Well if no one is going to answer this I will. The answer is we don't know for sure. We speculate that there should be earth rocks on mars but until we 'see' one and analyze it we will not know for sure.  The comments here all point to this answer.
Mars hit with thousands of Earth rocks possibly containing life following asteroid impacte talks about the Chicxulub impact and how it probably spread rocks to all the terrestrial planets in our solar system and the moons of all the planets.

The statement comes from Penn State University researchers who have calculated the approximate number of rocks from our planet large enough to possibly carry life that have made their way into space over the past few billion years.
Said the paper’s lead author Rachel Worth: “We find that rock capable of carrying life has likely transferred from both Earth and Mars to all of the terrestrial planets in the solar system and Jupiter. Any missions to search for life on Titan or the moons of Jupiter will have to consider whether biological material is of independent origin, or another branch in Earth’s family tree.”

The article cites its source as The BBC's Dinosaur asteroid 'sent life to Mars' which cites a 2013 paper published in Astrobiology Seeding Life on the Moons of the Outer Planets via Lithopanspermia:

The asteroid that wiped out the dinosaurs may have catapulted life to Mars and the moons of Jupiter, US researchers say.
They calculated how many Earth rocks big enough to shelter life were ejected by asteroids in the last 3.5bn years.
The Chicxulub impact was strong enough to fire chunks of debris all the way to Europa, they write in Astrobiology.
Thousands of potentially life-bearing rocks also made it to Mars, which may once have been habitable, they add.
"We find that rock capable of carrying life has likely transferred from both Earth and Mars to all of the terrestrial planets in the solar system and Jupiter," says lead author Rachel Worth, of Penn State University.


South-eastern sky above the tree line just now, NE Cape Cod, I saw a very bright, twinkling "star"

I just did the following:

opened in-the-sky.org,
manually chose the location New Bedford, MA (close to Cape Cod) because I'm somewhere else (you wouldn't have to do that)
selected planetarium

and looked in the southeast and found that the brightest star in the sky Sirius is currently only 20 degrees above the horizon.  At that low altitude, and being a star (not a planet) it would definitely be bright and twinkle, and not appear to move like a plane.

It is rising, and moving diagonally to the right and up. Check back in an hour or so to confirm that if you are interested, or check in the next several days to confirm it's in the almost the same place each night.


Is any meteorite known to come from a comet?

This page -- http://www.amsmeteors.org/fireballs/faqf/ -- suggests that it's very unlikely to find cometary-origin meteorites, even though the majority of observed meteors are cometary in origin, because the latter are too fragile to survive all the way to the ground:

  Based upon photographic fireball studies, cometary meteoroids have
  extremely low densities, about 0.8 grams/cc for class IIIA fireballs,
  and 0.3 grams/cc for class IIIB fireballs. This composition is very
  fragile and vaporizes so readily when entering the atmosphere, that it
  is called “friable” material. These meteoroids have virtually no
  chance of making it to the ground unless an extremely large piece of
  the comet enters the atmosphere, in which case it would very likely
  explode at some point in its flight, due to mechanical and thermal
  stresses.

It goes on to claim that cometary meteoroids make up about 95% of observed meteors, 38% of observed fireballs, and 0% of fresh meteorites. (Since a hypothetical cometary-origin meteorite would weather faster than meteorites of asteroidal origin, they would be even less likely to be found later, and so it's not surprising that the page says 0% of all known meteorites are of cometary origin.)


What is the fifth moon-like object I saw around jupiter through my telescope?

A seventh magnitude star.  See this link:

http://www.cloudynights.com/topic/533968-jovian-events-414-1516/

Further info: the star in question was probably HIP 54057 (also known as SAO 118636 or HD 95848.) On April 12th it was occulted by Jupiter (that is, Jupiter passed in front of it) for about three hours. So the star would have been in the area of Jupiter for a few days.


Why don't planets and asteroids have (cometary) tails?

See Jeans Escape. If the average velocity of the volatile molecules is above escape velocity, volatiles will escape. And the with the shallow gravity wells of comets, escape velocity is very low.

The earth and Mars have lots volatile gases and ice. But with their deeper gravity wells, sublimated volatile ices aren't hurled into space as they are with comets.

A lot of asteroids have high enough temperatures that any volatiles would exceed their tiny escape velocity. Which is why a lot of asteroids in our neighborhood have little or no volatile ices. If they ever had ice, it would have long since boiled off. For example, after Comet Wilson Harrington outgassed much of its volatiles, it'd no longer have a tail when coming near the sun. They thought they had lost the comet when they stopped sighting it. Then later they discovered asteroid 1979 VA. Then they noticed they were the same object.

It's speculated there are more "dead" comets like Asteroid 1979 VA. Former comets who've outgassed most their volatiles.


What methods exist to calculate the ellipticity of galaxies

The first two methods mentioned in the question are really similar. They both use moments of the light distribution, only the weighting is different.

Another family of methods is 2D profile fitting to the image, with more or less sophisticated luminosity profile models. A simple case would be to fit a Sérsic profile. But instead, one could make a composite model with two or more elliptical profiles (say one exponential disk and a de Vaucouleur bulge). This would allow to separate the ellipticity of the disk from the ellipticity of the bulge. A widely used software for this is GALFIT by Chien Peng.


Yes, the dark side of the moon is mostly hidden from earthbound observers. This is due to tidal locking:

  A tidally locked body takes just as long to rotate around its own axis as it does to revolve around its partner. This causes one hemisphere constantly to face the partner body.

In other words, one moon-year is just as long as one moon-day.

The left-hand side shows the actual configuration of earth and moon: The orbit is locked and we can only observe one side of the moon. On the (fictional) right-hand side, the moon does not rotate around itself and we can observe different areas of the moon at different times.

I should note that we can actually observe not 50%, but around 59% of the moon's surface because of libration and parallax effects that cause the moon to slightly "wobble":

You can find all those images and further explanation in the linked wikipedia articles on tidal locking and libration.


How far away is the nearest compact star remnant likely to be?

There can be no closer white dwarf. The coolest, oldest white dwarfs (3000K), would be rare, but are still luminous enough $6\times10^{-6} L_{\odot}$ to have been easily detected at distances closer than Sirius. At the distance of Sirius, such an object would have a visual magnitude of around 12-13 and would be brighter at near infrared wavelengths where all sky surveys such as 2MASS would definitely have spotted it from its parallax.

Neutron stars and black holes could be almost undetectable but are expected to be roughly 10 and 100 times rarer respectively. Calculated as follows:

Let us assume that $N$ stars have ever been born in the Milky Way galaxy, and given them masses between 0.1 and 100$M_{\odot}$. Next, assume that stars have been born with a mass distribution that approximates to the Salpeter mass function - $n(m) \propto m^{-2.3}$. Then assume that all stars with mass $m>25M_{\odot}$ end their lives as black holes, all stars with $8<m/M_{\odot}<25$ end their lives as neutron stars and about half the stars with $0.9<m/M_{\odot}<8$ end their lives as white dwarfs (the other half are still alive as main sequence stars, as are all stars born with lower masses).

So, if $n(m) = Am^{-2.3}$, then
$$N = \int^{100}_{0.1} A m^{-2.3}\ dm$$
and thus $A=0.065N$.

The number of black holes created will be
$$N_{BH} = \int^{100}_{25} Am^{-2.3}\ dm = 6.4\times10^{-4} N$$
i.e 0.064% of stars in the Galaxy become black holes. NB: The finite lifetime of the galaxy is irrelevant here because it is much longer than the lifetime of black hole progenitors.

In a similar way, the number of neutron stars
$$N_{NS} = \int^{25}_{8} Am^{-2.3}\ dm = 2.6\times10^{-3}N$$
and the number of white dwarfs
$$N_{WD} = 0.5\times \int^{8}_{0.9} Am^{-2.3}\ dm = 0.027 N$$

Now we use these results as scaling factors to apply to the local stellar population. There are about 1000 "normal" stars in a sphere of 15 pc radius, thus a density of 0.07 pc$^{-3}$. Thus one uses the results above to calculate the density of compact remnants and then take $(3/4\pi n)^{1/3}$ as an estimate of the average distance to one of them. This gives an expectation value of 18 pc to the nearest black hole, 11 pc to the nearest neutron star and 5 pc to the nearest white dwarf.

Thus the distance to the nearest white dwarf is roughly as expected. For reasons discussed in my answer to this related question the distance calcuated to the nearest black hole and neutron star remnants is likely to be an underestimate because many escape from the Galaxy or have very high velocity dispersions and much larger Galactic scale heights than normal stars. So whilst it is possible that an unseen one exists closer than Sirius, it is highly unlikely.

How could such an object be detected? An old, cold neutron star or black hole could be completely undetectable at all wavelengths of electromagnetic radiation - though it could be fruitful to examine carefully any candidate detections [see below] for signs of X-ray emission due to accretion from the interstellar medium). But your question has I think the correct suggestion. The objects would likely have a substantial proper motion and so there is a decent chance that you would see a "moving" gravitational lensing signature. This would still be very small unless the object just happened to pass directly in front of a background star - but such a microlensing event would be transient and may not be observed. More likely is that Gaia would pick up the subtle shifts in the positions of background stars changing over the 5 years of its mission. As per your other question: Will Gaia detect inactive neutron stars?


How to determine the formula to calculate this position of reflected Sun?

When the "sun angle" is zero, the position of the "sun" doesn't depend on the season angle, so the lines will cross at the point that is the image of the sun at "sun-angle 0"
One important aspect in your model is that the season doens't change the "angle" but the position of the sun.  In summer, the sun rises and sets in the north East and north west and describes a small cirle around the north celestial pole. In winter it rises and sets in the southeast/west. And only at the equinox does it describe a great circle.


Is communication with proxima b possible with todays technology?

This is a pretty complex question.  You might have better luck at space exploration than here.

Since you asked about current technology, how NASA does it now is they have very large microwave receivers, 3 stations, 70 meters each and the transmission is in microwave frequencies.    Microwaves are likely the best wavelengths to send long range space communications because it's a comparatively low background noise and because it travels through Earth's atmosphere, so receivers can be built on the ground, which is much easier than building a 70 meter (or larger) dish in space.   I'm using this as a source, it seems reliable, though it's apparently a hobbyist, not anything official.

