Are there any common characteristics of habitable planets?

I think I will answer your question from two vantage points. Firstly, what do we have to measure about a new exoplanet to consider it potentially habitable? Keep in mind that in detecting these 1000 or so new exoplanets, we've only gone so far as to show that they (statistically) exist and to measure some very very basic properties of them. It is on these very basic properties that we make our judgement of habitability and further research and observation is necessary to confirm these suppositions. When a new exoplanet is announced, we generally know the following information, which we use to determine habitability:

Distance from the star. This is a crucial piece of the puzzle because it tells us generally how hot or cold the planet is. We presume that life cannot exist on planets which are boiling hot or freezing cold. This information can be ascertained from either the radial velocity method or the transit method.
Size of the planet. Note here, size generally refers to diameter or radius. This is important because we want to generally know what type of planet we've detected. Gas giants are, well ... giant and inhospitable to life as we know it. We want to make sure the planet we've found is small enough that it is likely terrestrial. Although size is not a guarantee that it is terrestrial, its a piece of the puzzle. We can measure size by the transit method, however there is some degree of uncertainty to this process because we need to know the size of the host star to a good degree of accuracy, which is not always possible.
Mass of the planet. This is a crucial component in determining habitability. Size alone may tell us that a planet is what we call a "Super Earth" or "Sub Neptune" in that it is slightly larger than our own planet. The problem with that is, if the planet sits on the size border between Earth and Neptune, how are we to know whether it is rocky or gaseous? This can be resolved by looking at the mass, which is ascertained by the radial velocity method. The mass combined with size, gives us density and density is a surefire way to know if the planet is made of dense material or light gases.

That's really about it for basic information gleaned from simple planet detection methods. If you have distance from star, size, and mass (all of which you can get from the RV and transit methods) then you can presume that the planet is Earth-like and at a good temperature. That's enough to claim potential habitability, but not enough to actually show it can support human life. For that, you need to investigate the planet further and show it meets a variety of other conditions. As of know, some of those conditions are measurable with enough work, some aren't. These conditions may include, but are not limited to:

An atmosphere with water in it. We have the technology to measure this fact, but I'll say that it is very very difficult to do so and I don't know if anyone has really successfully shown water exists in an exoplanetary atmosphere (some have convincing evidence though). Generally, the process to do this is to measure the radius of the planet as a function of wavelength. Different wavelengths of light will refract through the atmosphere differently (just like in a prism) and so you measure a different radius of the planet (radius being the physical size of the planet plus the extent of the atmosphere) at different wavelengths. From this information, you can model what the atmosphere must be made of to cause the radius to change depending on wavelength. The issue with this process is that is requires removing all possible false detections and having a great signal to noise which is very hard.
A magnetosphere. We believe this is a necessity for life. Every second of every day our planet is being bombarded with high energy particles from our Sun. Most of these particles are harmlessly deflected by our magnetosphere and so we're protected. Planets without this protection would have a hard time hosting life as that life would be destroyed by radiation (or at least life like us). There is currently no way to detect whether a planet has a magnetosphere, although I believe there are potential ideas on the chalkboard.
The presence of a moon. This one is a bit iffy. Some people will argue our Moon has been instrumental to our development and for life in generally. The argument generally goes that the Moon has helped stabilize our tilt and thus kept our seasons and global weather well regulated. While we might be able to exist on a planet without such conditions, some believe life cannot begin on such a harsh planet without that regulating moon. There is also the benefit that the Moon has shielded us from a large number of impacts by taking the blow itself. Again, we could live on the planet, but the lack of a moon might not foster growth of life naturally. Moons, at this time, are out of our reach to detect.

There are numerous other factors for what helps define whether a planet is habitable, be it by us or in hosting new life. The list could go on and on, but these are the major points I think.


Open source code for the maths behind a heliostat?

I wrote the source you need some years ago:
https://jumpjack.wixsite.com/progetti/sorgenti-ipsun
The Arduino/Processing version was just a demo program to manually control a TENVIS camera by multiple buttons, it lacks the "astronomical algorithm".
The Javascript version contains astronomical calculations and a demo page which connects to a local TENVIS camera.
I also built a prototype...

https://jumpjack.wixsite.com/progetti/ipsun
...then I found, at MakerFairRome 2016, a maker which sold on Kickstarter a ready-made heliostat, initially named "Lucy", then renamed "Caia"... but 5 years passed and it has not yet been born... so I think I'll have to go back to my project and develop my own heliostat.
But in the meantime I found this page about a "static heliostat", and now I am trying to reverse-engineer it to figure out if it could be easier/cheaper:

http://cleardomesolar.com/solareflexpanels.html
Additional resource with dozens of links to heliostats, coelostats, and whatelse...:
http://www.redrok.com/main.htm#gsc.tab=0


If the earth stopped spinning what would happen to the moon

Earth's current rotation is faster (~1 day) than the Moon's orbital period around Earth (~28 days). This leads to tidal acceleration -- the tidal bulges raised by the Moon rotate ahead, pulling on (and being pulled by) the Moon. Over a long period, this converts Earth's rotational momentum into Lunar orbital momentum. The Earth's rotation slows, and the Moon's orbit expands. Eventually, if nothing else interfered, Earth and Moon would become tidally locked, with the length of a day and the length of a month equal.

If Earth weren't rotating, this effect would happen in reverse. Tidal acceleration (deceleration?) would transform Lunar orbital momentum into Earth rotational momentum. The Earth would slowly start to rotate again, and the Moon's orbit would contract, again until the two were tidally locked.

I'm not in a position to run the numbers, but it should be possible to estimate the final configuration -- how close the Moon would be, and how long the day/month would be.


Light that travels eternally because of inflation?

Infinities are a type of singularity. As soon as you introduce an actual infinity into any kind of mathematical model you can't expect it to make useful predictions. Mathematical models can't deal with infinite time. Similarly, using a frame that moves at the speed of light, with infinite rapidity, is singular.

In the particular case you describe, in a regular frame, the photon becomes progressively redshifted by the expansion of spacetime, its energy (to an observer in a inertial frame) becomes lower and lower.


Understanding The Turnover Point of Angular Diameter Distance

On the one hand an object spans a smaller angle the farther away it is, as expected. On the other hand, due to the expansion of the Universe and the finite speed of light, very distant objects were closer to us when they emitted the light we see today. At that time they spanned a larger angle.

The following paragraph is hard to understand without a drawing , so have a look at the drawing in this answer, which may help. The light from the edges of a galaxy that happened to travel toward where we were at the time it was emitted did not reach us, because by the time it arrived to that place, we were no longer there. But the light that happened to be emitted toward some point behind us reached that point at the same time as we did, so that's what we see.

The exact turnover, the threshold between "looking smaller because far away" and "looking larger because closer in the past" depends on the expansion rate history of the Universe, as well as on the way that light propagates in the Universe, which in turn depends on the densities od the various constituents of the Universe. The interrelationship of these quantities is given by the Friedmann equation.


How can a planet capture a moon?

There are 178 moons in the Solar System, according to the NASA Planetary Fact Sheet, so it seems to be a common event. The following sections will show that moon capture is actually unlikely, but when a planet has one or more moons capture becomes easier.

Initial Conditions

Starting from the initial conditions, the planet is in orbit about the sun, and an asteroid is in a different orbit about the sun.

In order for capture to become possible, the asteroid and the planet must come into proximity. When the asteroid comes inside the Sphere of influence of the planet, the gravity of the planet is the main factor in determining the path of the asteroid.

Possible Outcomes

Relative to the planet, the asteroid will be following a hyperbolic trajectory, and hence has sufficient kinetic energy to avoid capture. A large variety of outcomes may occur, but the ones that lead to capture are those where the asteroid somehow loses enough kinetic energy for its velocity to fall below the escape velocity of the planet while retaining enough energy to achieve a closed (elliptical) orbit. The main (not the only) possible outcomes are

the orbit of the asteroid is perturbed, by a greater or lesser extent, and it continues on its way out of the sphere of influence of the planet.
the orbit of the asteroid is perturbed, and the asteroid impacts the planet surface. That would usually be the end of the process, but current theories on how Earth captured the Moon are that a body named Thea impacted the Earth, and the Moon formed from some of the collision debris.
the orbit of the asteroid is perturbed, and the path of the asteroid intersects the atmosphere of the planet, losing kinetic energy as heat in the atmosphere (similar to aerobraking).
the orbit of the asteroid nears an existing moon of the planet and is accelerated (in the sense that deceleration is just acceleration with the opposite sign) by the existing moon, as used by the MESSENGER spacecraft to slow its speed before orbiting Mercury.

