All telescopes have in common that they gather and focus light from far away objects. They use a primary opical element, such as a concave mirror or a (planar- or bi-)convex lense (or lense system), and they use an eyepiece with another lense system (for viewing) or a camera in their primary focus.

A refractor telescope does not sharpen the image per se. The convex lense concentrates the light rays, not unlike a magnifying glass. To actually focus the enlarged image on your retina, you need an eyepiece, which is another bi-convex lense (in its most simple form). This will re-align the light rays after they passed through the primary focus. See this image for a visual explanation:

Source: Wikipedia

The above image also explains why the image of a refracting telescope appears upside down. You don't need (or want!) any prisms in this kind of setup.

On the other hand, a reflector telescope uses a concave mirror plus an eyepiece. There are different configurations, but one of the most simple and most common ones is the Newtonian telescope:

Source: Wikipedia

So instead of refraction of light by a lense, we use the reflection of light on a mirror, to enlarge the image. Focusing on the retina is again done by an eyepiece in the same way as with the refracting telescope.

The advantage of refracting telescopes is that there is no obstruction in the optical path inside the telescope. This is not the case with reflector telescopes. They usually have a secondary mirror in the middle of the optical paths, hence reducing light gathering performance.

On the other hand, reflector telescopes are often much lighter, and cheaper to assemble. Also, very compact models of reflector telescopes can be built.

Also, simple refractor telescopes will produce colourful fringes on object edges, called chromatic abberation, which is due to the glass used in the lenses. This can be compensated by multiple lenses, but this will make the refractor even heavier and more expensive.


Is the Ordovician meteor event detectable in the lunar cratering record?

The very, very short answer to your question is: "Not that's detectable by any current methods or data that we have."
The longer answer has to do with how we know how old anything is on the Moon or other planetary surfaces.  What we do is we use craters themselves to date a surface: The more craters, the older the surface because it's been around longer to accumulate those craters.  We can tie to those to an absolute age by saying that X rock sample collected by Apollo or Luna or Chang'e is Y years old (dated via radiometric methods) and it is from a surface that has Z craters.  Do this a lot and you can build up a function that relates the number of craters to an age.
The problem comes in using this to date any specific craters to try to find a non-uniform function (e.g., an age spike):  You have already assumed a crater chronology function that does NOT have that spike, so using it to find a spike becomes a circular argument that is very difficult to get out of.  There are some proposals on how to do that, but they're based on models and not actual data (such as, the large objects that form large craters would show those spikes, but the small objects that form small craters are a steady "clock" of impacts).
Right now, the lunar crater chronology has roughly a dozen points on it, so we don't have the resolution to say anything about a spike ≈468 million years ago.  It is certainly possible (likely, even) that the lunar crater record would show that spike if we had the resolution to measure it by dating many more craters through methods other than superposed crater measurements, and, indeed, it is generally assumed that the lunar crater chronology really should have lots of little spikes in it when asteroids break up / asteroid families form, but again we just don't have the data to really measure that ... yet.


Does the Hubble constant depend on redshift?

Yes, definitely.

The Hubble constant describes the expansion rate of the Universe, and the expansion may, in turn, may be decelerated by "regular" matter/energy, and accelerated by dark energy.

It's more or less the norm to use the term Hubble constant $H_0$ for the value today, and Hubble parameter $H(t)$ or $H(a)$ for the value at a time $t$ or, equivalently, a scale factor $a = 1/(1+z)$, where $z$ is the redshift.

The value is given by the Friedmann equation:
$$
\frac{H^2(a)}{H_0^2} =
\frac{\Omega_\mathrm{r}}{a^4} +
\frac{\Omega_\mathrm{M}}{a^3} +
\frac{\Omega_k}{a^2} +
      \Omega_\Lambda,
$$
where
$\{
\Omega_\mathrm{r},
\Omega_\mathrm{M},
\Omega_k,
\Omega_\Lambda
\} \simeq
\{
10^{-3},0.3,0,0.7
\}
$
are the fractional energy densities in radiation, matter, curvature, and dark energy, respectively.

For instance, you can solve the above equation at $z=0.1$ and find that the expansion rate was 5% higher than today.

Since everything but dark energy dilutes with increasing $a$, $H(a)$ will asymptotically converge to a value $H_0\sqrt{\Omega_\Lambda} \simeq 56\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$.

The figure below shows the evolution of the Hubble parameter with time:

As noted by KenG, the fact that $H$ decreases with time may seem at odds with the accelerated expansion of the Universe. But $H$ describes how fast a point in space at a given distance recedes. Later, that point will be farther away, and so will recede faster. From the definition of the Hubble parameter, $H\equiv\dot{a}/a$, multiplying by the scale factor shows the acceleration $da/dt$:


Are we moving ever closer to the center of our Galaxy due to a super massive black hole?

That's not really how gravity works.

When a galaxy forms, all the dust that falls into it retains angular momentum round the centre (and this pattern repeats locally round stars, and round planets round those stars etc) which leads to the relatively stable structures we see today with spiral arms etc.

Collisions and near collisions can change the direction of stuff orbiting a centre, but we are quite far out, so while we could change orbits we would be as likely to move wider than closer in the short term.

Over the longer term, it's a balance between the gradual loss of angular momentum of the Milky Way as a whole due to drag, gravity waves etc. and the slow growth of the black hole at the centre.


As has already been mentioned, stars within a galaxy do not generally expand. The stars in a galaxy are gravitationally bound together.

But as you observe more and more distant galaxies, you see a general trend that distant galaxies are moving away from us at a speed that's proportional to their distance.  (There are some small variations due to the random motion of individual galaxies; for example, the Andromeda galaxy is approaching us, and will collide with the Milky Way in about 4 billion years.)

This does not indicate that everything is expanding from some central point.  An observer in a galaxy a billion light-years away would see the same thing, all other galaxies receding at a speed proportional to the distance from the observer.  This can be interpreted as space itself expanding.  Going into more detail would require General Relativity, which I'm not competent to explain.

Your question is about the rate of expansion.  The answer to that is what's called the Hubble Constant, which describes the ratio between a galaxy's speed of recession and its distance from us.  The current best estimate of this ratio is 67 kilometers per second per megaparsec.  A parsec is about 3.26 light-years; a galaxy one million parsecs away will be moving away from us at about 67 kilometers per second.

Source: https://en.wikipedia.org/wiki/Hubble%27s_law#Determining_the_Hubble_constant


Why isn't the day backwards between leap years?

Leap seconds keep the clock in sync with solar time of day. Leap years keep the calendar in sync with the seasons. The two (leap seconds and leap years) have nothing to do with one another. We have leap seconds because a day is now a tiny bit longer than 86400 seconds. We have leap years because a year (tropical year) is about 365.24219 days long.

If we didn't have leap years but did have leap seconds, solar noon and noon by the clock will continue to occur at more or less the same time (ignoring cyclical changes such as the equation of time and daylight saving time). The calendar would become out of sync, becoming almost a month off from the seasons in just a century. Adding an extra day every four years doesn't quite do the trick; this is why 2100, for example, will not be a leap year.


How does V838 Monocerotis look today? Any subsequent light echo images, or is the party over?

I'm not an observational expert, but here's what I found.

I found this object (V838 Monocerotis) on the HST data archive.

https://archive.stsci.edu/hst/search.php

I input the 'Target Name' as "V838 Monocerotis", clicked all 'Imagers', 'Spectrographs', and 'Other,' and searched it.

(You can also do the exactly the same thing I did because this site is opened for everyone.)

And there are 199 images of this object. And the last observation was made in 2012. Unfortunately, I couldn't see any echoes in the picture taken in 2012. So I looked from the beginning. And I found this!

This picture was taken from 2006-09-09 20:04:57 to 2006-09-10 01:14:57. Dataset number is J9SF01020. You can see that the shape of the echo. There are other pictures in the archive, but I couldn't see significant echo shape anymore after that.

I'm sorry to say that the party is over, the echo was faded away.

I also love this picture, it means a lot to me because this picture made me study Astronomy. Thanks for making me remind this memory.

P.S.: If you want other pictures, especially color pictures, check out this link too.

http://hla.stsci.edu/hlaview.html#Images|filterText%3D%24filterTypes%3D|query_string=V838%20Monocerotis&posfilename=&poslocalname=&posfilecount=&listdelimiter=whitespace&listformat=degrees&RA=106.020093&Dec=-3.847399&Radius=0.200000&inst-control=all&inst=ACS&inst=ACSGrism&inst=WFC3&inst=WFPC2&inst=NICMOS&inst=NICGRISM&inst=COS&inst=WFPC2-PC&inst=STIS&inst=FOS&inst=GHRS&imagetype=best&prop_id=&spectral_elt=&proprietary=both&preview=1&output_size=256&cutout_size=12.8|ra=&dec=&sr=&level=&image=&inst=ACS%2CACSGrism%2CWFC3%2CWFPC2%2CNICMOS%2CNICGRISM%2CCOS%2CWFPC2-PC%2CSTIS%2CFOS%2CGHRS&ds=


Are we so sure global warming is a result of humans burning fossil fuels?

We know a lot about the sun. It isn't the main cause of global warming, as we can monitor exactly how much energy it is producing. The sun's output can fluctuate though, and we have yet to measure how badly this affects the climate.

So what is the main cause of global warming?

The main cause of global warming is emissions of C02 (carbon dioxide) into the atmosphere. C02 is odorless, colourless and non-toxic, however, it is the main cause of global warming. C02 is a greenhouse gas and  can stay in our atmosphere for about 20 years.

C02 causes something called the greenhouse effect. The greenhouse effect is the idea that long wave energy which is meant to go back to space, doesn't go back to space, and is instead captured by greenhouse gases such as C02. This causes lots of heat that should go off into space to be trapped inside the earth's atmosphere causing your temperature to rise.

Are there any other greenhouse gases?

Other greenhouse gases do exist, carbon dioxide is just the main one that causes the global warming. Other greenhouse gases include water vapor, methane, nitrous oxide and ozone.

Does global warming happen only on Earth?

