As pm-2ring already commented, the direction of the axis has a period of about 8.85 years.

It's worth noting, though, that in reality the moon's orbit does not correspond very closely to an ellipse, because it is rather heavily perturbed by the Sun.

Relative to this reality, the concept that the moon's orbit around the earth has a 'major axis' results from treating the real orbit approximately as if it were really an ellipse.  The approximations can be made in several different ways, and so in principle there can be a number of different approximating ellipses.

In effect the question presupposes the (idealised) rotating mean ellipse which results (in concept) when nearly all of the perturbation effects are treated as if they were averaged out -- except for the mean rate of axial rotation and the current position of the mean axis.

This idea of the mean ellipse is the one that underlies the 'mean elements' given in tables that express the fundamentals of the moon's motion, such as the 1954 Improved Lunar Ephemeris, which was slightly adapted from Brown's theory and 1919 Tables of the Motion of the Moon.

A more recent estimation of the position and rotation of the idealised mean ellipse, derived from lunar laser ranging via the JPL ephemeris DE200 (official basis of the Astronomical Almanac from 1984-2002), is found in J L Simon et al. (1994) "Numerical expressions for precession formulae and mean elements for the Moon and the planets", Astronomy & Astrophysics 282, 663-683.

According to this source, the mean longitude of the moon's perigee in arc-seconds (counted from the mean equinox of date) (see Simon 1994, p.670 at b.3)

$    =  300071".6752   +14648449".0869 t   -37".1582 t^2 $
$           -0".044970 t^3   +1".8948*10^-4 t^4 $

at any time defined by t, where t is the time-interval counted from the J2000 epoch of 2000 Jan 1 at 12h TT (= JD 2451545.0), in units of Julian centuries (36525 days) of Terrestrial Time).

The negative quadratic term in the expression above shows that the rate of rotation of the mean axis is very slightly slowing down; in 20 years it now rotates by about 1 minute of arc less than it used to do in 20 years around AD 1600 in the age of Tycho and Kepler.

Besides the mean approximation to the moon's orbital ellipse just described,
there is also the (variable) osculating ellipse. The largest periodic variations in the osculating longitude of perigee, relative to the mean perigee, are given in the same 1994 paper by Simon et al. at p.671, Table 4, 5th column.

Last but perhaps not least is the librating-and-rotating ellipse approximation which results when the rotating ellipse is taken to include the effect of the largest of the solar perturbations, now known as the evection.  This is described in older textbooks such as Godfray's Elementary Treatise on the Lunar Theory, see esp. pp.69-71 in the 1871 edition here.


How could the Moon have arisen from a collision - what would make for an elliptical orbit?

How could a mass of ejecta as large as the Moon all end up travelling together in a single direction and therefore a single orbit?

The Moon did not form this way. The giant impact hypothesis instead proposes a collision between the Earth and an object that was significantly larger than the Moon, roughly ten times the mass of the Moon. Per this hypothesis, large amounts of material ejected from the surface of the Earth. The energy released from the collision would have heated the impactor and the surface of the Earth immensely. The proto-Earth would have been enshrouded in an atmosphere of vaporized rock. Meanwhile, a handful of Earth radii out, there was an orbiting collection of bits, a few somewhat large, but most rather small.

Over the course of decades to centuries, these bits coalesced to form what would become our Moon. The Earth cooled and formed oceans. The close-by Moon raised huge tides on the Earth. Like today, but amped up by orders of magnitude by those huge tides, the tidal interactions between the Earth and the Moon made the Moon's orbit recess from the Earth. Now we see our Moon at about sixty Earth radii from the Earth, compared to the roughly one tenth of that when the Moon first formed.


What causes horizontal and vertical lines coming out of pictures of stars?

They are called diffraction spikes, and they're artifacts from a supporting structure inside a reflector-type telescope.


Comet Formation
From Altweg et al., Fred Whipple first came up with the idea of comets being "dirty snowballs" and Oort came up with the idea that comets originated in a very distant cloud around the Sun (now aptly called the Oort cloud).  Matter in the cloud gradually came together through gravity and other processes to form comets.  Now we know some comets also come from the Kuiper belt.
Comets are incredibly important in astronomy because they contain pristine matter from the early formation of the galaxy through the accretion of the protoplanetary disk, as shown in this illustration, also from Altweg et al.

More recent space mission analysis of comets has caused the community to more accurately refer to comets as "icy dirtballs."
Asteroid Formation
Asteroids are not as old as comets in some sense since, according to Raymond and Nesvorny:

The asteroid belt was dynamically shaped during and after planet formation.

The asteroids can differ much more significantly from one another than the comets.  Raymond and Nesvorny go on to say:

A number of processes may have implanted asteroids from different
regions of the Solar System, dynamically removed them, and excited
their orbits. During the gaseous disk phase these include the effects
of giant planet growth and migration and sweeping secular resonances.
After the gaseous disk phase these include scattering from resident
planetary embryos, chaos in the giant planets' orbits, the giant
planet instability, and long-term dynamical evolution.

Here is a cartoon from their paper comparing and contrasting three competing solar system development models for the creation of the asteroid belt:

Why do comets and meteors move so fast?
Comets are very slow moving until they get bumped into eccentric orbits with the Sun.  When we see them from Earth, they are moving at near their fastest speeds as they pass closest to the Sun, as an example of Kepler's 2nd law.
Meteors and meteorites are mostly moving around the sun at similar speeds as Earth before colliding, but in somewhat different directions.  This makes it appear that they are travelling very fast indeed with respect to the Earth as they enter the atmosphere.  In addition, they are sped up by Earth's gravitational pull.


Indeed space is not a physical entity (as far as we know). Saying space expands is another way to say that galaxies recede away from each other at a rate proportional to their distance.

The "expanding space" picture is unfortunately a source of endless confusion and misconceptions, as can be seen on the page of this very question: mpv mentioned that "The real expansion of space would manifest itself in the following way: you have 2 rockets in empty space, stationary with respect to each other. None of them starts its engines, but yet, they start moving away from each other."

This is plain wrong, the "expansion of space" is not some magical force that pulls objects away from each other. If you mean empty space as in no dark energy, no dark matter and no visible matter then the two rockets would move towards each other due to their mutual gravitational attraction (if you neglect their mass then they would not move at all). In the general case what will dictate the motion of the rockets is their gravitational attraction, the matter nearby (visible and dark) and dark energy (whatever that might be), the combination of which might pull them closer or further away from each other. The expansion of space is not an additional force. Galaxies don't keep moving away from each other because some undetected substance is created between them, but because they have an initial velocity which they acquired at the time of the big bang: without anything to stop them they keep moving away.

Joan.bdm said: "As I said in a large scale all objects recede from us, if it wasn't for the space expansion it would mean we are at the center of the universe."

The second part of the above sentence is wrong as well. As long as all galaxies recede from one point at a rate proportional to their distance, it follows that from ANY point all galaxies will be seen to recede at a rate proportional to their distance. This is easily seen with a little drawing (each * corresponds to a galaxy):

Before

After

If all galaxies are seen to recede at a rate proportional to their distance from the point of view of the galaxy at the center, then from any galaxy all other galaxies will be also seen to recede at a rate proportional to their distance.

I don't think we have any satisfying explanation as to why galaxies move away from each other at a rate proportional to their distance in the first place. The "expansion of space" is not an explanation, it is an analogy. A related and interesting read: Expanding Space: the Root of all Evil?


How does angular resolution of a telescope translate to its parallax precision?

Parallax measurement in practice is not as is explained above using the popular diagram you have used. The parallax causes the star to prescribe an ellipse in the sky, the semi-major axis of whose is equal to the parallactic angle.

The telescopes generally measure the shift in co-ordinates of star(RA and Dec) and then translate the information to that of the ellipse prescribed and determine the parallactic angle. Now, the ability of telescopes to note this change in RA and Dec is based on their least count, obviously which is dependent on their angular resolution.

Hence, distance measurements using parallax method have their range based on the angular resolution of the telescope being used.

Image source: http://documents.stsci.edu/hst/fgs/documents/handbooks/ihb_cycle16/c01_intro4.html


What is a "fundamental location", and why is Mosting A defined as one for the moon?

Every coordinate system needs one point defined as exact and unambiguous - the fundamental location which all other location measurements in the vicinity are relative to.
This can be the origin of the coordinate system (like 0° longitude and 0° latitude) - but it can also be any other point instead. You choose that (or one) point which you can revisit the easiest in a reproducable manner.
The given point in the center of the Mösting A crater previously was one of the fundamental points for the Moons reference frame as it was/is easy to identify: it is very round, has a sharp edge, thus the mid-point is easy to identify. There seems to be some original correspondance on the topic available from the late 19th century on ADS and discussions on improving the map accuracy prior to the moon landing which all refer to these properties as being the reason of choice.
Currently it's defined more accurately via the retroreflectors placed on the moon by the Apollo missions. As such the coordinate for that were calculated such that 0°/0° is at the center of the on average visible disk of the Moon and then the location for that retroreflector was defined to be the resulting coordinates. Without such fundamental point, different measurements might use (slightly) differently-defined coordinate systems and comparison of data would become difficult. The retroreflector on that location is the largest one, so the easiest to find.
It's basically similarily arbitrary that the 0° latitude on Earth pass through Greenwich observatory.


How can I find out the maximum magnification of my telescope?

The magnification is given by (focal length of telescope)/(focal length of eyepiece).  The spec I saw said the telescope focal length is 800mm, and the maximum magnification is 160x, so to get that would need a 5mm eyepiece.

Barlow lenses have the effect of increasing the telescope focal length: the 2x barlow would therefore give an effective focal length of 1600mm, and using the 9.7 mm eyepiece would give magnification of 160 (approx).  However be aware that the occasions when 160 magnification gives really clear views will be quite rare, as it is at the limit of what an 80mm scope can achieve.  The rule of thumb for maximum useful magnification for a good scope is 50x the diameter of the objective lens (in inches).


Non-baryonic dark matter (ie dark matter made of some particle that doesn't interact with light) had an important role in forming the large scale structure of the universe. The early universe was dominated by light radiation, and this would have prevented the early clumps of matter from growing. But dark matter isn't affected by radiation, so it can form clumps, which then attracts the normal matter towards it. Without dark matter, galaxies wouldn't have been able to form. You can read about structure formation on Wikipedia.