The last two cases admit the possibility of capture.

Possible Capture

After losing energy in the planetary atmosphere, if the asteroid has lost enough energy it may enter a closed orbit around the planet. The problem is that the orbit will intersect the atmosphere again, losing energy each time it does so, until it impacts on the planetary surface. Capture can occur when an existing moon is present and is in just the right place for its gravity to reduce the eccentricity of the orbit of the asteroid.

So, the most likely case where a planet can capture a free asteroid is when there is already one or more moons present. The incoming asteroid must avoid entering the Hill sphere of the existing moon - the region where the moon would dominate the path of the asteroid.

Gravity assist can accelerate an asteroid when the asteroid is passing outside the orbit of the moon, but can decelerate the asteroid is passing inside the orbit of the moon. In this case some of the kinetic energy of the asteroid is transferred to the moon. As is the case with aerobraking capture, gravity assisted capture requires the existing moon to be in just the right place.

Another mechanism

A rather elegant paper published in Nature (mentioned below) shows how two bodies orbiting each other as they approach the planet could have led to one being captured by Neptune. This mechanism could apply in other cases also. This Dissertation (pdf) discusses a similar process for Jupiter.

Irregular bodies

It turns out that irregular shaped bodies can be captured more easily than spherical bodies. Orbiting within the Hill sphere of the planet is not enough for capture to be permanent. Only orbits in the lower half of the Hill sphere are stable. Bodies in higher orbits can be perturbed by nearby planets, and the body can eventually be ejected. But irregular shaped bodies exert minute fluctuations in gravitational attraction on the planet, and actually orbit in a chaotic manor. When other moons or rings are present these chaotic orbits gradually transfer energy to the bodies in the lower orbits, causing the new body to orbit lower, and hence become immune to external perturbation. [citation needed]

Prograde vs retrograde orbits

The same analysis of chaotic orbits, and earlier work also concluded that retrograde orbits are more stable than prograde orbits. Whereas prograde orbits are only stable in the inner half of the Hill sphere, retrograde orbits can be stable out to 100% of the Hill radius. Hence retrograde capture is more commonly observed (this is not the whole story, it is still a matter if research).

Multiple existing moons, rings, and the early Solar System

While the probability of a single moon being in the right place at the right time is low, when there are multiple moons the probability of an initial helpful interaction rises linearly. But the probability of additional interactions rises geometrically, so the more moons a planet has the more likely it is to capture more. The existence of rings also aids capture by exerting a drag on the new moon, taking it's energy and lowering it's orbit, in much the same way that uncaptured gas would do in the early Solar System.

The biggest planets have the most moons

It may be obvious, but the biggest planets have the most moons. This is because they have deeper gravity wells, and sweep in more objects. Even though the probability of capture is low (most objects are just pulled into the planet), a steady trickle have have captured over millions of orbits.

Conclusion

Each capture mechanism requires a fortuitous set of conditions, and so is actually a quite rare event. One mechanism is a that a pair of co-orbiting asteroids become separated when one enters the planetary Hill sphere. The odds for an individual asteroid are improved when the asteroid arrives with low kinetic energy that must be given up to other bodies orbiting the planet, and when there are already many moons or a ring system.

See also

Could Earth's gravity capture an asteroid? - earthsky.org
Dynamics of Distant Moons of Asteroids - Icarus (pdf)
Hypothesis: New evidence on origin of the Moon support cataclysmic collision theory
Neptune's capture of its moon Triton in a binary-planet gravitational encounter - Nature
Planetary Fact Sheet - NASA
The use of the two-body energy to study problems of escape/capture (pdf)
Wikipedia
Aerobraking
Elliptic orbit
Escape velocity
Hill sphere
Hyperbolic trajectory
Gravity assist
Retrograde and prograde motion
Sphere of influence
Thea


How can a star's rotation speed be measured?

The projected equatorial velocity of a star (commonly written $v \sin i$, where   $i$ is an inclination angle for the rotation axis to the line of sight) is measured by observing the Doppler broadening effect on spectral lines.

There is no need to resolve the star. Light from one limb is approaching at $+v\sin i$, whilst light from the other limb is receding at $-v\sin i$. The net effect is that the spectral lines are convolved with a broadening function that has a full width of $2 v\sin i$.

Providing that this rotational broadening makes a significant contribution to the overall line width, then it can be measured (by direct fitting, Fourier transform or cross-correlation methods).

True rotation speed requires a measurement of rotation period and knowledge of the stellar radius. Whilst the former is known, the latter usually isn't.

Note in your article, I suspect the rotation speeds are actually the inverse of rotation periods. Angular velocities are often referred to as rotation rates and can be derived without spectroscopy by measuring periodic light variations from surface inhomogeneities on a star.


Could the dinosaurs have seen the asteroid that killed them?

The answer is yes; for a few nights prior to the impact (assuming they had eyes with a similar sensitivity to our own and could look up!). It could also be much longer than this if the object was a cometary body.
Details:
Impacting solar system objects would have relative closing speeds from around 11 to 72 km/s.
We could take the optimal case that the asteroid approaches whilst fully lit by the Sun (which probably precludes the minimum and maximum speed in the range quoted above) and then scale from another similar body - say the asteroid Vesta. This has a diameter of around $a=520$ km, gets as close as $d=1.14$ au from the Earth and has a maximum brightness of about $m=5.2$ apparent magnitude (and is hence just visible to the naked eye) and an observed flux $f = f_0 10^{-0.4m}$, where $f_0$ is a zeropoint for the magnitude scale.
Thus the flux $f_n$ received by a near-Earth asteroid of diameter $a_n$, at a distance $d_n$ from Earth (in au) and with the same reflectivity would be
$$ f_n = f\left(\frac{a_n}{a}\right)^2 \left(\frac{1+d}{1+d_n}\right)^2
\left(\frac{d}{d_n}\right)^2$$
The magnitude of the dinosaur killer would then be
$$m_n = m -2.5\log (f/f_n)$$
To be an at all conspicuous naked eye object, $f_n \geq f$. If we assume the dinosaur-killer had $a_n=10$ km, then
$$ d_n^2(1+d_n)^2 \leq 0.0022$$
An approximate solution is obtained by assuming $d_n \ll 1$ and thus we find
$ d_n \leq 0.047$ au or 7 million km.
Moving at say 30 km/s, then it gets closer by 2.6 million km per day, thus hitting the Earth about 3 days after becoming a naked eye object. Obviously this would be longer for a slower approach speed or for a larger or more reflective asteroid. But shorter for a smaller, faster asteroid or if the asteroid approached from a direction not fully illuminated by the Sun or had a smaller albedo than Vesta.
Another possibility is that the object is of a cometary origin with an icy composition. If that were so then it could be much brighter as a result of sublimation, outgassing and having a bright cometary nucleus and tail. he answer would still be yes, but the visibility period could be weeks (comets are rather unpredictably bright).
It thus seems to me that there is a plausible range of parameters and trajectories where a dinosaur-killing asteroid could be observed and then observed to grow brighter over a few nights, but probably not much longer than that unless it was a comet.


Do we know how old the matter that makes us is?

The material (heavier than helium) that makes up the solar system was made in millions, if not hundreds of millions of stars that lived and died  in the ~7 billion years between the formation of the Galaxy and the birth of the Sun. This material has been thoroughly mixed in the interstellar medium and so the heavier elements arise from countless individual stars of different ages. You can say something probabilistically, based on what we understand about the build up of chemical enrichment in our Galaxy.

Starting with hydrogen - the vast majority of hydrogen is primordial. That is, it hasn't been inside another star.

The carbon in the solar system is made in a variety of stars. Some will come from high mass stars and dispersed either when they exploded as supernovae or via winds from their surfaces during the Wolf-Rayet phase; some will have been manufactured in lower mass asymptotic giant branch stars and expelled in their winds. The distinction is important - massive stars enrich the Galaxy very promptly (see below, when discussing oxygen), near the beginning of the life of the Galaxy, long before the Sun was born. The enrichment rate by supernovae and massive stars tailed off considerably at later times. On the other hand, enrichment by lower mass stars takes longer because of their longer lifetimes. Thus the carbon in the solar system will have a spread of ages from about 13 billion (the oldest stars in the Galaxy), right up to the age of the Sun.

Elements like oxygen, phosphorus, magnesium, sulphur and silicon are primarily made and dispersed by massive stars and supernovae. These are likely to be at the older end of the range discussed above. On the other hand, nitrogen is also made in a mixture of stars, both high and low masses, with a consequent range of ages.

Iron and nickel are mostly dispersed into the interstellar medium by type Ia supernovae. These are exploding white dwarfs, which in turn are the remnants of comparatively long-lived stars of a couple of solar masses or so. Most likely, these elements were made not long before the Sun was born.