This picture below is a page from NASA, showing the microwave window, which is used for communication.

And the microwave receivers, called the Deep Space Network (DSN), operated by NASA's Jet Propulsion Laboratory (JPL)

The precise band used is a little tricky (and I'm not an expert), but bands for satellite communications are divided up by countries.  Article on that here.   I would think they might try to use a band in the Water Hole, 1.42 to 1.666 GHz, which is in the L band, but most deep spacecraft communication and photos sent from distant space crafts is done in the S band.    I'm not sure why that is.

As distances double, the strength of the signal is reduced by four and the signal has to stand out against background radiation, which is constant, while the signal weakens by the square of the distance.   This isn't an easy problem over vast distances.   But when you say "communicate", there's different kinds of communication, Morse Code or Binary Codes are very simple, only a few bytes per letter and messages can be transmitted that way at far greater distances.   NASA is generally interested in photographs, and our spacecraft relay pictures back to NASA by ones and zeros via microwave messages, this can be compared to internet connections for speed.  New Horizons, which recently flew past Pluto, takes about 1 hour to send 1 picture.   Very good write-up on that here.

The Mars Reconnaissance Orbiter (per Quora answer), can communicate with as fast as 2 million bits per second, but it's much closer.    How far and/or how fast (the 2 questions are related) a space craft can communicate with Earth depends on how big a transmitter they give it, how much power it has and how long it's power source is required to last.

Now, I'll try to answer your question badly, using Pluto and New Horizons as a baseline.

New Horizons is a relatively recent launch, (at least, compared to the Voyagers), it was launched in 2006, so it's using 10 year old technology.   It's currently a bit over 35 astronomical units from earth and it's still sending pictures because it sends them very slowly.   This article is 5 months old, but it says that it took New Horizons about 8 months to send half of the pictures it took of Pluto and Charon (and Nix and Hydra and . . . the others), by the equivalent of a tediously slow internet connection, (from the article)

  New Horizons is transmitting data at a poky 2,000 bits per second (March 2016), and by summer that will drop to 1,500—a fraction of one percent of
  broadband data rates here on Earth. It takes about an hour for an
  average image file to arrive, from the first bit to the last.

It's also relatively low power.   I couldn't find specifics on the wattage of it's transmitter/receiver unit, but the entire New Horizons power usage is currently about 200 watts Source and there are several components on it that share that energy.   Source.  Obviously we have the means to build much more powerful transmitters than 30 or 50 watts, or whatever it uses, but that's all that was needed for this specific mission.   New Horizons is expected to operate for 20-25 years, till about 2030, at which point it will be over twice as far from Earth as it is now, and transmitting data to us at about 500 bits per second.   Source.

A final bit of information is the angle of the beam.   (from the first New Horizons link above):

  The system includes two broad-beam, low-gain antennas on opposite
  sides of the spacecraft, used mostly for near-Earth communications; as
  well as a 30-centimeter (12-inch) diameter medium-gain dish antenna
  and a large, 2.1-meter (83-inch) diameter high-gain dish antenna. The
  antenna assembly on the spacecraft’s top deck consists of the high,
  medium, and forward low-gain antennas; this stacked design provides a
  clear field of view for the low-gain antenna and structural support
  for the high and medium-gain dishes. Operators aim the antennas by
  turning the spacecraft toward Earth. The high-gain beam is only 0.3
  degrees wide, so it must point directly at Earth.

Now, onto (approximately) Proxima Centauri.   It's about 268,000 AU from Earth, which is about, 7,600 times further from Earth than New Horizons currently is.   That requires a beam, everything else being equal, some 57 million times stronger.   That's not going to work.

I couldn't find any good information on the technology.   If a beam could be tightened from 0.3 degrees wide to 0.1, or 0.01, the first would provide 9 times as strong a signal.  The 2nd, 900 times.   A .01 degree wide beam from Proxima Centauri would spread out to about 46 astronomical units by the time it reaches Earth, so I would think, the tighter they could make the beam the better, but at .01 degrees, the signal strength would still be about 1/60,000th the strength of New Horizons current transmission.   .001 degrees, and I have no idea if that's possible, but that tight a beam, you're getting into the "this could work" range, about 1/600th the strength of the New Horizon's signal, and with greater power behind the signal and, perhaps, a larger satalite dish built on Earth, then it's in the range of workable.  (but I have no idea if microwave beans can be sent over just .001 degrees)

Presumably with a much more energetic transmitter, say a few thousand watts and a very tight beam and they might be able to get a good enough transmission for Morse code, but, I would think, no pictures, only a letter at a time, like the old telegraph communication.

Do we have the technology to do it today?   Well, to me, it looks pretty touch and go.   I think, maybe, if there was a human colony already there, with nuclear power plants and equipment to build a large transmitter/receiver then, if they can get the beam tight enough, we probably could exchange dots and dashes with today's technology.

Can we build a spacecraft that could communicate to us from 4.2 light years away - no, I don't believe we could do that with today's technology, not at that distance.

All that said, communication is a far less difficult problem than actually getting there.

I hope my layman's answer isn't too far off.


Calculate the true diameter of stars from photographic plate

Let's pretend for the moment that this is possible (it's generally not, as I'll explain below) and see how we would go about it.

In principle, this is basic trigonometry: you have a measured angle ($\alpha$, the diameter of the star), you know the distance to the star ($D$, from the six-month-separated parallax measurements), so to determine the linear diameter of the star ($d$, in the same units as the distance) it's a simple matter of $\sin \alpha = d/D$. In practice, the angles are small enough that you can use the small-angle approximation, so it's really just $\alpha \approx d/D \rightarrow d = \alpha D$ (assuming $\alpha$ is in radians).

As your book notes, you need to know the distance to the star; for nearby stars, this can be obtained by stellar parallax, which involves the same basic trigonometric argument (only this time you know the "diameter", which is the size of the Earth's orbit).

So why is this, as I said, not generally possible? In practice, the diameter of the telescope sets a lower limit on observed angular sizes -- anything with a true diameter smaller than the telescope's angular limit will appear to have an angular diameter $\approx$ that limit. In addition, any telescope underneath the Earth's atmosphere will suffer the effects of blurring due to atmospheric turbulence, which sets an angular diameter limit of $\sim 0.5$ arc seconds (full-width half-maximum) or worse at the best sites. Since the largest stars in angular terms have diameters of about 0.05 arc seconds, the result is that all stars in the image (photographic plate, CCD camera, etc.) will have the same observed angular diameter (e.g., 0.5 arc seconds at a really good site), regardless of their true diameters.

It is possible to do better for bright, nearby stars using large telescopes in space or special interferometric techniques -- see the second link in the previous paragraph for an example -- but this is going beyond the "stars on photographic plates" scenario your book seems to envisage.

And if for instance, using a 100'' telescope, you measure that a star has an apparent diameter of 0.66mm, can you directly compute the true diameter of the star?

Leaving aside the problems mentioned above, you don't have enough information. You need to convert the linear diameter in mm to an angular diameter, which means you need to know the plate scale (e.g., arcsec/mm) of the particular telescope + camera setup. You also need to know the distance to the star.


What will the universe be like in a googolplex years time?

That is nearly long enough to reach heat-death, which is estimated as about $10^{10^{120}}$. What that means is rather speculative, since it depends on various events that we have never observed, such as the spontaneous formation of black holes by quantum tunnelling. Such events are utterly rare, but are predicted to occur at very long timescales.

Any matter, will have at some time over a googolplex years quantum tunnelled into having enough mass to collapse as a black hole. So no matter will exist

Any black hole will have evaporated in Hawking radiation.  So no black holes exist.

Redshift will reduce the energy of photons to the point where their wavelengths are comparable to the scale of the universe.

Nothing is happening in the universe.

See https://en.wikipedia.org/wiki/Timeline_of_the_far_future

Now there is a: immovable object/unstoppable force paradox here, for in a universe that is spatially big enough, there are pockets of order among the chaos of heat death.


What are the real conditions for the creation of an atmosphere/loss of an existing atmosphere?

Technically speaking, every planet has an atmosphere (though most drop Mercury from the list because it's so thin). Let's start with the big guys, which are pretty straightforward. Gas giants just keep their atmosphere from when they gathered the remnants of gas from the nebula that their parent star formed from. If you're like the gas giants in our solar system, you usually keep this atmosphere for a very long time because you have a strong magnetosphere and you also are fairly far from your star. Your star also isn't particularly powerful. Some exoplanets are categorized as hot Jupiters because they are similar to jupiter in composition but are orbiting very closely to their star. If this is the case, despite the strong magnetosphere, their star may quite literally boil their atmosphere off, either adding it to the star's mass, or blowing it away, sort of like a comet. Eventually they may become what we call chthonian planets, which are hypothetical planets that are the rocky cores left over from boiled gas giants.

How about rocky planets like Earth? They get their atmosphere from a process called outgassing, which basically means that volcanoes burp out gases. Earth is unique in that we believe our atmosphere was at least partially created from comets and wet asteroids. Venus burped out too much greenhouse gases and got a hellhole for an atmosphere. At least it rains there...rains sulfuric acid...and even that boils before it reaches the surface! Mars probably had a thicker atmosphere than today, as evidenced by big volcanoes like Olympus Mons. What happened to it? Well first, Mars doesn't have a very strong magnetosphere because it is simply smaller than the Earth. Because it's smaller its core cooled down faster, and its mantle no longer supported a good magnetic field. More charged particles means a higher chance of atmospheric stripping, which Mars suffered from at least partially. However, we think that a large impact probably knocked a lot of Mars's atmosphere off. Finally, that leaves us with Mercury. Mercury is pretty small and very close to the Sun. It just can't retain an atmosphere because its gravity is weak. The same can be said for most moons in our solar system, as they are typically smaller than Mercury. They are simply too small to retain an atmosphere.

Wait, then how come some moons have atmospheres and others don't? Well the ones that don't, as I said before, are too small to retain one, and they also have to deal with interactions with the solar wind, atmospheric stripping, and interactions with their planet. Some moons have volcanic activity, so their thin atmosphere can be replenished with burped up gases. In fact, some may have weather! I believe Io and Triton both have weather systems, despite having very thin atmospheres. Titan has the thickest atmosphere of the bunch, with one that is about 50% denser than Earth's. Thus, it also has complex weather systems. But why? Titan is similar to Ganymede in structure. Why does Titan get to have an atmosphere? Short answer: we don't fully know why. We think it might've been comets, but we really aren't 100% sure on anything at this point. Go check out the wikipedia article for more.

