Look at the stars. Are they tiny and round in the center, and large and fuzzy at the edge? Those are aberrations from various sources (eyepiece, primary mirror, etc).

Coma

Of course, at f/4 even the best eyepieces out there cannot do anything about coma - an aberration coming out of any parabolic mirror, which becomes pretty obvious around f/5, very obvious at f/4, and a major problem at f/3. Again, coma is zero in the center of the image, and increases towards the edge.

A coma corrector is used in some cases, such as the TeleVue Paracorr, but I strongly recommend that you DO NOT use one - I suspect your instrument is aberrating in ways that overwhelm coma anyway. Jupiter would not be too blurry even at full f/4 coma at the edge. This paragraph is for informational purposes only.

Coma should become a concern with large telescopes, using high quality optics, with a focal ratio of around f/5 and less. E.g., you have a 20" dob with an f/4 mirror, then you should worry about coma - provided that collimation and so on are taken care of.

Optics quality

An f/4 parabola is not super easy to make at any size. I've made my own optics, and the lower the f/ ratio, the more difficult the process is. Many small, cheap telescopes are made in a hurry, and the difficult focal ratio poses additional problems - as a result, many manufacturers do a poor job. There are even cases where the primary mirror is left spherical, with disastrous results.

This is something you can do nothing about. If the primary mirror is bad, then that's just the way things are. An optician might try to correct it, but it's a difficult process, and quite expensive. I only added this here so you are informed.

This is what I would do in your case:

I would take the scope out 1 hour before observing, every time.

I would try and learn how to collimate the scope. I would try to figure out a few simple collimation techniques, and a few simple tests. I would spend a few days / weeks practicing that. I would keep reading about collimation.

When collimation is at least partially under control, I would learn how to properly focus the scope. Seems simple, but it can be tricky. Use a bright star, and try and make it as small as possible. Use the Moon when it's visible, and try and make it crisp and clear. Do not try this with a miscollimated scope, since it's pointless.

After a few months, when I gain confidence that the scope is in better shape, very well collimated, very well focused, I might try to borrow a better eyepiece from a friend. I said borrow, not buy. Something like a 3 ... 4mm eyepiece, good quality, that would give me a comparison for the existing eyepieces. This ONLY makes sense with a scope that is in perfect collimation, perfect temperature, perfect focus. If an improvement is seen, then get a better eyepiece - but do not spend hundreds of dollars for an expensive eyepiece that will then be used in a tiny cheap scope. Second-hand eyepieces often work exactly as well as new ones.

If you know someone in your area who makes mirrors, see if they agree to put your primary mirror on the Foucault tester, and assess its condition. But beware: the results might be very disappointing. Or not. You kind of never know with these little scopes.

EDIT: After the scope is collimated and so on, you could try to increase magnification by using a 2x barlow with your eyepieces, but do not expect miracles - the image will be bigger, but probably rather "mushy". More magnification is not always better, there's always a trade-off.

Good luck, and clear skies to you!


There are several reasons dark energy cannot be pressure due to the Pauli exclusion principle.  First of all, pressure does not cause expansion of the universe, because pressure is not a force-- pressure gradients are a force (per unit volume), and the cosmological principle precludes them.  Indeed, the pressure that exists everywhere only appears as a force when it is put into gravity, and there it is an attraction if it is positive, as it is for what you mean by Pauli pressure.  So what you'd need for dark energy is negative pressure, appearing in the gravity.

The second reason is that not only is what you mean by Pauli pressure positive, not negative, it is not any different from normal gas pressure, in the sense that it appears as a consequence of kinetic energy density in exactly the same way as any gas pressure does.  This is widely misunderstood, many people think there is some kind of extra pressure that comes from the Pauli exclusion principle, but this is not the case if you simply track the kinetic energy involved.  If you are tracking the kinetic energy, then gas pressure is 2/3 the kinetic energy density for nonrelativistic particles, and 1/3 for relativistic, and that's all-- it's just the same for "Pauli pressure," there's nothing at all different about it when you track kinetic energy density-- which is generally exactly what you want to do when you have an adiabatic expansion such as for the universe.  Hence, in order for us to understand the pressure from neutrinos, we only need to understand their kinetic energy density, which we think we do because it's pretty easy to track given some assumptions about how neutrinos couple to matter and what their mass is.

You might be wondering, if there's nothing different or extra about "Pauli pressure," then when can it be treated as an extra source of pressure?  Only when you are talking about the temperature, rather than the energy.   If for some reason it made sense to think of temperature as the fundamental constraint, rather than energy, then it would make sense to think of the Pauli exclusion principle as adding pressure.  But even then it's kind of a strange way to think about it, because all the Pauli principle does is constrain how the energy gets partitioned, which in turn affects the temperature.  The chain of causality in any gas is energy + partition -> temperature  and energy -> pressure independently of temperature, so it's kind of weird to do what is often done, and pretend that temperature + partition -> pressure.  It's only that last way that makes degeneracy look like an added pressure, and it would only make sense in a system whose temperature is, for some reason, being controlled.  That is rarely the case in an expanding universe, where the temperature is just whatever it needs to be given the energy and the contraints on its partitioning.


What is the projected range of the JWST to be able to detect exoplanet atmospheres?

According to this article, the TRAPPIST-1 system is supposed to be studied by JWST with the potential to detect atmospheres' composition in infrared band. So at last 40 ly (approx. 12 pc) should work. Somewhere else I read there is an assumption to study earth-like exoplanetary atmospheres at distance of 10-20 pc. But it might be better to see it from different angle: the real limit is due to the produced light energy/intensity, not the distance itself. For example, see an article on exoplanet transit spectroscopy.


The reason that some telescopes have long eye-piece tubes is that the focal plane of the objective has been placed far enough outside the main tube assembly to allow erecting prisms or binocular adapters to be fitted, if desired.


How can the equation of state for cosmic strings and domain walls be derived?

My topological defect cosmology is a little rusty, but I'm pretty sure this is how it goes.
Start with the fluid equation,
$$
\dot{\rho} + 3 {\dot{a} \over a} \left( \rho + p \right) = 0,
$$
and the equation of state,
$$
p = w \rho.
$$
Plug the equation of state into the fluid equation, assume a constant $w$, and you'll find
$$
\rho \propto a^{-3(1 + w)}.
$$
Now we'll find $\rho(a)$ for strings and sheets, and read $w$ off of them.
For cosmic strings, $\rho$ is
$$
\rho_{\rm string} = \sum^N_i {\lambda L_i \over V},
$$
where $N$ is the number of strings in our cosmic horizon, $\lambda$ is the linear density of the strings, and $L_i$ is the length of each string.

Here's the key: we've got to assume the length of any cosmic string scales with the expansion of the universe, since they're topological defects. Given that assumption, we can get the dependence of $\rho_{\rm string}$ on $a$:
$$
\rho_{\rm string}(a) \propto {a \over a^3} = a^{-2}.
$$
Thus,
$$
2 = 3(1 + w_{\rm string}),
$$
and $w_{\rm string} = -{1 \over 3}$.

Similarly, for domain walls, $\rho$ is
$$
\rho_{\rm wall} = \sum^N_i {\sigma A_i \over V},
$$
and since $A_i \propto a^2$, we get $w_{\rm wall} = -{2 \over 3}$.

Hope that helps!


This link gives you all the papers that have cited the instrument description paper.
The list of papers shows that it has been used for studying: the centres of AGN, close binary systems, discs around young stars, the atmospheres of AGB stars and interferometric imaging of exoplanets at least.
Here is a paragraph from the instrument description paper

Inspired by the potential of phase-referenced interferometry to zoom in on the black hole in the Galactic center and to probe its physics down to the event horizon (Paumard et al. 2008), we proposed in 2005 a new instrument named GRAVITY as one of the second generation VLTI instruments (Eisenhauer et al. 2008). At its target accuracy and sensitivity, GRAVITY will also map with spectro-differential astrometry the broad line regions of active galactic nuclei (AGN), image circumstellar disks in young stellar objects and see their jets evolve in real time, and detect and characterize exo-planets especially around low mass stars and binaries – in short we will “Observe the Universe in motion” (Eisenhauer et al. 2011).