See also Parent stars of our Sun - Where are its remains? and How can there be 1,000 stellar ancestors before our Sun?

A more technical description of this differential enrichment history can be found in Pippino & Matteucci (2008).


Which quantum reality will be seen in a future Earth

All of them.

If you sit on Earth then (in the many worlds interpretation) each quantum mechanical observation creates multiple universes, and there is a copy of "you" in each one. So if I were to ask, which universe will I be in tomorrow, the answer is "all of them" (except the ones in which I die)

If I go into space, float around and come back I will come back in all of the universes. Because I have been moving (relative to Earth) there will be a slight clock discrepancy in all the universes.

If I go into space, orbit a black hole, then come back. I will come back in all the universes (except for the ones that I die in). There will be a large discrepancy in clock time but that is irrelevant. I've not done any movie time-travel. It's just that my watch doesn't match the clocks on Earth.

In the Copenhagen interpretation you come back in the one universe (with a certain probability that is close to 1). As the mathematics is identical, there is no experiment that can distinguish between different interpretations of QM.


It's very easy to create a star: you just put an amount of hydrogen gas quite near to each other and gravity will do the rest.

The problem is that the amount of gas you need for that is so large that even replacing the whole earth with hydrogen atoms is far from enough: you need far more hydrogen atoms for that and as such you can't do something on the earth when you need more space than the whole content of the earth :-)


Is the Dwarf Planet class really necessary?

A lot of the naming conventions were originally "because they remind us of things we already called this", or simply "tradition".  How we name things has slowly but surely adjusted with time as more objects were found, and a more robust classification system was needed.

Imagine it like having bins to sort your toys into.  If you have a small number of balls, perhaps you'd stick them all into a single bin called "balls".  But as you get more and more balls, you find they no longer fit into that bin.  Looking at them, you notice they are all either quite small, or quite large.  So you make two bins: small balls and large balls.  Or maybe you notice they are all either very soft (stuffed or Nerf versions, perhaps), or very hard (baseballs, for example).  So you could have a "soft balls" and "hard balls" bins.  Keep collecting ever more balls, and you may need three, four, etc. boxes to fit them all in.  And each time you will naturally try to order them so that each box contains balls that are similar to each other: these are all soft and green, these are all hard and white, these are all big and bouncy, etc.  This applies equally well to all sorts of collections: baseball cards may start all together, but then get sorted into teams and years or even individual players as a collection gets very large, and so on.

This is basically what's happened as we've classified and reclassified objects we've found in our solar system.  We started with a small number of objects, but as we found more and more of them it became too unwieldy to stick them all together, so we started to separate them more and more.

Ceres was labeled a planet because it seemed like a tiny version of, but otherwise very similar object to, existing planets.  Then we eventually discovered there are a lot of things in the same general orbital region as Ceres, and if those were to also be planets then we'd soon have a gigantic list of planets.  Since all these objects seemed pretty similar, and came from the same general region, they were given their own class of objects: asteroids.

Pluto was labeled a planet due in part to some initial errors in the data that suggested it was larger than it actually is.  Despite a growing number of oddities about it that made it look increasingly un-planet-like, tradition left it as a "planet"; it wasn't particularly problematic for this one little odd-ball to be lumped with the others.  Finding a host of new, similar, and sometimes—in the case of Eris, at least—bigger objects forced a reconsideration.

Incidentally, there are also comets, which are not the same as asteroids.  Asteroids are rocky objects in the inner regions of the solar system.  Comets are icy objects from the outskirts of the solar system: the Kuiper belt and Oort cloud, in particular.  They are vaguely similar to each other: generally very small, lumpy, potato-like objects, but we noticed that asteroids were mostly rocks and metals between Jupiter and Mars, and comets were mostly ices beyond Neptune.  So we separated them into their own groups.

Pluto is more comet than asteroid, and likely originated from either the Kuiper belt or Oort cloud.  In this case, it likely got knocked into a closer orbit via gravitational interactions with the gas giants, or possibly a star that got close enough to the Oort cloud.


Jupiter's magnetosphere interacts with the one from Saturn?

The simple answer is yes, because magnetic fields extend to infinity.

Your question should ask if a significant interaction is present. My question would be, what is a significant interaction?


How to find the moons maximum culmination at a location

Let's suppose, that moon is always located on the ecliptic. Ecliptic ranges from declination $\delta=-23.5°$ to $\delta=23.5°$. That means, that the maximum apparent height of the Moon in Calgary will be $90°-\phi+23.5°=90°-51.04°+23.5°=62.5°$
But moon isn't always located on the ecliptic. Its inclination is around 5.14°. That means, that it can deviate from the ecliptic for such amount. So the maximum height changes to $62.5°+5.14°=67.6°$. But note that this won't happen every month, but rather every year.
This is just a simple geometrical problem. On figure:

Also, the culmination (the highest point) on northern hemisphere always occurs on the south.


Can we detect exoplanets with perpendicular plane of ecliptic?

There are (at least) three main methods to detect planets that orbit so that the orbital plane is at right angles (or thereabouts) to our line of sight.

The first of these is direct imaging. This is already responsible for the discovery of a number of planets. However, in order to be visible against the glare of their host star, the exoplanets need to be in wide orbits and be massive. It also favours the detection of giant planets around young stars, since these planets are still contracting, are still warm, and emitting their own infrared light. A list of about 20 directly imaged exoplanets can be found here; arguably some of these should more rightly be called brown dwarfs.

A second method is the so-called astrometric detection of exoplanets. This works because the star and planet orbit their common centre of mass. This means that the star executes an orbit with the same period as the exoplanet, but with a semi-major axis that is smaller by the ratio of the planet mass to the star mass. This motion of the star can be detected, in a similar way to which parallax is measured, by precise measurement of position with respect to background objects. The method is most sensitive to wider binaries and massive stars, but is limited by the requirement to see appreciable motion of the star within the duration of the observational dataset. In practice then, this means it is difficult to detect planets with orbital periods of decades, unless one is prepared to wait for some time!

There may only be one accepted exoplanet discovery by this method so far (HD 176051b), but the Gaia satellite should be capable of detecting many thousands of exoplanets with masses at or above that of Jupiter and with orbital radii of about 1--10 au (see Perryman et al. 2004).

Thirdly, gravitational microlensing is quite insensitive to the plane of the orbit. This works when a foreground object gravitationally lenses and magnifies the light from a background star. An exoplanet is marked by a "blip" in the lensed light curve. There are also about 20 planets detected by microlensing. The disadvantage of this technique is that the event is a non-repeatable one-off, so not much more can be learned about these systems.

Notably, all three of these techniques are less sensitive to planets that have small mass (as indeed are transit experiments and the radial velocity technique). However, whilst small planets (in terms of mass) have been found by microlensing, it seems unlikely that the other two techniques will discover Earth-like planets for quite some time.


Is their any clue where gravitons come from?

There is no fully satisfactory quantum theory of gravity. The electromagnetic field is quantized, the quantum excitations are photons. It would be reasonable to believe that the gravitational field would also be quantized, and the quanta would be particles, called gravitons. In fact it would be very hard to explain how gravity could not be quantised.

The image of a particle emitting photons is rather misleading. Instead, two particles exchange a photon in an electromagnetic interaction. And the particles are not classical, but quantum mechanical.

Photons carry the electromagnetic force between any charged particles: not just electrons, but also muons, tauons, quarks, and the W particles. By this analogy, anything that can create a gravitational field would be exchanging gravitons: that includes anything that has mass or momentum. (ie more or less anything).

However, there is neither a theoretical, nor an experimental basis for the existence of gravitons.


Are we made of the stuff of one star or more stars?

The precise presolar history isn't known. I'll try to tell a likely story backward in time.

Our sun probably has brothers and sisters scattered throughout the Milky Way. Further back in time our sun likely was part of on open star cluster like the Hyades. Open star clusters are not stable and eject stars over hundreds of millions of years. (More about attempts to find Sun's siblings.)

Such star clusters typically form in nebulae like the Orion Nebula. Dense nebulae typically contain short-lived stars with multiple solar masses, which typically end as supernova, adding heavy atoms to the nebula and surrounding interstellar medium. Our sun is rich in heavy atoms, easy to detect in spectral lines of sunlight or in solar wind; those atoms cannot have formed in the sun, because its core is not yet hot enough to form these elements; some will never form in the sun. Earth mainly consists of chemical elements heavier than helium. Stars like our sun, rich in heavy elements, are called population I stars (astronomers call all chemical elements besides hydrogen and helium metals). Population I stars (and Earth) can only have formed, at least in parts, from supernova remnants. Old stars, born in a time, when the immediate predecessor(s) of our sun were born, can still be found in the Milky Way. They are called population II stars. They contain less material made of heavy elements than our sun, but still more than provided by the big bang. Therefore there must have existed even older stars, called population III stars.