Global warming happens all over the  solar system, planets like venus have many greenhouses gases in their atmosphere too, infact, on venus greenhouse gases alone make the planet 465 degrees celsius hot, which is way over the boiling point of water - 100 degrees celsius.

Do I disagree?

Yes, as there is too little evidence to prove that the sun fluctuates enough energy to gradually increase our climate dramatically, and we have also proven the effects of greenhouse gases in a planets atmosphere.


If you want just the "shape" you need one number: the eccentricty.
The shape of an ellipse is determined by the eccentricty. It is a measure of how "uncircular" an ellipse is.  Conventionally this is written as $e$
If the ellipse is centred at the origin, and its long axis ends at (1,0), then the focus of the ellipse is at (e,0)
A very simple geogebra construction that does just this: https://www.geogebra.org/classic/ydp9cthr
If you want the "shape" and the "size" of the ellipse, you need a second number, and conventionally this is half the length of the long axis of the ellipse, this is written as $a$.
If the half length of the short axis is $b$, then these are related by the formula
$$e^2 = \frac{a^2-b^2}{a^2}$$
So if you have two, you can work out the third, and plot the ellipse at a given size using parametric equations
$$\begin{align}
x &= a \cos(t)\\
y &= b \sin(t)\\
\end{align}$$
Now, this draws the ellipse in 2d space and oriented along the x-axis.  If you want the ellipse drawn in 3d space and at an arbitrary alignment, you need three more pieces of information, usually these are the "argument of periapsis" (the angle of long axis of the eclipse) "inclination" (the angle of the plane containing the eclipse to the x-y plane) and the "longitude of the ascending node" (the bearing of the point at which the object crosses the x-y plane)
A sixth number is required to define the position of the planet on the ellipse at a given time. (usually given as the "mean anomaly" or "true anomaly")
These six numbers are called the orbital elements
All this assumes Kepler's laws.  Real orbits are perturbed by the gravity of other planets (and slightly deviate from Kepler due to relativity). The real paths of the planets aren't actually periodic, but Kepler's laws are a very good model.


How can I work out longitude and latitude through my telescope?

I used to have this exact telescope. It took a lot of practise, but I was eventually able to reliably find most targets I was after. I used this scope for about 2 years learning my way around the night sky before moving on to another scope and mount. You should practise finding bright targets before moving on to dimmer ones, just to get used to the scope. However, you can use the setting circles on the mount to help you too. The setting circles are the 2 silver disks with all the numbers on the head of the mount. The Declination one is just underneath where the telescope is fitted (upper red circle in the picture). The Right Ascension is on the other rotational point of the mount head (lower red circle in the picture)

First, you need to reliably polar align the scope, which will make nailing down other objects easier. This may take some time so for a "roundabout" method, you can just point the mount towards polaris and it'll be good enough to keep objects centered when adjusting the RA knob, but I'll still go through the "drift" method for a more reliable alignment.
Drift aligning on a manual EQ mount can take a while, but it starts with a rough polar alignment by eye. Align the mount with Polaris as best you can. Now, move the telescope and find a bright star to the south. Center the star in your eyepiece then get the highest power eyepiece you have (will probably be the 9mm one in your accessory kit) and move the star so it is sitting nearly top center of the eyepiece. Wait 2 minutes or so, then move the RA knob slowly to bring the star back into view. If the star has drifted up, then your alignment is too far west. Give it a small adjustment to the east. If it drifts down, you are too far east. Adjust it westwards. Repeat until you have minimal drift.
Next, find a bright star in the east. Use the same method of centering a star then moving it to the top of the eyepiece. If this start drifts up, your latitude setting is too high, so move it down slightly. If it drifts down, your latitude settign is too low, so lift it up slightly. Repeat till minimal drift and you have a pretty good polar alignment!
Now to actually find a target!
Find yourself a starting star, which is the brightest star in the nearby area of your eventual target. Center it in the eyepiece and look up the co-ordinates on Google or any astronomy app you may have. Adjust the setting circles so that the 'pointers' point to the corresponding values in the RA and DEC axis. Do another quick check that the star is centered in the eyepiece and the setting circles are correct. Now find the co-ordinates of your target. Loosen the declination adjustment on your mount and move the telescope s that the pointer is pointing at the declination of your target. tighten the clutch. Now loosen the RA and move the scope to the correct RA co-ordinate. Tighten the clutch once there to secure the scope. Using a low power eyepiece (your 20-25mm one that came with the scope will likely be the lowest power), have a look through the scope and your target should be somewhere in the FOV.
It should be noted that this is isn't a 100% reliable method, and the setting circles do have markings very close together so accuracy can be difficult, which is why it is best to use a low power eyepiece to find your target, but it should get you very close. It may be best to practise by starting on a bright star and using this method to find another bright star. Once you get the method, you'll start getting more accurate and speedy with it.
I hope this helps.


Fractional contribution of secondary star

It just means that in a star that is an unresolved binary system (i.e. the light you receive comes from both stars), the secondary contributes a fraction of that light.
This fraction could be expressed as a fraction of the total luminosity or a fraction of the flux at a particular wavelength or in a particular waveband.
How do you work it out theoretically? Well you would have to know exactly what kind of stars the primary and secondary were. e.g. If you knew they were both main sequence stars and you knew their masses, you could use a main sequence mass-luminosity relation to turn a mass ratio into a luminosity ratio. If you knew their temperatures, you could model their spectra to estimate the primary Vs secondary contribution at any wavelength.


Is it possible for a moon to continuously have a side facing its star whilst orbiting a planet?

I worked out the math, it's pretty straight forward.   If we take the Hill sphere, which is an estimate for the furthest possible orbit, and I'm going to just run the math on circular orbits.  Eliptical orbits are harder to put in resonance anyway.

the simple Hill Sphere formula.  $3\frac{r^3}{a^3} = \frac{m}{M}$, where $a$ is the semi major axis planet to sun and $r$ is the distance moon to the planet, little $m$, mass of the planet, big $M$, mass of the star.

To calculate tidal forces, the force changes with the 3rd power of the radius, so for equal tidal force on a moon from both the star it's planet orbits and the planet it orbits, $\frac{r^3}{R^3} = \frac{m}{M}$, and that rather neatly puts the equal tidal forces outside the outer edge of the Hill Sphere by a factor of the cube root of 3, or about 1.44

The outer half of the Hill Sphere isn't long term stable anyway in real systems, so for the Moon to experience a stronger tidal force from the star it's planet orbits than it experiences from it's planet, it would need to be well outside the stable orbital region around the planet, by at least a multiple of 2.88.

There's no way for any moon in any system to be tidally locked to it's star and not it's planet.


Rotation of far away and near by galaxies

It depends a bit on what you means by "far away" and "the same", but:
Galaxy formation
Galaxies form from collapsing and colliding clouds of gas and dark matter in the early Universe. The first structures began to form a few hundred million years after the Big Bang, with masses of the order of $10^5$ Solar masses (e.g. Mo et al. 2010). As time went, progressively larger structures built up in a hierarchical manner from smaller clumps, with more galaxies forming as time went, until galaxy formation peaked around 2–3 billion years after the Big Bang (e.g. Thomas et al. 2017).
Because the Universe expands, everything was closer in the early Universe. Hence, early galaxies formed under constant harassment by other galaxies coming from all sides. This made it difficult for galaxies to settle into an ordered, rotating disk. Indeed, some galaxies in these early epochs had all their gas blown out by merging with other galaxies and feedback from supernovae and quasar activity. Without gas it is difficult to settle into a disk, and to form new stars. These galaxies are seen today as elliptical galaxies without an overall rotation (e.g. De Lucia et al. 2006).
When did rotating disks form?
Galaxies that survived such a bombardement eventually tended to form a rotating disk. So when did galaxies start to rotate? This question is still not entirely resolved. Galaxies don't just suddenly start rotating, but rather progress from being dispersion-dominated (i.e. stars and gas follow randomly oriented orbits) to being rotation-dominated.
Previously, it was thought that (most) galaxy formation took place in a "hot mode", where accreting gas from the intergalactic medium is shock-heated to high temperatures ($T\sim10^6\,\mathrm{K}$), eventually "drizzling" down on the central galaxy (e.g. Rees & Ostriker 1977; Fall & Efstathiou). Consequently, disks would form quite late in the history of the Universe (at redshifts $z\lesssim1$, some 6 billion years after the Big Bang).
However, partly due to numerical simulations it was eventually realized that galaxies could accrete gas through cold, narrow streams of gas which reach the central galaxy faster, and is cold enough to form stars (e.g. Dekel & Birnboim 2006; Dekel et al. 2009). These streams also carry angular momentum which aid the rotation.
Simulations now routinely see rotating disks forming already ~1 billion year after the Big Bang (e.g. Grand et al. 2017; Pillepich et al. 2019). Observationally, disks at these high redshifts are detected (primarily I think) through carbon and carbon monoxide with large radio telescopes (e.g. ALMA), and rotation-dominated disks have been detected out to redshifts of at least $z=4.55$, corresponding to 1.3 billion years after the Big Bang (Fujimoto et al. (2020). The distance to this galaxy is 25 billion lightyears. I think the "record" for seeing rotation is Smit et al. (2018) at redshift $z=6.8$, seen only 800 million years after the Big Bang, at a distance of 28 billion lightyears, though its rotation is perhaps a bit up to interpretation.
A very recent (yet non-refereed) study by Kretschmer et al. (2021) find galaxies 1.8 billion years after the Big Bang to be rotation-dominated for several orbital periods, before being disrupted by mergers and counter-rotating streams.
Although these galaxies are not as evolved as the Milky Way, some of them already show not only rotation, but also sign of spiral structure. So I'd say that the answer to your question is "Yes, for the last 90% of the history of the Universe, some galaxies do, although most early galaxies don't."


How can we be certain that Ryugu is a remnant of the material that did not form planets?

Most of my information is from this link. So here's a watered-down version.