What is exactly the "longitude of the perigee"

Perigee is the Earth-specific name for periapsis. People use longitude (which is a composite angle rather than an angle) because this solves the problems of circular and equatorial orbits.

The reason you need to use an epoch time to specify the Moon's argument of perigee and longitude of ascending node is because the Moon's orbit about the Earth precesses. Neither the argument of perigee nor the longitude of ascending node is constant. Instead, they are functions of time (and hence, so is longitude of perigee).

In the case of an inclined, non-circular orbit, "longitude" is the sum of the right ascension of the ascending node (which is measured on the fundamental plane of the reference system) and some angle (or angle-like measure such as mean anomaly) on the orbital plane. Thus longitude of periapsis (or longitude of perigee in the case of an object orbiting the Earth) is the sum of the object's right ascension of ascending node and it's argument of periapsis.

The use of "longitude" as orbital elements are for two reasons: To be able take advantage of Hamiltonian mechanics (this was done by Delaunay, resulting in the Delaunay orbital elements), and to address problems related to orbits with very small inclinations and/or very small eccentricities (this was done by Poincaré, resulting in the Poincaré orbital elements). Anomaly and longitude are synonymous in the case of an orbit with zero inclination.

While it is invalid to add two angles on different planes and treat the sum as if it were an angle, it is not necessarily invalid to treat that sum as what it is, a composite angle or dogleg angle, which is exactly what the Poincaré orbital elements do. ("Dogleg angles" aren't angles. Analogous terms include "dwarf planets", which aren't planets, and "red herrings", which oftentimes are neither red nor herrings.)


The article you link to refers to Borsato et al. 2019, which attempted to rectify discrepancies in the measured properties of planets in the Kepler-9 system between transit timing variation measurements and radial velocity measurements. They arrived at $\rho\sim0.31^{+0.05}_{-0.06}\text{ g cm}^{-3}$ for Kepler-9c. However, Borsato et al.'s Figure 10 shows that there are other exoplanets in this mass regime with substantially lower densities, e.g. WASP-107b, which comes in at about $\rho\sim0.19\text{ g cm}^{-3}$:

Even WASP-107b, however, doesn't hold the record for the least dense exoplanet. The three planets in the Kepler-51 system, Kepler-51b, Kepler-51c, and Kepler-51d, may hold that record. Multiple groups (Masuda 2014, Roberts et al.) have found densities of around $\rho\sim0.03\text{ - }0.06\text{ g cm}^{-3}$ for both planets.


Venus transiting behind the Earth’s moon December 7, 2015

Well, depending on where you live, the probability is either 0 or 1, modulo cloudy weather.
From the picture below (from this site), you see that you'll need to be in the US to see the occultation. However, only in the regions outlined in cyan (West Alaska, East Siberia, East Canada, and in Caribbean islands east of the Dominican Republic) will it be during darkness (just before sunrise / after sunset).
But with binoculars (or without, if you have very sharp eyes), you should be able to see it over all of the US.

(cyan=occultation at moonrise/moonset; red dotted=daytime occultation; blue=twilight occultation; white=nighttime occultation)

Miscalculation
After your edit, I see that you are not referring to the probability of seeing the occultation, but the probability of a mis-calculation. I'd say that those odds cannot really be calculated, but it can be said that they are extremely small. No physical theory can ever be proved, but after a sufficient amount of verifications, we usually accept a theory as "true for all practical purposes, until disproved". Many, many factors go into a calculation like this (all the way down to mathematical axioms).
But a calculation like this has to do with celestial mechanics, which is very well understood, and which continuously makes accurate predictions and hence is continuously empirically verified. So the odds you request are definitely much, much smaller than, say, the odds of calculating the weather tomorrow, or the mass of a galaxy cluster.
Uncertainties
Of course there are always uncertainties associated with such calculations, which gives an uncertainty in the position of celestial bodies. But if you take a look at this video, you'll see that the predicted path of Venus is more or less through the middle of the Moon. If the actual path should not cross the disk of the Moon, the calculation would be off by $\sim15$ arcmin, the probability of which is virtually zero.


Would you recommend this telescope to a beginner?

The Celestron SCTs at all apertures have been honed by 30 years of critical amateurs and a market-attentive manufacturer, so you will almost certainly get high-quality images from the OTA, the telescope itself. The Kellner-design eyepieces are another matter. Their apparent field diameter (AFOV, the size of the circle as you look through the eyepiece) is only about 55°, or roughly the diameter of the circle if you close your index finger to your thumb and hold it about 12.5 cm (5 in.) from your eye as you look at the sky. Moreover, these are mass-manufacture products sold as original equipment supplied to many telescope makers. Their quality assurance is not guaranteed. Most people find the star images are generally not as crisp as more modern designs. You would be wise to budget another €200 to €300 for two wider 70°–82° AFOV eyepieces to your budget. A good focal-length suggestion would be 25-30mm and 8-10mm. Quality wide-field eyepieces vastly improve the viewing experience. Objects stay in view longer, and generally are crisper and "deeper" (more pinpointy stars in the field than with mediocre optics). There are numerous telescope equipment retailers which you can locate with search terms such as "1.25 telescope eyepiece 82° 25mm". It is best to patronise a recognized astronomy-specialist than a drop-ship outfits like Amazon. Professional dealers will know most of the problems you encounter and have a vested interest in answering your queries promptly. Since you represent possible future business, they will go out of their way to give you info and service you don't get from sell-&-forget outfits.

The scope's 127mm aperture is a good starter choice. With the exception of M31 the Andromeda Galaxy, galaxies will not be impressive; "faint fuzzies" aptly summarises it. However, nearby star clusters and nebulae more than make up for it; you will find plenty of eye candy just roaming the Milky Way band. If your observing location is Italy, you may find that "rough air" (atmospheric turbulence) degrades image quality much more than any optical problems in the telescope or eyepieces. If you just can't get a good focus on any of the stars in a given field, it means the "seeing" is not great. In bad seeing, the moon will look like the seething shimmer of a distant village on a hot day. Except for really bright inner-city or moonlit skies, light pollution will be less of a problem in telescope fields compared with naked-eye impressions. You are effectively looking at a star cluster through a soda straw with a magnifying glass, so the inherent brightness of the stars will easily override the slight grayness of the background field. Yes, you do see a lot more under truly dark skies, but if you can see the MW visually where you live right now, your telescope will give you years of pleasure before you have to move on to that cabin in the woods. Finally, have a look at the amateur astronomy community www.cloudynights.com. Just about any telescope or astronomy info you need can be answered by the enthusiasts on that website. Moreover, there are four amateur astronomy forum/discussion groups in Italy. Check them out, you have a lot of friends out there you never knew you had.


Meade ETX-90 with computerized mount slews too far during auto-alignment

It seems unlikely that it would have non standard gears.  On the ETX forum that Bonnie refers to there is a thread about a problem similar to yours - its one of the earliest posts.  The problem turned out be grease on an encoder, which when cleaned off, fixed the problem.


Yes, the same equation as is used for other matter.
Marc Postman, Ray Villard and Donna Weaver of the Space Telescope Science Institute posted the results of a CLASH study: "News Release 2011-25":

"This image of galaxy cluster MACS J1206.2-0847 (or MACS 1206 for short) is part of a broad survey with NASA's Hubble Space Telescope.
The distorted shapes in the cluster are distant galaxies from which the light is bent by the gravitational pull of an invisible material called dark matter within the cluster of galaxies. This cluster is an early target in a survey that will allow astronomers to construct the most detailed dark matter maps of more galaxy clusters than ever before.