The second question is just about definitions. Generally, when we talk about a "vacuum" we mean a space with few molecules of gas or other atoms. This is because other particles, such as photons or dark matter particles don't contribute to pressure. A vacuum is a place with a very low pressure, not necessarily a place with no particles at all. So you can talk about a vacuum without needing to make a fuss over dark matter. However the actual density of dark matter is still very low: about the same as one hydrogen atom for every 3cm³. Even in the best vacuums that we can create on Earth, and even in interplanetary space, regular matter dominates.


mm-wavelengths Moon map, scientific case

Millimeter and sub-mm observations (110-300 GHz = 2.7-1.0 mm) are sensitive to the thermal emission and provide a brightness temperature of airless bodies like the Moon or asteroids. The radio emission is from just below the surface/skin of the body (down to $\sim10\lambda$ so ~1 cm in this case) and so is also sensitive to the thermal inertia and conductivity of the surface and regolith. A paper discussing sub-mm observations of the Moon (which is freely available) is Pardo et al. 2005.

A major issue with observation in the mm/sub-mm is atmospheric absorption due to precipitable water vapour (PWV) as seen in the plot below which shows the atmospheric transmission in the mm/sub-mm at the summit of Mauna Kea (~4100 meters):

Note that the worst case shown (in red) with 5mm PWV is already a very dry atmosphere that is rarely found even at other high mountain sites; sea level sites could easily have 50+mm of PWV. This could make these type of measurements very challenging.


How to calculate the orbital velocity, the high-orbit (quasigeostationary) holding force and the shadow of the disk on Earth?

Let's just pick up the second point. The disk is 42000 km (more or less) from the centre of the Earth, do the gravitational acceleration is $$9.81\times \left(\frac{6371}{41922}\right)^2 =  0.22 ms^{-2}$$ The component of that along the Earth's axis (Southwards) is $$0.22\times\frac{4505}{41922} = 0.024 ms^{-2}$$

The mass of the disk is $$5000^2\times\pi\times 4.5 = 3.53\times 10^8 kg$$ so the total force which must be counteracted by thrust is $$3.53\times 10^8\times 0.024 = 8.5\times 10^6 N$$

This is roughly the thrust of 4 of SpaceX's new Raptor engines burning continuously, each consuming about 1 ton per second of propellants. The best ion engines I can see quickly on wikipedia produce about $2\times 10^{-5}
 N/W$ (thrust/power consumption), so you'd need over $10^{12} W$ of power to achieve this thrust with  ion engines. The disk collects at most about 200GW of incoming sunlight, so the numbers don't add up that way. You'd also need quite a lot of Xenon and the ion engines would add a lot of mass.

Moving on to the shadow. Seen from Earth's surface the disk subtends an angle of about $\frac{10}{41922} = 2.5\times 10^{-4}$ radians. The Sun subtends an angle of about $0.01$ radians (half a degree). So the disk would not come close to blocking the Sun at all. That is there would be no umbra at all. In the penumbra the amount of sunlight reaching the Earth would be reduced by about 0.1%


I'm pretty far from an expert at ancient astronomy, but as I understand it, the ancient astronomers looked for the date when the planet was highest in the sky.   This is called being in opposition or opposite the sun.

Source of image

See here for further explanation.

I've also read that the synodic period can be calculated between periods of retrograde.   In neither case, the position of the sun is necessary.

The synodic period, because it was initially based entirely on observation, works for both Geocentric and Heliocentric models.

"Position of the sun" infers our current understanding of the heliocentric system.  The triangle, between the 3 points, Earth, Sun and observed planet and that only works for the heliocentric model.  Most of the ancient astronomers didn't use that model.  Their synodic period was based entirely on the motion of the planet relative to the fixed background stars.


The sun still rises in the East and sets in the West. So you quickly identify the cardinal directions.

By observing the point around which the stars rotate each night you can find the altitude of the pole, which gives you your latitude. You can't find your absolute longitude, but with good time keeping you can find your longitude relative to your starting point. The planets are still in the same orbits. You might find that Saturn doesn't have rings (it's not certain how old the rings are)

The Galaxies would still be in the same positions, The Milkyway, Andromeda and the Magellanic clouds are all visible. Many more with a moderate telescope, The Milkyway would look different, in detail, the LMC and SMC would be in different positions.

However, the stars in the background have all changed. There are no constellations that have not undergone radical changes. There would be no stars that you recognise: Stars are visible either because they are close (and so have a large proper motion) or powerful and short-lived. All the stars you know would either have moved, or not be born yet.


Why more matter still left after collision with anti-matter?

I presume you're asking about the mystery of the missing antimatter. This features extensively in the popular and scientific media. For example the CERN webpage on the matter-antimatter asymmetry problem is fairly typical. It says the Big Bang should have created equal amounts of matter and antimatter, but everything we see today is made almost entirely of matter. It goes on to say “something must have happened to tip the balance. One of the greatest challenges in physics is to figure out what happened to the antimatter”.

I think it's something of a "clickbait" mystery I'm afraid, to try to curry favour with the paying public. Why do I think that? Because there’s baryon asymmetry, where we see lots of protons but hardly any antiprotons, and there’s also lepton asymmetry where we see lots of electrons but hardly any positrons. And they match. Take a look at positronium. It’s “a system consisting of an electron and its anti-particle, a positron, bound together into an exotic atom”. The electron is matter, and the positron is antimatter. Positronium however is neither matter nor antimatter. It’s both. It’s an exotic atom. Now see page 314 of photon and electron collisions with atoms and molecules edited by Philip Burke and Charles Joachain. It says “the positronium atom thus created is in essence a very light hydrogen atom”. So positronium is both matter and antimatter, and it’s like light hydrogen.

The clear inference from that is hydrogen is also a combination of matter and antimatter. We say the electron is matter, and the proton too. But why? Surely since the proton is more like the positron than the electron, we ought to say the proton is the antimatter. Like this:

That would mean antimatter is a lot more common than people say. The proton mass is 1836 times the electron mass. Which would mean that weight for weight, 99.95% of everything around you is made of antimatter. Where’s all the antimatter gone? Nowhere. Because it’s hiding in plain sight. Because truth be told and by and large, you are made of it.


How do I calculate B-V to make a Hertzsprung Russell diagram?

Unless you know all the stars in the diagram suffer the same extinction, then you should de-redden them individually.

If you do know they have the same extinction (e.g. they are all nearby stars and it is zero; or they are all stars well above the Galactic plane in a similar direction; or stars in a cluster at the same distance) then you can plot a diagram using the observed magnitudes. In the latter scenario of a cluster, this diagram can be used to estimate the reddening of the cluster.

If you want to compare with models, to estimate temperatures or masses, then you must de-redden (or redden the model!)


Technically, yes, the moon does have an atmosphere that gets ionized by the sun.
NASA: Is There an Atmosphere on the Moon?

  our moon does indeed have an atmosphere consisting of some unusual gases, including sodium and potassium, which are not found in the atmospheres of Earth, Mars or Venus. It's an infinitesimal amount of air when compared to Earth's atmosphere. At sea level on Earth, we breathe in an atmosphere where each cubic centimeter contains 10,000,000,000,000,000,000 molecules; by comparison the lunar atmosphere has less than 1,000,000 molecules in the same volume.

Pretty thin soup. Sodium and potassium get kicked up by photons hitting the surface. There's also argon, helium and other minor components.
As shown in this figure from the first link:  The sodium in that atmosphere does glow; (Rayleigh unit definition). At least some of the glow is due to full fledged ions rather than charge-neutral fluorescence:

  It is confirmed that the ions include oxygen, sodium and potassium.

So, whether by ionic or fluorescent emission, the moon's atmosphere does glow. Unless one wants to argue about the glow not being restricted to the poles by a magnetic field, or airglow, that glow counts as an aurora.


Your hypothesis is correct: it is the tidal interaction between neighbours which generates spin.

Think of your proto halo (galaxy, stellar disc etc) as an ellipsoid (set by it's inertial tensor). Should the matter around it not be spherically distributed, it will apply a torque onto that ellipsoid and make it spin. The assumption in cosmology is that even though the primordial angular momentum you mention may be null, tidal interaction between neighbouring proto structures is sufficient to allow then to exchange (hence acquire individually) angular momentum, while preserving a zero sum.

A Reference on this subject is given by this paper

This diagram may enlighten the issue: it represents the tidal torquing of the blue ellipsoid by the tides represented by the pink ellipsoid.

PS: the expression spin up is unfortunate it seems because it is in fact spin conservation at the later stage of collapse of the proto halo which makes it rotate faster while preserving it spin (= total internal angular momentum).


What I ask myself is: Couldn't there be "negative" bundles of mass just the other way that pushes matter away instead of invisible dark matter that pulls it?

The galaxy rotation curve indicates a (positively massed) dark matter distribution that is close to spherically symmetric; cf. dark matter halo. I take it that you are asking whether instead of modeling it as a positive-mass distribution around the galaxy, we can model it instead as a negative-mass distribution surrounding the galaxy pushing the stars inward.

No, that's not possible. Since the effect is spherically symmetric, the hypothetical negative-mass distribution would need to be spherically symmetric as well. However, Newton's shell theorem guarantees that everywhere inside a void surrounded by such a distribution with this symmetry, whatever the sign of the mass is, the gravitational force vanishes. So a negatively-massed dark matter distribution outside the galaxy would do nothing at all.

In general relativity, the analogous statement is guaranteed by Birkhoff's theorem, and although GTR is nonlinear (and there are stars in the way), in our galaxy gravity is weak enough to be handled by the linearized theory.

  What is the proof for the fact "Gravity is always attractive" beside that fact that it seems to be?

In general relativity, gravity doesn't have to be attractive. For example, inside a perfect fluid with density energy $\rho$ and pressure $p$, $\rho+3p<0$ would imply that gravity is locally repulsive in the sense that a ball of test particles initially at rest expands rather than contracts under gravitational freefall. More generally, the strong energy condition characterizes whether or not gravity is attractive in this sense.

However, there is no known material that violates the strong energy condition, besides possibly dark energy, which is a cosmological constant in the standard ΛCDM model, as Gerald says.

  Has someone tried to find something like that?

There are many studies and simulations that consider dark matter with a nonzero pressure component, sometimes even anisotropic pressure components. For example, "hot" dark matter would have $p\sim\rho/3$, while "cold" dark matter would have $p\sim 0$, "warm" dark matter somewhere inbetween. The CDM in the ΛCDM model stands for "cold dark matter", naturally.