When the Solar System first formed, numerous concentrations of mass, called 'planetesimals', formed over time from slowly accreting material. Some of these would go on to combine with other planetesimals and form planets in their own right, but some would simply float in space. Over time, these planetesimals would shatter apart into primordial asteroids from impacts with smaller bodies. However, these remnant asteroids would originate from a single body, so they would have about the same physical properties. In addition with our modern orbital calculation techniques*, we would able to identify groups of these primordial asteroids, and flag them as such. However, an asteroid created from a planetary impact would likely be ruled out as non-primordial, since there would be no "brother asteroids" to compare it with.

*To clarify, we can trace orbits of asteroids back over billions of years to see where the bodies might have been in the distant past. Asteroids that originated from a single parent body would be observed to converge to a single point.


Why do rocks on other solar system bodies that have an atmosphere seem to be flat?

I would say that your initial observation is flawed, so the question is moot.

Huygens landing site, Titan:


TL;DR: Not all life requires atmospheric oxygen or carbon dioxide. There are many organisms here on Earth that use chemosynthesis to process chemicals like methane or hydrogen sulphide for energy.

The search for life on Mars has concentrated on finding evidence of autotrophic microorganisms – organisms that convert simple chemicals in the environment into complex organic compounds. Most autotrophs use water as the reducing agent, but some can use other chemicals like hydrogen sulphide.

The energy source for these autotrophs can either be sunlight, in which case we call them phototrophs (which includes algae and green plants here on Earth), or obtained through the oxidation of electron donors in their environment, in which case they're called chemotrophs.

There's a wide variety of chemotrophs here on Earth, typically living in completely dark and often hostile environments such as next to deep-sea volcanic vents. For example, iron-oxidizing bacteria even colonise new lava beds in the deep ocean, and in effect use the oxidation of available ferrous iron as their respiratory process.

There's no reason to think that autotrophs couldn't find niche environments on Mars in which to exist, but the biggest hurdles are the significantly lower temperature and pressure, and the significantly greater radiation. The latter can be overcome by colonising sub-surface environments, which eliminates the phototrophs from contention. While UV radiation penetrates soils only in the millimetre range, one study found that viable dormant cells would have to be at least 7.5m below the planet's surface to survive the cumulative damage from cosmic radiation.


If we look at the planet's cores and I'm going to ignore liquid vs solid and focus on size overall.

Mars:  Core estimated 1,794 +/- 65 km radius.  The planet is 3,390 km radius.   About 53% of the planet's radius is its core.   Mars also has more sulfur in it's core and more Iron in it's mantle than Earth, suggesting that it probably didn't mix as well as Earth did, but I'm not sure that would significantly effect the size of it's core.

Venus:  Core estimates have some uncertainty, but by this article, it's core is thought to be about 3,000 km, about 49.5% of it's 6,052 km radius.

Earth's core is about 3,400 km and it's radius 6,371 km, about 53.4% of it's total radius.    If we use the 49.5%-53.4% as a guideline, Mercury's 2,440 km radius (85% core, so it's core is about 2,074 KM), so a rough estimate using the other 3 planets as a guideline, 3,880 - 4,190 km radius.    That puts it 500-800 km larger than Mars in radius, roughly 2,000 km smaller than Venus in radius.

That's obviously just a rough estimate, but assuming it was similar to the other 3 inner planets, it's probably in that range.   I should probably also account for compression.  The more massive the planet the greater the compression of it's core, but that would probably only vary a couple percentage points off the estimate and it wouldn't make a big difference.


How would the temperature on earth be if the sun was slightly more or less powerful?

Don't have high enough reputation to comment, but someone should mention the Stefan-Boltzmann Law: https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law
When a planet (or anything) gets warmer, the amount of radiation it emits increases with the 4th power of temperature (measured on an absolute scale like Kelvins).
A planet's temperature reaches equilibrium when it receives about as much radiation as it emits back into space.
This should give you a better approximation than the linear model.


Scientific evidence against CMB originating from “Oort Soup”?

The idea that belts or spheres of dust might be responsible for (some) microwave emission is not crazy. Indeed we know that dust does emit microwaves and indeed the contribution of such dust has to be removed from the CMB signal before it can be interpreted cosmologically.
There is some debate about some of the larger scale anisotropies (particularly the "quadrupole anisotropy" (of about 1 part in $10^5$) might be the result of badly-subtracted emission from within the solar system because the anisotropy is roughly aligned with the ecliptic plane.
However, the bulk of the article you refer to is nonsense and betrays a basic lack of understanding of physics and a lack of knowledge about the microwave background.
Just to address a couple of things.

The CMB is an almost perfect black body spectrum. A blackbody spectrum is emitted by perfect blackbodies. The thermal emission from dust is not blackbody radiation. Thermal radiation $\neq$ blackbody radiation. To become blackbody radiation the emission has to come from something that is "optically thick" to that radiation - i.e. it should have a surface that we can see. In other words if we had an opaque sphere of dust around us that had somehow come into perfect thermal equilibrium at 2.7K then this would produce something that looked like the CMB. Unfortunately we would then be unable to see anything else beyond that! For a cosmological interpretation that isn't a problem. We cannot see (at any wavelength) "beyond" the CMB at a redshift of $\sim 1100$. It is a fatal problem for any model where the CMB is optically thick emission from a sphere surrounding the solar system, since we wouldn't be able to see anything beyond the solar system!

The CMB is anisotropic. The most obvious anisotropy is the large (0.1%) dipole anisotropy caused by the motion of the solar system with respect to the rest frame of the CMB. Why should the Sun be moving at about 370 km/s with respect to a spherically symmetric Oort cloud (in the direction with Galactic coordinates $l=264$, $b=48$? Even if the Oort cloud were at a distance of 10,000 au, the Sun would travel beyond it in only $\sim 100$ years! If the sphere is supposed to be further away, what is preventing gravity disturbing its almost perfect uniformity and how does it maintain a perfect thermal equilibrium?


Could liquid water have existed in open space 15 million years after the Big Bang?

Let's interpret your question to be about whether the conditions would permit blobs of water to remain liquid, whether or not water existed yet. And the answer is No, because the pressure was by then far too low. Basically, space was already a vacuum, just not as hard a vacuum as intergalactic space is now.
It is appealing to imagine an era when the universe was simultaneously dense enough and cool enough for liquid water (and thus perhaps humans) to exist. But alas it is not so. At the time of the creation of the cosmic microwave background, around 370 thousand years after the Big Bang, the temperature was around 3,000K, but the pressure was around $10^{-17}$ atmospheres (see the Wikipedia article Chronology of the Universe, and search for "Recombination").


A clock travelling "faster than the speed of light"

Let's restrict to special relativity, meaning two inertial frames moving in a Minkowski spacetime.

A clock in the first inertial frame ticks slower, when seen from the second. A clock in the second inertial frames ticks slower, when seen from the first.

Now assume, that you are fixed to one of the two inertial frames. Usually you measure velocities within your own inertial frame, meaning measuring distance and time in your inertial frame to calculate the velocity of the other frame. In this case you'll get a velocity below the speed of light.

If you measure the distance in your inertial frame, and divide this distance by the time you observe on the clock of the moving inertial frame, you'll get higher velocities, which may exceed the speed of light. But this result has only the dimension of a velocity; it is not a velocity with respect of any of the two inertial frames.

The moving observer would observe a slow-down of external events, not a speed-up.
The observer would observe a speed-up only in the sense of a relativistic Doppler blue-shift when moving towards the observed object.

The people on the destination planet age the same as on the Earth, provided they don't move relative to the Earth. It's just you, since you travel that fast, the distance gets shorter, and therefore the time for the travel also gets shorter. By the Doppler shift and space contraction you'll see them age faster.

From Alpha Centauri, it's again the space contraction for the traveller, making the distance shorter. Again we have a combined space contraction/Doppler effect looking the people on Earth aging rapidly for about 4.5 years.

For people on Earth observing the travel takes 4.5 years. For the traveller it may take just a few days. With an acceleration of just 1g you could traverse the Milky Way in about 20 years, seen from the spaceship. The same travel seen from Earth would take about 100,000 years.

More on the twin paradox on Wikipedia.


A question about star names... E.G. CW Tauri, DZ Bootis, IQ Lupi, EU Eridani

The Bayer catalogue uses Greek letters, then lower case Latin letters and if needed, upper case letters. But it never gets beyond "Q"

In 1855, the German astronomer Argelander proposed naming a particular variable star "R" (as using upper case letters at the end of the alphabet would avoid clashing with the Bayer catalogue.

In 1867 it was agreed by the German astronomical society to name variable stars R,S,T... Z, then RR, RS, RT up to ZZ, followed by the constellation name.

This system provided enough names until 1907, when they reached ZZ Cygni.  It was decided to start again with AA, up to AZ, Then BB to BZ etc.  That system provides 344 names and worked only until 1929, and after that, when more variables than this were found in a single constellation, they switched to a 'Vxyz " where "xyz" is a number starting with 345 e.g. V345 Cyg.

So the "2-letter designation" is particular to variable stars http://cdsarc.u-strasbg.fr/afoev/var/edenom.htx and is used in the General Catalogue of Variable stars, but the scheme doesn't originate in the at Catalogue.


When we look at spectral lines in a star's spectrum, we're actually looking for absorption lines, not emission lines. A star's spectrum usually resembles that of a black body, continuous and smooth. However, there are elements in the star's atmosphere that absorb some of the emitted light; these create characteristic absorption lines in the spectrum we observe.

There's a reason that a star's spectrum closely resembles that of a black body. H- opacity means that some of the solar photosphere is optically thick for a wide range of frequencies. This in turn means that the emission no longer comes from just a few frequencies corresponding to a few elements. This spectrum appears continuous, and therefore like a black body.


Can the moon turn red even without an eclipse?

Yes, pollution from (for example) wildfires can cause the moon to appear red. It is normally not noticed unless the moon is low in the sky.

Earthsky magazine is reporting that red sunsets and red moons are being reported across the USA. caused by wildfires. Even when the moon is quite high in the sky. I believe that fires from logging in the jungles of Borneo and the Philippines can have the same effect.