On the NASA website: "A Clash of Clusters Provides New Clue to Dark Matter":

"... this composite image, made using data from the Hubble Space Telescope Chandra X-Ray Observatory, researchers have discovered clues to increase our knowledge of dark matter. This image is a powerful collision of clusters. These clusters show formations of dark matter, where it separated from the ordinary matter. The different color disparities, along with the interesting evidence for dark matter ...".

Another article with the same photo: "heic0818 — Science Release -- Clash of clusters provides new dark matter clue":

"New Hubble and Chandra observations of the cluster known as MACSJ0025.4-1222 indicate that a titanic collision has separated dark from ordinary matter. This provides independent confirmation of a similar effect detected previously in a target dubbed the Bullet Cluster, showing that the Bullet Cluster is not an anomalous case.".

Wikipedia webpage on CLASH - Cluster Lensing and Supernova survey with Hubble.


Is the expansion of the universe proof of the big bang?

In and of itself, no the expansion of the universe is not proof of the big bang. Other theories could be constructed that would also be explained by the expansion. For example Fred Hoyle and others proposed the Steady State Theory which proposed that new matter was constantly being created thus causing the expansion. A the time there was no other evidence for either theory so either could have been true.

However, when you couple the expansion with other evidence such as the cosmic microwave background radiation discovered in 1964 then it does strengthen the case for the big bang being the cause.


Googling "hyperinflation" mostly returns articles about Zimbabwe and the Weimar Republic, so I'm going to assume that you are referring to what is usually just called (cosmological) inflation.

We are still pretty uncertain about when and for how long inflation occured, but we do however have some sort of constraints. In general, inflation is thought to be caused by some sort of scalar field, usually written as $\phi(\mathbf{r},t)$, and its associated potential energy $V(\phi)$. Various forms of how this field changes as a function of time exist, but a popular one is, or has been at least, the slow-roll field, which makes it behave like a cosmological constant, enabling it to drive an exponential inflation.

Inflation can be initiated by a symmetry break when the unified forces of electromagnetism and the weak and strong forces are separated in the strong and the electroweak force (the "GUT" era, for Grand Unified Theory), at temperatures of $T\sim10^{28}$ K. It will continue until the field has "rolled" from its metastable false vacuum state to a true vacuum. During this time, the Universe expands something like $e^{100}$ times, and its temperature drops by the same factor, i.e. to a freezing temperature of $T\sim10^{-15}$ K. This doesn't last long, though, because the energy of the field is converted to photons and other relativistic particles. This is called reheating.

The factor by which the Universe expands during inflation is essentially set by the minimum factor needed to explain in particular three cosmological "problems", viz. the flatness of the Universe, the causality of the horizon, and the missing monopoles, and the maximum factor allowed in order not to blow stuff too far apart that no structure would have had the time to form before gravitational attraction was too small, so that we wouldn't be here to ask the question. This requires at least 60 e-foldings (i.e. expanding by a factor of $e^{60}$), but could be 100 or maybe more. The factor is discussed e.g. in Galaxy Formation And Evolution by Mo, Bosch & White, Sec. 3.6.2.

Since the size evolution of the Universe during inflation is exponential, the scale factor can be written as
$$a(t) \propto e^{Ht},$$
where $H$ is the Hubble constant during inflation, so for an expansion of, say, $e^{100}$ we have
$$e^{100} = \frac{a(t_{\mathrm{end}})}{a(t_{\mathrm{begin}})}
          = \frac{e^{H t_{\mathrm{end}}}}{e^{H t_{\mathrm{begin}}}}
          = e^{H(t_{\mathrm{end}} - t_{\mathrm{begin}})},$$
or
$$\Delta t \equiv t_{\mathrm{end}} - t_{\mathrm{begin}} = 100/H,$$
where $t_{\mathrm{begin}}$ and $t_{\mathrm{end}}$ marks the time of beginning and end of inflation, respectively.

During inflation, the Hubble constant should not evolve, but have a constant value of $H=(\Lambda/3)^{1/2}$, where $\Lambda$ is a cosmological constant. If inflation started when the GUT era ended, which was around $t_{\mathrm{GUT}} \sim 10^{-36}$ s, the Hubble constant was of the order $H \sim t_{\mathrm{GUT}}^{-1} \sim 10^{36}$ s. Then a hundred e-foldings will take you to $t_{\mathrm{end}} \sim 10^{-34}$ s. Wikipedia states that inflation should last somewhat longer, something like $10^{-33}$ to $10^{-32}$ s. This might be if you do a more careful analysis, or maybe a different model. For instance, H is not exactly constant during inflation, but does evolve a little.

Various observations can constrain the physical parameters that enter the equations, e.g. cosmic microwave background and baryon acoustic oscillations. The latest results from the Planck satellite can be found here. Wikipedia also has a summary of the observational status of inflation.

So, I guess the answer to your question is, "We don't, really, but we have several theories that together with several observational constraints tell us that the duration of inflation should be of this order of magnitude. If it were much shorter, it wouldn't solve the cosmological problems that gave rise to the idea of inflation, and if it were much longer, the Universe would be blown apart".


Plotting Equatorial coordinates to X,Y plane (simulating telescope/camera view)

This isn’t code for a full implementation, but you could do something like this:

Transform your RA, Dec coordinates (which are spherical polar coordinates, just all with the same $r$) to Cartesian $x, y, z$ coordinates.   (The $r$ value you would use here is arbitrary, for example $r = 1$, and should be the same for all stars, not their real distances; you're just projecting them onto a sphere so you can do the transformation.)

Apply a coordinate frame rotation using your field-center RA, Dec coordinates (or perhaps their complements) as the rotation angles, so that the field center direction is the new $z$ axis.

Plot the resulting new $x, y$ coordinates, ignoring $z$.

You might have to do a further rotation of those $x, y$ coordinates to get the plot oriented the way you want, or maybe there’s a clever way to do that with the proper choice of angles in step 2.


What is the minimum mass of a celestial object so that it can have a moon?

Planetoids can have moons and the minimum size is "pretty small". For example 2003 SS84is a small Near-Earth asteroid, with a diameter of 120m and a moon of about 60m in diameter, which orbits at a distance of 270m ever 24 hours.  It probably didn't form by "attracting the moon" but the moon probably formed as a result of impact splitting a "rubble pile" asteroid.  The moon, in this case, is similar in size to the main object.  There is a wide distribution of size ratios from close to 1:1 to very small objects orbiting much larger ones.
However one aspect of your question is different:  you ask for a "round" shape. That makes things harder, since a round shape requires a fairly large size to pull the object into a sphere.  There is probably only one object except for the major planets, with a spherical moon, and that is Pluto with its moon Charon (and even Charon isn't perfectly rounded, but it is close)
So it is quite easy for a planetoid to have a moon.  But it is hard for a planetoid's to be big enough to pull itself into ellipsoidal equilibrium


Do massive bolide entries coincide with meteor showers?

In historical times we do not have enough data to do serious statistics about meteors of the size of the Chelyabinsk meteor.

In Wikipedia's list of meteor air bursts there are only six events within two orders of magnitude of the Chelyabinsk meteor.

$$\begin{array}{lll}
\hline
\text{Date} & \text{Location} & \text{Energy [TJ]}\\
\hline
\text{1490-04-04} & \text{Qingyang district, China} & unknown\\
\text{1908-06-30} & \text{Tunguska event} & 42000–63000\\
\text{1930-08-13} & \text{Curuca River Area, Brazil} & 38–20920\\
\text{1963-08-03} & \text{Prince Edward Islands} & 740–1490\\
\text{2009-10-08} & \text{South Sulawesi, Indonesia} & 130–210\\
\text{2013-02-15} & \text{Chelyabinsk meteor} & 2100\\
\hline
\end{array}$$

Note that for the Tunguska event it is not proven to date that it had been caused by a meteor.


Actually, most of the surface of the Moon is covered with craters like that. The exception are the large dark 'seas', which are in fact basaltic plains. The seas are mostly present on this side of the moon; the far side of the moon is almost entirely composed of craters.