Supernovae also can destabilize the interstellar medium due to shock fronts, eventually leading to local collapse especially of nebulae.

Presolar grains are still raining down on Earth; "the elements in these particles were made at different times (and places) in the early Milky Way".

By this, the likely answer to the question is, that we probably consist of remnants of several supernovae, and of several generations of supernovae.


Generally, astronomical bodies rotate around their common center of mass, not the one or the other body.

If one of the bodies is much heavier, the center of mass is much nearer to its center, sometimes even inside the larger body. For example, Earth and Sun both rotate around a point that is much nearer to the Sun's center than to the Earth's center, as the sun is much heavier. Looking from the outside, it seems as Earth rotates around the Sun, while the Sun just wobbles a bit in sync. But the reason for that is exclusively the masses of the bodies, not their being-a-star or not-being-a-star.

Your theoretical example of a super-heavy planet and a lighter star would result in the star seemingly rotating around the planet; however, here you need to look at why a star is a star:
Stars are created by gravity that is compressing the mass to a level that starts fusion, and thereby a star. In your example, either the 'light star' wouldn't make it into a star (and be a light body/planet), or the 'super-heavy planet' becoming a much brighter star (or both of it). That reverses the roles effectively.

Looking at it this way, the bodies in a system rotate always around the center of the mass, which is near the largest mass, which will be the first star of the system (if any).
So your first constellation is not possible.

The second is well possible - if neither of the bodies has enough mass to become a star, they will be still rotating around each other the same way, with the heaviest body being nearest the center. Such a system would be dark and basically invisible from some distance. We wouldn't be able to find them.


I did not find any Richter-scale numbers for Io, probably since there are no direct measurements yet, or how atmosphericprisonescape puts it:

Nobody put a seismometer on Io yet, so we don't know. The only other bodies except for Earth, where seismometers have been placed so far are the Moon and Mars.

But I found some interesting comparison at on NASA's SpacePlace: High Tideo on Io!

On planet Jupiter's moon Io ("EYE-oh"), the ground itself moves up and down like an elevator taking people to the top and bottom of a 30-story building!

This would mean amplitudes for the tidal waves (and therefore also for regular Io-quakes) of ${\rm 60 \dots 90 m}$! I am not sure though where that number comes from, but it compares with ${\rm 0.2m}$ tidal waves of the Earth's solid crust which is not really measurable on Earth, except with interferometers like e.g. GEO600.
References

Wikipedia on Tital Heating of Io
Are planet-destroying quakes possible or observable?


Luminosity L(t) for a homologously contracting star

Fairly sure this is standard bookwork if you are talking about a PMS star on the Hayashi track.

Differentiate your GPE wrt time, assuming that the mass is constant. Take half of this as the luminosity (via the virial theorem). This gives $L$ in terms of $dR/dt$.

Then if you are on the Hayashi track you can assume surface temperature $T$ is constant and Stefan's law gives you $dR/dt$ in terms of $L$, $T$ and $dL/dt$.

Substitute that in and integrate the resulting differential equation to get $L(t)$.

Using this approach, I managed to get
$$ L = \left(\frac{\alpha GM^2}{6}\right)^{2/3} \left(4\pi \sigma T^4\right)^{1/3} t^{-2/3},$$
where $T$ is the temperature of the Hayashi track in question.

If you don't want to make the Hayashi track assumption, you can say that $L = AM^B T^C$ defines tracks in the HR diagram and this can be used in addition to Stefan's law to eliminate $dT/dt$ and give your result in terms of constants $A$, $B$, $C$. These can come from polytropic theory or by fitting the $L,M,T$ function to numerical calculations. (E.g. see Jackson & Jeffries 2014 https://arxiv.org/abs/1404.0683 ).


How can the Earth and the Moon be in synchronous rotation if the Moon won't be in geostationary orbit?

Although the Earth's rotation is slowing due to tidal interactions with the Moon, the timescale for the Earth to reach synchronous rotation with the Moon's orbit is quite long, even by astronomical standards, and certainly won't happen during the red giant phase.
The current change in Earth's rotation is 1.7 ms/century over the past 100 years, and 2.3 ms/century since the 8th century BCE.  (See Wikipedia for references.)  To be conservative (i.e. to give the Earth the best chance to get to synchronous rotation), let's take the larger rate of spindown.
Converting units, this is
$$ \frac{2.3 \times 10^{-3}\ \mathrm{sec}}{10^2\  \mathrm{yr}} \times \frac{1\ \mathrm{day}}{86400\ \mathrm{sec}}=\frac{2.66\times 10^{-10}\ \mathrm{day}}{\mathrm{yr}}=\frac{1\ \mathrm{day}}{3.76\times 10^{9}\ \mathrm{yr}}$$
So at the current rate of change, the day length increases by one (current) day every 3.8 billion years.  So by the time the Sun reaches the red giant phase in 5–7 billion years, the day will be at most 2 days longer, i.e. about 3 of our current days.  The orbit of the Moon will be longer than its current 27.3 days (it is spiraling outward due to the same interactions), so we won't be anywhere near synchronous rotation at that point.
Assuming the Earth survives being engulfed by the Sun during the red giant phase, how long will it take?  At the rate calculated above, changing from the current 1 day to the Moon's current orbital period of 27.3 days requires a 26.3 day change, which takes about 100 billion years (about 7 times longer than the current age of the universe) at the current rate of change.  We still aren't synchronized at that point, since the Moon will have moved outward at that point, and thus (by Kepler's third law) will have a longer orbital period. (To calculate the right period to compare to, you would have to work out what orbital period gives synchronous rotation given the current amount of angular momentum in the Earth-Moon system.)
There are two major reasons why the above is an underestimate of how long this would take:

As already noted, as the Moon gains orbital angular momentum from the Earth's rotation, the Moon moves into a larger orbit.  This weakens the tidal interactions between Earth and Moon, which makes the rate of change slow down over time.

During the red giant phase (or well before that), Earth's oceans will evaporate.  Without a liquid ocean, the tidal bulge raised on the Earth by the Moon will be much smaller. Thus the exchange of angular momentum, and resulting rate of change of day length, is again slower.

There are various other effects you could speculate about (i.e. will the Earth even have a Moon, assuming it survives the Sun's red giant phase?), but this gives a sense of the long timescales you'd have to consider for this specific orbital evolution.


Do the cometary gas ions which make up the plasma tail of a comet reach the velocity of the Sun's solar wind?

Yes, the ions are said to be "swept out" by the solar wind, such that they reach the speed of the solar wind.  That speed varies a fair amount, however, because there is both a "fast" (about 600 km/s) and "slow" (about 300 km/s) solar wind.  It is really the magnetic fields entrained in the wind that sweep out the charged ions, since each ion has such a low mass it is easily affected by those fields.  Note the dust particles from a comet, which is the tail that we see, are much more massive and are not affected by the solar wind, so they are moving out much more slowly and often make a noticeably curved shape.


Finding angle between a given sky position and ecliptic pole

The angle between an ecliptic pole and another sky position is complementary to the ecliptic latitude of that position.
To find ecliptic (λ, β) coordinates, one method is to convert the equatorial (α, δ) to Cartesian form:
$$ \begin{align}
x_\mathrm{equ} &= \cos \alpha \cos \delta \\
y_\mathrm{equ} &= \sin \alpha \cos \delta \\
z_\mathrm{equ} &= \sin \delta
\end{align}$$
then rotate by ε = 23.44° on the x axis:
$$ \begin{align}
x_\mathrm{ecl} &= x_\mathrm{equ} \\
y_\mathrm{ecl} &= y_\mathrm{equ} \cos \varepsilon + z_\mathrm{equ} \sin \varepsilon \\
z_\mathrm{ecl} &= z_\mathrm{equ} \cos \varepsilon - y_\mathrm{equ} \sin \varepsilon
\end{align} $$
and convert to spherical form:
$$ \begin{align}
\lambda &= \mathtt{atan2}(y_\mathrm{ecl}, x_\mathrm{ecl}) \\
\beta &= \arcsin z_\mathrm{ecl}
\end{align} $$

If you are not interested in the ecliptic longitude λ, you can express the ecliptic latitude as
$$ \beta = \arcsin (\sin \delta \cos \varepsilon - \sin \alpha \cos \delta \sin \varepsilon) $$
Then the angle off the north ecliptic pole (α=18h0m, δ=+66.56°) is 90° - β.