This has also been answered on https://earthscience.stackexchange.com/questions/12223/why-were-both-the-sun-and-the-moon-red-today


The formation of new stars as depicted in Feynman's lectures

imo, all the 5 stars are already present and with a little separation in the left image.
Their distance to Earth should have been investigated on the image of the right. The ones nearer could travel, proper motion, much faster than the ones on the background.
In short: they could be at large distances one from the others and it is a coincidence the fact that in the left image they are aligned. With zillions of stars out there the probability is not null.
There are very fast stars (VFS) for ex. US 708 (1200 km/s on WP e 708 km/s heliocentric radial velocity - Hirsch et al. 2005) not known at Feynman time.
The time of discovery of VFS was 2005 and 10 years later, Warren R, Brown et al, reported on the PROPER MOTIONS AND TRAJECTORIES FOR 16 EXTREME RUNAWAY AND HYPERVELOCITY STARS (2015).

EDIT ADD:
One thing that can be done is to use software enabled with Point Spread Function (PSF) to discriminate the number of sources and decide how many stars are in each of the images.
I will not do it because it takes time that I do not have.

There are several packages available, fi, DS9 (video: find a moving star), MAXIM DL, SExtractor. The images have to be scaned, eventually formated as FITS (python code) and if using SExtractor look at the procedure used by E. Bertin and M. Dennefeld (1997).
Search on youtube 'Detecting asteroids'


Asteroids don't have control of their orbit. They are influenced by gravity; they obey the laws of physics.

There aren't any traffic lights for them in space to obey so they know to slow down and stop to avoid a collision with a planet. They'll just carry right on going like someone driving through a red light. If it goes right in front of a planet at the wrong moment then it gets t-boned.

Or it might miss. But here is where the analogy falls apart. Because if someone drives through a red light and misses everyone, they carry on without any subsequent repercussions (except maybe a ticket). An asteroid passing close by a planet will be influenced by gravity between the planet and the asteroid, and it's path will change. That might mean it is flung far away, or it may mean it is captured, putting it on a looping path that eventually draws it in to crash onto the planet.


Just because a planet is in the "Habitable zone" doesn't mean it is habitable. A planet is said to be in the habitable zone if it is possible for liquid water to exist on the surface. A planet may be in the habitable zone, and yet not be habitable if, for instance, it's atmospheric pressure is too low (like Mars) or too high (like Venus), both of which are in the habitable zone of the Sun.

A star may be unsuitable for life for other reasons: it may be prone to extreme flares, for example. A blue giant is a star that burns bright and dies young, only lasting a few million years. Planets around such a star would have only just formed (they may still have molten surfaces). Blue giants tend to be unstable, being variable over a range of timescales, and ejecting large amounts of matter into space.

So while there would be a region which is "habitable" in the sense that the surface temperature of a planet could be in the range 0-100, you wouldn't expect such a planet to actually be suitable for life.


Are we sure there are no planets inside Mercury's orbit?

Yes, Vulcanoids larger than 5.7 km diameter are considered to be ruled out:
A Search for Vulcanoids with the STEREO Heliospheric Imager.

An object sufficiently large to be called "planet Vulcan" would change Mercury's orbit in a detectable way by gravity.

So, there are two lines of evidence, that such a planet doesn't exist.

More to read in Wikipedia:
https://en.wikipedia.org/wiki/Vulcan_(hypothetical_planet)


It does not have to start from the big bang. There are also different universe models such as the big bounce. Where the universe has an infinite past, such that it expands and then contract.
See
https://www.quantamagazine.org/big-bounce-models-reignite-big-bang-debate-20180131/
https://www.wired.com/story/what-if-the-big-bang-was-actually-a-big-bounce/
For more detailed explanations you can look some cosmology textbooks


Are Wikipedia's sun ecliptic-coordinate formulae accurate?

Assuming that you used the time of 170900 EDT(-4 UT), which is 230900 UT, the following is my math:

Let the Julian Date for 25Mar2016 at 230900, which is 2457473.464583, be set to JD;

Given the formula (on Wikipedia:n = JD - 2451545.0 ), your n is 5928.464583.
Then, taking the second formula ( g = 357.528° + 0.9856003° * n ) you get g=357.528°+5843.0964715441749 = 6200.3544715441749 mod 360 = +80.35447°.  Were you to round up this would give +80.4.

Without how you handled your date/time in c++ and your rounding, I would wager that is were the error occurred.


You are right that most of the dots of light that you see in the at image of Andromeda are stars in our galaxy that happen to be on the same alignment.

However we can resolve stars in the Andromeda galaxy. (Here I am understanding "resolve" means "to see as individuals and not just a haze")

https://www.spacetelescope.org/images/heic1502a/zoomable/

Even without the Hubble telescope, individual stars had been identified in the Andromeda galaxy, and (since the brightness of some of these stars can be determined) the distance was calculated. This was also done by Hubble, but by the man, not the telescope.  He found that the Andromeda galaxy was not part of the Milky Way, but much further away.

Hubble's photograph from 1923 can be compared to a modern skysurvey image.

The light in the image that you show is the combined light from billions of stars. They are too dim to be seen at the resolution of your image, but combine to make what looks like a shining cloud.  Just like a real cloud is made of lots of dots of water that are too small to see.

There are background galaxies. Some can be seen on the high resolution view I link above. One surprising thing is how small the area of the sky in the Hubble Deep field image is. It would be only a few pixels across on the scale of the image you have of the Andromeda galaxy.

You could expect that many galaxies to be in every few pixels, but too dim to be seen in the image of Andromeda.


"Mysterious radio signals" - how could a geostationary satellite and Ross-128 line up with Arecibo?

The original team suggested three main possibilities: "(1) unusual stellar activity, (2) emissions from other background objects, or (3) interference from satellite communications"  Clearly radio astronomers know about satellites.

They noted "in the absence of solid information about the signal, most astronomers would think that [radio interference or instrumental failures] would probably be the most likely explanation."

The satellite hypothesis has some problems it "does not yet explain the strong dispersion-like features of the signals (diagonal lines in the [weird] figure [in the question])" But on a balance of probablities this is the most likely explanation.

There are satellites in the region of Ross 128:

Location of Geostationary satellites operating between 4-8 GHz in the same region of the sky as Ross 128 (yellow dot).
Credit: Enriquez et al. (SETI Berkeley), http://seti.berkeley.edu/ross128.pdf

Your analysis of the location of a satellite, based on the geometry of satellite orbiting over the equator is not supported by this image.

The observers did not claim to have received a SETI signal. Indeed if you look at their original report they note that "aliens" is at the bottom of their possible explanations. The signal looks different from typical satellite bursts. They hoped the signal was astronomical in nature so there may have been a little wishful thinking.

So the most likely explanation is that we are observing multiple reflections from geostationary satellites transmitting on the 4-8 GHz band.

All quotations from the press release from the planetary habitability observatory.


What's the origin and culture of funny astronomical terminology?

Any causal claims or associations relating to language use are never going to be able to be proven, so the best answer to your overall question is: "nobody knows".

You can use Google's ngram viewer to visualise the appearance of a specific term or terms over time.

Perhaps astronomers generally have been better as a field in making their research more approachable by not pitching their discoveries in superior-sounding high-falutin sesquipedal pomposity. But it isn't universal; for every approachable 'Hanny's Voorwerp' there is an ancient-language-originating label like 'heliosphere'.

Geologists also use non-latin terms like "slip" and "fault" and "hanging wall". Is that to make geology more approachable for mass-media public reach-out PR?


Are we seeing a conjunction these mornings?

Strictly speaking, a
conjunction
is when two objects appear at the same
right ascension or  ecliptic longitude.
These are not exactly the same but are usually close since the major planets remain near the ecliptic.

When Venus or Mercury is in conjunction with the Sun, we distinguish between inferior (near side) and superior (far side) conjunction.
The outer planets can have another type of alignment, opposition, in which they are 180° away from the Sun around the equator or ecliptic.

When two solar system objects appear only a few degrees apart, it's fair to say they are near conjunction.
If they don't quite reach conjunction, their minimum angular separation is called an appulse.

Mars is in conjunction with Jupiter on March 20 and with Saturn on March 31.
Other planet-planet conjunctions are listed here.
On 2020-12-21 Jupiter and Saturn will appear only 0.1° apart at conjunction and should be interesting to see together in a telescope shortly after sunset.


The sun is the nearest star to Alpha Centauri (unless you count Proxima Centauri, which is really part of the same system).

There is a very small and dim pair of brown dwarfs, called Luhman 16 that are closer, at about 3.6 light years from Alpha Centauri. Brown dwarfs are not true stars, but they do glow from their own heat. They were only discovered in 2011. It is possible that there are other very dim objects, but these would not be stars either in the strict sense.


What are shooting stars and how are they formed?

A shooting star is simply the visible path a meteriod takes burning up in the atmosphere. As it flies through the atmosphere, it leaves a trail of fire, creating the streak of light you see.

Before they enter the atmosphere, they are simply chunks of rock. They could form in one of many ways - be broken off of a planet, etc. After they do this:

they look like this:

As for the reasons for meteor frequency varying, it's because of where the Earth is in it's orbit around the sun. Since almost all meteors are in orbit around the sun, and there tend to be clumps of them, this makes a 'donut' of meteors in orbit around the sun.

When the earth intersects one of these donuts, there is a spike in meteor activity. Ultimately, it's about how many space-rocks happen to hit Earth.

One annual 'high', for example, is the Perseid peak, which tends to occur mid-late summer each year. Sometimes, this even delays launches of space vehicles.


How could eclipse happen on the same day in a different calendar? (45 AD)

From the documentation for your program, http://www.alcyone.de/plsv/documentation/overview.html,

  Dates are in the Julian Calendar through 4 October 1582, and then in the Gregorian Calendar, which begins on 15 October 1582. (Compute for 1582, and you will see that 1. oct to 1. nov. is short.)


Why is the Oort cloud presumed to be spherical?

The latter.  Long-period comets appear to come from random directions.


Why does the adiabatic exponent decrease at the ionization zones?

There are two things going on. (1) When you add heat to a gas that is on the threshold of ionisation or partially ionised, some of that heat goes into ionisation. This means that it takes a much larger amount of energy to produce a rise in temperature. (2) However, as the gas becomes ionised the number of particles per unit mass also increases and so the pressure increases at a given temperature. The net effect is that $C_V$ (heat capacity for a fixed volume) increases by more than $C_P$ (heat capacity for a fixed pressure) and so the adiabatic index decreases.