There are more exotic possibilities with anisotropic pressure components, etc. But I don't know of any paper that looked specifically for $\rho<0$ dark matter, and for the above reasons I'd be surprised if one existed.


Star Map Vs Star Catalogue vs Planetarium Software

A catalogue is just that what one would understand usually: it's a table of stars with their coordinates and possibly other properties like relative or absolute brightness, proper motion etc. The accuracy depends on how it was created, thus mostly is expression of the available instrumentation, processing pipeline and its limitations. They usually are output of a particular mission or survey and represent the observations in tabular format and help to make sure different people talk about the same thing. Often the stars in a catalogue are thus also assigned a number to make it easy to refer to.
The most recent star catalogues are those from the Gaia mission, the older yet still often used catalogues are like Hubble Guide Star Catalogue which hold MUCH fewer stars, but the brighter ones only. There's also specialized ones which focus on particular type of stars like the General Catalogue on Variable Stars - but all those are only a selection.
Star maps and planetarium software need catalogues as input - they simple read the table and then display it somewhat differently. Both need to know where to place stars. As such their accuracy depends for one on the accuracy of the catalogue(s) they use - and on the accuracy of their projection which is at least limited by the point or pixel resolution but also the need to draw stars with more than one dot or pixel, especially to also indicate different brightnesses.


Is it a coincidence that both the Sun and the Moon look of same size from the Earth?

The coincidence isn't so much that they appear very similar sizes from Earth, but that we are alive to see them at the point in time in which they appear very similar sizes.  The moon is slowly moving away from the Earth, and at some point in the future the moon will be unable to totally eclipse the sun and conversely, if you could step far into prehistory, you would be able to see the moon with a much greater angular diameter than you see it now.

Most research I've found on the topic seem to be unavailable through my institute, however I did find one paper, "Outcomes of tidal evolution", which references results from Goldreich's research on the subject.

  This qualitative description of the eventual disruption of the Earth-Moon system is confirmed by the results of Goldreich's numerical integration, which showed that the moon will recede to 75 Earth radii, when spin-orbit synchronism will be reached; then the Moon's orbit will decay steadily inward because of the influence of the Sun.

For reference, the Moon is currently at a distance of approximately 60.3 Earth radii.  As such, the moon will steadily move away until synchronism would be reached, and from that point begin to recede towards the Earth due to the tidal affects of the Sun on the Earth disturbing the synchronization. It would seem that at some eventual point in the far distant future, it will return to this coincidental position once again.

  Counselman III, Charles C. "Outcomes of tidal evolution." The Astrophysical Journal 180 (1973): 307-316.


The short answer is that the redshift of stars that are close enough to put into H-R diagrams (or "color magnitude diagrams", if we're being precise, given that you're talking about using colors) is so small that the effects are minimal.
The correction to color you're thinking of (called "K correction") depends on the redshift: the smaller the redshift, the smaller the effect.
But in order to distinguish and measure individual stars (so as to construct an H-R diagram), the galaxy has to be close enough -- the upper limit (for observations with the Hubble Space Telescope) is about 10 Mpc, or a redshift of $V \sim 1000$ km/s. For that kind of redshift, the K corrections will be $\sim 0.01$ magnitudes or smaller (you can test out different values using this online calculator -- note that it uses $z = V/c$ for redshift). This means the effect will be smaller than typical observational errors.
Put simply, galaxies that are far enough away to have large enough redshifts to worry about the effects on colors are too far away for you to observed individual stars and make H-R diagrams.
For stars within our own galaxy, the redshift (or blueshift!) effects will be due to relative velocities $< 200$ km/s or so, which means the effects will be really small.


What makes a flying object keep moving with the Earth's rotation?

They aren't.  To first order, airborne objects are moving with the Earth because they were moving when on the ground.  However, what they retain is their angular speed, and since the circumference increases with altitude, an object that doesn't accelerate appropriately will appear from the ground to fall behind to the West (the Earth is rotating to the East).  This is a well-known phenomenon whose magnitude is latitude-dependent, falling to zero at the poles.


List of Moving stars on the northern hemisphere? [Stellar kinematics]

What you're looking for is stars with "High Proper Motion".  A Google search of "High proper motion stars" will provide a number of such lists.  A quick look at Wikipedia gives this link:

https://en.wikipedia.org/wiki/Proper_motion#Stars_with_high_proper_motion

You can follow the links to individual stars if you wish to determine their latitudes.


Why are distant objects observed in the near infrared?

Yes.

When light moves through the expanding space, it loses energy and becomes redshifted. As galaxies form, they a emit a lot of UV light, and are often detected from there emission at the so-called Lyman $\alpha$ light, which has a wavelength of 1216 Å. Galaxies began forming just a few hundred million years after the Big Bang. Since then, the Universe has a expanded by more than a factor of 10 in all direction. The wavelength of the light expands by the same factor, and thus a Lyman $\alpha$ photon emitted in this epoch today has a wavelength of $\sim12,000$ Å, or 1.2 $\mu$m, which is in the infrared.


Gamma ray detection for SETI. How far away could we detect an alien nuclear powered rocket?

It's unclear if nuclear energy would be a choice propulsion for a space faring civilization.   It's possible, but it's not the only option.   That said, if it was the method of choice, it's not at all clear to me that it would be easily detected.

Let me preface this with stating that I'm not an expert.   There's two kinds of basic nuclear engine that could be used for space travel, closed or open.  An open nuclear propulsion is much simpler.   Basically explosions are made outside the ship that propel the ship by force from the explosion.   Tiny nuclear pelots are detonated and that's what pushes the ship.

The other method would be an internal nuclear engine, well insulated that generates electricity and electricity generates magnetic fields and presumably ion thrusters that push the ship.   We wouldn't detect nuclear energy if it's internal and insulated.

The Orion ship that you mention in your question wouldn't make one megaton explosions, but many smaller explosions.  It would carry more fuel than 1 megaton, but it would use it for tiny explosions over it's journey.    Space is also full of Gamma radiation, so I don't think gamma radiation would stand out.

You can't look for gamma-rays from Earth's surface because the Atmosphere blocks most of them, so you need a space telescope and our best space telescope for that is the Fermi Gamma-ray Telescope.  https://fermi.gsfc.nasa.gov/  (having trouble with linking, not sure why).

The Fermi scans the entire sky, so it's really more of a survey than a telescope, and gamma ray sources are still being studied, with 31% (per chart) listed as unknown sources.   https://www.nasa.gov/mission_pages/GLAST/news/gamma-ray-census.html

I don't think our current equipment is close to detecting a 1 megaton nuclear bomb a few star systems away, much less an orion type spaceship.   I think it's too few gamma rays at several light years distance.  Perhaps with better equipment we'd have a shot, but I don't think we're anywhere close to that now.    That's largely a guess though, I'm not an expert on the subject.

Edit

There's something wrong with my account, I can't comment either, so I'll post it here.

Your assumption is that 1 megaton is large enough to be visible from lightyears away - by gamma ray telescope.   I'm not sure that's true.

1 Megaton = 4.18 x 10^15 Joules.
1 Joule = 6.24 x 10^18 electronvolts.

So, 1 Megaton in explosive energy (and not all of it would go to gamma rays, but a decent fraction).   But ignoring that fraction, 1 Megaton = 2.61 x 10^34 Electron volts.

The surface area of a sphere 4.1 light years in radius (the distance to the closest star), 4 Pi r^2, where the radius is 3.88 x 10^16 meters, is 1.89 x 10^34 square meters.

So a single megaton blast 4.1 light years away would average out to a bit over 1 electron volt on a 1 meter gamma-ray telescope.    A typical gamma ray is thousands of electron volts, so unless you have a very large telescope, there's a good chance you wouldn't pick up a single gamma ray from a one megaton blast 4.1 light year away, or maybe you'd pick up one if you were lucky.

4.1 light years is very far.  Far enough that a telescope, unless it was enormous, would have a hard time seeing a nuclear blast that far away, and that's the closest star system.

Corrections are welcome but that's what the math looks like to me.


Musk's Roadster is now Russell's Teapot?

Small objects in space are hard to detect and characterise. Compare the tesla with another small solar system body: 2015TC25, It is currently about 0.3AU (3/10 of the distance from Earth to the sun) and it has a magnitude of about 30. That is perhaps detectable if you know exactly where to look. It would not be "discoverable" at this distance, with current technology. We know about because it made a close pass of Earth in 2015, and it was discovered during that pass.

The Tesla could be discovered if its orbit takes it close to the Earth. Observations of its spectrum would indicate it is not a regular asteroid, The cherry red paint would mark it as being utterly different from natural asteroids.

How close depends on what technology you have. By comparision with 2015 TC25, it might be observable from 0.3AU, if you have enough time and money to spend.


Where does the dust on 67P/Churyumov-Gerasimenko come from?

Your assumption that solar wind should have blown the dust away is only valid for comets that have a close perihelion. But you are correct in that there is a process which regenerates dust.
From Wikipedia

A comet will experience a range of diverse conditions as it traverses its orbit. For long period comets, most of the time it will be so far from the Sun that it will be too cold for evaporation of ices to occur. When it passes through the terrestrial planet region, evaporation will be rapid enough to blow away small grains, but the largest grains may resist entrainment and stay behind on the comet nucleus, beginning the formation of a dust layer. Near the Sun, the heating and evaporation rate will be so great, that no dust can be retained. Therefore, the thickness of dust layers covering the nuclei of a comet can indicate how closely and how often a comet's perihelion travels are to the Sun. If a comet has an accumulation of thick dust layers, it may have frequent perihelion passages that don't approach the Sun too closely.
A thick accumulation of dust layers might be a good description of all of the short period comets, as dust layers with thicknesses of order meters are thought to have accumulated on the surfaces of short period comet nuclei. The accumulation of dust layers over time would change the physical character of the short-period comet. A dust layer both inhibits the heating of the cometary ices by the Sun (the dust is impenetrable by sunlight and a poor conductor of heat), and slows the loss of gases from the nucleus below. A comet nucleus in an orbit typical of short period comets would quickly decrease its evaporation rate to the point that neither a coma or a tail would be detectable and might appear to astronomers as a low-albedo near-Earth asteroid.


Most stars form in clusters, so it is very likely that the Sun was part of a star cluster when it formed.