The reason you see a ring near the terminator (the boundary between the illuminated and dark part) is that the shadows make the craters much more visible.

  Due to the angle at which sunlight strikes this portion of the Moon, shadows cast by craters and other geological features are elongated, thereby making such features more apparent to the observer. This phenomenon is similar to the lengthening of shadows on Earth when the Sun is low in the sky.


Is the James Webb telescope going to orbit around the earth?

Your first question - is JWST going to orbit Earth - is a little complicated. It will follow a mission profile that will send it to the Sun-Earth $L_2$ Lagrangian point. It will take the telescope about three months to achieve its orbit in $L_2$. Now, $L_2$ is unstable, and so some station-keeping - essentially, course corrections by thrusters - will be needed. It won't be orbiting Earth, but orbiting the Sun with Earth, in a nearby location.

You're not quite right that JWST will only be observing objects outside the Solar System. Some of the most exciting targets - young, evolving galaxies and exoplanetary atmospheres - lie far beyond. However, the telescope's capabilities for atmospheric analysis (thank you, NIRSpec!), useful for determining the composition of exoplanetary atmospheres, will also be used for observations of the atmospheres of Mars and the giant planets, as well as compositional data for other bodies.

Now, it's not really helpful to send JWST elsewhere in the Solar System, given that most of its observations will be of objects outside the Solar System. Okay, maybe you could get higher-quality data on Mars if you sent it to Mars orbit, but that would be quite costly, and not effective for a spacecraft designed to be used by many scientists for a number of purposes. Keeping it at $L_2$ simplifies the orbital trajectories immensely, although it's still much too far away to repair, unlike Hubble. You also lessen the communication time.

Well, you might say, why not just launch JWST into Earth orbit? We have scientific instruments there; why do we go to the trouble of putting so many things at $L_2$? It turns out that debris high above Low Earth Orbit (LEO) would be too dangerous for the sensitive optics on the telescope. The best instruments are often also the most delicate, and JWST would be damaged if it was truly orbiting Earth, at LEO or elsewhere.


I have a question about the wording in a paper about a planetary system and comets

What they have done is take each of the 580 Jupiter-Family Comets (JFCs) from our solar system (from the JPL Small Body DataBase; set object kind=comets,
Comet Orbit Classes to Jupiter-family Comet) and kept all of the orbital elements the same, except for the semi-major axis which they have scaled by the ratio of the semi-major axis 55 Cnc d (5.74 au) to Jupiter's (5.2 au). So a comet with a semi-major axis of 3 au in our solar system would be $3\times(5.74/5.2)=3.312\,au$ in the 55 Cnc system - this is what they are referring to as comet analogs
All of these 580 comet analogs are being inserted into the 55 Cnc system, For each comet analog inserted, they are creating 100 clones of each analog, varying the mean anomaly. This creates almost 60,000 "comets" which are then integrated along with the 55 Cnc system planets and the outcomes of the comet clones are tracked to see how many survive, how many are ejected from the 55 Cnc and over what timeframe. To illustrate this, they have picked the 3 comet analogs of 2P/Encke, 9P/Tempel 1 and 31P/Schwassmann-Wachmann 2 as examples of JFCs which have high, medium and low eccentricity respectively (their Table 2) and find that comets which have eccentricity tend to preferentially ejected and in a shorter time due to a greater number of encounters with the planets of the 55 Cnc. This occurs much quicker in the 55 Cnc system than their corresponding counterparts in our Solar System.


Why is Saturn the only large planet without any trojans?

They are pulled out of stable orbits by Jupiter.
The details are in https://ui.adsabs.harvard.edu/abs/2012jsrs.conf..225B/abstract
Full text https://syrte.obspm.fr/jsr/journees2011/pdf/baudisch.pdf

The planet Jupiter is solely responsible for the hole of instability for short time integrations
($T < 10^{7}$) compared to the age of the planetary system. On the long term scale this planet also destabilizes the whole region around the Saturnian libration points. If we find in the future Trojans of Saturn,
these Trojans could only be captured asteroids, in orbits in the 1:1 MMR for a short time


Venus's surface temperature is hot everywhere at about 460⁰C. There may some small differences due to latitude, but not nearly as pronounced as on Earth or Mars. The supercritial CO2 that forms the lower atmosphere is a good conductor of heat, keeping surface temperatures very stable.

The article you refer to notes that models of the Venusian atmosphere predict somewhat higher temperatures at high altitude and latitude than can be inferred from measurements made during aerobraking. This tells us that the models of the atmosphere need to be tweaked. It doesn't mean that there is a region of cool temperatures at the poles.


Is Mars brightest from Earth at opposition, perihelion or somewhere in between?

JPL HORIZONS gives these values for Mars as seen from Earth.
r is the Sun-Mars distance in AU, delta is the Earth-Mars distance in AU, and S-O-T is the Sun-Earth-Mars angle in degrees.
I added [] to indicate minimum or maximum values.

Date__(UT)__HR:MN   APmag         r     delta    S-O-T

2018-Jul-23 16:00   -2.75   1.40246   0.38911   172.30
2018-Jul-25 14:00   -2.77   1.40106   0.38730   173.27
2018-Jul-27 12:00  [-2.78]  1.39971   0.38600  [173.51]
2018-Jul-29 10:00  [-2.78]  1.39840   0.38522   172.94
2018-Jul-31 08:00   -2.77   1.39714  [0.38497]  171.73
2018-Aug-02 06:00   -2.75   1.39592   0.38523   170.11

2018-Sep-14 12:00   -1.74   1.38147   0.50640   128.93
2018-Sep-16 12:00   -1.69  [1.38144]  0.51602   127.49
2018-Sep-18 12:00   -1.64   1.38147   0.52588   126.08

Mars is at opposition on Jul 27, closest to Earth on Jul 31, and at perihelion on Sep 16.
Mars is at peak brightness for several days between opposition and closest approach to Earth.
A difference of 0.01 magnitude is imperceptible to the eye and smaller than the uncertainty of most ground-based CCD photometry.
Albedo features and dust storms probably contribute at least that much variation in magnitude anyway.


What would it take to capture a passing asteroid or other celestial body into Earth's orbit?

There have been proposals to capture a small (boulder-sized) asteroid into Earth orbit (more or less) as several other answers have discussed. This needs a lot of planning, a big rocket and a great deal of patience, but all of the asteroids being considered already orbit the Sun, and generally are among the relatively small number of asteroids that periodically come close to Earth.

Capturing an extra-solar object like Oumuamua would be much harder. Firstly we'd need to detect it a long time  (decades, ideally) before it gets anywhere near the Sun, which means spotting something that is far away, very dimly illuminated, and coming from a completely unpredictable direction.  We'd want to get to it as quickly as possible, break a lump off if it was too big to redirect as a whole, and then use something, probably nuclear explosives to change the course of the lump enough for it to come close to one of the giant planets, ideally Jupiter. That encounter would have to be arranged so that the planet's gravity deflects our (now slightly radioactive) lump of rock into a closed orbit around the Sun.

Once that is done, the time pressure is off, and we can use less drastic measures (rockets, light-sails,...), over possibly many decades, to steer our target lump, via whatever gravitational slingshots seem useful until it encounters the Earth and Moon at a fairly low relative velocity. Then a combination of more rockets and the Moon's gravity and maybe aero-braking in the upper atmosphere can be used (over several passes if necessary) to actually capture it.


What distance is considered 'close' for the 9th nearest neighbor of a galaxy?

If galaxies were randomly distributed (a spatial Poisson process), then the probability of having $N$ galaxies inside a radius $r$ sphere is $\Pr[N]=\lambda^N e^{-\lambda} / N!$ where $\lambda = (4\pi /3)\rho r^3$. So the cumulative distribution function of the distance to the $N$th neighbour is $$\Pr[R_N<r]=1-e^{-\alpha r^3}\sum_{n=0}^{N-1} \frac{\alpha^n r^{3n}}{n!}$$ ($\alpha=4\pi\rho/3$ for brevity). Taking the derivative to get the PDF gives $$f(r) = e^{-\alpha r^3}\left [(3\alpha r^2) + 3\sum_{n=1}^{N-2} \frac{\alpha^n r^{3n-1}}{(n-1)!}\right ].$$ While one can try taking the derivative of this to get the mode (or setting the CDF to 1/2 to find the median) this becomes algebraically messy so a numerical solution is likely best. I get that the mode for the 9th neighbour is 2.44 times the mode of the first neighbour.