It is a tricky and detailed computation to work out how $C_V$ and $C_P$ change as ionisation proceeds. An example calculation for pure hydrogen gas is give on pp 123-126 of "Principles of stellar evolution and nucleosynthesis" (1983, D. Clayton). Both $C_V$ and $C_P$ increase drastically (by factor of 30-40) as ionisation increases, both peaking at around 50% ionisation before falling back to exactly twice their unionised values once the gas is completely ionised (when it just behaves like a monatomic perfect gas but with twice the number of particles per unit mass). However, the behaviour of $C_V$ and $C_P$ are slightly different, $C_P$ increases by less than $C_V$, and so the ratio of $C_P/C_V$ changes and goes through a minimum at around 50% ionisation before regaining the standard monatomic gas value once completely ionised.


I think you may be looking at the wrong date? -21.9964 is the first value in the current bulletin B but that is for 3/2/2016. The current bulletin B only seems to go up to 5/1/2016, with -134.3179 as the value. I think you may instead want bulletin A, which shows -0.154825s for today (5/12/2016)?

http://maia.usno.navy.mil/ser7/ser7.dat


I compared C type asteroids and comets for this question. C type asteroids have around 30% of empty space, meaning that that much space can be for water.

Comets have around 40% of empty space.

Obviously, just because they have more empty space does not mean they have more water. However, evidence for water on Earth originally comes from comets. Not only that, but comets are occasionally balls of ice with some rock. Asteroids are known for being just rocks, and only have a limited amount of water.

In conclusion, comets are more water rich.

Hope that helps...?


But if yellow light is what's getting through the atmosphere - why doesn't everything have a yellow tinge to it?

Everything does have a yellowish tinge to it, but only for roughly the hour after sunrise and the hour before sunset. These a photographer's golden hours:

As the Sun rises, its color changes from reddish to orangish to yellowish -- and then to nearly white, even when the Sun never does get very high.

You must not (and cannot) look at the Sun, even for the most fleeting of moments, once the Sun gets above 10 to 15 or so degrees above the horizon. The only time you can look at the Sun, even briefly, is when the Sun is low on the horizon and air mass is high. We think the Sun is yellow (or in some civilizations, orange or even red) because we cannot see the Sun for what it truly is, which is nearly white.

What bit of off-white color sunlight does have when the Sun is 10 to 15 or so degrees above the horizon is offset by the blue light from the sky, aka "diffuse sky radiation". The color of the light from the Sun and the sky when the Sun is high in the sky is about as white as light can get.

There's a bit of physiology going on here as well. To those who have lived their most of their lives inside buildings, bright sunlight is a bit yellowish and a bit too harsh. To those who spend much of their time outdoors, the fluorescent lighting that now dominates the urban landscape is a bit bluish and a bit too harsh. What city dwellers who rarely ventures outdoors thinks of as white light is not very white to those who spends most of their waking hours outdoors, and vice versa.


The positions of many galaxies are recorded, so astronomers can point their telescopes to the exact point in the sky that they wish to observe. In addition,the relatively nearby galaxies can be recognised by their appearance,& these are the galaxies most intensively studied. Professional astronomers have memorised most of the stars you will see in the night sky, and the constellations (Sagittarius, Ursa Major, Orion etc) guide them to their target. Those galaxies near the limit of observation are of lttle interest except when something unusual & very luminous happens within them,such as a supernova or gamma ray burst.


How was the mass of Venus measured for the first time?

In the mid 19th century, Urbain Le Verrier's predicted of the existence of a then unknown planet beyond the orbit of Uranus. He even predicted this planet's orbit. The discovery of Neptune based on his predictions was perhaps his greatest accomplishment.
Le Verrier then went on to investigate Mercury. He used observations of Mercury, Venus, the Sun (as a stand-in for the Earth) and Mars and calculated that Mercury should precess by 532 arc seconds per century based on Newtonian mechanics. Along the way, he had to (and did) estimate the mass of Venus. There was a problem here; the observed precession of Mercury's orbit is 575 arc seconds per century, 43 arc seconds per century greater than his calculated value. This led Le Verrier to conjecture that there was a planet even closer to the Sun than Mercury.
Despite the failure to discover the non-existent planet tentatively named Vulcan, Le Verrier's estimate for the mass of Venus was fairly close to the correct figure, within a couple of percent. Once the cause of this 43 arc second per century discrepancy was discovered by Einstein, the mass of Venus was determined with even greater accuracy. Of course, once probes were sent into orbit about Venus, its mass was determined with greater accuracy yet.
Reference:
Leverrier, M. "On the masses of the planets, and the parallax of the Sun." Monthly Notices of the Royal Astronomical Society 32 (1872): 322.


How was the acceleration calculated for this CME (Coronal Mass Ejection)?

The average velocity in the interval 8:05 to 8:36 is 331.2. So, use $v_1=$ 331.2 at $t_1=$ 08:20:30 under the assumption that acceleration is constant.

The average velocity between 8:36 and 9:27 is $v_2=$ 306.5, so use $t_2=$ 9:01:30

Acceleration is $(306.5-331.3)/(41\times60)=-0.01\ $km/s$^2$.

The same calculation gives me $0.029\ $km/s$^2$ and $-0.038\  $km/s$^2$.

Is this just arithmetic error on the first value (using 51m instead of 41m), a small rounding issue on the second and the third I agree with.

Other than that, then some unstated alternative interpolation scheme has been used.


How are the atomic clocks synchronised between worldwide VLBI telescopes?

They don’t have to synchronize the atomic clocks at each telescope.  Instead, they synchronize the collections using GPS (so the telescopes observe the same target in the same frequency range at the same time), as shown in the diagram from this paper.

When the data is brought together, they use differences between the arrival times of quasar data to calculate relative antenna locations.  From this paper from NASA:

During geodetic VLBI observations, signals emitted by distant sources
of radio frequency energy (quasars) are received and recorded at
several antennas. At each antenna (VLBI station) a very stable
frequency standard (hydrogen maser) provides a reference signal that
enables time tagging of the quasar signals as they are being recorded.
For each VLBI experiment, correlation of the time-tagged, recorded
information from the participating antennas yields the differences
among the arrival times of any specific quasar radio wave at the
antennas. These time differences are used to calculate the locations
of the antennas with respect to each other.


Calculating Air Mass at Zenith for observing sites above sea level?

I would use the wikipedia variable density atmosphere model

https://en.wikipedia.org/wiki/Air_mass_(astronomy)#Variable-density_atmosphere

Which gives airmass as a function of zenith distance for an exponentially decaying atmosphere of scale-height $H$. This formula is valid for all heights above sea level but the airmass given will be in terms of airmass at zenith from that height. To make your adjustment to an equivalent sea-level airmass then you would multiply by the ratio of the density at your observing altitude to the density at sea level.


Do we know of any tidally-locked planets with atmospheres?

As we haven't measured the atmosphere of many rocky planets yet, there aren't many examples. But I found 55 Cancri e:
It is a super-earth with very close tidally locked orbit and an atmosphere that has been measured by Hubble. A year on the exoplanet lasts for only 18 hours and temperatures on the surface are thought to reach around 2000 degrees Celsius. The detected atmosphere seems to consists mainly of hydrogen and helium. So you do have a constant Twilight Zone, just a rather hot one.

Sources:
- Beeing tidally locked
- Detection of the Atmosphere


How to convert factors of time dilation into something relatable?

I am curious on how to turn those factors into relatable terms such as: At 1 km from the event horizon ____ time passes on earth in an hour.

The key here is to understand what the formula relates to.  The (all important) observer is in this case a notional observer so distant that curvature (and hence gravitational effects) are negligible.

That observer is the "base clock" we reference.  Everything is dilated relative to that clock.

That means that the dilation on Earth (not strictly speaking an ideal distant observer) has to be taken into account to get an answer in the form you mean.  So starting with the formula you had :

$$t = t_\infty \sqrt{1 - \frac{r_s}{r}}$$

We want this relative to Earth so we need :

$$t_e = t_\infty \sqrt{1 - \frac{r_s}{r_e}}$$

And combining youy get :

$$t = t_e \sqrt{ \frac {1 - \frac {r_s} r} {1 - \frac {r_s} {r_e}} }
= t_e \sqrt { \frac { r_e(r-r_s) } { r(r_e-r_s) } }$$

Now in the specific case of Earth and Sagitarius A*, $r_e$ is huge so the dilation factor for Earth is as close to one as makes no difference in practice, so you don't really need to go to all that trouble and you can use $t_\infty \approx t_e$.


Why is 3GHz the best frequency at which to measure the sun's microwave output?

From "The 10.7 cm solar radio flux ($F_{10.7}$)",

It has become clear that wavelengths in the region of 10 cm are best for monitoring the level of solar activity because solar emissions at these wavelengths are very sensitive to conditions in the upper chromo- sphere and at the base of the corona.

This is not saying that exactly 10 cm wavelength radiation (3.0 GHz) is optimal. Ten centimeters is being used in this context as a nice round number. 10.7 cm radiation is definitely in the "region of 10 cm". Schmahl and Kundu reported in "Microwave proxies for sunspot blocking and total irradiance" that anything in the range of 1000 MHz to 9400 MHz works quite nicely. 3.0 GHz (3000 MHz) is more or less in the middle of that range, logarithmically.
Measurements of the F10.7 radio flux go back to 1946. This was a serendipitous choice as a proxy for measuring variations in the Sun's output, not just in the microwave / radio frequencies, but also in very high frequencies (ultraviolet and X-rays).

References:
Tapping, K. F. "The 10.7 cm solar radio flux ($F_{10.7}$)." Space Weather 11.7 (2013): 394-406
Schmahl, E. J., and M. R. Kundu. "Microwave proxies for sunspot blocking and total irradiance." Journal of Geophysical Research: Space Physics 100.A10 (1995): 19851-19864


If we were to see the Sun with our naked eyes from the Orion belt, would all planets be encompassed inside the star? Is this calculable?