But in On the Dynamics of Open Clusters, the relaxation time of a cluster is calculated to be in the order of $\tau=4\times10^7 \textrm{yr}$. During that time, about one hundredth of the stars will escape from the cluster (i.e. reach escape velocity). The dissipation time of a star cluster is therefore in the order of a few billion years. These figures were calculated for an idealised star cluster containing stars of equal mass. For more realistic clusters the dissipation time may be much shorter. Especially dwarf stars seem to dissipate faster.

It is, therefore, likely that after 4.5 billion years, the Sun has escaped from its star cluster. It would be very difficult to identify the stars with which it was formed, unless you can trace back the position of the Sun and other stars before dissipation, which is impracticable.

As far as I am aware, no sibling stars have been positively identified. The sibling stars should be similar in composition (element abundances), although they do not have to be solar twins as the masses of these stars may vary widely.

EDIT: @adrianmcmenamin mentions an article in Sky & Telescope, where a sibling has been proposed. This refers to an article on arXiv by Ramirex et al. in which the element abundances have been analysed of stars that were earlier identified as possible siblings using galactic dynamics (i.e. tracing back the movement of these stars). Of the 30 stars identified as possible solar siblings using dynamical methods, only 2 stars had similar element abundances, which in itself says something about the accuracy of these dynamical methods...


LOFAR does not go 'through the ionosphere' as it's ground based. Rather, it is able to receive signals from outside the ionosphere due to the very low frequencies involved. These frequencies (naturally) have very long wavelengths, which means that LOFAR must be very large to obtain a decent level of resolution. The number of antennae will affect sensitivity and reliability.


How did UV from the earliest stars 'alter the state of the 21 cm line' such that it shows up in CMB today?

The Intergalactic medium at the relevant redshift is made of neutral hydrogen. What we can measure is the brightness temperature (the temperature that the IGM would have if it emitted as a blackbody) relative to the CMB. This quantity depends crucially on the following expression:

$\frac{T_S - T_{CMB}}{T_S}$

where $T_S$ is called spin temperature.

The spin temperature is just a measure of the ratio between the number of hydrogen atoms in the first excited hyperfine state (spin parallel) and ground state (spin antiparallel).
It turns out that the spin temperature can be modified in three ways: one of them is Lyman resonant scattering (the other two are collisional coupling and scattering of CMB photons).
Using atomic physics it's not difficult to see that $T_S$ is a weighted mean of the temperature of the CMB and gas temperature.

UV photons can change the spin temperature because a hydrogen atom in the lowest state, n=1 (if you are familiar with chemistry, a s state) with antiparallel spins, can absorb an UV photon and jump in a p (n=2) state, and then fall again in a s state but with parallel spins.
This mechanism is known as Wouthuysen–Field coupling.

Before the emission takes place, the gas is at the same temperature of the CMB and both are equal to spin temperature, so the relative brightness temperature is zero.
When Lyman alpha emission begins the spin temperature decreases resulting in the observed absorption peak.
At some point the peak stops because the first stars emit X-rays and the gas becomes hotter than the background radiation.


Thinking about the Twin Paradox, what would a Twin Traveling at .8c, calculate as the amount his EBT aged at the mid-point in a 4 lightyear trip?

I find it easier to just write down the coordinates of the events of interest.
Given coordinates $(t, x)$ in the frame of the Earth and choosing units such that $c=1$, the Lorentz transform to the frame of the traveller is $(t', x')$, where
$$\begin{align}
t' &= \gamma(t - \beta x) \\
x' &= \gamma(x - \beta t)
\end{align}$$
In your scenario, $\beta = \frac{v}{c} = \frac{4}{5}$, giving $\gamma = (1 - \beta^2)^{-1/2} = \frac{5}{3}$. The relevant events, in the reference frame of the Earth, are then:

Traveller leaves Earth at $(0, 0)$.
Traveller reaches halfway point at $(\frac{5}{2}, 2)$.
Earth is at the origin for all $t$, giving coordinates $(t, 0)$.

Transforming these into the traveller's frame, these become:

Traveller leaves Earth at $(0, 0)$.
Traveller reaches halfway point at $(\frac{3}{2}, 0)$ — as expected, the space coordinate is zero because the frame is moving with the traveller.
Earth is located at $(\frac{5}{3}t, -\frac{4}{3}t)$ — note this is expressed in terms of $t$, the time coordinate in the reference frame of the Earth. As expected, the Earth falls behind the traveller as time advances.

Equating the time coordinates in the traveller's frame gives $\frac{5}{3}t = \frac{3}{2}$, which gives $t = \frac{9}{10}$ years for the time that has passed on Earth, as viewed in the traveller's frame (who would have experienced a duration of $\frac{3}{2}$ years up to this point).


Dark matter hypothesis and visible light

Nice thought, but as the saying goes, space is really big. Stars have a tiny angular size when viewed from even their nearest neighbours (except perhaps in clusters & galaxy cores), so not much starlight actually gets intercepted by other stars, relative to the amount of light that gets absorbed by interstellar (or intergalactic) gas and dust.

And even that absorbed light isn't actually lost: it increases the kinetic energy and the temperature of the gas & dust. So some of the energy is re-radiated at a lower frequency, typically as infrared or microwave radiation. By looking at the spectrum of that radiation we can determine its temperature, and use Doppler techniques to estimate the kinetic energy of the gas & dust.

When a star intercepts light from another star, the light energy isn't destroyed. Some of the light is actually reflected, the rest of the energy is assimilated by the receiving star's atmosphere, where it will contribute to the receiving star's own light emissions.

Astrophysicists take these things into account when evaluating the brightness of galaxies. Plenty of people have tried all sorts of alternative hypotheses to dark matter, but so far dark matter is the best explanation that we have.


What does "the depth of the 13.0 Å feature differs by 7 sigma from zero" mean?

In this case itseems to mean that the depth of the line is 7 times its error bar below the continuum level.

Impossible to answer. You say it can't be done, but the authors say that they fitted a Gaussian.

You either use a rough estimate (attributable to Cayrel de Strobel 1988) of
$$\Delta {\rm EW} \sim 1.5\frac{\sqrt{RP}}{{\rm SNR}},$$
where $R$ is the resolution element (the FWHM) of your instrumental wavelength response, $P$ is the pixel size (both in the same wavelength units) and SNR is the signal-to-noise ratio per pixel. Or you use an integration method and combine the SNR of each pixel used to find the uncertainty in the integral. Or you fit a Gaussian but using the area under the Gaussian as one of the fitting parameters.


How did astronomers first come to understand that the Sun's core was a gas?

Well simply said: All data fits the premise that it should be a gas.

But what data?

First of all, in the late 1960's the 5-minute solar oscillations on the surface of the sun have beend discovered. Those have been recognized as surface features of internal pressure waves with penetration depth $d$, wavenumber $k$ and their relation roughly $d \sim  \frac{1}{\sqrt{k}}$.
So from the measurment and simple existence of certain wave-ripples on top of the photospheric plasma-ocean we can learn about the interior of the sun. To relate the strength of certain wavemodes to the interior structure of this plasma-ball we use usually a reverse-engineering approach called modelling. We take what we know about gases, energy transport etc. and calculate from this how a solar-oscillations-spectrum on a modelled sun would look like. Then compare, scratch your head, change the modell and repeat until it fits. More mathematical details, if you are interested can be found in the excellent lecture notes on Helioseismology from Christensen-Daalsgard.
So all this gave us a pretty good look into the sun's interior and it was quite clear, that the observed solar parameters like, size, energy output, surface temperature must lead to certain Pressure, Temperature and Density values for the core corresponding to a fusing plasma.
But those theoretical predictions would imply the core-plasma to create neutrinos from nuclear reactions, which famously (wiki: Solar_neutrino_problem) were off by a factor of about 3. With the discovery of neutrino oscillations (wiki: Neutrino_oscillation) this problem was also solved in that the sun didn't produce too few neutrinos, but the produced neutrinos would simply change their types. After some distance from the sun those changes would redistribute the numbers from 3:0:0 for electron-myon-tau neutrinos to 1:1:1.
This problem persisted until approx. the year 2000.


Does light from the sun reach the surface of Venus

Yes, a surprising amount of sunlight does reach the surface, so you very nearly answered your own question. The Venera spacecraft used visible light to take its pictures. Bear in mind that being nearer to the sun than we are, Venus gets more light from it than we do. What surprised me is that like Earth and Titan, there is a substantial gap of clear air between the ground and the cloud base. The atmospheric pressure is 90 atmospheres, and temperature about 460 C. On Earth, sulphur dioxide from volcanoes and other sources is oxidised in a few weeks to become sulphur trioxide, which forms sulphuric acid on contact with moisture. This process is impossible on Venus, where there is only a minute trace of oxygen in the atmosphere, so the origin of the clouds of sulphuric acid said to be there is a bit of a mystery. Photographs of the surface show no sign of acid erosion.


How can we estimate if a certain astronomical object will be visible from a place on earth?

A simple estimate can give you a rough idea what area of the sky will be visible at different times of the year, using only a bit of mental arithmetic.

I just remember one number, the right ascension (RA) at (or nearest to) the meridian (between due S and the zenith) at midnight on Jan 1st. This is (roughly) equivalent to local sidereal time. Actually, I just remember the hour, 06h, at my longitude of just under 72 degrees E. This changes very slowly from year to year due to precession (I may have to use 05h by around 2050 or so, if I am still alive!). Since the difference between sidereal time and UTC is about 4 minutes per day, RA on the meridian at midnight shifts about 2 hours later per month. Now around Jul 15th, about 19h RA is on the meridian at midnight here. Latitude determines what range of declination can be seen. At ~42 degrees N, I usually cannot see much below -30 degrees declination, depending on the horizon.

Knowing what RA is near the meridian (I can easily subtract or add hours corresponding to time before or after midnight), by looking at a sky chart I can easily estimate what objects should be visible, or well placed for observing. I used this simple technique for many years, before digital everything! Now I can just use SkySafari on my iPad. :-)


List of galaxies with their dark matter halo

This paper by Wechsler & Tinker (2018) should give you a better direction. Depending on the model, different estimates of DM halos come up. You just need to go through the sections on observational evidence and the underlying papers.
https://arxiv.org/pdf/1804.03097.pdf


Is there any plausible astronomical source of energy that could wipe out humanity without rendering the planet inhabitable?

The cockroach and the tardigrade can withstand just about anything.  Next, any radiation blast consisting of wavelengths reasonably attenuated with water wouldn't affect, directly at least, all the deep-sea creatures.


Is a General Relativity approximation available in Universe Sandbox (1 or 2)?