Working your example (α=5h52m, δ=-63°),
I get (λ=75.63°, β=-86.34°),
and the angle off the south ecliptic pole (α=6h0m, δ=-66.56°) is β + 90° = 3.66°.


Problem with Tychonian model of geocentric universe?

Tycho was concerned that the Earth must be too "heavy" and "sluggish" to move.  So his system solves this problem. It turns out not to be a problem at all.  The problem it solves is "we know that the Earth can't move, so how can we describe the motions of the planets".  As the Earth can move, this isn't a problem. So the whole purpose of Tycho's system is gone.
It is a kinematic, not a dynamic model.  It describes the motion of the planets but does not give any reason for that motion.  This is also the case for Kepler's and Copernicus model.  But Newton's model, on the other hand, describes planetary motion in terms of forces.
Now you can do Newtonian mechanics in any coordinate system.  It is just more convenient to do it in an inertial system.  When you do Newtonian mechanics in a barycentre based inertial frame you derive Kepler's orbits.  But if you do Newton's mechanics in an Earth based non-inertial coordinate system, you get Tycho's model. (Or something very like it, but with elliptical orbits)
There are other problems:

It treats the Earth as "special". This treatment is unjustified. So it isn't isotropic and homgenous:  The Earth is still, everything else moves about it.
It requires the stars to orbit the Earth once a day, this would require them to move faster than light.
It does not predict things like Foucault pendulum, which demonstrates a rotating Earth.
It does not predict stellar parallax.

If you start from the axiom "The Earth cannot move" then you must get Tycho's system, or something like it.  But if that axiom is rejected then you can get a much simpler model with an inertial coordinate system in which the sun is almost stationary.
Some of these can be "fixed", for example by allowing for a rotating central Earth.  But remember the whole point of Tycho's model is to have a stationary Earth.  If you can have a rotating Earth, then why not a moving Earth too?


Why don't we build an interferometer between Earth and the Moon?

Put simply: controlling the phasing would be impossible (not just difficult).  Not only do you have to deal with a constantly changing physical distance but also with the dynamics of the atmosphere.  It's difficult enough to maintain transverse phasing across an earth-based receiver (think adaptive optics);  I can't imagine a reliable way to track the radial variation.  Without that, unless you are sticking with multi-meter wavelength radio systems, you cannot hope to phase up the receivers.


Bright light in the sky, in the '90s, what could it have been?

Gamma ray bursts are very rarely bright enough in visible light to be seen with the naked eye. This is why they weren't detected until there was a network of satellites looking for atom bombs. You didn't see a GRB.

If what you saw was Bright, and Fast it was likely to be a bright meteor, sometimes called a fireball. These appear as a very bright, fast-moving "shooting star". They are rare, but not so rare that it would be possible to identify a particular event from the information that you give.

A reflection from a satellite can look like a star that brightens and then fades. The Iridium satellites are well known for this though other satellites can also produce "flares". Iridium flares have been visible since the late 90s.

There are other mundane possibilities, but these would seem to be the most likely. Remember memory is a funny thing. After 20+ years thing that you "remember" might be different.


Is the age of the universe relative to an observer's location in that universe?

You are labouring under the misapprehension that how far we can see directly gives the age of the universe. Whilst it is true that the oldest light we can see was emitted some 13.7 billion years ago, the stuff that emitted that light is now roughly 46 billion light years away, thanks to expansion of the universe.

The universe itself probably extends significantly beyond this and may be infinite. That does not conflict with the big bang model or relativity. If the universe is infinite then it has always been infinite. Places that are separated by more than the oldest light ray as you put it, are not in causal contact and, depending on the cosmological parameters, may never be.

In a homogeneous, isotropic universe (and don't forget we can only make assumptions about regions beyond our observable universe), all observers would agree on the age of the universe and that the universe was once very much smaller and all would see (on average) the same sort of universe in all directions.

To ensure that places that are outside of causal contact now are homogeneous requires that they were in causal contact in the past. This is the nature of the "horizon problem"".

I took the picture below from the "slideplayer" website, I am unsure who the originator is, but it serves its purpose here and I think encapsulates your question. It appears in this diagram that A and B can never have "communicated" and can never have originated from the same place. The solution provided by inflation is to have a massive exponential expansion of space in the first fraction of a second. In short, the distances between points in space (or galaxies if you like, although there were no galaxies at the epoch of inflation) are suddenly increase by many, many orders of magnitude. This give the appearance of faster-than-light motion, although the cosmic speed limit only applies to local measurements and not to the expansion of space itself. The end result is a universe that appears homogeneous well beyond the limits of a radius of the age of the universe in light years.


Kuiper belt (and beyond) objects do sometimes come into the inner solar system: we call them "comets".  However comets have a large proportion of volatiles (water etc) and the non-volatile matter tends to be in the form of small dust particles, neither of which form useful meteorites  Ceres is also covered mostly in ice, so even if some was ejected into space and reach the Earth, it wouldn't survive the descent.
An older paper wonders if very old dead comets might be a source of meteorites, but notes that none have been found. A much more recent paper describes a small (100 µm) particle of cometary dust found embedded in another meteorite.  This dust particle would have come from the outer solar system.


If a captured moon has prograde, circular and equatorial orbit, can we know it is captured?

If a satellite has a prograde, circular(ish) orbit, it will certainly be much more difficult to determine if said satellite has been captured or not. However, there are other indications we could look for in determining if that body was captured or formed in place.
Composition
As a general rule, the composition of any given body in a stellar system is a function of its distance from the star. The reason for this has to do with the temperature of the primordial disk as a function of distance from the Sun. Early on, as all the objects in our solar system were forming, the began coalescing out of the stellar disk surrounding our Sun. However, in order for objects to solidify out of this disk, they had to be cool enough to do so. Very close to the Sun, it was of course extremely hot and molecules like water or methane could not precipitate into solids. Only materials like iron and silicate could do so. Much father out, where it is colder, things like iron and silicate can still solidify, but now its cold enough that you can pack on ice and methane. This process is evident by the fact that Mercury, our closest planet to the Sun, has a core of almost complete Iron which makes up 42% of the planet's volume (compared to the Earth's 17% by volume Iron core). A graph of this concept is shown below.

What this means for your question is that you can look at the composition of a given satellite and determine if that composition matches what we'd expect for an object forming at that location. If for example, you see a satellite around Saturn which has a lot of Iron and very little ice and methane, you can likely presume it formed closer in to the Sun and was captured by Saturn (how this might happen, I don't know but its just a simple example).
In fact, the Moon's nearly identical composition to the Earth is a strong indicator that our Moon formed in place (and from debris from Earth) rather than being captured.
Now of course measuring composition is no easy feat. You can determine something like that by analyzing any emissions the planet produces (such as with the water plumes from Europa) or else by flying a satellite very close to the body and measuring gravitational effects.
Cratering
This is a less reliable method, but can be telling. The number and distribution of craters on a body tells us a lot about its history. If we assume that the surface of a given satellite is not constantly changing (through some geological process) then the craters on the surface represent a record of past interactions with that satellite. One approach I can imagine is to statistically predict the number of craters that should exist on the body given that it formed at its current position, then compare to what we see. If the surface hasn't been changed for billions of years, then a difference in the expected cratering indicates the body hasn't formed in the way we thought it might and possibly formed elsewhere or under different conditions. I admit though that this method is much less telling and more of a guess than using composition.


Does more ocean on an Earth analog produce a different rotation period?

Would an exoplanet that has more ocean than Earth rotate at a
  different speed as a result of this?

Basically no.   Rotation speed, or angular velocity doesn't measurably change with your proposed example.

To explain this in more detail, there's two important concepts.  First, is Angular Momentum, (Short explanation or longer) and the 2nd concept is Moment of Inertia.

The basic formula is that Angular velocity (time for 1 full rotation) = Angular Momentum divided by Moment of Inertia.  Formulas and explanations are in the link(s) above, and if I was to explain it, it would get wordy, but your question seemed to be more general, less about doing the mathematical calculations, so I'll skip the formulas.

In nutshell, talking about a planet, the state of matter doesn't affect the angular momentum.   Angular momentum is conserved and that, divided by moment of inertia determines rotation speed.  Now if you get something spinning fast enough the angular rotation can overwhelm the gravity and when this happens, the planet can begin to fly apart, which happens more easily with water than a rocky surface which has some cohesion, but ignoring crazy super fast rotations, a water world, a deep ocean world, a shallow ocean world and a rocky world all obey the the angular velocity law and composition doesn't matter.  The angular momentum is conserved.  The moment of inertia of a planet can change but for the most part, doesn't change much.