The problem here is of course that actual galaxies are highly clumped. That will cause a (potentially large) reduction in the median distance.

In fact, most galaxies have satellite galaxies, so unless one has a definition in the survey excluding them the answer will be that the 9th nearest neighbour is very close (in the case of the Milky Way the distance would be about 30 kpc). So any answer will depend on the criteria on inclusion in this particular catalogue.


If I eliminate ec and z, and do this:

    eq = ephem.Equatorial(ra, dec, epoch = d)
    zg = ephem.Galactic(eq)

the crossings I find in gzt.csv agree with Stellarium within a minute of time.

If you're open to another package by the same developer, skyfield.almanac can help with such calculations.
In this case you would implement a function which takes a time argument and returns an integer indicating which side of the galactic equator the zenith is on.
Then you would pass this function as an argument to find_discrete.


How does a supernova occur when a massive star collapses into a black hole?

You are right. A direct collapse of a heavy star into a black hole can lead to a failed supernova without a detectable explosion.
However, a black hole can also form with an SN explosion. One possibility would be the initial formation of a neutron star and bounce off of the outer layers but subsequent fall back of enough material to collapse the neutron star to a BH.
In rapidly rotation stars the supernova can also be fueled by the energy released during the accretion of matter around the newly formed black hole. This would than be a Collapsar.
There is also the possibility for rapidly spinning stars that a magnetar forms during the supernova. Here, the explosion is not primarily driven by the bounce off and the neutrino wind but by the release of magnetic energy and a subsequent spin-down of the neutron star. As the rapid rotation can stabile a NS against its collapse, this spin-down can also lead to a delayed black hole formation. Here a paper about that.

So, SN explosions can become pretty complex and a black hole can form with and without a supernova explosion. All depending on mass, spin and metalicity of the progenitor star (& whether these is a binary companion).


How much heat is generated from waxing and waning of reflected radiation from the Sun?

The article doesn't go into specifics and appears, not to be written by Dr. Evans at all, but I'll pull some quotes.

  When it is completed his work will be published as two scientific
  papers. Both papers are undergoing peer review

and

  He has been summarising his results in a series of blog posts on his
  wife Jo Nova’s blog for climate sceptics.

  He is about half way through his series, with blog post 8, “Applying
  the Stefan-Boltzmann Law to Earth”, published on Friday.

(footnote, I'm guessing in Australia, summarizing and skeptics are spelled differently than in the US, cause that's a copy-paste)

I've only read one of his summaries, and in Mathematics, you have to look at the details, which he's not provided, so, he's basically saying little more than "this is true, here's generally why, I've done the math, please take my word for it, I'll publish the numbers later".   Now Michael Mann is also presenting non peer reviewed work to Paris, so, what's good for the goose is good for the gander I suppose.

Now, I'm just a guy who likes science, but my understanding is that ideas have been presented before they are finished or peer reviewed fairly often.   Einstein did this in fact regarding general relativity and another scientist actually beat him to publishing the theory (though, the other scientist was gracious about it and gave Einstein full credit), so, I don't think it's necessarily bad to present a summary prior to peer review.

I think it is, however, unusual to post summaries of an idea on a blog saying "I figured it out, the majority of research on this subject is wrong".   Dr. Evans is saying he has proof, but he's acting like a junk science blogger.

Dr. Evans "trust me I did the math" claim kind of requires that we look
at his track record, and his track record isn't very strong, though he will say, he's being attacked by the establishment.

According to this site, he hasn't published anything peer-reviewed since the 1980s and he's on a their Climate Denier List.  He's also on Skeptical Science's Climate Misinformers List.

And here's a list of debunked claims he's made (and if this list is accurate, he's not a scientist at all, just a guy on a fishing expedition.   No good scientist would agree with pretty much every counter argument against climate change cause that's not how the scientific method works.  You can disagree with something, that's fine, but to agree with every counter argument - that's silly.  Here's another more detailed explanation of what he's gotten wrong, from 2011.    Evans seems to be in this debate to disagree with it much more than he's in it to do scientific research.

It's pretty much impossible to make a true scientific argument against Evans "proof" without his specifics, and, that kind of proof/disproof can get a little long and complicated, but for now, disproving him is impossible.   But should we listen to him?

It says above he's published 8 blogs related to his recent research.   I'm not going to dig up all 8, but here's the most recent one. Applying the Stefan-Boltzmann Law to Earth.   Now, I'm just a layman, but even I can see problems with his argument here (and he should too given that he's an engineer with a PhD).   The Stefan-Boltzmann law is an approximation.   It's a physical model to calculate radiation into space.

The problem with his approach is, the best way to measure how much heat/energy leaves the earth by radiation is to measure it directly, by satellite.   The amount of energy that radiates from the Earth into space varies with temperature, snow cover, cloud cover, even humidity, and probably 1 or 2 other things I'm overlooking.   If you try to calculate this energy leaving the earth into space by playing with with the Stefan-Boltzmann Law instead of relying on direct measurements, you're allowing yourself a lot of fudge factors and inviting a far greater error than direct measurements would give you.

On the quoted article, let me pull out an example:

  Dr Evans has a theory: solar activity. What he calls “albedo
  modulation”, the waxing and waning of reflected radiation from the
  Sun, is the likely cause of global warming.

OK, so, which is it, solar activity or albedo modulation, cause they're not the same thing.  The first takes place on the sun, the 2nd, the earth.   This paragraph makes no sense to me.

  He predicts global temperatures, which have plateaued, will begin to
  cool significantly, beginning between 2017 and 2021. The cooling will
  be about 0.3C in the 2020s. Some scientists have even forecast a mini
  ice age in the 2030s.

Now, this paragraph is particularly devious.   El Nino's tend to warm the earth, La Nina's cool it.  The effect is temporary and not huge, but enough to cause yearly variation.   A strong El Nino drove the big spike in global temperature for 1998 and we're in an El Nino now (edited my answer, since 2014 they've been talking about entering an El Nino, I gather it's officially started now).

We had more La Nina years than El Nino 2006-2013 with the only small El Nino coinciding with 2010, which set records for temperature.  A lot of the hiatus in warming that is often talked about is related to there being only 1 small El Nino over 7 years.

Predicting 2017 as the time when the cooling will "begin" is devious because that could be around the time the El Nino has ended and the oceans could switch back to a La Nina (which usually follows El Nino).   This will create a temporary cooling for a year or two, which he, no doubt, will take credit for if it happens.   Now, he also predicts 2021 which could go either way and he gives an amount, but that doesn't change the fact that he's making a prediction and hoping the El Nino of 2015 will end and make it prediction look good.

Real global warming or cooling can't be measured in 1 year anyway, unless, maybe, if it's ocean current and the occasional mega-volcano adjusted - then, maybe you can get some measure of warming/cooling based on one year, but it's still only one year.  That's a really really short period to make any predictions on and not something I'd trust very far at all.

and on the "scientists have predicted a mini ice age in 2030", that's not actually true.   There was a study on sun-spots and they predicted that we could see a sun-spot low period around 2030, perhaps similar to the Maunder Minimum that may have caused the mini ice age, but the scientists who predicted that were very clear that they were not predicting a new mini ice age, they said the effect would be smaller than the effect of CO2.

Here's the mini ice age prediction, which a few people made (but not the scientists who did the research).

Here's an article that explains why it isn't true.

So, there's a lot of bad and a handful of false statements in that article you quoted, which, granted, wasn't written by Dr. Evans himself, but still, it's hard for me to take it seriously.