As noted by @MikeG in this comment, the Universe Sandbox 2 FAQ addresses the question, but not in a satisfying or helpful way:

Does it account for relativity?
No, the physics in Universe Sandbox² is currently only Newtonian.
Why? The short answer is that you need a supercomputer to accurately simulate general relativity. (emphasis added)
Jenn, astrophysicist and Universe Sandbox² developer, explains more in a blog post: "General relativity requires simulating the spacetime itself. That is, taking your simulation space, discretizing it to a hi-res 3-D grid and checking the effect that each and every point in that grid has on all neighboring points at every timestep. Instead of simulating N number of bodies, you are simulating a huge number of points. You start with some initial data of the shape of your spacetime and then see how it evolves according to the Einstein equations, which are 10 highly non-linear partial differential equations."

Why do I say this is not very satisfying or helpful?
Because it's quite easy to nearly completely account for General Relativity (GR) effects in the orbits of solar system bodies just by adding some terms to the ODEs that are integrated to propagate orbits.
Answers to the question How to calculate the planets and moons beyond Newtons's gravitational force? (including mine) show how this is done all the time. I reproduced what's normally done in Python rather than in a compiled language for transparancy, it would be much faster if included in a reasonable simulator implementation.
As explained here, to the Newtonian acceleration given by:
$$\mathbf{a_{Newton}} = -GM \frac{\mathbf{r}}{|r|^3},$$
You can just add this expression:
$$\mathbf{a_{GR}} = GM \frac{1}{c^2 |r|^3}\left(4 GM \frac{\mathbf{r}}{|r|} - (\mathbf{v} \cdot \mathbf{v}) \mathbf{r} + 4 (\mathbf{r} \cdot \mathbf{v}) \mathbf{v} \right),$$
To get close, you certainly don't need to evaluate these GR terms for every pair of objects. Interaction between the Moon and Ceres for example could still be Newtonian.
All approximations are approximate. The FAQ's suggestion that they don't do it because you need a supercomputer to be accurate is slightly misleading. Using this straightforward approximation is much better than not using it in this context.
Yes for Gravitational Waves, collapses, and other effects "requires simulating the spacetime itself. That is, taking your simulation space, discretizing it to a hi-res 3-D grid and checking the effect that each and every point in that grid has on all neighboring points at every timestep" but just adding the major GR effects to the propagation of solar system bodies certainly doesn't need anything more than  whatever computer you are already running your copy of Universe Sandbox 2 on already.
EDIT:
The FAQ answer continues:

We are, however, interested in adding in a few features which would address some effects of relativity. One example is setting gravity to travel at the speed of light, instead of instantaneously taking effect as it currently does. You can read more about these in Jenn's blog post: Gravitational Waves & Universe Sandbox².

I had thought so as well, but according to @DavidHammen's thoughtful answer this is probably not so. I haven't tested this yet, but since this comes from a reliable source, I tend to believe it. Retarding gravity alone, without these other terms, seems to be the wrong thing to do in this case.


It seems Walter Baade and Fritz Zwicky may have coined the term in 1934 (see here) and the name stuck around.

I've known some astronomers to label them 'stellar remnants' for the reason you gave; they are not stars in the sense they do not produce energy via nuclear fusion, and that they are not made entirely of neutrons as the name would suggest.

They seem to be called stars for lack of another layman's term (from our point of view on Earth, not all neutron stars are pulsars), and that they still share some characteristics of stars: they emit their own light, they have masses greater than the Sun and can even have planets orbiting them.


Are the Venusian "continents" likely to have existed before the global resurfacing event?

Ishtar Terra belongs to the tessera type of terrain, "one of the most tectonically deformed types of terrain on Venus", representing ~7.3 % of Venus' surface (Ivanov & Head 2011, Global geological map of Venus). Tessera are "dated" (relatively, from stratigraphic relationships) from the Fortunian, the oldest period:

The most important features of the Fortunian are large tessera-bearing
crustal plateaus (e.g., Fortuna, Ovda, etc.) that form a specific
class of first-order (a few thousands of kilometers across) highs on
Venus [...] The lowest stratigraphic position of tessera and its
relative abundance show that this unit is of extreme importance as a
probable ‘window’ into the geological past of Venus.

Stratigraphy of defined and mapped geologic units on Venus. The
visible geologic history of Venus consists of the Fortunian,
Guineverian, and Atlian Periods. The major portion of the surface of
Venus (about 70%) was resurfaced during the Guineverian Period. See
text for discussion. T is the mean model absolute age of the surface

Further works by Ivanov & Head detail the history of volcanism (2013) and tectonism (2015) on Venus. They consider tessera themselves as a resurfacing process:

The globally observed stratigraphic relationships among the tectonic and volcanic units that make up the absolute majority (~96%) of the surface of Venus divide the observable portion of its geologic history into three different episodes, each with a specific style of resurfacing. These are as follows: (1) Global tectonic regime, when tectonic resurfacing dominated. Exposed occurrences of these units comprise about 20% of the surface of Venus. (2) Global volcanic regime, when volcanism was the most important process of resurfacing and resurfaced about 60% of Venus. (3) Network rifting-volcanism regime, when both tectonic and volcanic activities were about equally important. During this regime, about 14% of the surface of Venus was modified.

So, to summarize, "continent-like" (tessera) terrains like Ishtar Terra have existed for a long (yet unknown) time, and have survived the resurfacing from volcanism that covers most of the surface today. But they are themselves considered to represent a (tectonic) resurfacing process that has affected older (and unknown) pre-Fortunian terrains.

A global correlation chart that shows the three major regimes of
resurfacing on Venus: global tectonic regime, global volcanic regime,
and Network-rifting-volcanism regime.


Where can I find a database of galactic spectra?

The Sloan Digital Sky Survey Data Release 15 contains over 4 million spectra of both galactic and extra-galactic origin from the multi-fiber spectrographs. Of these spectra, 0.7 million came from the original spectrographs during the SDSS-I/II Legacy Survey and the remainder from the upgraded spectrographs as part of the BOSS survey during SDSS-III (see SDSS surveys details pages). These data went through an automatic pipeline that determined the redshift automatically (where the data quality was good enough) and the details of the process are given here.

The spectroscopy data are available through the links to the SkyServer Quicklook which lets you visualize and download the spectra and through the SkyServer Spectroscopic Query Form which lets you search for galaxies. (Try setting the 'Parameters to return' to 'typical' under the 'Spectroscopy' section' and setting the 'Classification=GALAXY' under the 'Spectroscopy constraints' section to only return objects that have been classified as galaxies).


How frequently are other planets in the Solar system hit by meteorites?

Moon
Even though the Moon obviously isn't a planet, it's a good place to start as most of the observations of meteorites hitting other bodies has been focussed on the Moon. NASA runs the Meteoroid Environment Office which in part monitors lunar impacts. They state;

  The lunar impact rate is very uncertain because observations for objects in this mass range are embarrassingly few. Source

This statement from what I can tell hold true for most objects in the solar system aside from the Earth. However according to this source reckons the moon receives around 2800kg of meteor material each day, which seems about right. The LRO and increased amounts of astronomical observations have led to this increased knowledge on the issue.

Jupiter
Aside from the Moon, Jupiter is probably the second most studied object in regards to meteorites. This is because it regularly has some fairly energetic collisions due to its large gravitational pull. According to this paper Jupiter can receive around 12-60 meteorites with comparable strength to the Chelyabinsk airburst each year.

Also noteworthy was the Shoemaker-Levy impact in 1994, which gave a lot of importance to the study of collisions and NEO studies.

Comet Shoemaker-Levy 9 Impact

Source

Mars
Noteworthy planet as its only planet where we've found a meteorite on the surface and studied it up close. Opportunity, Spirit and Curiosity all found candidates for meteorites. Curiosity studied up close the iron meteorites Lebanon A and B.

Source

Earth

  On average, 33 metric tons (73,000 lbs) of meteoroids hit Earth every day. Source

Now while meteoroids don't count as meteorites, but this is useful in gaining a sense of the relative scale. The atmosphere of the Earth means that the meteoroids don't reach the surface, which separates Earth from objects without an atmosphere. Most atmosphere-less objects in the solar system have a fossil record of the amount of collisions they have received, in the form of all the craters on the surface.

Summary;
There's still a lot of unknown information in the area, but observations are beginning to ramp up and with that we can began to gain an idea of just how frequently objects in the solar system are hit by meteorites, so in the future a number could be put forth for each planet in the solar system.


Would a future hole in the Cosmic Microwave Background show us the edge of the universe?