So to answer your question, atmospheres are usually formed from volcanic activity or, for gas giants, retained gas from nebula remnants. Atmospheres are lost due to low gravity, high temperatures, solar wind, atmospheric stripping, large impacts, and planet-moon interactions.


From CMB anisotropy data observed in 1992, did astronomers figure out that the universe should be accelerating before its discovery in 1998?

The only CMB anisotropies that were well determined in 1992 (and by the date, I assume you refer to the results from the COBE satellite) were those at small angular frequencies. This revealed the overall scale of the CMB structure (which tells you about a need for dark matter in order to get the large scale structure seen at the present epoch) and the dipole anisotropy, which tells you about the Galactic motion with respect to the CMB rest frame. There were no constraints on any possible dark energy - or rather, the effect of dark energy on the low frequency anisotropies is degenerate (i.e. cannot be distinguished from) the effects of the overall curvature of the universe.

To measure the curvature of the universe and constrain the mix of dark matter and dark energy (and constrain other cosmological parameters) needs information from higher angular frequencies - in particular information from the location and amplitudes of the first three "acoustic peaks" at angular scales of . This really only became available with the WMAP data (launched in 2001) and with improvements in ground based CMB measurements that took place at the end of the 1990s. The results and conclusions from this emerged over several years from 2000 onwards (e.g. Durrer et al. 2003; Spergel et al. 2003).

In actual fact, the evidence for $\Lambda$ from the CMB alone is rather weak. As explained in one of Wayne Hu's cosmology tutorials, the strongest evidence from the CMB is that the total density of the universe $\Omega = 1$ from the CMB and this can then be combined with other data (e.g. primordial abundances of D, He; galaxy cluster dynamics; gravitational lensing) to suggest that the matter contribution to this is only $\simeq 0.3$.


Why don't we feel the speed of earth moving

In a car, you have a perception of speed because of (a) the "wind" passing by as you rush through the air which is not moving at the same speed as the vehicle, and (b) you perceive the stationary objects nearby as "moving" off into the distance behind.

As the earth moves in its orbit, you don't notice any "wind" from the planet rushing through space, as the atmosphere of the earth is moving along with the planet. There isn't any substance to the space around the planet to create an impression of "wind". The car is surrounded by air, so when you move through the air, you have an impression of moving because of the air rushing past. Because the planet isn't moving through an especially gaseous medium, there is no impression of movement, no "wind".

Also when it comes to our planet, there is nothing "nearby" to create a reference point that would make it appear as if we are moving relative to that reference. Even the closest of objects, such as the moon, are sufficiently far off that the only obvious impression is that these move and we are still. Hence the reason why the geocentric view of our world was dominant for so long.

Even sitting still at home, your entire body is moving at the same velocity as the rest of the planet, so your brain isn't being jostled or 'gyrated' at all. If someone shook you, then your velocity would be changing and your body would feel the effects of acceleration, and then you would be jostled and gyrated. But that is because you are perceiving sudden and rapid changes in velocity, accelerations.


What are some theories that explain the 11 year solar cycle?

It is linked to the solar dynamo and the dynamical behavior of the so-called $\alpha$-effect. One way to understand easily the solar dynamo and the the $\alpha$-effect with the hand is the following: you can generate a magnetic field by differential rotation (the $\Omega$-effect, $\Omega$ being the classical notation for the angular velocity). But this sole effect, twisting the magnetic field, would fold it so much that it will reconnect and dissipate. If you don't want to lose the magnetic field, you need a process to "regenerate" it, and a good way to do it is to create a poloidal component of the magnetic field. This poloidal component can be generated by convection in the solar interior; the magnetic field is, in the same time, twisted and pulled and that explains the solar dynamo.

Now, you can show that dynamo waves will propagate from the equator to the poles, and when the poles are reached, a reversal is observed. This propagation of dynamo waves is well observed by the sunspots patterns during the solar cycle.

Sources:

Parker (1955)
Proctor (2006); you can find it freely available here: http://damtp.cam.ac.uk/user/mrep/solcyc/paper.pdf


How to chose a comfortable 2 inch eyepiece for a Skywatcher Dobsonian 150 mm f/8 reflector?

Short version:
The reason that 2" eyepieces will offer more comfortable views is because they are lower powers and therefore have larger exit pupils.

Long version:
In a telescope, the objective is the lens or mirror which collects light and focuses it at a distance equal to the focal length, creating an image at that point.
(You can test this out most easily on the moon: leave the dust cap in your focuser and use your finder to point the telescope at the moon. You should be able to see a clear (after focusing!) image of the moon projected on the dust cap)
When you insert an eyepiece, you're using it to inspect the image created by the objective.
This is why eyepieces with smaller numbers actually magnify the image more: with a 5mm eyepiece you're looking at a smaller central portion (imagine 5mm, though that's not correct, as it also depends on the FoV of the eyepiece) of the image which the objective has collected; and with a 30mm eyepiece you're inspecting a larger part of the image.
And this results in the exit pupil - a function of telescope and eyepiece - which is the size of the image you're looking at. It's the beam of light which enters the pupil of your eye. At lower magnifications (higher eyepiece focal lengths) the exit pupil is larger, and therefore easier to view comfortably.
You can test this out: cut a series of small holes (from 0.5mm up to 7mm) in a piece of paper and try looking through them. You'll find that the larger holes are much easier to see things through.
To calculate the exit pupil, divide the focal length of the eyepiece by the F-ratio of your telescope.
(And, related, to calculate the magnification you'll get from an eyepiece, divide the focal length of your telescope (in mm) by the focal length of the eyepiece (in mm))
For example, your telescope is f/8, so a 30mm eyepiece would give an exit pupil of 3.75mm - nice and sufficiently wide for you to be able to find the view and not lose it.
But if you want to look at a planet then you'll want more magnification, and you'll have to use a shorter focal length eyepiece. Take a 5mm for instance. That would give an exit pupil of only 0.625mm.
Another effect of decreasing exit pupil is decreasing image brightness. You'll find that, apart from the brightest objects, you probably won't want to go below a 1mm exit pupil very often.
Here's an image from the Wikipedia article on the subject which nicely illustrates the concepts:

(Image credit: By Evan Mason - Own work, CC BY-SA 3.0)
You'll see another term in the image: eye relief.
Eye relief is the distance from the surface of the top lens of the eyepiece to the point where the image is formed (the point where you position your eye). This distance increases as the diameter of the lens increases.
If you wear glasses while observing then it's important to buy eyepieces with an eye relief of at least 15mm (actual figure varies, maybe ±5mm, depending on the shape of your face and glasses you wear) so that you can see the whole view.
(Also, if you wear glasses while observing, are you sure that you need to? Simple long- or short-sightedness can be corrected by focusing, and mild astigmatism isn't visible at higher powers.)

You ask

how can I choose a 2" eyepiece that would be more comfortable for
observing the Moon and planets? What should I look for in terms of
types or specifications?

The Moon and planets are objects which you'll generally want to use a high magnification to view. High magnifications are served by 1.25" eyepieces (there's no need to make them as large as 2" - they're looking at a smaller part of the objective image - and making them in 2" barrels costs more to make) and give a smaller exit pupil.

In your telescope, a good starter set of eyepieces, in my opinion, would be:

A low power wide field eyepiece, with a focal length of 30-40mm, and a FoV of 68°-82°.
A medium power wide field eyepiece, with a focal length of 10-20mm, and a FoV of 76°-100°
A high power wide field eyepiece, with a focal length of 5-10mm, and a FoV of 76°-110° (This one's a bit more tricky, as the maximum magnification which will be usable will depend on your local conditions - the "seeing". If seeing is generally good you might be able to go higher than 5mm, but if it's bad then you might not be able to go much higher than 8 or 9mm)

(Notable brand names which come to mind are APM, Baader Morpheus, Explore Scientific, Pentax, Tele Vue)
You don't really need more than three good eyepieces.
You could also get away with only one high-power, an 8-24mm zoom, and a Barlow lens. But I think the wide field eyepieces are the better option in your telescope, because with a wider FoV you don't have to nudge the telescope so often to keep the object in view.
(The "good" is quite important: almost all eyepieces look good in the dead centre. Then the view starts to degrade towards the edges. Generally, the more money you pay, the less the view degrades (or starts to degrade further away from the centre). Also, and fortunately for you, the degradation of the view is greater as the F-number gets lower. At f/8, you won't need to spend loads of money to get great views, but you'll have to do your research and find out what other people with the same telescope like to use)

Finally, a couple of tips for comfortable viewing:

Make sure you're sitting down while observing. It makes a huge difference to your ability to keep your head still. If you must stand, then try and have something to hold on to or lean against.
Practice makes perfect: the more you use your telescope and eyepieces, the better you'll get at using them, and you'll find it easier to maintain a steady view.

These related posts Eyepiece diameter for small focal length planetary viewing on a large dobsonian? and What are the advantages and disadvantages of a 2-inch eyepiece versus a 1.25-inch eyepiece? have good answers to similar questions.


Can a habitable planet be smaller than 0.58 Earth radii?