The Earth's moment of Inertia, for example, changes as glaciers grow or shrink, or when there's an earthquake.   Even, every time we a tall building is built the Earth's moment of inertia increases a teeny tiny bit, similar to a skater extending their arms to slow down coming out of a spiral.

When there's an earthquake which, for the most part, settles the Earth, there's a small increases in the Earth's rotational speed.   The total angular momentum and total mass remaining the same but shifting of material changes the moment of inertia.   (Granted space dust and tidal effects change the Earth's moment of Inertia, but quite slowly).

  Would the amount of water impact the weight, gravitational pull,
  and/or tidal forces and cause a difference in the exoplanet's rotation
  period?

This is harder question to answer precisely because adding water changes the mass of the planet and changing the mass changes the moment of inertia, but sticking to the principal of your question, there's no measurable effect.

Lets take a somewhat simpler example without changing the planet's mass.  Ice ages.  When the Earth is in an ice age there's less liquid oceans and more ice at the poles but the total mass is unchanged.   More mass at the poles and less mass in the oceans decreases the Earth's moment of inertia because the bulk of the Earth's moment of inertia is around the equator,  so, as a result, the Earth rotates slightly faster during an ice age and slightly slower after an ice age.   Over time, the Earth's crust has a tendency adjust for this effect but that takes tens of thousands of years.  Parts of the Earth's crust is still rebounding from the last ice age.

Gravitational pull isn't relevant.  Neutron Stars with enormous gravitational pull can rotate very fast and the planet with the fastest rotation in our solar-system is Jupiter and the one with the slowest rotation is Mercury.   Angular velocity has no direct correlation to mass or gravity though there is an indirect correlation.   As a star, for example condenses it's rotation speeds up, because the angular momentum is conserved but as it settles the moment of inertia decreases.   That's why young stars, White Dwarfs and Neutron stars can spin very fast.

Tidal forces can create drag on rotation but the effect is slow, taking millions or hundreds of millions of years.  With enough time, tidal forces cause a planet or moon to stop rotating and become tidally locked but there's no short term affect.  (I'll say a bit more on this later).

So, a planet with large oceans wouldn't rotate any slower than a planet with no oceans because liquid or solid can have equal angular velocity, but over time, tides will slow a planet with oceans more quickly than a planet without them.

Because the Earth has oceans, the Moon's gravitation on the Earth's tidal bulge does slow down the Earth's rotation, but this has been happening for 4 billion years and the Earth still rotates every 24 hours - one of the faster planets.   If the Earth had more water the Moon's tidal tug would slow the earth down a bit faster, but it would still be very gradual.

Jupiter, which is basically a ball of gas, is the fastest rotating planet and Mercury, basically a rock, the slowest, so those are 2 examples of composition not being a factor, though Mercury's slow rotation is in large part due to the strong tidal forces it receives from the Sun.

Now, I Understand the logical approach to your question, as there's something apparent about water resisting rotation - touched on in this question, but the fact that water doesn't spin with a glass when you spin a glass is an example of conservation of angular momentum, not an argument against it.  On a planet, the oceans are rotating with the planet and the angular momentum is already there.

Hope that wasn't too long, but that's the gist of it.   I can try to clean up or clarify if needed.


Could neutrinos be at least a part of Dark Matter?

Ethan Siegel has a good discussion of this question here.  Read that for more detail, but briefly, the main issue is that neutrinos move too fast - they are a form of what cosmologists call "hot dark matter".  The high speeds mean that they wouldn't settle into dense areas (e.g. galaxies, and clusters of galaxies) as dark matter is observed to do.   Evidence from structures we see in the universe tell us that most dark matter must be cold.
There certainly can be some hot dark matter in the form of neutrinos, but if there were a lot of it, it would leave an imprint on the cosmic microwave background that is not observed, constraining this component to be about 1% of the mass in the universe.


Are meteors and meteorites considered "Small Solar System Bodies"?

Your question is really well answered by this article at the Encyclopedia Britannica website, written by Prof. Ed Tedesco of the Planetary Science Institute.

Two quotes that resonate well are...

The small bodies populate the solar system in vast numbers and include ... the fragments of these bodies—commonly called meteoroids—over a continuum of sizes down to microscopic grains known as interplanetary dust particles or micrometeoroids.

And...

The term small body is sometimes restricted to objects that can be observed telescopically while still outside Earth’s atmosphere. In practice, this results in a lower limit of a few metres on the diameter of a small body.

So in general you can assume that anything other than the Sun, planets, minor planets, and their moons is a small solar system body. But in practice it makes sense to limit the minimum size for a SSSB, and the "observable telescopically" caveat provides that limit so that we don't have to include single molecules or one grain of sand in the list.


What would Jupiter look like from a Galilean moon?

You can simply startup Stellarium and have a look yourself. Choose any locations you are interested in. The Galilean satellites of Jupiter are inside the default list of locations.

The attached image shows Jupiter as viewed from Io at the given time, Europa is the bright object to the right; with -9 mag it is considerably brighter than Venus when viewed from Earth, but slightly less so than Earth's Moon as seen by us.

Interesting to note is also the huge extend of Jupiter in the field of view, if you choose the right hemisphere (Io is in a tidally-locked state). With roughly 15° you need about your entire hand at arm's length to block it while the sun is a lot smaller (~30x larger than Sun or Moon when viewed from Earth). Also the daily eclipses of the Sun by Jupiter might be noteworthy.


Calculate time when star is above altitude 30°

Yes, this is the correct approach.  The $h$ in the equation is the altitude above the horizon of the object at which you consider it to be rising or setting.  This is typically non-zero, because of atmospheric refraction, and, in the case of the Sun or the Moon, because of their finite diameters.  In your case, the object 'rises' when it climbes above $h=30^\circ$ and 'sets' when it drops below that altitude.

If $\left|\cos(LHA)\right|>1$, there is no solution, because your object never crosses the $h=30^\circ$ line.  $LHA$ is the local hour angle, and you can find the local sidereal time $\theta$ using

$LHA = \theta - \alpha,$

where $\alpha$ is the right ascension of your object.


How do disk-shaped galaxies then form at the sites of spherical galactic halos? Do the halo stars start orbiting or just sit in place?

The key to understanding this is that dark matter and, effectively, stars are collisionless, whereas gas is collisional.

Structure formation

As written in the second quote, structure forms in a hierarchical, bottom-up fashion. That is, small halos of gas and dark matter (DM) collapse first on small scales, and these halos then merge to form larger halos. This is in contrast to what was previously thought, namely that galaxies formed in a so-called monolithic collapse from a huge, primordial cloud (Eggen, Lynden-Bell & Sandage 1962).

On the very smallest scales, thermal motion will tend to wash out structure and prevent collapse. The first structures to collapse are thought to have masses of some $10^5\,M_\odot$, but the exact value depends on your assumed initial conditions. During the collapse, these clouds then fragment to form stars.

Collisionless matter

Because DM is collisionless, it is difficult for it to "relax", i.e. settle down into a dense, "virialized" structure: A DM particle falling into the combined potential of gas and DM will tend to go right through through the center. However, because the potential isn't static, but contracts, the particle doesn't go as far to the other side, and eventually a DM halo will relax (thanks to Peter Erwin for making me aware that I underestimated this effect).

The same is effectively true for stars: Of course, stars can smash together, but the chance of two stars getting close enough for this to happen is minuscule because they're so far apart.

Collisional matter

The story is different for gas: The collapse is dominated by DM (because there's much more), but when gas gets dense enough, hydroforces become significant. The atoms will collide, electrons are excited, de-excite, and emit photons, removing energy from the system. In other words, gas can cool.

Formation of the disk

This causes the gas to basically smash together in the bottom of the potential well, and eventually settle down into the center of the halo. Although there are ways to shed itself of some of the initial angular momentum, it will usually maintain enough to form a rotationally supported disk, with centrifugal forces impeding the collapse in the plane of the rotation.

The halo

The first population of stars are formed before the disk has settled, and so have random motions around in the potential. The massive stars quickly die, leaving the longer-living, dispersion-dominated (as opposed to rotation-dominated) population of relatively low-mass, metal-poor stars, known as Population II stars.

That is, the stars in the halo do not just "sit there", but orbit the galactic center like the disk stars. However, the orbits are quite unordered.


Why is Mars considered the outer edge of the "goldilocks zone"?

The circumstellar habitalbe zone can be defined as the distance range around a star, where the mean temperature of a rotating planet would be between 0 and 100 centigrades, if radiation (heat) received from the star and thermal radiation emitted by the planet form an equilibrium.