Until he publishes his results, he can't be proved or disproved but based on what I've read, I find it hard to take him seriously.   My hunch is, he's not trying to reach scientists at all, but he's trying to reach his target audience.  Those who question climate change and he gives them a name and an alternate argument that they can stand on.   An argument doesn't have to be correct, it only needs to sound correct and with that, you can usually convince a percentage of people to agree with you.

Not sure how much that helps, but that's my take and I went through and tried to clean up my long answer a bit.   If you'll forgive me, it reminds me of the the old joke.   How can you tell Dr. David Evans is lying?   He's talking or writing.  :-)


What is this gap between hot and cold Jupiters?

Giant planets like Jupiter form through accretion in the outer part of the disk, beyond the frost line where the material is cool enough for volatile icy compounds to remain solid. These planets cannot form closer to their host star, since there is there is not enough matter in the protoplanetary disk at smaller radii.

Due to interactions with other planetesimals and the protoplanetary disk itself, the orbit of the young planets changes. This is planetary migration. Interactions with the protoplanetary disk cause planets to migrate inwards. This migration occurs rather fast, so we don't see many Jupiter mass planets at intermediate distance from their star.

The protoplanetary disk is somewhat doughnut shaped. When a migrating planet reaches the inner edge of the doughnut, the sum of forces exerted on the planet change which halts the migration.

This is why there are two clumps in the diagram, for Jupiter-mass planets. One clump (i.e. cold Jupiters) where these planets form, and one clump (i.e hot Jupiters) where they end up after migrating, but nothing in between.


Supposed SN 1006 was only 4 lightyears away what would the blasts impact be like on earth?

Last week, a paper was posted on arxiv (https://arxiv.org/abs/1707.04253v1) which studies the "The Resilience of Life to Astrophysical Events", in particular it deals with the temperature change of oceans (as boiling oceans would evaporate the water of a planet which seems to be bad for biological life forms...). For a assumed SN explosion of Proxima Centauri the temperature change of Earth oceans would only be 0.1 degree Celsius - not enough to extinct life by boiling the oceans.

(extract from Sloan, Batista and Loeb, 2017)


What time and where on earth is the latest solar noon?

That should be in western China, since all of China uses Beijing's time zone.

Reference: https://en.wikipedia.org/wiki/Time_in_China

--- Edit below ---

In response to the comment by @adrianmcmenamin: I'll leave this as guesswork since I simply don't know the peculiarities of every timezone there is. Here's a back-of-the-envelope calculation for local noon in western China.

The Sun covers 15 degrees in a period of 1 hour. All of China is on Beijing time (UTC + 8 hours). With the above, it's centred on $15 \cdot 8 = 120^{\circ}$ east. As a consistency check, Beijing is at $116^{\circ}$ east. The westernmost part of China is at about $73^\circ$ east (See here). Local noon there is delayed by $(120^\circ - 73^\circ)/(15^\circ/h) \simeq 3h$.

So, local noon is a little later than 3pm. This is discounting daylight savings time, which is not currently observed in China.


It would be easier if you could include an example of the kind of image you mean, but I can guess that you're thinking of those drawings where there's a sine wave drawn in space and suggesting that that's what a wave "looks like".

This plot from EMS 2- Wave Properties of Light shows "crest", "trough" and "rest position" when explaining light waves, and this might be a good example of the problem. All three imply that there is some up-and-down-ness and that's not what a light wave is.

I've seen much worse, floating sine waves with arrows at one end, or hitting people or plants, and those drawings are crazy-wrong! ;-) When we draw those sine waves, it's on a graph or a plot of something versus time or versus space on the x-axis, and the y-axis axis on the plot is pressure for sound or usually electric field for light or electromagnetic radiation.

Light or photons, or sound waves for that matter don't physically go up-and-down like that.

The plot below is somewhat better. The arrows are the amplitudes of the electric and magnetic fields. There's nothing that is actually moving up and down, it's just a mathematical plot.

 Source

  English: Electromagnetic waves can be imagined as a self-propagating transverse oscillating wave of electric and magnetic fields. This diagram shows a plane linearly polarized wave propagating from left to right. The electric field is in a vertical plane and the magnetic field in a horizontal plane.

Found in Transverse wave


Seeing things upside-down is normal with an astronomical telescope.  You have to take that into account when pointing.

Sounds like you have at least four other distinct problems here:

1) You don't know how to align your equatorial mount.

For someone just starting out, you don't have to do this precisely, just enough so it doesn't get in your way too much.  First, adjust the tilt of the polar axis to match your latitude.  If you live at 40 degrees north, you want your polar axis to be tilted 40 degrees up from horizontal.  Then, turn the mount so the upward-tilted part of the polar axis points north.  Again, your first few times out with the telescope you won't need to get this precisely right.  You should find that if the polar axis is correctly aligned, it should point near (although not precisely at) the North Star.  (This assumes you live north of the equator; if you live south of the equator, you'll want the up-tilted part to point south).

2) You don't know how to use an equatorial mount.

It is tricky to use a German equatorial mount (like the one in the picture) on objects near the meridian (the imaginary line connecting the north and south poles in your sky) since as you move the telescope around, it will need to swap places with your counterweight.  This just takes getting used to.

3) You don't know how to align your finderscope.

It is actually easiest to do this during the day.  Select the lowest power eyepiece you have (the one with the biggest lens will be lowest power-- don't use a barlow if they gave you one).  Point the telescope at a distant, but distinctive, terrestrial object on the horizon.  A tall tree sticking up, a church steeple, a smoke stack, something like that.  Do this just by sweeping the telescope along the horizon line while looking through the low-power eyepiece.  Adjust the focus on the eyepiece so distant objects appear sharp.  When you have your target centered in the eyepiece, leave the telescope there (the telescope should stay where you point it if it is balanced properly; if the telescope wants to move around, you might try locking the axes--make sure to unlock them again when you want to move the telescope).  Then look through the finder scope-- you may have to adjust the focus on the finderscope.  Be very gentle with the finderscope, you don't want to move the telescope around.  You should be able to see your target through the finderscope.  Loosening and tightening the screws on the finder mount will change the aim of the finder slightly.  You want to adjust the finder aim until the finder crosshairs are on your target.  Then doublecheck that the target is still centered in the eyepiece.

4) You don't know how to use your finderscope.

When the finderscope is properly aligned, you can use it to point the telescope.  Center the target you want to look at in the finder cross hairs.  It should appear somewhere in the view of the telescope through the low power eyepiece.  Again, it is easiest to practice this during the day on distant terrestrial objects on the horizon, before you start looking for the moon and planets at night.


How to differentiate elements that have same spectral lines in a star?

If you only had that single line, you would be unable to differentiate them.  This is especially true of redshifted objects where we couldn't tell the actual wavelength of an isolated peak.  But a good spectrograph will give you information over a range of wavelengths.

Each element has multiple lines.  It is the overall pattern of relative strengths that is definitive.

As an example, the stronger lines for a few elements are shown below:

(From https://cnx.org/contents/LnN76Opl@10.1:H5KhIDcK@3/Spectroscopy-in-Astronomy)

You can see there's a cyan line that is similar in both hydrogen and calcium.  But the other lines are so different, you could tell if one or both were present, even if they were overlaid on a single chart.


What is the equation of state for a relativistic fluid/gas?

Your equations seem correct. Note, you can also get $v$ without having to differentiate, from $E=\gamma mc^2$ and $p=\gamma mv$.

Here are some notes on relativistic fluids as related to stellar interiors, including derivation of these equations and their solutions in the non- and ultra-relavistic limits http://www.jb.man.ac.uk/~smao/starHtml/equationState.pdf

Correct, the integrals cannot be solved analytically (only in the two limits) so there is no general analytic formula


What is the farthest we can see into the universe?