Short Answer:
In part Two of the long answer below, it is stated that a planetary mass object needs to have a mass of at least 0.1, 0.12, 0.23, or 0.25 Earth mass to be habitable.  Worlds with those masses and with a radius of 0.58 Earth radius should have densities of 2.827, 3.392, 6.502, or 7.067 grams per cubic centimeter.  Such densities are possible, though it would be unusual for such small mass worlds to have such high densities.
If such a world is made more dense, it could have such a mass with a radius a bit smaller than 0.58 Earth radius, though a small world with such a high density would even rarer and more unlikely.
Taking planetary density to a theoretical extreme, if a planetary mass object was made out of osmium, rough calculations indicate it could have those minimum masses with radii of 0.29 to 0.3945 of the radius of the Earth.
And of course several solar system bodies with no atmosphere and surfaces of solid ice are believed to have liquid water in subsurface oceans, and thus possibly be habitable for alien life in their subsurface oceans.  The smallest of them of them is Enceladus, a moon of Saturn, with a radius of only about 252.1 kilometers, or 0.039 the radius of the Earth.
Long Answer:
When you say "habitable" do you mean "habitable for carbon based liquid water using organisms in general" or "habitable for human beings, other intelligent beings, and multicellular land animals that need an oxygen rich atmosphere in particular"?
Part One of Two: Habitable for multicellular land animals that need an oxygen rich atmosphere.
As far as I know, the only scientific study of what planets would be habitable for multicellular land animals that need an oxygen rich atmosphere in general, and for humans in particular, is Habitable Planets for Man Stephen H. Dole, 1964, 2007.
https://www.rand.org/content/dam/rand/pubs/commercial_books/2007/RAND_CB179-1.pdf[1]
Dole believed that there are formulas for calculating the relationships between
the mass of a terrestrial type planet and its radius, volume, density, surface gravity, and escape velocity.  Thus the mass of a terrestrial type planet would determine its radius, volume, density, surface gravity, and escape velocity. And that is largely true for the few terrestrial type worlds in our Solar System, which formed from the same protoplanetary disc with similar percentages of different elements with different densities.
But even in our Solar System Mercury is much more dense than such a small planet should be.  It is believed that billions of years ago Mercury collided with an even smaller planet, and most of the lighter elements in Mercury's crust were lost in the impact, leaving only the denser elements behind.
And in the outer Solar System, where the protoplanetary disc included a higher percentage of the lighter elements,  many of the large, planetary mass, moons and other objects have, are made of mixtures of ices of various liquids with rock, and so are less dense than the terrestrial planets.
Astronomers have discovered thousands of exoplanets orbiting other stars, and have measured or calculated the masses, radii, densities, surfaces gravities, and escape velocities of some of them. Planets formed out of the protoplanetary discs of other stars with other percentages of various elements are likely to have somewhat different densities than terrestrial planets in our solar system.
A few large exoplanets are believed to have very high densities, and it is theorized that they are the remaining dense rocky cores of giant planets that most most of their light elements in various cataclysms.
So it is possible for a science fiction writer to justify a terrestrial type planet with a density somewhat different than what would be calculated based on the terrestrial type planets in our Solar System.  But not too much different.
On page 53 Dole sets a rough maximum possible mass for a planet habitable for humans.

"Now it will be recalled that, to be considered habitable, a planet must have a surface gravity of less than 1.5 g. From figure 9, it may be seen that this corresponds to a planet having a mass of 2.35 Earth masses, a radius of 1.25 Earth radii, and an escape velocity of 15.3 kilometers per second.

Of course an exoplanet which formed from a protoplanetary disc around a different star, might be made of matter with a higher or lower average density, and so might have a slightly greater mass and radius than Dole calculated, while still having a surface gravity of 1.5 g or less.
And of course lifeforms evolving on an alien exoplanet could be capable in surviving in a much higher surface gravity than 1.5 g.
So what did Dole calculate for the minimum mass of a planet with an oxygen rich atmosphere habitable for humans, and would the minimum mass of a planet with liquid water on its surface be any higher or lower?
On page 54 Dole says:

However, if we take as a rough approximation that maximum exosphere temperatures as low as 1000 degree K are not incompatible with the required surface conditions of a habitable planet, then the escape velocity of the smallest planet capable of retaining atomic oxygen may be as low as 6.5 kilometers per second (5 X 1.25).  Going back to figure 9, this may be seen to correspond to a planet having a mass of 0.195 Earth mass, a radius of 0.63 Earth radius, and a surface gravity of 0.49 g.

Dole believed that a planet that small could retain an oxygen rich atmosphere, but didn't think that it produce an oxygen rich atmosphere in the first place.  Thus a planet with 0.195 of Earth Mass would probably not form a natural oxygen rich atmosphere but could be terraformed and given an oxygen rich atmosphere by an advanced civilization.
Dole used 2 different possible rules to calculate 2 different possible minimum masses for a planet with an oxygen rich atmosphere.  Those calculations produced a minimum mass of 0.25 and 0.59 Earth mass respectively.  So on pages 56 and 57 Dole decided that the true value would be intermediate and chose a value of about 0.4 Earth mass.

This corresponds to a planet having a radius of 0.78 Earth radius and a surface gravity of 0.68 g.

So a planet with the mass of 0.4 Earth mass would have a radius of 0.78 earth, which is 1.34 times a radius of 0.58 Earth radius, and even a planet with a mass of 0.195 Earth would have a radius of 0.63 Earth radius, which is about 1.08 times a radius of 0.58 Earth radius.
And I guess that 0.63 Earth radius is close enough to 0.58 Earth radius, that a planet with a lesser density and more light elements could possibly produce an oxygen rich atmosphere and replace as fast as it loses oxygen from the exosphere.  Or maybe a planet made of denser materials might have a high enough escape velocity and produce enough oxygen to retain an oxygen rich atmosphere.  and possibly that might stretch the minimum radius for habitability down to 0.58 Earth radius.
Part Two of Two: Habitable for carbon based liquid water using lifeforms.
Such a planet would not need to have an oxygen rich atmosphere, just a dense enough atmosphere to have sufficient pressure for water to be liquid at most temperatures on the planet's surface.  The lower the atmospheric pressure of a planet is the lower the boiling temperature of water is, until with a low enough atmospheric pressure ice will turn directly into water vapor without ever being a liquid.
"Exomoon Habitability Constrained by Illumination and Tidal Heating", Rene Heller and Roy Barnes, Astrobiology, Volume 13, number 1, 2013, is a comparatively recent discussion of the possible habitability of worlds in other solar systems, in this case as yet undiscovered exomoons orbiting exoplanets in other solar systems.  Heller and Barnes do not mention any reason to suppose that exomoons could be habitable if they were outside the mass range for an exoplanet to be habitable.  Therefore they summarize scientific opinions about the mass range for potentially habitable worlds.
https://faculty.washington.edu/rkb9/publications/hb13.pdf[2]
In section 2. Habitability of Exomoons, the last paragraph before section 2.1 2.1. Formation of Massive Exomoons, on page 20, discusses the mass range for habitable worlds, including exoplanets and exomoons.

A minimum mass of an exomoon is required to drive a
magnetic shield on a billion-year timescale (MsT0.1M4;
Tachinami et al., 2011); to sustain a substantial, long-lived
atmosphere (MsT0.12M4; Williams et al., 1997; Kaltenegger,
2000); and to drive tectonic activity (MsT0.23M4; Williams
et al., 1997), which is necessary to maintain plate tectonics
and to support the carbon-silicate cycle. Weak internal dynamos have been detected in Mercury and Ganymede
(Gurnett et al., 1996; Kivelson et al., 1996), suggesting that
satellite masses > 0.25M4 will be adequate for considerations
of exomoon habitability. This lower limit, however, is not a
fixed number. Further sources of energy—such as radiogenic
and tidal heating, and the effect of a moon’s composition and
structure—can alter the limit in either direction.   An upper mass limit is given by the fact that increasing mass leads to high pressures in the planet’s interior, which will increase the mantle viscosity and depress heat transfer throughout the mantle as well as in the core. Above a critical mass, the dynamo is strongly suppressed and becomes too weak to generate a magnetic field or sustain plate tectonics. This maximum mass can be placed around 2M4 (Gaidos et al., 2010; Noack and Breuer, 2011; Stamenkovic´ et al., 2011).

Their source for a minimum mass of 0.1 Earth Mass for a magnetic shield is

Tachinami, C., Senshu, H., and Ida, S. (2011) Thermal evolution
and lifetime of intrinsic magnetic fields of super-Earths in
habitable zones. Astrophys J 726, doi:10.1088/0004-637X/726/
2/70.

Their source for a minimum mass of 0.12 Earth mass for a substantial, long lived atmosphere is

Williams, D.M., Kasting, J.F., and Wade, R.A. (1997) Habitable
moons around extrasolar giant planets. Nature 385:234–236.

Their source for a minimum mass of 0.23 Earth mass for plate tectonics is

Williams, D.M., Kasting, J.F., and Wade, R.A. (1997) Habitable
moons around extrasolar giant planets. Nature 385:234–236.

Their sources for the weak magnetic fields of Mercury and Ganymede are:

Gurnett, D.A., Kurth, W.S., Roux, A., Bolton, S.J., and Kennel,
C.F. (1996) Evidence for a magnetosphere at Ganymede from
plasma-wave observations by the Galileo spacecraft. Nature
384:535–537.
Kivelson, M.G., Khurana, K.K., Russell, C.T., Walker, R.J.,
Warnecke, J., Coroniti, F.V., Polanskey, C., Southwood, D.J.,
and Schubert, G. (1996) Discovery of Ganymede’s magnetic
field by the Galileo spacecraft. Nature 384:537–541.

Their source for the importance of plate tectonics for habitability is:

Williams D.M. Kasting J.F. Wade R.A. Habitable moons around extrasolar giant planets. Nature. 1997;385:234–236. [PubMed] [Google Scholar]

Their sources for that maximium mass limit of 2 times the mass of Earth are:

Gaidos, E., Conrad, C.P., Manga, M., and Hernlund, J. (2010)
Thermodynamics limits on magnetodynamos in rocky exoplanets. Astrophys J 718:596–609.

Noack, L. and Breuer, D. (2011) Plate tectonics on Earth-like
planets [EPSC-DPS2011-890]. In EPSC-DPS Joint Meeting 2011,
European Planetary Science Congress and Division for Planetary Sciences of the American Astronomical Society. Available
online at http://meetings.copernicus.org/epsc-dps2011.

Stamenkovic´, V., Breuer, D., and Spohn, T. (2011) Thermal and
transport properties of mantle rock at high pressure: applications to super-Earths. Icarus 216:572–596.