Such observations are possible, see Fomalhaut b.
It depends mainly on the diameter of the objective of the telescope (diffraction limit) and contrast. Telescopes can (at least in theory) be combined to an effective telescope of larger diameter (aperture) by interferomentry.
Masks can help to occult bright stars to overcome the contrast limitatios.

There are indirect methods to "see" faint planets, as e.g. applied in the Kepler mission, or by Gaia.

And yes, it is calculable. With naked eye we wouldn't be able to distinguish the sun from the planet orbits from a distance of about 1000 lightyears, as the orion belt stars: 1000 lightyears are about 300 parsec. This means, that the earth orbit would be about 1/300 arc second. The resolution of the human eye is about 4 arc minutes, the 72,000-fold. For Pluto it's still more than 1000-fold.


Power requirement for transmitting communications further than 1 ly?

It's possible to transmit an intelligible signal to a target roughly 4.3 light-years away using 0.1 milliwatts. The trick for that particular level of power is that both the sender and receiver have to be carefully positioned several hundred AU from a star, with those stars lined up between them, in order to take advantage of a gravitational lensing effect. If only one of the two radios can be positioned at a star's gravitational focus, then the necessary power rises, but can be under 40 watts, according to one of the diagrams reprinted at https://www.centauri-dreams.org/?p=10123 .


What are the maximum and minimum impact speeds for an asteroid that would strike the Earth?

Although most objects orbit the Sun in the same direction — having emerged out of the same rotating gas cloud that spawned the Solar System — some asteroids and other minor planets do move in opposite, or retrograde, orbits (see this Wikipedia article for a list of such objects).

The minimum speed for an asteroid is achieved if it has more or less the same velocity around the Sun as the Earth. In this case the gravitational attraction of Earth will accelerate the object to the escape velocity of Earth, i.e.
$$
v_\mathrm{min} = v_{\mathrm{esc,}\oplus} = \sqrt{\frac{2GM_\oplus}{R_\oplus}} \simeq 11\,\mathrm{km}\,\mathrm{s}^{-1}.
$$
Here, $G$, $M_\oplus$, and $R_\oplus$ are the gravitational constant and the mass and radius of Earth, respectively.

The maximum speed is achieved at a "head-on" collision. Earth's speed around the Sun is
$$
v_{\mathrm{orb,}\oplus} = \sqrt{\frac{GM_\odot}{d}} \simeq 30\,\mathrm{km}\,\mathrm{s}^{-1},
$$
where $M_\odot$ and $d$ are the mass of the Sun and the distance from Sun to Earth (1 AU).

If the asteroid travels on the same orbit, but in the opposite direction, the impact will then be at 60 km/s. However, if the asteroid comes from far away (e.g. the Oort Cloud), it will be accelerated by the Sun and achieve a velocity equal to the escape velocity from the Sun at the location of Earth. As is seen from the two equations above, the orbital speed and the escape velocity differ by a factor of $\sqrt{2}$. That is, an object falling from infinity toward the Sun, will have a speed equal to $30\,\mathrm{km}\,\mathrm{s}^{-1}\times\sqrt{2}=42\,\mathrm{km}\,\mathrm{s}^{-1}$ when it reaches Earth.

Hence, the maximum impact velocity is
$$
v_\mathrm{max} = 30\,\mathrm{km}\,\mathrm{s}^{-1} + 42\,\mathrm{km}\,\mathrm{s}^{-1} = 72\,\mathrm{km}\,\mathrm{s}^{-1}.
$$


How can orbital resonance sometimes have a stabilising effect, whilst other times, it has a destabilising effect?

According to my understanding, for an orbital resonance to be stable, there must be at least three circumstances present:

The forces on the bodies in resonance have to cancel out over time,
The momentary forces shouldn't be that high that the orbits are changed significantly,
after a perturbation, the bodies should pull each other back into resonance.

For example, the Trojans are in a 1:1 resonance around the Lagrange points of Jupiter, and the Galilean moons of Jupiter are in a 4:2:1 resonance. A 2:1 resonance wouldn't be stable at it's own because the gravitational forces would always apply at the same point, but the perturbances are cancelled out by the third moon, and the outer and inner moon are in a stable 4:1 resonance.
In general, a resonance can be stable, if the encounters of the objects are equally distributed, so imaginary lines between these points have point symmetry / a regular shape (for example a straight line across the orbits (3:1; 3:2), an equilateral triangle (4:1; 4:3), a square (5:1; 5:2), a pentagon or pentagram, etc.) But, the closer the orbits get to each other, the more unstable they get, because the deflection gets bigger, so those resonances only work if the bodies aren't too massive. 2:1 resonances can be relatively stable too, but without any other resonance preventing it, the eccentricity will rise over time and one of the bodies may be ejected. However, orbital resonances are only stable in certain circumstances, because if the eccentricity and/or the inclination are too high, the forces don't cancel each other out enough. But there still are some configurations in which highly inclined and eccentric orbits are stable.


There is a few things one has to consider when talking composition:

It is usual to talk about relative abundances, and usually the solar composition is taken as the measure for this in the assumption that its composition is mostly prestine with a few exceptions. Now you can compare observations of different bodies and look at ratios of different materials, say nitrogen vs. carbon.

Now, materials have a different condensation temperature, and that is one of the main points of the paper. In the sun everything is in gas phase and thus has about pristine ratio.

Planets on the other hand form from solids. Thus - assuming a planet forms at distance R from the sun - it forms only from the materials which already solidify at the temperature at the given distance. (This is simplified due to radial mixing in the PPD and planetary migration, but the point is still valid). While the overall chemical composition of the PPD is the same, the chemical composition of the solids is different, the high(er)-temperature condensates are enriched, thus they are also enriched in the planets forming there

Carbon dioxide condenses from gas phase at slightly higher temperature than nitrogen. That means that you enrich the solid phase of the PPD in carbon and oxygen. Planets form from solids and only accrete gas when they have grown sufficiently - a threshold which Uranus and Neptun only barely reached. At the same time the gas of the PPD has to be in pressure eqilibrium, thus a flow towards the condensation zone exist, leading to slightly further enrichment of the condensing matter in the solid fraction - which is only partially balanced by the radial drift of the solids inward where they evaporate again.

In essence, due to different condensation temperatures, and the region the two outer gasous planets formed, they are enriched in carbon when comparing to nitrogen due to carbon dioxide ice condensing in the solar nebular. It doesn't mean that anyone took away the silica nor the water ice in the formation - as such.

I don't find there any support in the paper that the ratio of C/O can be enhanced beyond 1: the most abundant solids are slica, water ice - and the point of the paper at that distance CO which is increasingly enhanced during the formation process due to transport processes of solids and gas and condensation in the vicinity of the CO snowline (see figure 4 therein).


Why does the Solar System have no (natural) satellites of satellites?

This is due to the mechanics of capture.

For one object (moon) to be captured by a another (planet), some energy has to be removed from the system. If the incoming moon has an existing satellite then it would be ejected, carrying a lot of kinetic energy.

If a small body were to be captured by the planet/moon combination, it would usually be captured by the larger object. Again some energy has to removed from the incoming object by being transferred to one the objects already present. One way that could occur is for the existing moon to be accelerated into a higher orbit.

In summary, it could happen, but with a very low probability.

For much more information, see this answer to the question How do moons get captured?.


Why the Circumstellar Habitable Zone is defined as it is, if life could be possible outside of it?

From your first link, the definition is:

  "The circumstellar habitable zone (CHZ), or simply the habitable zone, is the range of orbits around a star within which a planetary surface can support liquid water given sufficient atmospheric pressure.[1][2][3][4][5] The bounds of the CHZ are based on Earth's position in the Solar System and the amount of radiant energy it receives from the Sun."

but as you will see further down in that same article:

  "In subsequent decades, the CHZ concept began to be challenged as a primary criterion for life, so the concept is still evolving. Since the discovery of evidence for extraterrestrial liquid water, substantial quantities of it are now thought to occur outside the circumstellar habitable zone. The concept of deep biospheres, like Earth's, that exist independently of stellar energy, are now generally accepted in astrobiology given the large amount of liquid water known to exist within in lithospheres and asthenospheres of the Solar System. Sustained by other energy sources, such as tidal heating or radioactive decay or pressurized by non-atmospheric means, liquid water may be found even on rogue planets, or their moons."

So really, it's just historical - it is accepted now that it is a limited view, and there is just as likely to be liquid water elsewhere. That doesn't necessarily mean life is as likely, as with lower solar radiation, there may be insufficient energy to support life even with liquid water.


Why has the orbit of Venus such a low eccentricity?

It has a low eccentricity, but there may not be a particular reason.

Image by Kheider on wikipeda using Gravity Simulator by Trevor Dunn

In a simulation of the solar system, both Earth and Venus had orbital eccentricities that were much below those of Mars and Mercury (note the two axes. Mars and Mercury are on a scale that is 10 times bigger). But in 25000 years, Earth will actually have a smaller eccentricity, and 20000 years ago Venus's eccentricity was much larger.

So there may not need to be a special explanation for this. Planetary eccentricities do change over 1000s of years due mostly to perturbation of other planets.


Can we, with arbitrary precision, calculate what's the place on earth that's closest to the sun on a given day?

Calculation of the subsolar point is relatively straightforward. The actual point moves at nearly 500m/s in a circle, as the Earth spins. The linked website does the calculation in a few lines of python using a free library, The value is probably correct to within a few 10s of kilometres.

But you are looking for a different order of accuracy:  If you (for some reason) need 1cm accuracy, you would need to take into account the actual, non-spherical shape of the Earth.

The Earth is not a sphere, it can be better modelled as an oblate ellipsoid. But for 1cm accuracy, you would need to consider all of the mountains and hills. And you would need a very accurate time, not just a date. That is not a practical calculation. I can think of no practical application.


Are all satellites thought to be the result of collisions with the planets?

There are three main formation scenarios for planetary moons.