First, let me clear up a misunderstanding:
Particle horizon
The "edge" of the observable Universe is called the particle horizon, and lies roughly 47 Gly (billion lightyears) away. It is always receding, both because the Universe expands and because light from increasingly large distances eventually reach us. In comoving coordinates (the coordinate system that expands along with the Universe), it asymptotically reaches a finite size (some 63 Gly), but in physical coordinates it increases forever$^\dagger$.
Event horizon
The event horizon is the boundary between the part of the Universe that contains the galaxies (and other stuff) from which light emitted today will be able to reach us, and the part where it won't. Currently, it is roughly 17 Gly away, but his horizon is always shrinking because the expansion of the Universe accelerates. In other words, stuff outside the event horizon is so far away that the expansion carries it away too fast for the photon amitted today to ever reach us.
Thus, at Big Bang (BB), the event horizon was 63 comoving Gly away, because that was the maximum distance from which we would ever receive information.
The CMB will always lag behind the particle horizon
The farther away we look, the more we look back in time. If there were galaxies a second after BB, and if we had the technology to look that far, we would see them right after BB, and they would be 47 Gly away. However, galaxies weren't born until several hundred Myr (million years) after BB. And now to your question: The CMB wasn't emitted at BB, so it doesn't origin from a region that is now 47 Gly away. Instead, it was emitted 380,000 yr after BB, and it was emitted from all of the Universe, in all directions.
This emittance wasn't instantaneous, but almost (took something like 10,000 yr), so let's assume for now that it was. Let's call that time $t_\mathrm{CMB}$. In that case, because light travels at $c = 300\,000\,\mathrm{km}\,\mathrm{s}^{-1}$, if you were present one second after $t_\mathrm{CMB}$, you would see the CMB that came from all regions that were 300,000 km away. That is, the CMB would be the light from a thin shell of radius 300,000 km.
Two seconds after BB, the CMB would be the light from a shell of radius 600,000 km, and so on. However, because the Universe is expanding, it is actually at little more than 600,000 km in radius.
And now, 13.8 Gyr after BB, and hence 13.7996 Gyr after the CMB was emitted, this shell is not almost 14 Gyr away, but in fact 45 Gly away which is almost, but quite, at the particle horizon.
Because of the expansion, the distance to the CMB shell always increases. But so does the distance to the particle horizon, and because the CMB was emitted after BB, the particle horizon will always "be ahead" of the CMB. As the CMB shell recedes, it becomes more and more redshifted, slowly fading into darkness but never disappearing. This is irrespective of whether the Universe is finite or infinite, except in the possible, but rather fantastic, event that the Universe is larger than the ~47 Gly we can see today, but smaller than the 63 comoving Gly that is the maximum size we will ever see.
Finite or infinite?
We don't know if the Universe is finite or infinite, but if the un-observable part of it looks like the observable Universe, then it is infinite. What we can say is that, at least in the observable Universe, the CMB was emitted everywhere in space, and there is currently no reason to believe that it was different elsewhere (although it could bee, if it expanded differently in other parts). To the best of our knowledge, space is homogeneous, meaning that (on average) there are as many baryons, photons, dark matter particles, and bicycles in one place as there is in another. In other words, to the best of our knowledge, there is no edge. Even if it's finite, it is difficult to (and most, if not all, astronomers don't) imagine an edge; rather it is thought that it would "curve back on itself", just as the surface of a ball is finite in area, but doesn't have an edge.
2D analogy
As an analogy, imagine living on a flat, inifinite Earth, populated evenly by people with loud voices. At some point in time, everbody yells "CMB". First you'll hear your neighbors yell, after 1 sec you'll hear those living 340 meters away, after 1 minute those living 20 km away, after a year those 10 million km away, and so on and so on forever.
Now imagine the same on a round Earth. After 16 hours, you would hear the folks living at your antipode, and if I understand correctly, this is when you expect to "hear a hole in the sound". But now you would start hearing the ones that are closer, but whose sounds have been traveling more than halfway around Earth. After 32 hours, you'd hear your own yell. And so on.
But, our Earth analogy is also expanding. If it were expanding at a steady rate, we would eventually hear our antipode friends. You might think that if it were expanding fast enough, their sound would never reach you, but as explained by the ant-on-a-rubber-band puzzle, that's not the case).
However, our fictional Earth is not only expanding, but in fact this expansion is accelerating. And in that case, there is a limit to how far away you can hear people.
Back to the real Universe, there is a limit to from how distant regions you may receive CMB photons (or any other information).
Moreover, if the Universe were finite, and were not accelerating, you might think that you would eventually see the CMB that was emitted from our own neighborhood and had been traveling around the Universe.
However, it can be shown that, in this case, the Universe will reach a maximum size, start collapsing, and end in a Big Crunch exactly at the time where you expect to see your antipode.
Only in a finite Universe with a rather special mix of gravitating matter and accelerating dark energy might it be possible to delay the collapse or the eternal expansion to a point where we eventually look back at ourselves. But that's sort of fine-tuning.

$^\dagger$Unless the Universe contains a yet unknown substance that is able to halt the acceleration. This is entirely possible, but in this answer I consider only a well-behaving Universe that acts according to current knowledge.


How strong are the tides raised by Io on Jupiter relative to the ones raised by the Moon on Earth?

Both expressions are incorrect. The first should be
$$\frac{GM_{\text{moon}}}{(R_{\text{moon}}-r_{\text{planet}})^2} - \frac{GM_{\text{moon}}}{{R_{\text{moon}}}^2}\tag{1b}$$
or
$$\frac{GM_{\text{moon}}}{{R_{\text{moon}}}^2} - \frac{GM_{\text{moon}}}{(R_{\text{moon}}+r_{\text{planet}})^2}\tag{1a}$$
where $R_{\text{moon}}$ is the distance between the center of the planet and the center of the moon. Equation (1a) pertains to the point on the surface of the planet closest to the moon while (1b) pertains to the point on the surface of the planet furthest from the moon.

Note that the results have units of acceleration rather than force. The tidal force on a small mass on the surface of the planet is the product of the tidal acceleration and the mass of that small mass.

Since $R_{\text{moon}} \gg r_\text{planet}$ in the case of Io and Jupiter, and also in the case of the Moon and the Earth, the two expressions are nearly equal to one another, and both in turn are nearly equal to
$$2\frac{GM_{\text{moon}} r_{\text{planet}}}{{R_{\text{moon}}}^3}\tag{2}$$
This is the correct version of your second expression.

On plugging in the numbers one finds that the tidal acceleration exerted by the Moon at the surface of the Earth is 1.068 μm/s^2 while the tidal acceleration exerted by Io at the surface of Jupiter is 9.014 μm/s^2 -- over 8 times the tidal acceleration caused by the Moon at the surface of the Earth. This does not quite show the extent of the tides raised by Io on Jupiter (or by the Moon on the Earth).

It is important to note that equations (1) and (2) pertain only to two special points, the submoon point and its antipode. The tidal acceleration at the point where the moon is on the horizon is half the value of equation (2) and is directed inward. What would be nice would be a potential model whose gradient results equation (2) and also in the inward acceleration at the horizon. This potential function is, to first order,
$$V = \frac{GM_{\text{moon}} {r_{\text{planet}}}^2}{{R_{\text{moon}}}^3}\frac{3\cos\phi^2-1}{2}$$
where $\phi$ is the angle between the line segment between the center of the planet and the center of the moon and the line segment between the center of the planet and the point in question on the surface of the planet.

If the planet consisted of a fictitious frictionless fluid, the fluid would form a prolate spheroid whose height at some angle $phi$ differs from what would be the planet's moonless spherical shape by $V(\phi) / g_\text{planet}$. The height of the tidal bulge is the difference between the maximum at $\phi=0$ and the minimum at $\phi=\pi/2$, or
$$h = \frac32 \frac{M_{\text{moon}}}{M_{\text{planet}}} \frac{{r_{\text{planet}}}^4}{{R_{\text{moon}}}^3}$$

On plugging in the numbers one finds that the tidal bulge created by the Moon on a fictitious frictionless fluid Earth would be 0.5327 meters while the tidal bulge created by Io on a fictitious frictionless fluid Jupiter would be 22.52 meters -- over 42 times the size of the tides raised by the Moon on the Earth.

While neither Jupiter nor the Earth is composed of such a fictitious fluid, the tidal bulge height nonetheless does provide a nice picture of how large the tides that result from a moon can be.


What are the smallest star clusters affected by Galaxy Rotation Curve?

Yes, dwarf galaxy rotation curves are affected -- in fact, they tend to require relatively more dark matter than is required to explain the rotation curves of giant galaxies like the Milky Way. The smallest known systems are so-called ultrafaint dwarf galaxies (UFDs), which can have as little as $\sim 10^{4}$ solar masses worth of baryons (in the form of stars), but apparently have as much as 100 times this in the form of dark matter -- versus the roughly 10:1 (DM:baryons) case for galaxies like the Milky Way. This means that some of these systems have fewer stars than typical globular clusters.

But globular clusters, on the other hand, do not show signs of dark matter/non-Newtonian motion: you can explain the motions of their stars as being due to the normal, Newtonian gravity of all the baryonic mass (i.e., the stars, plus a plausible amount of stellar remnants like neutron stars and stellar-mass black holes).


The sun does not orbit another star. A star, even a very small brown dwarf, would have been seen, if not in visible light then in the infra-red by the WISE survey.

However there is no mystery about precession. The Earth is non-spherical, it is wider around the equator, and because the Earth is tilted with respect to the ecliptic, there is an asymmetry in the force of the sun, and the moon, on the Earth. This force pulls the axis of the Earth's rotation in a circle. The basic maths of this is not so hard, and is in the Wikipedia page on Axial precession. This explanation of axial precession was known to Newton. There is nothing controversial about it.

The sun does orbit the galaxy it takes about 230000 years to complete one orbit. This has nothing to do with axial precession.

The page you link to is worthless. It offers an "alternate explanation" of a phenomenon that is very well understood, and pretends that there is some mystery, where there is none. It the uses this to justify a lot of astrological nonsense.

To repeat.

There are not "many theories" explaining Axial precession.
Axial precession is a well understood phenomenon, due to the Earth not being a sphere and the axis of the Earth not being perpendicular to its orbit
The sun does not orbit another star. We know this because we could have seen it.
No astrology based on the presence of another star is valid.


Helioseismology of the Sun, what is actually measured?

The two main methods to detect the solar p-mode oscillations, which have a period of about 5 minutes (so frequency of ~3 milliHertz) are line of sight Doppler shift and change in irradiance.

The Doppler method works by measuring the intensity of light in (normally) two narrow regions either side of a prominent spectral line in the Solar spectrum and using this change to measure the velocity shift caused by the oscillations (once the solar rotation is removed). The ground-based GONG instrument uses the Ni I at 676.8nm, BiSON network and the GOLF instrument on the SOHO spacecraft both use the sodium D lines. After isolating the light around a suitable spectral feature, then you can either use a Michelson interferometer to get a velocity image of the shifts across the surface (GONG and MDI on SOHO) or a disk-averaged velocity, treating the Sun as a star (BiSON and GOLF on SOHO). The difficulties are in keeping everything stable so that the measured shifts only come from the Sun and not the instrument. For example GONG note that their optical elements are stabilized in an oven to the order of 0.00001 K.

The irradiance method simply involves measuring the disk-integrated intensity of the Sun and watching for changes caused by the oscillations. However fluctuations in the Earth's atmosphere make this very difficult to do from the ground and the most successful has been the VIRGO instrument on the SOHO spacecraft.

The imaging instruments have a resolution of about 2" (MDI) to 5" (GONG) so return velocity data at approx. 800,000 to 100,000 locations across the Sun. This allow the study of oscillations with very high degree (up to 1000) which have very small scale on the surface. The disk-integrated "Sun-as-a-star" instruments by comparison are sensitive to the low order oscillations.


Question related to Sun rise and Sun set times

I understand 'Sun rise' and 'Sun set' is caused by 'earth rotation'.
  I read 'Sun light' takes around 8 minutes to reach earth
  Every city in this world has its own 'Sun rise and sun set' timings.

This is all correct. The apparent motion of the sun in the sky is mostly due to the rotation of the Earth.