The density of the planet Earth is 5.513 grams per cubic centimeter.  A sphere with a radius of 0.58 Earth radius would have a volume of 0.195 of the volume of Earth.
So putting a mass of 0.1 Earth mass within a volume of 0.195 Earth volume would result in a density of 0.512 of Earth's density or 2.827 grams per cubic centimeter.
So putting a mass of 0.12 Earth mass within a volume of 0.195 Earth volume would result in a density of 0.615 of Earth's density or 3.392 grams per cubic centimeter.
So putting a mass of 0.23 Earth mass within a volume of 0.195 Earth volume would result in a density of 1.179 of Earth's density or 6.502 grams per cubic centimeter.
So putting a mass of 0.25 Earth mass within a volume of 0.195 Earth volume would result in a density of 1.282 of Earth's density or 7.067 grams per cubic centimeter.
In a comment, Xi-k says that since the relatively common element iron, a major component of the cores of most planets, has a normal density of 7.874 grams per cubic centimeter, higher than the figure of 7.067 grams per cubic centimeter quoted above, and the iron in the cores of planetary bodies is compressed to greater densities than 7.874 grams per cubic centimeter, a mostly iron planet could be habitable with a small radius.  Thus a planet 80 to 90 percent iron with 10 to 20 percent silicate crust could have an average density of about 7.067 grams per cubic centimeters.
This seems like a plausible way to get the radius of a habitable planet down to about 0.58 Earth radius. But an expert on planetary development could probably give a better estimate about the probability of planets being that small and dense.
At at the extreme of physical possibility, and very, very, very, very improbable, is a small planetary object made mostly of Osmium, a very, very rare element.  Such a planet might have to be created by a civilization collecting osmium from several solar systems and building a planet from it.

Osmium is the densest naturally occurring element, with an experimentally measured (using x-ray crystallography) density of 22.59 g/cm3.

https://en.wikipedia.org/wiki/Osmium[3]
So a sphere of solid osmium, with a thin layer of lighter materials necessary for life, could possibly exist - without discussing the probability of such a sphere forming.
Ignoring the compression and increased density of a planetary sized osmium sphere, and assuming that the hypothetical planet has an average density 4.0915875 that of Earth, It could have mass of 0.1 to 0.25 Mass earth with a radius of about 0.29 to 0.3945 Earth.


Why would the Earth's solar analemma would be still a figue eight even if Earth's orbit was circular?

It is correct that the Earth's axial tilt alone would result in an analemma of a figure of eight with equally sized lobes.  The axial tilt contributes to the equation of time.

Circular Orbit, no Axial Tilt

Let's look at an example case with a perfectly circular orbit and no axial tilt, to observe what happens with no equation of time.  In this example, it's noon at the equator and prime meridian.  The Sun is directly overhead.  Next, the Earth rotates exactly once.  This isn't a normal solar day, but a sidereal day, about 23 hours, 56 minutes.  The Earth has rotated exactly once with respect to the stars.  Where is the Sun directly overhead now?  The Earth has moved a little bit in it's orbit around the Sun, so that our observer at the equator and 0 degrees longitude has observed that the Sun has moved in the sky with respect to the stars.  The Sun is now directly overhead at a location to the east, a little less than 1 degree longitude away (360 degrees in a circle, usually 365 days in a year).  The Earth must rotate a little more so that the Sun is directly overhead again (which completes the 24 hours in a day).

After each rotation, the Earth must rotate a little more each day to bring the Sun back overhead at 0 degrees latitude, 0 degrees longitude.  After a quarter of a year, the Earth must rotate another quarter of a rotation (90 degrees) for the Sun to be overhead at that spot.  Without that extra quarter of a rotation, the Sun is overhead at 90 degrees east longitude.  That makes sense; the Earth has moved 90 degrees in its orbit.  With a perfectly circular orbit, the rate at which this point moves eastward per day is constant, a little under a degree per day.

These points define a great circle around the Earth that coincides with the equator.  A great circle is a circle on the surface of a sphere whose center is also the center of the sphere; it's the biggest possible circle that can exist on the surface of a sphere.

A Large Axial Tilt

Let's keep Earth's perfectly circular orbit, but let's give it a very large axial tilt, 80 degrees, for the purpose of this explanation.  This Earth now begins at noon on the northward equinox at 0 degrees latitude, 0 degrees longitude.  After one Earth rotation, where on Earth is the Sun overhead now?  It's still a little less than one degree of angle away, but the direction of the displacement has changed.  Instead of this point moving due east, it is mostly northward and only a little bit east.  The Earth needs to rotate far less than 4 more minutes for the Sun to reach local noon for our observer at 0 degrees latitude, 0 degrees longitude.  These points are west of where they would be without axial tilt.  Solar noon has occurred before 24 hours have passed.  As days continue to pass after the equinox, solar noons continue to occur earlier each solar day, as these points continue mostly northward and only a little bit eastward.  The analemma shape as seen in the Northern hemisphere sky is moving up and to the right (west is to the right when looking south towards the Sun in the Northern hemisphere).

A quarter of a year later, it's the northern solstice, so let's see what has happened to our sun-overhead plots as the sidereal days have piled up.  These points have continued to move mostly northward and a little eastward, but now they are moving purely eastward, because at the solstice the sun has stopped moving northward in the sky.  It's still moving at a little less than 1 degree per day along that great circle, but because longitude lines are spaced much more closely together at 80 degrees latitude, one degree along a great circle covers many degrees of longitude.  In other words, the eastward movement of these points has been "catching up" with the longitude of the corresponding no-axial tilt points.  At the solstice, both the no-axial tilt point and the axial tilt point have the same longitude, 90 degrees east.  The analemma shape as seen in the Northern hemisphere sky is moving up and to the left (eastward), where it reaches its highest point.

These points will continue to move through longitude lines rapidly while the Sun's overhead point remains far in the north.  Now, the sun's overhead point is further east than it would be with no axial tilt.  The analemma shape as seen in the Northern hemisphere sky is moving down and to the left (eastward).

As the southward equinox approaches, the sun-overhead points are moving mostly southward and only a little eastward.  The analemma shape as seen in the Northern hemisphere sky is moving down and to the right (westward).  This allows the non-axial tilt points to "catch up", and the longitude differences begin to decrease again, until the time of the southward equinox comes, when the two points coincide again.

Here is a crude ASCII drawing of these points so far, with longitude along the horizontal direction, and latitude along the vertical direction.

                                   (2)
                   K       L        M        N      O
            J                                              P
         I                                                    Q
      H                                                          R
    G                                                               S
(1)F                                                                 T(3)
  E                                                                   U
 D                                                                     V
 C                                                                     W
B                                                                       X
A  B  C  D  E  F  G  H  I  J  K  L  M  N  O  P  Q  R  S  T  U  V  W  X  Y

A = northward equinox
M = northern solstice
Y = southward equinox

The points around (1) are further west than their corresponding points on the equator. At (2), the points have caught up in longitude.  At (3), they are further east than their corresponding points on the equator, but the equator points are catching up.

After the southward equinox, the longitudes of the sun-overhead point diverge once again, with the same mechanism as described above for the northward equinox.  Just reverse north and south, and the eastward movements are the same.  The analemma shape as seen in the Northern hemisphere sky first moves down and to the right, then down and to the left where it reaches its lowest point at the southern solstice, then up and to the left, then up and to the right where it reaches its starting point at the northward equinox.

Back to Reality

With an axial tilt of 80 degrees, the equation of time would show some extreme values, approaching almost 6 hours of divergence from the no-axial tilt case.  With the true axial tilt of about 23.5 degrees, our equation of time difference values are far less substantial, but the effect is real.

The true analemma shape we see on Earth is the combined effect of the axial tilt that we have seen plus the fact that Earth's orbit is elliptical and it slows down during the parts of its orbit when it's farther from the Sun and it speeds up during the parts of its orbit when it's closer to the Sun.

The site Analemma has good explanations about the individual effects of the elliptical orbit and the axial tilt and how they combine to create our analemma.


Is it accurate to compare comets to clouds and rain?

Dating cometary material is tricky, just because you have to go to space to get it. But the dates from the "stardust" mission show that comets are very very old. As old as the oldest asteroids, and date back to the origin of the solar system.

Now while there is evidence that material that formed in the inner solar system when the sun was young, and the planets still forming, was carried to the outer solar system. There is no suggestion that this process is ongoing. There is no astrononomical water cycle.

Although the planets might lose a little water to solar wind. It is very very little. Earth is protected by a magnetic field. Mars has lost all its  water already, as has Venus, and Mercury probably never had much.


Why didn't all the iron and heavier elements find their way to the center of the accretion disc in the early solar system?

The process of differentiation is how metals like iron and nickel can become separated from less dense substances. It generally occurs in large, partly solid bodies in the early solar system, whose interiors are melted either via accretion energy, radioactivity or some combination of the two. A solid iron meteor is likely a fragment of a differentiated larger body that has been disrupted by a collision.

I think what you are asking is why such a differentiation process does not occur in the primordial disc from which the solar system formed. The answer lies in turbulence and other mixing mechanisms that act against differentiation. If the mixing timescale in the disc is shorter than the differentiation timescale then the differentiation just won't occur. This is for example the same reason that stars like the Sun do not have all their heavy chemical elements at the core.


If you look on the main SIMBAD page, you will see the following in the box titled "Content":

The SIMBAD astronomical database provides basic data, cross-identifications, bibliography and measurements for astronomical objects outside the solar system.

(emphasis mine)
The Moon and the Sun are inside the Solar System, thus they are out-of-scope for SIMBAD.


How Can I interpret this particular light curve without reading data on the side

The radius of the planet is 0.17% of Jupiter's radius.  Under the "Planet" info on the left, it says $R_P$ which is the radius of planet, and since the units are shown as $[R_J]$ this is the value is in comparison to Jupiter. It also gives the $M_P$ (mass of planet) in relation to $M_J$ (mass of Jupiter) as well.
The interesting thing here is the size of the star - it's smaller than our Sun.
To get the radius of the planet, find the radius of Jupiter and multiply it by 0.17.
A "year" on that planet is shorter than one of our days and it's orbiting extremely close to its host star, as well.  The distance is given as 0.015 AU (astronomical unit which is the distance from the Earth to the Sun).