The giant impact hypothesis: The satellite forms as a consecuence of an impact between the planet an a large planetesimal. The Moon is an example, and one of the arguments is that the chemical composition of the Moon matches that of Earth with a significant accuracy suggesting that it is a in part our planet and in part the original impactor (Theia). We also know that the Moon has been getting farther away from Earth because we have evidence that it gained orbital potential energy by absorbing it from the rotational energy of Earth. We know this because days were not 24 hours long a few million years ago and we can keep track of those changes in the rotational period of Earth using rings in fossillized corals (those have rigns like the tree rings but ones that generate on a daily basis). We then can see that the moon was extremely close to the Earth some billion years ago (we have more evidence of this from the fact that tides were huge in those times and led geological evidence of daily floods across the recently formed planet). If you keep going back in time you see that the Moon was basically emerging from Earth. There are many more evidences of this scenario for our moon.

The accretion scenario: The satellite coalesced from a disc of material around the new-born planet (just as the planet accreted from the protoplanetary disc), the so called circumplanetary disc. As an example we have the four galilean moons around Jupiter (Io, Europa, Ganymede and Callisto). Since the disk was relatively flat the moons formed in the same orbital plane, also they move in the same direction as the planet rotates (which makes sense as both are generated from the same material spining with a certain angular momentum). This is the most frequent scenario for large moons. Our moon couldn't form like this because the expected size of the circumplanetary disc was in no way as massive as our moon is today (Earth is a tiny planet and it has a huge moon in relative terms).

Capture scenario: The satellite formed elsewhere in the Solar System as an independent minor body. With time some dynamical interaction might have led the object close to a planet and both got gravitationally binded. An example of this is Triton, Neptune's largest moon. The retrograde orbit is inexplicable in terms of the accretion scenario and the energy needed for a gian-impact scenario to work on Neptune is too large. Triton was captured (we think it formed as another planetesimal in the Kuiper Belt as it shares many chemical features of Pluto and other objects of the region). There are not so many moons on Neptune probably because they dissapeared (crashing into the planet or been ejected) just as Triton arrived to the system and dynamically destabilized their orbits. Another clear example are the tiny irregular satellites of Jupiter. This scenario is very difficult to immagine for Earth since capturing a massive moon like ours and making the orbit circular would have been a feat in terms of how precisely tunned would the orbit insertion parameters had to be. The giant-impact scenario leads to the current situation in simulations for a larger range of impact parameters, thus is much more probable in statistical terms.

There are some less frequent and some speculative scenarios also:

Eyecta fragments from other moons: Some satellites might have their origin on other satellites. A large impact might eject material into orbit. An example would be Hippocamp (a Neptunian moon) which is now regarded to be a fragment stripped out from Proteus (a larger moon).

Lagrangian/Trojan moons: This is similar to the circumplanetary disc scenario but here the accretion in the disc of the planet is further stimulated in certain regions because of a moon that formed a bit earlier. An orbiting body can generate five equilibrium points (Lagrange points) by sculpting the gravitational landscape. Two of those equilibrium points (L4 and L5) are stable equilibrium points; thus they are like gravitational traps where matter can accumulate until a new moon is formed. As a potential example we have Telesto and Calypso in the Saturnian system. They both lie on the L4 and L5 Lagrange points of Tethys (a much larger moon with great gravitational influence). They might have formed as regular objects and then got trapped on the equilibrium points or they might have actually formed there as matter coalesced on those gravitational traps.

Sprayed into existence by cryovolcanism from another moon: It sounds ridiculous because it is an hypothetical scenario, I just think might happen somewhere in the universe. Take a look to Enceladus (a large active moon of Saturn). Enceladus has water plumes and jets that shoot material from its interior into space through craks in the icy crust (because tidal stresses heat the interior like a pressure cooker and the pressure gets liberated that way). The entire E-ring on Saturn was created by orbiting ice and dust grains sprayed by Enceladus. We know that the E-ring has a mass of nearly $12\cdot10^8\; kg$ and we know that it is so far from Saturn as to make possible the coalescence of this matter (tidal forces wouldn't disrupt it: see Roche limit). Thus it is possible that the material from the ring could make a moon with the density of Aegaeon (another saturnian moon) and with a diameter of just $162\; m$ (a third of the size of Aegaeon). Enceladus prohibits this because of the gravitational influence it excerts on the E-ring, it also prohibits it to grow (the ring would be more massive if the material was not continuosly reabsorbed by Enceladus). But if Enceladus migrated to another orbit in relatively short timescales then at least I think is possible for this to happen. The new moon greated from the material of the ring could then end up coliding with Enceladus since both would probably still interact chaotically. The material once sprayed out of its interior would had returned to home.

Centrifugal breakup scenario: This is also hypothetical but we think this happens a lot on asteroids. Lightweight moons could be ruble piles just as many comets and asteroids are. Losely bounded material with little cohesion. If the moon starts to spin faster and faster (because of some mechanism like the Yarkovsky-O'Keefe–Radzievskii–Paddack effect), it could eventually break into two pieces due to the extreme centrifugal forces (described as seen from the co-rotating reference frame of the original moon). It is believed that asteroids like $1999 \;KW_4$ splitted in two because of this effect. I don't see any reason why a moon couldn't do the same, forming a new independent moon in the process.

Moon made of chunks of other disrupted moons: As crazy as it seems, it is one of the formation hypothesis regarding the formation of Miranda (one of Uranus satellites). Miranda's surface is so complex and varied that some speculate it might have formed as several pieces that orbited Uranus came toghether gently. These pieces might have been chunks from other moons or might have been chunks from an earlier iteration of Miranda itself, fragmented after a disruptive event. The geology on each chunk would have evolved independently until they re-assembled together. But this is also quite speculative.


1. Is material on Earth's surface not in free fall around Earth's center?

No. Material on the Earth's surface -- or inside it -- is not in orbit, and so is not in free fall. You can temporarily put yourself into an orbit (and thus into free fall) by jumping up into the air, or jumping off a higher surface. When you do this, you are briefly in a very eccentric orbit (one which would take you very close to the center of the Earth, if the Earth wasn't a solid body) -- but then you hit the ground and are no longer in orbit.

The Earth rotates in the same way that a spinning top rotates; this has nothing to do with orbits.

2. How are geostationary orbits even a thing? Seems like the only orbit that could be geostationary would be standing on the Earth's surface.

Again, the surface of the Earth is not orbiting. The Earth rotates as a rigid body, with (as AtmosphericPrisonEscape noted) residual angular momentum left over from its formation, like a spinning top.

Because your angular speed in an orbit decreases the further away you are from the Earth, there will be a point where it happens to match the Earth's spin rate. If you arrange the orbit so it is above the equator and in the same direction as the Earth's spin, then you will always be above the point on the equator: a geostationary orbit.

3. What changes as you orbit further above the Earth's surface? Does your angular speed increase or decrease? Does your tangential speed increase or decrease?

Both your angular speed and your tangential speed decrease the further away you get. (Your angular speed would decrease even if your tangential speed stayed the same, because the circumference of your orbit increases with altitude; but in fact the tangential speed decreases as well.)

4. Is magma near the center of the Earth not rotating faster than material in the crust, like in an accretion disk?

The Earth rotates approximately as a rigid body, so in general, no. The molten outer core (which is not magma) may rotate slightly slower, while the solid inner core might rotate a little faster, but we're talking about $\sim 0.1$ degrees per year differences, and this has nothing to do with orbits. (The Earth is nothing like an accretion disk.)

5. Can two objects be orbiting (circularly) at the same altitude but with different tangential speeds?

Ignoring minor deviations due to things like the non-spherical nature of the Earth, mass concentrations in the crust, etc., the orbital speed for a circular orbit is a function of the altitude only. So two objects in circular orbit at the same altitude must have the same tangential speed. (Note that they can have different velocities, because velocity is a vector quantity -- so you can have two object orbiting in different -- even opposite -- directions at the same altitude, at least until they run into each other.)


Where do Neutrinos Go if They Don't Interact?

It depends on the ultimate fate of universe, as with everything else in the universe. Fundamentally, neutrinos are not much different from “ordinary matter”. Yes, they do interact only via weak force and gravity, but this only quantitatively changes the frequency of interaction, but does not change the game qualitatively. As long as there’s non-zero chance for them to interact, and as long as there is non-zero density, there is non-zero probability they will eventually interact with another particle (or black hole), and given infinite timeframe, it will happen for sure.

Accelerating expansion of universe does however complicate things a bit, but again, eventually there is bound to be some random fluctuation which will knock the neutrino off from its track.

Eventually neutrinos would meet anti-neutrinos and annihilate each other. Since the chances for this are really low, I would suspect neutrino-antineutrino collisions being the very last blink of this universe - given heat death is the ultimate fate of this existence.


Dark enough to see stars during total solar eclipse?

One thing to double check is if you can see stars easily at night from the same location. If there are city lights nearby, combined with humidity or dust you can have a bright sky due to terrestrial effects.
Another thing to consider is the time needed for dark adaptation. If you walk from a brightly lit room out under a dark sky, you can't see many stars right away. There needs to be time for your eyes to adapt to the dark. If you were looking at the screen of a camera or phone or another light source, that would further slow down the adaptation process.
The astrometric data for the 1919 eclipse test of General Relativity was done with a telescope, not the naked eye. Stars are unresolved, which means that no matter what magnification, they are still small points of light who's size (visual or on film) is determined by other effects, like diffraction, aberrations and focus as well as the resolution of the retina or the film. In short, they stay small. However, sky brightness is reduced by magnification squared. If you expand a patch of sky by 10X then each point is 100 times dimmer.
Both of these are offset by the increase in light from the larger aperture of the telescope, but its mostly the sky brightness that decreases with magnification.
The interesting stars were the ones closest to the Sun, because they would have the largest deflection, which was really small, only of the order of one second of arc. There the problem was the brightness of the solar corona, more than the sky. If you look closely at the photographic negative below, you can see pairs of faint horizontal lines. In the middle of each pair is the position of a star to be observed. Even with the telescope and extended photographic exposure, the images are faint.

above and below: 1919 eclipse negative, from here. Original from: F. W. Dyson, A. S. Eddington, and C. Davidson, "A Determination of the Deflection of Light by the Sun's Gravitational Field, from Observations Made at the Total Eclipse of May 29, 1919" Philosophical Transactions of the Royal Society of London. Series A, Containing Papers of a Mathematical or Physical Character (1920): 291-333, on 332.