  Lets assume 'sun arises' at 6 AM in 'City A'.
  Does it mean that 'City A' sees / feels the light of Sun which was in the sky @ 5:52 AM ?

It means that at 6AM, light from the sun shines on city A. That light had left the sun at about 5:52. At 5:52, city A was in the shadow of the Earth. The time that the light left the sun is fairly irrelevant to the timing of sunrise. A 5:52 the light left the sun, in the 8 minutes that followed the Earth rotated by about 2 degrees and by 6 AM city A was just in the right position to catch a little of the light. We don't change the timings. We would not say that the sun rose at 5:52, because at 5:52 the sun appears to be below the horizon.


Giant variation of the proposed eLISA mission using reflectors on Earth and the Moon possible?

This would not be a giant version of eLISA, but a small (and complicated) one.

The distance between the LISA satellites will be 1 million kilometers, while the moon is on average 380.000km from Earth. So that arm would be less than half as long. Geosynchronous orbit is at 36.000km, not even 10% of the distance to the moon. The difference in length between your proposed detector arms is much shorter than the coherence length of any laser, so your interferometer would not work.

Further problems:

Positioning of mirror on Earth: Where do you want to put the mirror such that the moon is always visible in the sky?
Positioning of satellite: I doubt (without checking) that you can see the moon at all times from geosynchronous orbit. Maybe you could work with one of the Earh-Moon Lagrange points.
Constant mirror adjustment: Earth is rotating, and the Moon revolves around the Earth. So your interferometer arms would change angles constantly. (And you'd need to dig lots of tunnels straight through Earth, see above.)
Vibration: The mirrors have to be as free from vibration as possible. That is much easier to achieve with a free floating satellite than on earth or the moon. (Did you know there are moon quakes?)
Ellipticity: The distance to the moon varies between 360.000km and just over 400.000km during its orbit. So you'd be constantly adjusting the interferometer. A constant drift is not necessarily problematic, but the huge difference in distance might be.

Lastly, it is not clear to me how you want to replace precision measurement by calculation. What is PLL? How do you get meaningful interference patterns with lasers of different wavelengths? Also note that the "exact distance", if you want to count waves, must be better than half the wavelength of your laser, so a few hundred nanometers. At the distance of the moon, that is better than $1 : 10^{15}$ (hope that calculation is correct).


Cosmological redshift and comparing past galaxy sizes

Comoving separations go with the scale factor. Comoving volumes scale as $(1+z)^{-3}$.

At $z=3$, that volume was 64 times smaller.


Using setting circle for Right Ascension

I think I mis-read your question before. Let me rephrase it to make sure I understand. In step D (moving the scope from Sirius to M42), that process took 5 minutes to complete. Or maybe it took 30 seconds to complete, but you then took a 4.5 minute coffee break :-). So during that 5 minutes of time since you calibrated the setting circle, M42 has drifted another 5 minutes to the west.

Yes, the object will be 5 minutes west of the eyepiece. Since your setting circle does not update during that time, you need to position the scope 5m to the west, or to 5h 30m right ascension (not 5h 40m R.A.).
Yes. Before moving to the next object, you should reset the right ascension setting circle to the coordinate of the current object. Then move to the next object. Then compensate for the time it took to move the scope. Of course, switch to a low power eyepiece just before starting the move to increase the chance that the next object will be in the field of view.
Mechanical setting circles are useful for small jumps. I have used them on my scope to locate Venus near the Sun. I have also used the circles on the scope at my club's observatory to locate planets in the day. (That is slightly different that your setup since the club's scope is permanently mounted, professionally made, and has really large circles -- all of which help to reduce any errors.) If I remember correctly, the "rule of thumb" is that mechanical circles should be about the same diameter as the optics in order to be "accurate". A 200 mm reflector (the diameter of the mirror = 200 mm) should have circles that are about 200 mm in diameter. But anything should definitely get you close to the target.


The equation for calculating how many times Earth (or our solar system) has revolved around the Milky Way

I'm not 100% sure what equation you would like, but I'll try at an answer to see how it goes.

The observed speed of the Sun orbiting around the center of the Milky way is about 220 km/sec. The distance to the center is about 8.1 kiloparsec, which is 2.5E+17 kilometers.

The time it takes to travel around a circle is the circumference divided by the speed,

$$\frac{2 \pi R}{v} \approx \frac{6.28 \times 2.5\times 10^{17}}{220} \approx 7.17 \times 10^{15} \text{seconds} \approx 227 \text{ million years}$$

The speed is a measured quantity. There isn't a simple way to calculate the speed with an equation, because the mass distribution in the galaxy is complicated and dominated by dark matter, whose presence we can only infer from the orbital motion of our own star as well as a large number of other stars in our galaxy.

 Source

From Wikipedia's Milky Way:

  Galaxy rotation curve for the Milky Way. Vertical axis is speed of rotation about the Galactic Center. Horizontal axis is distance from the Galactic Center in kpcs. The Sun is marked with a yellow ball. The observed curve of speed of rotation is blue. The predicted curve based upon stellar mass and gas in the Milky Way is red. Scatter in observations roughly indicated by gray bars. The difference is due to dark matter


How Does IceCube Trace a Neutrino to its Source?

You ask a number of questions. I'll try an answer them.

Yes, galaxy clusters (except the one we are part of) are many millions of light years away and so the neutrinos from them (which travel at a speed only a tiny fraction below that of light) have taken millions of years to get here. However, the light from those clusters has also taken millions of years to get here, so if we look in a certain direction and see a galaxy, and we detect a high energy neutrino coming from the same direction, then the galaxy was there when the neutrino was emitted (or passed through it). It may have moved since, but that would affect neither the light nor the neutrino.

Neutrinos are affected by gravity, and so the neutrino images of distant objects can be distorted in the same way as the optical images when there is a massive object (like another galaxy cluster) between it and us. However, if that were happening, we would see the distortion of the optical image.

It is also possible (though extraordinarily unlikely) that a single neutrino might have been deflected by passing close to a very massive object  such as a black hole which is too small to make much of a difference to the optical image of the cluster. The chance of a neutrino from somewhere else being deflected by something we haven't detected, just enough to appear to come from a galaxy we can see, though, is absolutely amazingly minute. A bit like a bullet ricocheting off a falling meteorite in such a way that it seems to have come from that man over there holding a gun, even though it actually came from somewhere else.

Finally, on the question of detection. It is true that neutrinos interact weakly with matter, but these very high energy neutrinos interact less weakly than others. This paper and in particular its figure 1 suggest that neutrinos of the energy reported ($2\times 10^{15}$ eV according to the article you refer to) would interact about 10 million times more strongly than the ones from the Sun or from nuclear reactors. Even so, most neutrinos will go straight through ice cube without interacting with it at all, despite the fact that it consists of a whole cubic kilometer of ice. Of the few that do interact, the interaction will usually "kick loose" a charged particle (often a muon) from the ice, which will be moving at close to the speed of light (in vacuum) itself, and in the same direction that the neutrino was moving. In particular, it will be moving faster than the speed of light in ice. When a charged particle moves faster than the speed of light in whatever material it's passing through it produces a "shock wave" of light, called Cerenkov radiation. Detectors in ice cube pick up that radiation and track the movement of its source, thereby getting a good idea of the direction from which the original neutrino came.


Would an analogue of the definition for planets also work for moons?

In 2006 the IAU had a trilemma.

They could decide that Eris was a planet, and potentially allow for future discoveries of tens of new planets.
They could be inconsistent, declare that Pluto was a planet, but Eris (and Ceres) wasn't
They could come up with a definition of "planet" that would exclude Eris, and consequently also exclude Pluto.

Each option is problematic: We teach eight-year-olds to chant the names of the planets, would we want to make them chant the names of 20 or thirty Kuiper belt objects?  Being willfully inconsistent would lead to further argument. Re-describing Pluto would go in the face of 75 years of tradition. There was no option that would keep everyone happy, but they chose the third.

We don't have a problem with a more relaxed defintion of "moon" we are used to new "moons" being discovered. Eight-year-olds don't need to chant their names.  The casual definition of "if you find it, you can name it" works.

The issues with planets are particular to our expectations that "planets" are important. We have no such expectation with moons.


What is the thickness of the Solar System disk?

The thickness of the planetary disc is dominated by Neptune, due to its large orbital radius.
We can calculate a planet's maximum distance from the ecliptic $h$ from the inclination angle of its orbit $\theta$ and its aphelion distance $r$. We get a right triangle, with $r$ as the hypotenuse, so
$$h = r\sin\theta$$
The table below was calculated using data from the NASA Planetary Fact Sheet. Angles are in degrees, distances are in millions of kilometres.
Planet distance from the ecliptic plane.

Name
Inclination
Aphelion
Distance

Mercury
7.0
69.8
8.506

Venus
3.4
108.9
6.458

Earth
0.0
152.1
0.000

Mars
1.9
249.2
8.262

Jupiter
1.3
816.6
18.526

Saturn
2.5
1514.5
66.062

Uranus
0.8
3003.6
41.937

Neptune
1.8
4545.7
142.784

So the total thickness of the disc is $2×142.784 = 285.568$ million kilometres, which is almost $1.91$ au.

Here's the Python code I used to create that table:
from math import sin, radians

names = (
 'Mercury', 'Venus', 'Earth',
 'Mars', 'Jupiter', 'Saturn',
 'Uranus', 'Neptune',
)

# Orbit data from https://nssdc.gsfc.nasa.gov/planetary/factsheet/
# Inclination to ecliptic plane
inc = [7.0, 3.4, 0.0, 1.9, 1.3, 2.5, 0.8, 1.8]
# Aphelion in millions of kilometres
aph = [69.8, 108.9, 152.1, 249.2, 816.6, 1514.5, 3003.6, 4545.7]

print("|Name | Inclination | Aphelion | Distance|")
print("|-|-|-|-|")
for n, th, r in zip(names, inc, aph):
    # Perpendicular distance to eciptic
    h = r * sin(radians(th))
    print(f"|{n} | {th} | {r} | {h:.3f}|")

Here's a live version of the script running on the SageMathCell server.

As John Holtz mentions in the comments, the true $h$ value for a planet may be smaller than the value shown in my table. The table's $h$ value only occurs if the planet's argument of periapsis is ±90°. Fortunately, Neptune's argument of periapsis is currently ~272°, so my $h$ value should be fairly close to the true value.

James K has supplied a list of orbit inclinations to the Solar System's invariable plane. Here's the table using those values.
Planet distance from the Solar System invariable plane.