Understanding the angular peaks in the CMB power spectrum

My attempts at a short answer failed, so here is my best shot at an explanation.
When you look at what the photon-baryon plasma does in the dark matter background, you will bump into an equation that is almost the equation for a harmonic oscillator, but it contains a time dependent quantity $\chi = \frac{3\rho_B}{4\rho_\gamma}$ that spoils the fun. It is, however, possible to approximate the solution:
$$\Theta_k(\tau) + \Phi_k \approx A_k \cos\Big(k\int_0^\tau c_s(\tau')d\tau'\Big) - \chi\Phi_k$$
where $\Theta$ is the temperature fluctuation $\delta T/T$ (in Fourier space), $\tau$ is the conformal time, $\Phi$ is the potential caused by dark matter and $c_s$ is the sound speed in the plasma. The observable quantity is $\Theta_{0k} = \Theta_k + \Phi_k$. These are the intrinsic temperature fluctuations plus the background potential.
The prefactor $A_k$ is determined by evaluating this expression at $\tau = \tau_{LS}$, the moment of last scattering. This then yields:
$$\Theta_{0k} = \frac{1}{3}(1 + 3\chi)\Phi_k\cos(k\lambda_s) - \chi\Phi_k $$
with $\lambda_s = \int_0^{\tau_{LS}} c_s d\tau$ the sound horizon (i.e. the distance acoustic waves can travel in a certain amount of time). The quantity $\Theta_{0k}$ is called the monopole contribution to the CMB. The dipole contribution, that is due to the Doppler effect, is much smaller and I will not consider it here.
Let's take a look at an overdensity ($\Phi_k < 0$) and place our coordinate axes right in the middle of it. At the "time of the Big Bang", $\tau = 0$ the monopole is $\Theta_{0k} = \frac{1}{3}\Phi_k <0$. The argument of the cosine is 0, so the cosine itself reaches its maximum value.
Half a period of the cosine later: $\Theta_{0k} = -\frac{1}{3}(1 + 6\chi)\Phi_k$, which is a lot larger! In the standard cosmological model, $\Phi_k$ remains constant, so the only thing that could have gotten bigger is the intrinsic temperature fluctuation $\Theta$. This temperature fluctuation is directly linked to the density fluctuations of the photons and baryons.
What happens intuitively is the following: the baryons love falling into dark matter potentials. Baryons are approximately pressureless, so nothing would impede them to do so if it weren't for the photons to which they are coupled before $\tau_{LS}$. The photons are dragged down into the potential well by the baryons, which makes them hotter (photons do have pressure), so thermodynamics tells them to cool again by escaping the well. It's much harder to get out of a potential well than it is to fall into one, so the amplitudes of compression ($B$ and $\gamma$ falling into a well) and expansion ($\gamma$ pushing $B$ out of the well) are not equal.
Look again at the cosine at $\tau_{LS}$. It will reach maximal compression when $k\lambda_s = (2n + 1)\pi$ for $n = 0, 1, 2...$, which defines the wavenumbers of the fluctuations that reach maximal compression. It will reach maximal expansion when $k\lambda_s = 2n\pi$ for $n = 1, 2, 3...$.
The first peak reaches maximal compression at last scattering. It is huge because it is least affected by Silk diffusion. It corresponds to a large angular scale (compared to the other peaks) because its wavelength is $\lambda = 2\lambda_s$ and is therefore the largest fluctuation.
And finally, I think you already found Wayne Hu's website, but here is the link to a gif that shows you the oscillation of the baryon-photon plasma in a dark matter potential well.


Is the eventual heat death of the universe due to the expansion of the universe?

As far as I understand it, the heat death of the universe is a consequence of entropy, not expansion. All processes result in the shifting of some energy to higher entropy. Though the observable universe is an open system, the entire universe is an isolated system, so as more and more energy gets shifted to higher entropy over time the universe will eventually reach a state where effectively no work can be done.


The battle over Gliese 581 and Gliese 667C?

Going by multiple online exoplanet catalogues and a quick scan through the papers listed on the SIMBAD pages for the host stars:

Gliese 581

At the time of writing, the Extrasolar Planets Encyclopaedia (http://exoplanet.eu), the NASA Exoplanet Archive (https://exoplanetarchive.ipac.caltech.edu) and the Exoplanet Orbit Database (http://exoplanets.org) are in agreement and list only three confirmed planets for this system: in order of increasing orbital period these are e, b and c. It's not just Robertson et al. claiming that d is an activity signal, see for example Hatzes (2016). The most recent RV study of the system I have seen is Trifonov et al. (2018) who give a 3-planet solution, not sure whether they searched for evidence of a fourth.

Gliese 581 is known to host a debris disc with an inner edge around 25 AU (Lestrade et al., 2012), much further out than c (or indeed any of the claimed planets beyond c), so I wouldn't be too surprised if there are additional planets to be found.

Gliese 667 C

This one's a bit less settled because it doesn't seem to have attracted so many studies. Extrasolar Planets Encyclopaedia lists 6 confirmed planets, the NASA Exoplanet Archive lists 5 (they don't have planet d), while the Exoplanet Orbit Database only lists 2 (b and c). Again it is not just Robertson et al. claiming that there are only two planets, there's also the study by Feroz & Hobson (2014). The identification of planet d as being caused by the stellar rotation is supported by Suárez Mascareño et al. (2017) which is probably why this one was removed from the NASA list. Planet e is at roughly 2/3 of the rotation period and so is also considered suspicious. There doesn't seem to be nearly as much discussion about the outer candidates f and g.

Hope that helps as a summary. More data would definitely be useful for these systems if there is going to be any consensus, particularly for Gliese 667 C.

Update (June 2019): the recent study posted on the arXiv by Tuomi et al. brings back planet d, unfortunately they don't discuss this case in detail. Confusingly they also decided to swap the designations of planets d and e (at least they did in their tables, they kept the old name elsewhere in their text). They also include the 91-day planet d at Gliese 667C, also unfortunately without a detailed discussion.


How would one calculate the synodic period of the Earth and an elliptical orbit?

tl;dr:

The synodic period would be the same. For this object it will be about:

$$\left( \frac{1}{365.25} - \frac{1}{557} \right)^{-1} \approx 1061 \ days.$$

You can confirm that by reading this new ArXiv preprint about Roadster: The random walk of cars and their collision probabilities with planets. In the 2nd paragraph of Section 3 it says:

  The bodies reach the same orbital longitude on their synodic timescale of ∼ 2.8 yrs.

There's a slight difference, since the Roadster's orbit continues to be refined, and the paper is about evolution over long timescales.

However, I think what you really want to know are the times of closest approach when (at least) one orbit is significantly eccentric, and those  will not be periodic or come at regular intervals.

Well the Roadster is quite far away now and moving farther rapidly. See this answer for some details and some nice images from medium-sized tracking telescopes. It's already dimmer than +22 magnitude as seen from Earth, and as far as I know there won't be any really close approaches to Earth for a long time.

But to answer your question, the period is about 557 days. You can see this for example if you go to JPL's Horizons and look at the osculating elements. If you look a few weeks in the future when the Roadster is sufficiently far away from Earth not to be perturbed, the period around the Sun flattens out to 557 days.

above x2: Top: Distance data from JPL's Horizons plotted using Python from here. Bottom: Estimated apparent magnitude using distances from Horizons plus math shown in the text of this answer. Your milage may vary, but probably within +/-2 magnitudes of "official" predictions when they come out.

A really cool GIF can be seen in the Space.com article Observatory Spots Elon Musk's Tesla Roadster Zooming Through Space (Video). It is over 7 MB so I can't add it here. However, here are ten frames from somewhere in the middle. Still, you should go see the whole thing.

GIF:

DEMIOS image below, from here as also tweeted by Jonathan McDowell. The little dot near the center moving to the right and up is Roadster in reflected sunlight, probably mostly from the white FH 2nd stage still attached.

GIF:


A curious relationship between lunar periods and the solar year

The synodic month is the "average period of the Moon's revolution with respect to the line joining the Sun and Earth".  However, the Earth also moves in its orbit around the Sun during this month.  From our vantage point, the Sun has appeared to move in the sky with respect to the background stars, in the same direction as the Moon moves in the sky with respect to the background stars.

Your calculation deals with the sidereal month, so your result is the calculation of the sidereal year.

When the Sun returns to the same spot in the sky, that is the sidereal year, whose length is 365.256363004 days, very close to your calculation.  So why is that off from the tropical year by 20 minutes?  Because the tropical year (the year that keeps the seasons in place throughout the calendar year) is slightly shorter than the sidereal year.

  The tropical year is about 20 minutes shorter than the time it takes Earth to complete one full orbit around the Sun as measured with respect to the fixed stars (the sidereal year).

You just found the difference between the sidereal year and the tropical year, and that was no coincidence.

Addition

The number of sidereal months in a sidereal year is one more than the number of synodic months in a sidereal year.  That is because the Earth goes around the Sun once of year (of course), leading to one less synodic month than sidereal month.

Here, $P_{syn}$ is the synodic period and $P_{sid}$ is the sidereal period, and $Y_{sid}$ is the sidereal year, all in days.

$$\frac{Y_{sid}}{P_{sid}} = \frac{Y_{sid}}{P_{syn}} + 1$$

Dividing both sides by $Y_{sid}$ yields:

$$\frac{1}{P_{sid}} = \frac{1}{P_{syn}} + \frac{1}{Y_{sid}}$$

Solving for $Y_{sid}$...

$$\frac{1}{P_{sid}} - \frac{1}{P_{syn}} = \frac{1}{Y_{sid}}$$

or

$$P_{sid}^{-1} - P_{syn}^{-1} = \frac{1}{Y_{sid}}$$

Multiplying both sides by $Y_{sid}$ and dividing both sides by $P_{sid}^{-1} - P_{syn}^{-1}$ yields

$$Y_{sid} = \frac{1}{P_{sid}^{-1} - P_{syn}^{-1}}$$


You can easily convert flux to luminosity and vice versa using the spherical nature of the propagation of light. The equation you need is:

$$
L=4\pi D^{2}f
$$
Where $f$ will be your flux, and $D$ is the distance to the object, Tau Ceti in this case.

From Pijpers (2003) Tau Ceti's luminosity is $0.52\pm0.03L_{\odot}$. You should have everything you need there. $L_{\odot}$ is of course the luminosity of the Sun.


How is the entropy of a blackhole stored at its horizon?

Ultimately this is a highly theoretical question, and while appropriate here you may find perusing the Physics site helpful to get deeper answers.

But here's the basic idea for how I think about this.  It's based on asking "where else would the information be?"