Negative of the 1919 solar eclipse taken from the report of Sir Arthur Eddington on the expedition to verify Einstein's prediction of the bending of light around the sun. For positive version, see Image:1919 eclipse positive.jpg.
Original caption:
In Plate 1 is given a half-tone reproduction of one of the negatives taken with the 4-inch lens at Sobral. This shows the position of the stars, and, as far as possible in a reproduction of this kind, the character of the images, as there has been no retouching. A number of photographic prints have been made and applications for these from astronomers, who wish to assure themselves of the quality of the photographs, will be considered as as far as possible acceded to.

below: Screen shot from the March 2009 Physics Today article Testing relativity from the 1919 eclipse—a question of bias. The telescope on the right, in the square tube is the 4-inch instrument that provided the photographs used in the analysis. Heliostat mirrors driven by clocks were used to reflect light from the region near the Sun into the fixed telescope.


Do black holes pull in and keep spacetime itself or merely warp it but ultimately let it pass through?

Black holes orbit things and galaxies move within space, so it would be safe to assume that black holes move through space-time and do not hold it in any particular state or position. They simply drag their "depression" along with them as they go.


Is it possible to affect Earth's orbit so that, say, aphelion happens sooner?

Humans don't have this capacity.  The Earth has a kinetic energy of about $10^{33}$ joules, relative to the sun.  Even if we put the total energy that humans use per year into the Earth's orbit (about 10^{24} joules) we still only have one-billionth of the kinetic energy of the Earth.

Moreover, the annual changes in temperature have very little to do with the distance from the sun, they are caused by the tilt of the Earth to the sun. For people in the Northern Hemisphere, the greatest distance to the sun occurs in the middle of summer when the temperature is usually at its hottest.

While we can't change the orbit of the Earth very much, we have been much more successful (!) at changing the composition of the atmosphere. The build-up of CO2 and other greenhouse gasses is warming the Earth significantly, and we can expect significant harmful consequences from this warming. We might like "warmth" but we don't like the rising sea levels, droughts, floods, and other effects of global climate change.


What is the projection of Earth's axis on the sphere with the Sun in center (see explanation)

It will be two images of the nearly-circular ellipse of the Earth's orbit (one for the north pole, one for the south), distorted by being projected onto the interior of a sphere, which means that the side of each orbit that has the axis tilted "towards" it will appear enlarged and closer to the ecliptic, and the side of each orbit that has the axis tilted "away" from it will be smaller and closer to the pole.

To a first approximation the pattern will retrace itself every year. The Earth's orbit is stable with respect to the Sun and the fixed stars. Over a period of tens of thousands to hundreds of thousands of years, the Earth's orbit will precess (causing the figures to rotate around the sphere), and nutation will also cause small wiggles at a much higher frequency.


Is there any point on earth where the moon stays below the horizon for an extended period of time?

Depends on the interpretation of your question...
The best places not to observe the moon are the north and south pole. On the north pole you will only be able to see objects above the celestial equator. As the moon orbits the Earth in one month its orbit is inclined from the celestial equator. This inclination is almost the same as the inclination of the ecliptic (path of the Sun) with the celestial equator. The ecliptic crosses the equator at two opposite points on the celestial sphere. This means that for about half its orbit, the Sun, and as the moon's orbit is near the ecliptic, also the Moon, will be above the ecliptic and therefore visible from the north pole.

That being said, the Moon does not follow the ecliptic precisely as the moon's orbit is inclined from the ecliptic by about 5°. The inclination of the ecliptic is 23°, so during very special circumstances the maximum altitude of the Moon above the horizon on the North pole will be 18° during one month. The duration that the Moon will be above the horizon on the north pole will be about 10 days (a guess) about 13.6 days (edit:see comments below).

So if we interpret your question as: Is there a place where the Moon will be below the horizon for a long period of time (> 1 month), then the answer is NO.

But if this happens near June, then the Sun will also be above the horizon at the north pole (for six months), and as the Moon will be close to the Sun (as it follows more or less the ecliptic), it will be very hard to see the Moon during that time. So if you specifically ask whether the Moon will not be visible for an extended period of time, then the answer is YES.

And there are of course also places with perpetual cloud cover ;-)


TOTAL solar eclipses and TOTAL lunar eclipses are actually not that common in places outside of the solar system. For these TOTAL eclipses to occur, and I'm emphasizing total bc partial eclipses happen all the time, there needs to be the perfect distance so that an object perfectly covers the Sun.

In other words, you are correct - it is such a remarkable coincidence that this occurs in our Solar System. I'm not entirely sure about how many exoplanetary systems also experience this phenomenon, but I assume that it won't be a lot. I do know that it DOES HAPPEN in other places, but it just isn't common.

Just for a fun fact - the Moon is actually moving away from us, meaning that in a few million years, total solar eclipses won't be possible anymore - and we'll only see partial eclipses.

Hope that helps! :)


How would water waves behave in partial gravity?

Would an "indoor" pool of water on the Moon or on Mars behave differently than on Earth?
Depends on what you are observing.

Would the waves caused by a splash be higher or lower? The waves would be higher.

Would they propagate faster or slower? They would propagate at the same rate.

Would boyance be different? No, because buoyancy depends on the displacement of mass.

what would it be like to shower in slowly falling water? How can anyone here answer this since only a few people (<0.001%) have experienced this.


How can I see a star behind a mountain range? Do I move apart from it and up?

Moving away will help. If the mountain is very steep, and you are standing right next to it, you can only see objects that are directly above or behind you. Example: if the mountain is 100 ft tall, and you move 100 ft away, then you can see objects in the sky that make a 45 degree angle off the horizon. Angle = inverse tangent (mountain height / distance from mountain)


You can calculate this using the program "KAIROS", See here: http://www.raymondm.co.uk/
It will let you play through the variables: Is this a Julian or a Gregorian date? Is it using true or mean positions? And so forth.


It is a strangely worded question I agree.

I would start with the sun appearing directly overhead at the latitude of the tropic of cancer 23.5 degrees. As you go north, vertical shifts north and the Sun appears to be lower towards the horizon due south. To make the Sun appear to be at 73 Degrees from horizontal you would need to be 17 degrees north of the tropic of cancer. 23.5+17=40.5

Local differences in gravitational direction and the eccentricities of the circumference of the Earth due to the equatorial bulge make the answer a bit uncertain hence the range

30 to 45 degrees.


Is there any thing special with an axial tilt of roughly 25°?

Mars (the planet with an axial tilt closest to that of the Earth) has an axial tilt that varies greatly over time. For instance, 4 million years ago, the mean obliquity was $∼35 ± 10°$. From the same article:

Earth on the other hand has had its rotational features stabilized due to the effect of our large Moon.
So it is only by chance if the axial tilt of Mars is currently similar to that of Earth.


On Augustus' Actual Prescription to Restore the Julian Calendar to Accuracy

8 AD was definitely a leap year.
It was believed since Scaliger that the leap year sequence was (45), 42, 39 ... 12, 9 BC, AD 8, 12 ... This was based strictly on sources in literature. Chris Bennett claims that an astronomical papyrus published in 1999 (pOxy 61.4175) which gives lunar ephemeris in late 24 BC and implies that Scaliger was definitely wrong, and the correct sequence is actually 44, 41, ... 11, 8 BC, AD 4, 8, 12 ...
The major source for Roman chronology questions is A. K. Michels, The Calendar of the Roman Republic (Princeton, 1967).
If you just want the answers, Bennett's page has clear tables. It also has a very good list of primary sources.


Is it possible to define a state of zero absolute angular momentum anywhere in the universe?

Angular momentum does not change unless a torque acts upon a body. So your question is just whether the torque on a body could ever be exactly zero. The answer is of course that it can be arbitrarily small but we can never be sure it is exactly zero.

More interesting questions would have been "does the CMB exert a torque on an object?", or "is it possible to define a state of zero absolute angular momentum?".


Is there any orbit at which the Roche limit can be "felt"?

Roche limit happens where the gravity of the object, trying to pull the object together, becomes smaller than the tidal force (trying to pull the object apart).
But the astronaut is bound by not gravity, rather by the electromagnetic interaction between his/her atoms. The own gravity of the astronaut is negligible, compared to the electromagnetic interaction.
However, the tidal force affecting an astronaut, should require a little calculation. We can derive the formula of the gravitational acceleration around a point-like body ($F=\frac{GM}{r^2}$), we get
$$\frac{dF}{dr}=\frac{2GM}{r^3}$$
(We can ignore the sign on obvious reasons.)
Here $G$ is the gravitational constant, $M$ is the mass of the body, and $r$ is the distance.
Substituting the values of the Sun, we get $\frac{2\cdot 6.67 \cdot 10^{-11} \cdot 2 \cdot 10^{30}}{(7\cdot 10^8)^3} \approx 7.78\cdot 10^{-7} \frac{\mathrm{m/s^2}}{\mathrm{m}} \approx \underline{\underline{8 \cdot 10^{-8} \frac{g}{\mathrm{m}}}}$.
More clearly, if we are orbiting the Sun just above its surface, a roughly 2m long astronaut feel that his head and foot are pulled apart by around $1.6\cdot 10^{-7}g$ weight. In the case of a $70\:\mathrm{kg}$ astronaut, it is around the weight of $0.0112$ gram on the Earth.
The astronaut wouldn't feel it, but not very sensitive sensors could already measure it.

This calculation sometimes used $\mathrm{g}$ for "gram", as unit of mass, and $g$ as the (non-standard) unit of acceleration.


How imperceptible will the 4-July-2020 penumbral lunar eclipse be?

Completely imperceptible except to instruments.
Space.com reports

In general, most people don't notice the penumbral shadow projected on the moon until at least 70% of its diameter is covered. Some people who have very acute vision and better-than-average perception might notice an ever-so-slight shading when only 50% of the moon is inside the penumbra.

As your image shows, this is a lot less than 50%. Nothing will be observed by the naked eye.  "Visible" in your first source only means "the moon will be above the horizon at this time". It doesn't mean you will be able to see a change or dimming of the moon.