It depends. In some areas, far-away stuff is obscured by obstacles in the foreground. In other places, we can see almost all the way to the earliest (and furthest-away) galaxies, like in the Hubble Deep Field image:

The very earliest objects have so much redshift they disappear out of view for visible-light telescopes. That's where the JWST comes in: because it's an infrared telescope, it can capture objects that have too much redshift to be visible in a visible-light telescope.

Advances help us see more in other ways too. Gaia data revealed some small nearby galaxies which are mostly obscured by our own galaxy.


Find constellation over Earth coordinates on a specific date-time

Ahh got it. The coordinates are given as latitude, longitude.

Fortunately, the longitude hardly matters (depending what precision you need), but the latitude does. A star at the zenith will have a declination equal to the latitude on Earth.

As far as Right Ascension goes, well around the March equinox then stars with 12 hours in Right Ascension will reach their highest point (culmination) at "local midnight".

So one would expect roughly speaking for stars at around 21 hours, +40 would be overhead at 9h40m later.

Checking out Stellarium for that location and time it seems that this is pretty much spot on and that Cygnus (the Swan) is overhead, with the bright star Deneb very close to being at Zenith.


How many generations of stars can be formed in the Stelliferous Era?

One can make a theoretical upper bound by considering the most short-lived star possible $\tau_{short}$, and a large supply of initial hydrogen $M_H$. Then one could calculate the fraction hydrogen that is recycled $r$ after the star ends (with a supernova), and get a total number of generations as $ \log (M_{star}/M_H)/\log(r)$. If one uses the solar-mass $r\approx 0.5$ and a galaxy mass of gas $M_H=10^{10}M_\odot$ one could get 33 generations.
One could of course try tweaking this by using heavier stars that recycle better yet are not affected by the growing metallicity of the medium, but it would be rather wasted effort: it is not a realistic model.
The overlap between generations is growing: stars are made today from mixed gas recycled from all previous generations. They also last very different lengths of time depending on mass, so the gas reservoir does not conveniently fill up for the next generation to be created, but rather is topped up while star formation is ongoing. So speaking of generations becomes less and less useful, and instead it makes sense to mainly speak of when a star formed or how much metals it had at the start due to a highly evolved gas reservoir.
One can simulate star formation and recycling by using models of each, but it quickly turns rather complex and worse, dependent on assumptions about how much intra-cluster gas will fall in the future, something that is weakly constrained. Current trends in star formation suggest that we are close to peak star number and that peak star formation was in the past, so relatively few stars will form in the future until the gas levels go below a critical surface density and shuts off (see section D in this paper). But if intra-cluster gas cools and rains down slowly that may prolong things a fair bit. Still, in about $10^{14}$ years normal star formation will have ceased.


Is Earth's climate significantly different when the one solstice occurs at aphelion rather than the other?

Seasons on the Earth are driven primarily by axial tilt. I'm writing this answer in late August, when the Earth is a bit past aphelion. That event currently occurs in early July, a couple of weeks after the June solstice. Yet it's summer in the northern hemisphere. Precession have a secondary effect on the seasons. The current situation (perihelion close to the December solstice, aphelion close to the July solstice) was reversed ~12000 years ago when perihelion coincided with the June solstice. This made Northern Hemisphere summers hot but short, winters cold and long.

The brutally hot Northern hemisphere summers back then resulted in the end of the last glaciation (colloquially, ice age). Glaciations start when summertime temperatures at far northern latitudes (65° north is the standard proxy) are so mild that snow from the previous winter doesn't completely melt. Glaciations end when summertime temperatures in the Northern Hemisphere become hot enough to melt the ice that has accumulated over forty thousand to a hundred thousand years.

The reason we aren't we seeing a glaciation start now (conditions are right from the perspective of precession) is that the Earth's eccentricity is rather low. Precession, eccentricity, and other factors have to line up to trigger a glaciation.


Several options are possible:

Don't interpolate at all. Just assume the planet stays fixed until the next update. The planet Earth moves $10^{-7}$ degrees in 30milliseconds. In the vastness of space that is a rounding error.

Given two objects, their position and velocity and acceleration, and the values of their Keplerian elements at times t0 and t1, you want to estimate their position, velocity and acceleration at any time in between.

You can use simple linear interpolation of their position. This has the advantage of being very quick, but it assumes a constant velocity. Over short time-spans this is a reasonable assumption. For most orbiting bodies this approximation should be accurate enough for differences in time of many minutes, as the changes in velocity are relatively small.

You can assume constant acceleration and interpolate using the SUVAT equation. This does very well if the objects don't move too far. For example it can model the flight of a ball in the Earth's gravitational field very well, giving a parabolic shape. This should give a fairly good approximation of the position over hours, or even days.

You can interpolate the orbital elements, linearly, and solve Kepler's equation to get the position of the body at any given time. For most bodies the Orbital elements change only slowly with time (except for the Anomaly) That is the shape of the orbit stays nearly the same, only the position on the orbit changes. Approximating the orbit of a planet by interpolation of its elements should give a good approximation for several hundred years of orbital time. However the process of solving the equation is slow.

Finally you can numerically integrate the n-body problem for the solar system. Depending on the time step you choose this can be accurate for an indefinite amount of time

So these can be summarised as:

Constant position
Constant velocity
Constant acceleration
Variable acceleration following an inverse square law.
Variable acceleration n-body graviational field.

For a game engine, speed is good. If the gaps in time are on the order of milliseconds (of real time) then simply not interpolating should be undetectable.


How long do thunderstorms on Venus last?

Independent observation led to the conclusion that lightning is prevalent on Venus. Electromagnetic signals detected by the 4 Venera landers explained the main cause of lightning on Venus. Venera 9 spectrometer detected a lightning stroke. The Pioneer plasma wave instrument detected waves which explained lightning might be at low altitude. VLF observation suggested that lightning is a intra-cloud  phenomenon which is more common in the afternoon and evening time and the occurrence rate is more as compared to Earth. Venera 11, 12, 13 14 were equipped to detected magnetic signature(acoustic data were also obtained from V11 and V12).
Lightning on Venus occur from strong convective cumulus cloud and occasionally from dust storms and volcanic clouds. The clouds may vary in altitude from few km to 30 km . Lightning on Venus occur due to two distinct forms: cloud-to-ground discharge and intra-cloud discharge. Such discharge occur when electric field exceeds from 106 V/m in wet air to 3 $\times$ 106 V/m in dry air. Cloud to ground discharge carry current from 200-300 A which increase every 50-100m, travelling at a velocity of 1.5 $\times$ 106 m/s which followed with a return stroke with velocity 0.6 $\times$ 108 m/s carrying 10-20 kA which decays in about 20-50 us. There are 3 or more stroke in one flash separated by 40 ms and lasts 0.2 s.
Discharges within clouds are 2 to 4 times more frequent than that of cloud-to-ground discharge and carry currents upto 4kA but they generate less energy(4.5 $\times$ 108 J in cloud-to-ground discharge). Number of discharges on Venus is more than that of Earth. Over 7000 discharges occurs on Venus per day  You can check the details and observation in ref.1 and ref.3

Fig. 1 Venusian atmosphere where lightning is most likely to occur

Fig.2 Magnetic field due to intra-cloud discharge (altitude 50 km)
References

Venus lightning by Russell, C. T., Space Science Reviews, vol. 55, Jan-Feb 1991, p. 317-356. (link and the author's paper here)
Russell, C.T. et.al. (2011). Venus lightning: Comparison with terrestrial lightning. Planetary and Space Science - PLANET SPACE SCI. 59. 965-973. 10.1016/j.pss.2010.02.010.
Lorenz, R.D. Lightning detection on Venus: a critical review. Prog Earth Planet Sci 5, 34 (2018). https://doi.org/10.1186/s40645-018-0181-x


The material at the surface of a white dwarf is not degenerate.
The "visible" surface is defined as where the optical depth exceeds some threshold and this will occur at a low enough density that even at a few hundred kelvin, the ratio of the Fermi energy to the thermal energy is too low for significant degeneracy.