Imagine the universe is a 2-dimensional sheet.  A piece of paper, say.  And that's all there is for things inside this universe: no desk it's sitting on, no air around it, etc.  Whatever inhabitants of this universe wish to talk about must be contained within this 2-dimensional sheet of paper.  Now imagine there is a hole in this paper.  What can in-universe inhabitants possibly use to learn about this hole and whatever properties it might have?  What to us, as outside observers is "the hole" is inaccessible to them: it's not part of the paper (it's an absence of paper).  But the border of that hole is accessible to them.  So whatever information this hole conveys to this universe must be accessible through that border, and the border only: anything else either doesn't exist, or is some distance away.

The hole in this paper is our analogue for the black hole in our universe.  "The inside" of a black hole doesn't make sense to an external observer, because the universe—the sheet of paper— ends at the border of the hole.  There literally is no inside to speak of.

The analogy gets complicated a bit by the fact that if you are not an external observer, but are infalling, then you can in fact pass through what the external observer calls the border. There's the whole information paradox as well when we bring in quantum mechanics.  But this is the basic gist of things: the black hole is a topological/geometric defect (as general relativity is a theory about the geometry and topology of space), and the only "part" of it which is contained in our universe is its border, the event horizon.

The actual physics at and near the event horizon is still a matter of significant debate and research.  This little exercise does not explain why the information of the No-hair theorem is accessible, but nothing else is, for example.  It just offers some intuition about why whatever information is accessible must, in some sense, be stored "on" the event horizon.  As far as external observers are concerned, that's the only place it ever could be!


How to find if a set of objects lies within a given resolution (5 arcsec) on a telescope?

This would be your best resource. Just note that the variables are expressed in geographical latitude and longitude, and your values are Right Ascension and Declination. Longitude and RA are the same, but Declination is 0 at the equator and is positive toward the north pole, whereas latitude is 0 at the north pole and is positive toward the south pole.

Don't forget to mind your units (it'll probably be in degrees since that's what your RA and Dec are). If you're doing this with a computer, you'll probably want to convert to radians first since most computer trig functions are in radians. Whatever program you use, just be sure to check what units it prefers.

Finally, be sure to convert your separation to arcseconds and you should be all set.


Is this image captured by the Hubble Telescope an original image?

All images provided from large telescopes are a rendering of the data for viewing pleasure. Usually sensors are sensitive in a wide range of wave length, even outside what humans can perceive and color is added by using filters. Hubble uses infrared filters and the so-called Hubble palette (false color).

So, if you mean, do these pictures show what a human observer will see? No, they don't. They are a visualisation of the data collected.

All things observed at low light conditions are black and white, as humans lose color vision at low light.


This is really simple once you get it:

The usual analogy here is to look at the surface of a balloon (which is a surface in 3D space, rather than 3D space itself, but the analogy still works)

A partially inflated balloon has tight curvature, but as it is inflated, the curve decreases - the surface at any point becomes 'flatter'

From an article on this topic on scienceblogs.com:

The universe has expanded a lot since the big bang - any curves, ripples or inconsitencies have largely been flattened, so it is difficult to detect them now. This article on space.com quotes:

  The Harvard-Smithsonian study spotted gravitational waves as ripples in space-time possible left over from the rapid expansion of the universe (called inflation) right after the Big Bang nearly 13.8 billion years ago.


Any massive body (i.e., a body having mass) like the stars, planets, moons, and any atom in the universe have, what is called, a gravitational potential which results in its gravity.  This is described by Newton's Law of Gravity. Why masses cause gravity, we do not know.

So, in short, the earth's mass causes it to have gravity.


Is there any formulae to calculate whether the sun will depress 18 degrees or more below the horizon on a certain day / location?

An approximation of the Sun’s declination for a certain date is:
sin δ = 0.39795 ⋅ cos [ 0.98563 ⋅ ( N – 173 ) ]
where N is the number of days since January 1.
That, combined with the formula given by @user21, viz.:
$ a_{min} = \lvert \phi + \delta \rvert - 90° $
will give you the dates where the Sun is at least 18° under the horizon.
(Oddly enough, a simple instrument, invented by the Greeks almost 2,000 years ago and perfected by the Muslims over the next few centuries, the astrolabe, would give you the answer almost instantly, without having to do any calculations. Below is a photo of one I made in brass.)


Can a meteor shower happen simultaneously all around the planet?

Realistically, no, that cannot happen. Most meteor showers that we experience on Earth - the most famous being the Perseids and the Leonids - are a result of comets passing roughly through our orbit and leaving behind debris that was burned off as the comet passed by the Sun. We then come along, sweeping through this debris and from our perspective on Earth, we see a shower of meteors raining down on us.

That implies two important things. First, we primarily see meteors as we sweep through the debris from comets. They are not objects which have made a trajectory for Earth and hit us head on. Second, we only see meteors when we're on the side of the planet facing the orbital direction of the Earth.

That being said, for the sake of the story, one could envision an alien race of sufficiently advanced technology surrounding our planet with asteroid and comet debris and subsequently "dropping" these meteroids on us from everywhere all at once. I'm not sure if aliens are involved in the story you've linked, but aside from third party intervention, there's now way you're going to see meteors from every place on Earth all at the same time.


First, the nearest galaxy to ours is not Andromeda, according to the NASA based page "The Nearest Galaxies", it was until recently considered to be the Canis Major Dwarf Galaxy at approximately 42,000 light years away.  However, recently, there is suggestion that the closer Omega Centauri Globular Cluster may be a disrupted dwarf galaxy (Another source).

Now, in answer to your other query of how many galaxies there are - there is a global project called Galaxy Zoo which is attempting to catalogue and classify all observed galaxies, so far, according to the Galaxy Zoo for astronomers page, thus far they have

  the entire Sloan Digital Sky Survey spectroscopic sample and all existing Hubble Space Telescope surveys (around 1.5 million galaxies in total).

Pictured: The Hubble Ultra-Deep Field which contains over 10,000 objects, the majority of which are galaxies. The image is only a bit over 3 arcminutes across--a tiny sliver of the sky.


What causes random irregularities in the earth's rotation accounted for by dynamical time

On short scales, atmospheric phenomena can cause variations in mass distribution.

  On relatively short sub-annual to annual timescales, variations in the length-of-day have been clearly shown to arise primarily as a consequence of the exchange of angular momentum between the solid Earth and its overlying atmosphere (e.g. Hide et al., 1980) and oceans. On the inter-annual timescale an important recent discovery concerning l.o.d. variability has been the documentation of a significant excitation associated with El Nino-Southern Oscillation events (Cox and Chao, 2002; Dickey et al., 2002). Concerning the sources of wobble excitation, it is clear that on the timescale of the seasonal cycle of climate change the excitation of the annual component of wobble variability is due to the inter-hemispheric exchange of atmospheric mass. The Chandler wobble, however, a free oscillation of Earth’s spin axis in a body fixed frame of reference with a period close to 14 months, is apparently significantly forced by the dynamical state of the oceans (e.g. Gross, 2000) as well as by the atmosphere. These most recent analyses of the problem of Chandler wobble excitation, to be reviewed in what follows, appear to have finally resolved what had remained an unresolved problem for decades.

-- "History of Earth Rotation", W. R. Peltier, Department of Physics,University of Toronto, Toronto, Ontario, Canada

On longer time scales, glacier distribution, mantle convection, and effects from the larger planets on the orbit of Earth (which by altering the geometry of earth's orbit, would also alter solar tidal forces).


Do we know if there are asteroids leading or following Earth in Earth's orbit around the Sun?

There are several known Earth co-orbital asteroids. The first to be discovered was (3753) Cruithne, which is often mistakenly described as "Earth's second moon". Cruithne has a bean-shaped orbit when viewed in a reference frame rotating with the Earth.
The only known Earth Trojan is 2010 TK7, which orbits the L4 point located 60° ahead of the Earth.


How close would the sun have to get to Earth for there to be severe consequences?

I'm going to use Gm (1 gigameter = $1\times10^9$m) and degrees Celsius for this answer.

By getting closer to the Sun, probably the most major problem would be the increase in temperature.

A while ago I wrote a program that calculates the effective surface temperature of a planet. Having dug it back out, I've played around with the values a bit. It's important to note that the effective temperature is not the same as the actual surface temperature, since it doesn't account for Earth's atmosphere. If Earth had no atmosphere, it would be the correct actual temperature.

At our current distance of (avg.) 149.6Gm, the effective temperature is 257K, or -16°C (some websites cite other values ±2°C). NASA cites an actual average temperature of 15°C. Assuming a linear relationship between effective temperature and actual temperature, we can assume that $T_{actual} = T_{eff}+31 \pm 2$.

So, what would happen if we moved, say, 10Gm closer?

Well, the effective surface temp is now -7°C, and I'm guessing the actual temperature would be about 24°C. This is an increase of 9°C. We would probably still be able to survive, but with difficulties. For starters, the sea level would rise by over 40m, leading the world to look (at the best) like this. Not an easy situation, but nevertheless survivable.
Of course, there would be other effects like an increase in extreme weather, the mass extinction of species, and probably a whole load of things that are hard to foresee, but I'm not going to try to predict them.

So, what about a move towards the Sun of 20Gm?

The effective temperature is now ~3°C, and the surface temperature is about 34°C. Things start to get bad now. All of the ice caps have melted. Once fertile areas are now barren deserts. Surviving is hard, but possible, although famine is now a major problem across much of the world (especially around the equator). The Gulf Stream may have stopped, oddly enough cooling down some of western Europe and all of Britain. It's not looking good.

What about 50Gm?

The effective temperature is 42°C; the actual one is around 73°C. Where the Sun's habitable zone begins and ends is disputed, but now, at 0.65AU away from the sun, it's quite likely that we're not in it. It's very hard to predict what happens now. Humans would likely have to stay underground to remain alive, and food would be a major issue. The ecosystem would be pretty much destroyed, having not had a chance to adapt to the new temperature.

For fun, if we moved 100Gm closer, we'd be at about 205°C. Ouch. We're closer than Mercury now, and look how that planet's coping. Mercury ranges from -173°C to 427°C depending on various factors. Survival without massive life support is not possible.

Any closer, and things just get worse.

Hope this is a suitable answer for your question!


Can we draw a min(distance) and max(mass) relation based on current observation data of nearby astronomical objects within several parcsecs?