But that's only a rule of thumb. It has been redefined, and is still disputed.
Hence your criticism is justified to some degree.
There may exist exceptions of habitable planets outside the presumed habitable zone; and there exist certainy many planets within the habitable zone which aren't habitable.
But the probability for a planet similar to Earth to be habitable on its surface is thought to be highest in the habitable zone. When looking for habitable exoplanets, it's therefore more efficient to focus on the habitable zone.


What would a kHz gravitational wave detector look like? (mountains on millisecond magnetars) How would it differ from LIGO/Virgo?

In principle, not much different to LIGO, since LIGO is sensitive to kHz gravitational waves. But you need to find ways to increase the sensitivity at kHz frequencies, without compromising response at lower frequencies, or perhaps having a tuneable response. These improvements are likely to come with increased laser power and developments in the optics and mirror coatings.
Detecting gravitational waves (GWs) is limited at high frequencies by the length (or effective length, with some cunning deployment of mirrors) and at low frequencies by various sources of noise, including "radiation pressure noise" which is an inescapable consequence of using powerful lasers to do the detection.
Roughly speaking, to get maximum sensitivity at 1kHz (LIGO is most sensitive at 100Hz), you make your interferometer arm length equal to a quarter of the GW wavelength (or the total path length of the light to be $\lambda_{\rm GW}/2$). This ensures that perturbations to the arm length all "sum up" whilst the laser light is in transit. Thus for a GW frequency of 1KHz that means $L \simeq 75$ km (or a total path of 150 km).
That's fine - something of LIGO's size can be made effectively that long by bouncing the light up and down the arms about 10 times, which is much less than the $\sim 270$ times that LIGO uses in order to get sensitivity at lower frequencies.
Sources of noise at 1kHz are mostly limited to simple "shot noise" due to the finite number of photons in the laser beam. So you would need to boost the power at the beam splitter by using a more powerful laser to begin with or by employing clever recycling techniques to amplify that power within the interferometer arms.
In LIGO, they use a laser with a power of about 40W, split between two arms. The 4km arms form Fabry Perot resonators that boost this by a factor of 270, but that isn't the power at the beam splitter. From the beam splitter's point of view, the FP arms are about 0.97 times reflective. To amplify the power at the beam splitter you put a recycling mirror prior to the beam splitter, with a reflectivity of 0.97. This allows all power to transmit into the FP arm cavity, but also amplifies the power between the beam splitter and the FP input mirror by a factor of about 35.
At present LIGO is tuned to have a high amplification factor in the FP arms and a modest power recycling gain, which is limited by the effective reflectivity of the FP cavity due to losses at the reflective surfaces.
To get more power at the beam splitter, whilst maintaining sensitivity at 100 Hz, would need big improvements in the mirror and mirror coatings technology; improvements in laser stability at high power and the clever used of "squeezed vacuum" techniques which can to some extent shape the noise frequency profile, allowing modest improvements in the shot noise at the expense of radiation pressure noise at lower frequencies. For instance, by minimising losses in the mirrors, one could make the FP arms more reflective, which allows a bigger recycling gain (which goes as $(1-R)^{-1}$).
Even without these improvements, you could tune LIGO to work better at 1kHz, at the expense of its response at lower frequencies. The gain of 270 is not realised at high frequencies. You could drop this to around 20, which would increase the reflectivity of the FP arms  to about 0.997 by reducing the losses (which increase with how many times the light bounces around) and could then use a more reflective power recycling mirror and get a power recycling boost of 330. This would gain you a factor of $\sqrt{10}$ in SNR at 1kHz at the expense of losing a factor of 10 sensitivity at 100Hz.
There is lots more detail in the technical documentation for the Einstein Telescope. This proposed third generation GW detector operates on the same principle as LIGO, but with improvements. It will have several "V" shaped interferometers to improve its directional response and remove dead spots; it will use more powerful lasers; it will operate cryogenically to improve sensitivity at low frequencies; it will have 2 interferometers within each set of arms, tuned for low and high frequencies respectively, as I've indicated above; it will have 10 km arms and use more massive mirrors to reduce the effects of radiation pressure and other displacement noise at low frequencies; and it is proposing big developments in optics technology (e.g. mirror losses at 1 ppm) to boost power and get high frequency response.


How do sunspots move seen from the side?

The sunspots are actually depressions and low thermal pressure zones.

Incredible magnetic fields are piercing the surface of the sun and the magnetic force pacifies the movement and the convection at the zone of the sunspot. that's why the surface becomes darker clouds near those magnetic fields.

Elsewhere, the sun's surface has convection wells called solar granules a bit like a lava lamp, which are about 1000km diameter. You can check images of "solar convection" to study maps of convection.

At a sunspot the movement of colder plasma flows downwards, and there is a hot spot trapped under the sunpot where the thermal energy escapes radially away from the sunspot.

This is a video attempting to illustrate the view under the sun:

https://www.youtube.com/watch?v=zU9O8SmCHFQ

This is a virtual simulation of them. The versions are wildly different:

https://www.youtube.com/watch?v=W5dldC5dPMk

https://www.youtube.com/watch?v=UN_9HMVXjY0

https://www.youtube.com/watch?v=ARIJx0geaOE

here is a speeded up video of some sunspots:

https://youtu.be/BejRxZAa66E

https://youtu.be/l1VpMRMg4P4

https://youtu.be/T5MnZUINzjY


An orbit as you describe is not stable. Even if the two moons are intially exactly the same distance from the Earth, small perturbations from the sun and other planets will cause one of them to speed up and the other to slow down. Eventually either one of them will hit the other, or one of them will hit the Earth. Whichever happens, it's not good.


Can Doppler shift make colors of two stars at same distance different?

As you correctly mention, there are rare stars which move particularly fast. Check also hypervelocity stars section here, where a stellar population moving at around $10^3 \textrm{km/s}$ is described (by the way, these velocities were most likely produced in the galactic center). Note also, that these stars are too fast to stay in the disk for long times. As another sidenote, galactic center in not accessible in visual due to dust extinction.

Redshifts are calculated particularly easy:
$$z=\dfrac{v}{c}$$,
where $v$ is the velocity along the line of sight, $c$ is the speed of light, $z=\dfrac{\delta\lambda}{\lambda}$ and $(v/c)^2\approx 0$ (which holds even for hypervelocity stars).

Because hypervelocity stars are rare, they most likely can be found only close to regular stars. Yet, being very optimistic, take $v=3\cdot 10^3 \textrm{km/s}$, and you obtain that $\dfrac{\delta\lambda}{\lambda}=10^{-2}$ under very favourable conditions. In general it is not small, but when it comes to colors, the change is negligible.

Say, green light at $\lambda=555 \textrm{nm}$ would now have $\lambda=550 \textrm{nm}$ or $560 \textrm{nm}$ in wavelegth, i.e. would still be very green. So Doppler effect has no bearing on your problem.

What has possible bearing, however, is dust reddening. One of the two stars might be obscured by the dust, unlike the other one, and look reddish. So, both b) and c) are possible answers.

P.S. And actually, if a star "appears reddish", it is most likely dust extinction. Black body temperature colors are either blue, bluish, white, yelowish, yellow, orange or red.


Which telescope would be better with a DSLR?

Both of these look OK to me. The biggest plus in the Omegon looks like the extra 10mm. If you're on a budget, taking a cheaper refractor and later taking a decent budget tracker (e.g. SkyTracker Pro) is a good move. Gradually building up your set of gear is key. Good luck!

ps.
Don't think you're going to take "budget photos" because you're on budget gear. Good amateur astropics don't necessarily come from people with the best gear.


The current explanation is that "space" is continuously created in the intergalactic void, "pushing" the galaxies apart. From what i understood ( i am also ignorant ), the current rate at which "space" is created is expressed by the Hubble's constant 67.15 ± 1.2 km/s/Mpc ): for every million parsecs of distance from the observer, the rate of expansion increases by about 67 kilometers per second, which means that the distance between 2 points that are 1 megaparsec apart grows with 67km every second; between 2 points that are 2 megaparsecs apart with 134km/s, and so on so the relative speed between 2 points that are 5000 Mpc apart would be 335000 km/s which is more than the speed of light. In fact, two objects at this distance would not move, but 335000km worth of "extra space" would be "created" between them every second.

However in the last week there was an article depicting the shape of the universe, and it is hard for me to understand how it got to look like this : http://ifreepress.com/wp-content/uploads/2014/09/laniakea.png

http://en.wikipedia.org/wiki/Laniakea_Supercluster


Spiral galaxies may be the most commonly depicted probably because we find ourselves in one, however, galaxies come in many sizes and shapes. Wikipedia's page on Galaxy morphological classification explains the most common classifications.