Name
Inclination
Aphelion
Distance

Mercury
6.34
69.8
7.708

Venus
2.19
108.9
4.161

Earth
1.57
152.1
4.167

Mars
1.67
249.2
7.262

Jupiter
0.32
816.6
4.561

Saturn
0.93
1514.5
24.582

Uranus
1.02
3003.6
53.468

Neptune
0.72
4545.7
57.121

That reduces Neptune's $h$ considerably! Uranus may even be the current "winner", depending on their arguments of periapsis with respect to the invariable plane.


I think, that's a bit semantics, and a question of what do you call 'I see something': If you see the shadow cast by an object on a bright background - do you see the object? Do you see Mercury or Venus when it passes in front of the disk of our sun? Similarly, if you look an object through an atomic force microscope - do you see the object, or do you just see some force indicator giving a value?

So yes, in this general sense, you see something when you have an instrument to detect it which allows to deduce properties of the object. So you also can see an object when you just see its shape in front of a bright background like a black hole within its accretion disk - even when all light you see comes from the latter.


What's fixed in Ariel that is moving in other planet-hunting space telescopes?

I think this is a case of selective quoting/editing in the media reporting otherwise something major has changed in the design and plan for the mission that hasn't been released or published officially. The large and comprehensive ARIEL assessment report which was published less than 2 years ago discusses mounting all of the optical components on a large, thermally isolating optical bench to prevent movements of the optics introducing noise into the transit signals. However it also says (page 66) that secondary mirror (M2) will have the normal 3 axis position control to provide focus and tip-tilt. This is also mentioned in a Phase A study report published in 2018. The stability will be improved by the launch of Ariel into the Earth-Sun L2 Lagrange point which is more thermally stable and doesn't suffer from Earth eclipses compared to Kepler or TESS which are closer to the Earth.

This is the normal and sensible accommodation incorporated into satellite optical systems in order to compensate for anything moving slightly during the pretty violent shaking that occurs during launch on a rocket. Locking everything down pre-launch with no adjustment would be a very risky undertaking.

There are no more recent publications listed on the Ariel mission website that would suggest that the optical system has been redesigned to eliminate all moving parts although it is possible a study has been done and the result have not yet been communicated publicly.


You don't say what other information you have for the "several stars". Yes, you can use a mass-luminosity relationship if the stars are on the main sequence. In terms of mass uncertainties I would estimate that you might be at the level of 20% unless you can absolutely pinpoint them on a Hertzsprung-Russell diagram, because the luminosity of a (fixed mass) main sequence star changes, even whilst it is on the main sequence.

If the stars are in any other phase of their evolution then your problem is much harder, even if you can place them on the HR diagram. Not least because there is considerable disagreement between different evolutionary models as regards the positions of stars of a given mass at a given age, on both the pre- and post-main sequence. Furthermore, their luminosity and temperature evolution can be influenced by factors such as rotation, metallicity, magnetic fields and mass-loss.

Compact stellar remnants and brown dwarfs do not follow a mass-luminosity relation. In general you can only estimate their masses if they are in binary systems, or for white dwarfs, you can estimate their radii and either their surface gravity or gravitational redshift. Estimating a brown dwarf's mass needs both its luminosity and age.


Does gravitational lensing provide time evolution information?

Time delays between the light received from multiple lensed images is a well known and commonly studied phenomenon. A typical example can be found in Koopmans et al. (2003) where they study a quadruple lens system formed by a forefround object at z=0.6 and a background source at z=1.4. The time delays measured are 31-77 days and this seems to be typical for the objects studied so far.

The delay depends on the difference in path length taken by the multiple images. This in turn depends on the distance to the source and the distance between the source and the lensing object and on the mass distribution around the lensing object. It also depends on the expansion rate of the universe, which gives a route to estimating the Hubble parameter.

A rule of thumb appears to be that the delay is of order $r_s/c$, where $r_s = 2GM/c^2$, where $M$ is the mass of the lens interior to the path of the light ray.
$$ \tau \sim 10^{-5} \left(\frac{M}{M_{\odot}}\right)\ {\rm sec}$$
So for a $10^{12}M_{\odot}$ foreground lensing galaxy (a big spiral like the Milky Way), the delay is about 100 days.


How to see Saturn's rings through a pair of binoculars?

You will not be able to see Saturn's rings directly with a 15x70 telescope, but you will be able to notice that Saturn appears elliptical, not round. This is due to the shape of the rings around the planet, but you will not be able to resolve the hole between the rings and the planet (much less the Cassini gap).

Just as an impractical suggestion in the interest of giving to you a more complete answer, you might be able to mount a camera and use image stacking to see a smidge of ring and differentiate it from the sphere of the planet. If you do go this route, please post online about your setup and results, whether it succeeds or not.


What visual artifacts are expected from the JWST?

Are you asking about the PSF (point-spread function)?

There are some simulations at http://www.stsci.edu/jwst/software/webbpsf ; there are some basic images available at that site as well as a downloadable package you can use to compute the PSF for a particular instrument and wavelength.  Since the telescope hasn't been fully assembled yet these are based on simulations.


Has anyone back-calculated previous close encounters between the Apophis asteroid and Earth?

The JPL Small Body Database lists Apophis close approaches dating back 100 years before discovery.
Three fairly close ones were:

1907-04-13, 0.029 au
1949-04-14, 0.028 au
1990-04-14, 0.033 au

While it's possible to run a dynamical integrator arbitrarily far backward or forward in time, any given pos(t), vel(t) state is only a point in a cloud of possibilities.
Observations (MPC database, NASA news) constrain that cloud; close encounters accelerate its dispersion.
Specific predictions before ~1900 or after ~2050 should not be regarded with much confidence.


Regarding visible light, the nominal range is usually given as 400 nm (violet) to 700 nm (red). For light of the former wavelength to get Doppler-shifted to the latter means a z value of (700–400)/400 = 0.75, corresponding to a recession of 152,000 km/s, just over half the speed of light.


Finding the temperature of Earth from temperature of Mars and its distance from Sun

I might as well write up my comment into an answer.

One of my new favorite tools in my equations toolkit is the formula for effective temperature:
$$T=\left( \frac{L(1-a)}{16 \pi \sigma D^2} \right)^{\frac{1}{4}}$$
where $T$ is the temperature of the planet, $L$ is the star's luminosity, $a$ is the planet's albedo, $\sigma$ is the Stefan-Boltzmann constant, and $D$ is the distance between the star and the planet. To find the star's luminosity, simply solve for $L$:
$$L=\frac{T^4 16 \pi \sigma D^2}{1-a}$$
where all the terms are for the first planet (Mars). Now, knowing $D$ for Earth, solve for $T$ on Earth.

All you have to know for this that isn't in the question is albedo. That could be tricky, because it's not given. You could assume that it's the same for both planets - and so solve not for $L$ but for $L(1-a)$ - but that might not be better than an approximation.


Simon Littlefair of Sheffield University gives formulae for adjusting for precession

$$
Δα = M + N \sinα \tanδ
$$
$$
Δδ = N \cosα
$$

where M and N are given by

$$M = 1°.2812323 T + 0°.0003879 T^2 + 0°.0000101 T^3$$
$$N = 0°.5567530 T - 0°.0001185 T^2 + 0°.0000116 T^3$$

and $T$ is defined as

$$
T = (t-2000.0)/100
$$

where t is the observation date, in years. $Δα$ is the change in RA that you would need to add, in degrees.

(formulae quoted directly from the website)

So you would have T = 0.17, M= 0.21782075093,N=0.09464464234

The actual changes in RA and Dec can be calculated from the trig formulae above.

Alternately caltech provide an online calculator


What mechanisms limit the top speed of high altitude winds on Venus?

I found some recent papers which, if nothing else, discuss apparent mechanisms.
Quoting from Schubert, 2016,

  The large-scale circulation of the upper atmosphere from ~90 to ~200
  km altitude (upper mesosphere and thermosphere1 ) is a combination of
  two distinct flow patterns: (1) a relatively stable
  subsolar-to-antisolar (SS-AS) circulation cell driven by solar (EUV-
  UV) and IR heating, and (2) a highly variable retrograde superrotating
  zonal (RSZ) flow, in part a continuation of the lower-atmosphere RSZ
  flow discussed above

Then, ESA says, in part,

  In 2006, average cloud-top wind speeds between latitudes 50° on either
  side of the equator were clocked at roughly 300 km/h. However,
  detailed cloud tracking studies revealed that these already remarkably
  rapid winds are becoming even faster, increasing to 400 km/h over the
  course of the mission. The reason for this dramatic increase is
  unknown.
  [emphasis mine]

Alternatively,  these guys think they have a working model:

  The atmospheric circulation in Venus is well known to exhibit strong
  super-rotation. However, the atmospheric mechanisms responsible for
  the formation of this super-rotation are still not fully understood.
  In this work, we developed a new Venus general circulation model to
  study the most likely mechanisms driving the atmosphere to the current
  observed circulation. Our model includes a new radiative transfer,
  convection and suitably adapted boundary layer schemes and a dynamical
  core that takes into account the dependence of the heat capacity at
  constant pressure with temperature. The new Venus model is able to
  simulate a super-rotation phenomenon in the cloud region
  quantitatively similar to the one observed. The mechanisms maintaining
  the strong winds in the cloud region were found in the model results
  to be a combination of zonal mean circulation, thermal tides and
  transient waves. In this process, the semi-diurnal tide excited in the
  upper clouds has a key contribution in transporting axial angular
  momentum mainly from the upper atmosphere towards the cloud region.
  The magnitude of the super-rotation in the cloud region is sensitive
  to various radiative parameters such as the amount of solar radiative
  energy absorbed by the surface, which controls the static stability
  near the surface. In this work, we also discuss the main difficulties
  in representing the flow below the cloud base in Venus atmospheric
  models

.


As far as theories I've heard of, we can't see the edge of the universe (if truly exists) because the farther you look out into space through a telescope, the farther back in time you are actually looking. For example, if you are looking at a star 20 light-years away... you're actually currently observing what was happening on that star 20 years ago... the theory is that there is only so far you can look out into the universe because there is only so much history post Big Bang.

Image above: What is the furthest we can see? In 2003, NASA's WMAP satellite took images of the most distant part of the universe observable from Earth. The image shows the furthest we can see using any form of light. The patterns show clumps of matter that eventually formed into galaxies of stars. Credit: NASA/WMAP Science Team