First lets make sure we know the scale of the solar system. Neptune, the most distant major planet, has an orbit of 165 years. Sedna, the object with the longest known period, has an orbit of 11400 years.

Black holes come in different sizes. A black hole that has formed from the collapse of a giant star would have a mass more than three times the mass of the sun. If a black hole of this size were inside the orbit of Neptune, it would not be orbiting the sun, the sun would be orbiting it.

How about further out? Blackholes can be spotted in X-rays when matter falls into them forming a disc of matter orbiting at very high speeds, and heating up to enormous temperatures.  A black hole without an accretion disc would be very dark and hard to detect from a distance. But we can be reasonably confident that there are no in our immediate neighbourhood.

The motion of a blackhole in orbit with the sun would be described by Newtons laws of gravity. The sun and the black hole would orbit on elliptical orbits around their common centre of gravity. Planets of the sun or the black hole would be stable within their Hall radii. Close to the blackhole one would need General relativity to describe the gravitational field, but from further out, its gravity would be almost exactly as predicted by Newton. You could model it in any gravity simulator.

It is possible that micro black-holes exist in the universe. These could pass through the earth without doing much damage, as the are sub-atomic size and since they interact only with matter by gravity, and don't weigh much, they don't have much effect. They might be "hot" with Hawking radiation.

None of these are much of a risk to the survival of the solar system. Black holes are rare.


Is extraterrestrial mining more difficult or impractical for bodies without plate tectonics?

I vaguely remember my dad talking about this.

Uranium and other heavy elements are dense. When the celestial body was molten (early in it's life) all the heavy elements sank to the core of the body. Now, with tectonics, the heavy elements are brought back up to the surface. This is why we can mine Uranium on Earth near the surface: the Uranium was brought back up due to convention current on Earth.
In conclusion, it's not worth it at all to mine on a non-tectonically active planet/moon

If anyone can extend this, please do.


Are there any planned future observations of TRAPPIST-1 and its planets?

A website trappist.one has been created to detail information about this discovery. In the "Future" section, they specifically state:

In the short term, photometric follow-up using the repurposed Kepler satellite (named K2) along with with newer observations using Spitzer ought to reveal the period of planet 1h. We will also search for additional planets. Theses lightcurves, combined with ground-based measurements will increase the number of transit timing measurements for each of the planets. This will give us more accurate masses and orbital eccentricities. This will confirm whether the planets are mostly rocky or whether they contain a certain amount of volatiles, like water.
In the medium term, we can expect the first attempts at detecting the atmospheres of the TRAPPIST-1 planets, using Hubble, followed by deeper investigations thanks to the James Webb. The James Webb could in principle measure the temperature of the planets, and detect the chemical composition of their atmospheres. It will do so by collecting dozens of eclipses of the TRAPPIST-1 planets. The advantage of having seven planets in one system is that we will be able to compare them to one another.

It seems they're currently using Kepler and Spitzer and plan to use Hubble in the near future as well as JWST after it launches.


What visible star is closest to the ecliptic?

Regulus ($\alpha$ Leo) is the one to beat, with ecliptic latitude $\beta$ = +0.46$^\circ$ and apparent V magnitude +1.4.
Some dimmer stars are closer to the ecliptic, e.g.
$\alpha$ Lib ($\beta$=+0.33$^\circ$, V=+2.7),
$\delta$ Gem ($\beta$=-0.18$^\circ$, V=+3.5), and
$\delta$ Cnc ($\beta$=+0.08$^\circ$, V=+3.9).
For more suggestions, poke around in Stellarium with the Ecliptic turned on.

If you want linear distance from the ecliptic plane, multiply the distance from us by $\sin \beta$.


How large does refraction become in radioastronomy?

The refractive deviations in position are very similar for both radio and optical astronomy, until you consider very low frequency radio waves ($<200$ MHz) when the effect becomes rapidly larger.
For plane parallel refraction an approximation for the deviation you are talking about is
$$\Delta \theta \simeq (n-1) \cot \theta,$$
where $\theta$ is the observed elevation, $\Delta \theta$ is the change in elevation from its true value due to refraction and $n$ is the refractive index averaged over airmass.
According to this source from the Green bank radio telescope, they use something like this, with an added model for how $n$ varies with height, scaled by the atmospheric pressure. The largest value of $n$ quoted is 1.00031 at ground level. This is basically the same as the refractive index of air at visible wavelengths and amounts to about 60 arcseconds at elevations of 45 degrees.
So, to my surprise, the effects of refraction on radio telescope pointing are similar to those for optical telescopes. It simply turns out that the real part of the refractive index (that controls the phase velocity of light and hence refraction) is just as close to 1 for radiowaves as it is for visible light.
Here is another source that gives some algorithms to calculate the effective (small) real refractive index for radio waves, with similar results.
This source claims the calculations are valid for mm-wavelengths and upwards. Of course they cannot be valid as one approaches the ionospheric cut-off at around 40 MHz (wavelengths of metres), where the refractive index will depart sharply from unity and the deflections must get commensurately larger.
I did manage to find something about positional refraction at these low frequencies. The LOFAR radio array can apparently do work down to frequencies as low as 10MHz, but the practical limit may be a little higher. Anyway, they do need to account for refraction in the ionosphere and I found this presentation, which contains a section on refraction and in particular, the figure below.

Thus for low frequency radio astronomy (<200 MHz), refraction is certainly a bigger effect than in the optical. e.g. At elevations of 45 degrees, the refracted positions are shifted by about 0.1 and 0.4 degrees at 50MHz and 30MHz respectively.


The ones in the halo  have blue colors, which is the "old" part of the spectrum.  Each type of metallicity is usually measured using a form of logarithmic scale, since the flat percentage of metals (especially iron peak elements) of any star will be very small.  In this case, we have
$$[Fe/H] = \log(N_{\text{Fe}}/N_H)_{\text{star}} - \log(N_{\text{Fe}}/N_H)_{\text{sun}},$$
So we take the ratio of iron to hydrogen in the target star, measure it as well in our own sun, then take the ratio of those two and take the logarithm of that value.  Since the $\log$ function is monotonic and has $\log(1)=0$, a negative value of $[Fe/H]$ means our Sun has more iron in it (relative to the amount of hydrogen).  A value of $[Fe/H]=-1$ means our sun has an iron to hydrogen ratio that is 10 times greater than the given star's, for example.  The more negative this value, the more metal-poor the star is (or, at least, iron poor) than our own sun, which is considered of a fairly recent generation.  Lower metallicities are typical of earlier generations of stars, and so may be described as "old".  Note, however, that they need not necessarily be old stars.  In principle it is possible for a recently formed star to have formed in an unusually metal-poor region, or otherwise been unable to accrete the normal quantity of metals.


Does the relative motion between planets other than Venus and Earth form shapes such as the Flower of Venus?

It boils down to simple remainder mathematics.

If two planets have a 13:8 resonance, then, for every 1 outer planet orbit, the inner planet moves 1 and 5/8ths times around the sun.  Because both planets move in the same direction, that means the inner planet moves ahead 5/8ths of 1 orbit for every outer planet year.

5/8ths in 1 year, means the inner planet passes the outer planet every 8/5ths year, or every 1.6 years.  That's the catch-up time.  The formula is pretty simple:   8/(13-8) or small/(large-small).

That period is known as the synodic period of Venus viewed from Earth.

Throwing out the 13:8, because it's close but not quite right, Venus orbit is 224.701 days and Earth's orbit 365.256 days (sidereal orbits).

224.701/(365.256 - 224.701)
or, 224.701 / 140.555 = 1.59867 Earth years, or 583.92 Earth days.

Venus' synodic period is listed here as 584 days - close enough.

All you need for a repeating pattern is a synodic period that's very close to a divisor of an integer and you always get that when the planets are in resonance.    1.59867 is close enough to 1.6 that 5 Venus synodic years comes very close to 8 earth years.   That's what causes the pattern to repeat.   1.59867 times 5 is very close to 8.   I think it misses by maybe 2.6 degrees every 8 years, but it's still close enough to have gotten the attention and admiration of the ancients.

Venus and Earth aren't really in resonance, just close.  They don't have sufficient tidal forces on each other to maintain a true resonance.

The flower petal you get in your 2nd link, the 5 intersection points or meeting points, those happen in in that Spirograph like image when Earth and Venus are on opposite sides of the sun.  They also happen on a 1.6 earth years and move forward 225 degrees every Earth year.  It looks neat, but like a Spirograph it's a cool design based on nothing more than simple fractions.

As far as fractal, flowers are sometimes used as examples of fractal patterns, so I can see how, looking at the flower pattern, one might think it might be fractal.   It's really not.  Fractals are non-repeating.   Kepler Orbits are repeating.

Some of Earth's Milankovich cycles may be fractal in their variation.   That's another topic though.

@Chappo

using your 7:4

4/(7-4) = 1.333 years to catch up, so that corresponds to 3 passes every 4 years, or, as you said, a triangle.

For resonances and the number of points in the star-pattern, it seems to be a simple matter of subtraction as you said, because the denominator in your example is 7-4, or 3.   Divide by 3, you then need 3 cycles to get back to an integer.


How thick is the cosmic microwave background, including the part we cannot see within the observable universe?

If I understand you correctly, you want to know the distance from the point from which we observe the CMB, to the edge of the observable Universe.

During inflation, the observable Universe expanded from ridiculously small to some ten meters in radius, so that part can be safely ignored compared to the distances now$^1$.

The distance$^2$ to the "CMB shell"$^3$ is 45.4 billion lightyears ("Glyr"), and the distance to the edge of the observable Universe is 46.3 Glyr. Hence, the shell of the observable Universe that lies beyond the CMB shell has a thickness of only 0.9 Glyr.

Here's a sketch of how I interpret your question (not to scale):

If you want, I can add details on how to calculate these numbers.

$^1$The relative expansion during inflation was huge, however: The Universe expanded roughly by the same factor that it has expanded afterwards, namely a factor of $\sim10^{26}$.

$^2$Here, "distance" corresponds to the comoving distance, which is what you would measure if you froze the Universe right now, and started laying out meter sticks.

$^3$This shell is not infinitely thin, but has a thickness of some 60 million lightyears, so let's ignore that.