In addition, at these temperatures, the electrons attached to atomic hydrogen and helium would not be ionised and the material would mostly be in atomic and molecular states.

In terms of appearance a reasonable comparison would be with that of the very cool T- and Y-dwarfs; although some of the more complex chemistry that occurs in those atmospheres would not occur in very cool white dwarfs because the trace quantities of heavier elements would probably have sunk out of the atmosphere.

Edit: I note that there probably are white dwarfs cooler than 3000K. This is the upper limit to the temperature of an invisible companion of to the pulsar PSR 2222-0137. Kaplan et al. (2014) say that this most likely a massive ($\sim 1 M_{\odot}$) old white dwarf that has cooled so much that they cannot detect it at visible wavelengths - so probably you would call this a "black dwarf", though the term is not used in the scientific literature.


Are we really star-stuff from the interior of collapsing stars?

The straightforward answer is, "Yes, we are made of star stuff."

Some of it will be from the interior of collapsing stars, some will be from supernovas, some from normal everyday fusion, and some from other processes.

The answers from @HDE226868 and @RobJeffries on this question on where heavier elements come from gives good background, including this nugget:

  The split between r-process and s-process production of heavier than iron (peak) elements is about 50:50. ie They weren't mainly made in supernovae, which is a frequent, incorrect claim.

but of most importance is Rob's final point:

  The relative contributions of various sites to the r-process remains an unsettled matter. You could also read my answers on this topic in Physics Stack Exchange.

On following Rob's links I think this provides you with an excellent overall answer (and relative percentages)

  A more up-to-date visualisation of what goes on (produced by Jennifer Johnson) and which attempts to identify the sites (as a percentage) for each chemical element is shown below. It should be stressed that the details are still subject to a lot of model-dependent uncertainty.

Looking at C and N - the majority seems to be from dying low mass stars, and Ca and Fe are from exploding stars, which indicates that Carl is not far off the mark.


How much time passes in the outside universe when falling into a black hole?

You are essentially asking the following: if someone falls from the Earth from some way beyond the event horizon of a black hole, how long after they have left can an observer on Earth still signal to them with a light beam?

The answer of course depends on exactly how far the Earth is from the black hole. It is also often forgotten that it is not just light reaching them before the falling observer reaches the event horizon, but there is also a little bit more to add in the (proper) time it takes the falling observer to travel from the event horizon to the singularity.

The answer in all cases is that there is a finite time available, so no, the infalling person does not look back and see the cosmic aeons flying by.

The handwaving reason is that from the perspective of Schwarzschild coordinate time both the falling person and light beams approach the event horizon along an exponential asymptote. The light beams are always travelling faster, so their $dt/dr$ is always shallower and hence it is possible for them to catch the falling observer as long as the delay in sending the light beam does not exceed a finite critical value. Or to put it another way; you can contruct a past light cone from any point on the event horizon (or the space-like singularity) and it encompasses only a finite time anywhere outside the black hole.

The full, very gritty, details were worked out (for a Schwarzschild black hole) at https://physics.stackexchange.com/a/396157/43351

For the case of the person free-falling from a position $r_0$, then the span of time seen by the falling person when looking back towards $r_0$, before they hit the singularity, is given by
$$ c(\Delta t)_{\rm singularity} = r_s \ln \left(\frac{r_s}{r_0-r_s}\right) + \pi r_s\left(\frac{r_0}{r_s} -1\right)^{1/2}\left(1 + \frac{r_0}{2r_s}\right) -r_0,$$
where $r_s$ is the Schwarzschild radius.

Note that as $r_0$ gets bigger then of course $\Delta t$ get bigger and approaches $\Delta t \simeq r_s \pi(r_0/r_s)^{3/2}/2c$, which is just the freefall proper time (i.e. measured on a clock carried by the falling object) for an object to fall into a black hole. i.e. You could see that length of time into Earth's future, but only because you had spent that long falling into the black hole !

Edit: To address Jonathan's question in comments. Take a 10 solar mass BH with $r_s=30$ km. If it takes 1 hour (proper time) to "free fall" radially (formulae above are for radial infall), then $r_0 = 2.42\times 10^{9}$ m. The formula above then shows that $\Delta t$ is 1 hour to three sig figs. The reason for this is that on such a trajectory, the falling object spends a negligible fraction of its infall time anywhere near the black hole, so the GR effects are negligible.


Will an observer falling into a black hole be able to witness any future events in the universe outside the black hole?

The confusion about finite or infinite time dilation at the Schwarzschild radius began since Karl Schwarzschild in 1915 used a space-time coordinate system where coordinate time went to +∞ when a falling object approached the Schwarzschild radius. But t → +∞ for a world line that approaches the horizon is a thing different from t → +∞ far from the black hole. An object that is about to cross the horizon is in no position to synchronize clocks with an outside world.

No, completely opposite situation to the OP’s expectations. If a falling observer will transmit a powerful beam of light out of the hole, then photons will reach increasingly far future of the exterior universe. But the future seen by the falling observer will be very limited; see http://jila.colorado.edu/~ajsh/insidebh/penrose_schw.html for graphic depiction.


Betelgeuse in false colors, is this common?

This is a partial and perhaps temporary answer only, posted in order to get to the bottom of this. I think that the comments are correct, you're probably seeing an artifact of image manipulation and the original data is just a 2D intensity map.

I downloaded the original image at https://cdn.eso.org/images/screen/potw1726a.jpg, converted back to .png and analyzed in Python using  matplotlib and numpy. Right now I don't see these mysterious rings, but if you can provide pixel coordinates from the original image and explain more how you made the rings visible, that would be great!

Image breakdown into three color channels:

Line scans in x and y through center and maximum:

Same but multiplied by 256 (8 bits per channel) and plotted on a log scale:

Scatter plot of G and B versus R show that these are correlated. We're probably just looking at artificial coloring from a table, not different channels of data. It could be a similar thing going on here as well.

import numpy as np
import matplotlib.pyplot as plt

fname = 'Betelgeuse in false colors.png'

img = plt.imread(fname)
print(img.shape)

if True:
    plt.figure()
    plt.subplot(2, 2, 1)
    plt.imshow(img)
    plt.title('RGB')
    for i , char in enumerate('RGB'):
        plt.subplot(2, 2, i+2)
        plt.imshow(img[..., i], cmap='gray', vmin=0, vmax=1)
        plt.title(char + ' channel', fontsize=16)
    plt.show()

rgbx_cen = img[640]
rgbx_570 = img[570]

rgby_cen = img[:, 640]
rgby_570 = img[:, 520]

things = rgbx_cen, rgbx_570, rgby_cen, rgby_570
names  = 'rgbx_cen', 'rgbx_570', 'rgby_cen', 'rgby_570'

if True:
    plt.figure()
    for i, (thing, name) in enumerate(zip(things, names)):
        plt.subplot(2, 2, i+1)
        r, g, b = thing.T[:3]
        print(r.shape, g.shape, b.shape)
        plt.plot(r, '-r')
        plt.plot(g, '-g')
        plt.plot(b, '-b')
        plt.title(name, fontsize=16)
    plt.show()

if True:
    plt.figure()
    for i, (thing, name) in enumerate(zip(things, names)):
        plt.subplot(2, 2, i+1)
        r, g, b = thing.T[:3]
        print(r.shape, g.shape, b.shape)
        plt.plot(255*r, '-r')
        plt.plot(255*g, '-g')
        plt.plot(255*b, '-b')
        plt.yscale('log')
        plt.title(name, fontsize=16)
    plt.suptitle('scaled 0 to 255', fontsize=16)
    plt.show()

if True:
    r, g, b = [x.flatten() for x in np.moveaxis(img, 2, 0)[:3, ::8, ::8].copy()]
    plt.figure()
    plt.subplot(1, 2, 1)
    plt.plot(r, g)
    plt.title('green vs. red')
    plt.subplot(1, 2, 2)
    plt.plot(r, b)
    plt.title('blue vs. red')
    plt.show()