Answering the question on the title:

  The basic answer to your question is Conservation of Angular Momentum.
  Astronomers have been interested in this question for a long time and
  are still writing papers on the subject.   Even if you assume that at
  the time of the big bang there was no net angular momentum in the
  universe, one would expect that there were local fluctuations.  When
  galaxies initially formed by gravitational attraction these net local
  amounts remained.  Now angular momentum is conserved so when the
  initial collection of stars and gas in one galactic volume start to
  collapse the net angular momentum has to stay the same.  Since angular
  momentum $\vec{L} =m\vec{v}\times\vec{r}$,  for a given $v$ its easier to conserve $L$ if the $r$ is
  large.   In the orthogonal direction (along the axis of the accidental
  angular momentum) there is no such hindrance to collapse and so the
  collection turns out to be disk shaped.   By the way there has been
  several studies on the net angular momentum of a large collection of
  galaxies coming from the Sloan Digital Sky Survey and the answer is
  pretty close to but not quite zero.  Stay tuned.

Extracted from an University of Phoenix Department of Physics article


Roche Limit for Earth-like body around Earth-like body

According to wikipedia, the roche limit for an object with the same density as the Moon would be around 10,000 kilometres. Since the density of the Earth is about 166% of that of the Moon, if you don't take the atmosphere, or oceans into account, then the binary planets could be much closer to each other. However, when you take into account the oceans, and atmosphere, it would probably be a lot further away.


Why doesn't the sun pull the moon away from earth?

Short answer: Because the Moon is much closer to the Earth than it is to the Sun. This means the gravitational acceleration of the Earth toward the Sun is almost the same as is the gravitational acceleration of the Moon toward the Sun.

The Moon's acceleration toward the Sun, $-GM_\odot\frac{\boldsymbol R+\boldsymbol r}{||\boldsymbol R+\boldsymbol r||^3}$ is indeed about twice that of the Moon toward the Earth, $-GM_\oplus\frac{\boldsymbol r}{||\boldsymbol r||^3}$. This is irrelevant. What is relevant is the Moon's earthward acceleration due to gravitation compared to the difference between the Moon's and Earth's sunward gravitational acceleration,
$$\boldsymbol a_{\odot,\text{rel}} = -GM_{\odot}\left(\frac{\boldsymbol R + \boldsymbol r}{||\boldsymbol R + \boldsymbol r||^3} - \frac{\boldsymbol R}{||\boldsymbol R||^3}\right)$$
This relative acceleration toward the Sun is a small perturbation (less than 1/87th in magnitude) on the Moon's gravitational acceleration toward the Earth. Given the current circumstances, the Sun can't pull the Moon away from the Earth.

Longer answer:

The gravitational force exerted by the Sun on the Moon is more twice that exerted by the Earth on the Moon. So why do we say the Moon orbits the Earth? This has two answers. One is that "orbit" is not a mutually exclusive term. Just because Moon orbits the Earth (and it does) does not mean that it doesn't also orbit the Sun (or the Milky Way, for that matter). It does.

The other answer is that gravitational force as-is is not a good metric. The gravitational force from the Sun and Earth are equal at a distance of about 260000 km from the Earth. The short-term and long-term behaviors of an object orbiting the Earth at 270000 km are essentially the same as those of an object orbiting the Earth at 250000 km. That 260000 km where the gravitational forces from the Sun and Earth are equal in magnitude is effectively meaningless.

A better metric is the distance at which an orbit remain stable for a long, long, long time. In the two body problem, orbits at any distance are stable so long as the total mechanical energy is negative. This is no longer the case in the multi-body problem. The Hill sphere is a somewhat reasonable metric in the three body problem.

The Hill sphere is an approximation of a much more complex shape, and this complex shape doesn't capture long-term dynamics. An object that is orbiting circularly at (for example) 2/3 of the Hill sphere radius won't remain in a circular orbit for long. Its orbit will instead become rather convoluted, sometimes dipping as close to 1/3 of the Hill sphere radius from the planet, other times moving slightly outside the Hill sphere. The object escapes the gravitational clutches of the planet if one of those excursions beyond the Hill sphere occurs near the L1 or L2 Lagrange point.

In the N-body problem (for example, the Sun plus the Earth plus Venus, Jupiter, and all of the other planets), the Hill sphere remains a reasonably good metric, but it needs to be scaled down a bit. For an object in a prograde orbit such as the Moon, the object's orbit remains stable for a very long period of time so long as the orbital radius is less than 1/2 (and maybe 1/3) of the Hill sphere radius.

The Moon's orbit about the Earth is currently about 1/4 of the Earth's Hill sphere radius. That's well within even the most conservative bound. The Moon has been orbiting the Earth for 4.5 billion years, and will continue to do so for a few more billions of years into the future.


How to plot curvilinear coordinates in the frame of a Cartesian telescope field of view

Most CCD or photographic images are equivalent to a gnomonic projection.
I suppose this is true of the Kepler CCDs; while the Schmidt camera focus surface is curved, the CCDs have individual field flatteners.
If the projection is not exactly gnomonic, the difference is not very great in a small field.

As you found, a single gnomonic projection doesn't work so well for areas much wider than 90°.
TheSky user guide
says it uses orthographic projection for fields smaller than 60°.
The slightly curved RA lines and slightly oblique intersections near the corners of the chart in question are consistent with that.
Other projection choices are also valid, e.g. stereographic projection is common in maps showing a hemisphere at once.

In any case you'd need to transform the individual image tiles to the same projection as the composite overview.
This will change the tiles' shapes or sizes, but the relation between map coordinates and sky coordinates will be consistent.


Will NEOWISE be known as the Great Comet of 2020?

I'm inclined to say a hard no using the casual definition of a "great comet"

A great comet is a comet that becomes exceptionally bright. There is
no official definition; often the term is attached to comets such as
Halley's Comet, which during certain appearances are bright enough to
be noticed by casual observers who are not looking for them, and
become well known outside the astronomical community.

So, there's a dual definition of visible brightness and/or recognition outside the astronomical community.   With that in mind, I think it's pretty clear that Comet NEOWISE doesn't make the cut due to it not being easily visible   Granted, it peaks in 2 days from now (7-23-20) but so far it's hard to see by the naked eye and that's not likely to change significantly as it peaks.

Although it’s brighter than all but around 20 stars, its extended,
diffuse nature makes it a challenge for human eyes.

NEOWISE is the brightest comet since comet McNaught, but McNaught, with a peak mangitude of -5.5, was about 500 times brighter. (based on the estimate that Neowise is about as bright as the 20th brightest star, roughly 1.25 magnitude).  McNaught was actually brighter than peak Venus, though it probably didn't look that bright to the naked eye because it was more spread out.
Comet Lovejoy was a bit unusual.  It's "greatness" had nothing to do with it's visibility, which was unspectacular.   Lovejoy passed too close to the Sun and it lost too much material in passing, so it was much anticipated and I remember articles saying it was potentially a super-comet, but it lost everything on the other side of the sun and we never got a good view from Earth.   Perhaps if we'd been on Mercury or Earth's L3 trojan point, we'd have seen a rare and magnificent show.
Wikipedia doesn't list Comet Lovejoy in it's list of great comets.


Do all total solar eclipses have the ring of fire when observed from ground?

No, some solar eclipses are annular and some are total.
The difference is in the Moon's distance at that moment in its eccentric orbit around the Earth.
Hermit Eclipse has some illustrations which may be helpful.

In an annular eclipse, the Moon is farther than average and appears slightly smaller than the Sun.
Some of the solar photosphere remains exposed, briefly appearing as a ring for observers on the central path.
The phrase "ring of fire" applies to annular eclipses.
The eclipses of 2019-12-26 and 2020-06-21 are annular.

In a total eclipse, the Moon is nearer than average and appears slightly larger than the Sun.
Along the central path, the photosphere is completely obscured, briefly allowing observation of the solar corona.
There may be a "diamond ring" effect before and after totality as a bit of photosphere shines past low regions on the lunar limb.
The eclipses of 2019-07-02 and 2020-12-14 are total.

In the present few centuries, annular eclipses are slightly more frequent than total eclipses.


What does it mean for a black hole to "evaporate"?

It's due to Hawking radiation, a black body radiation due to quantum effects near the event horizon of black holes and named after Stephen Hawking that predicted this phenomena. From the same Wikipedia article:

  Hawking radiation reduces the mass and the energy of the black hole
  and is therefore also known as black hole evaporation. Because of
  this, black holes that lose more mass than they gain through other
  means are expected to shrink and ultimately vanish. Micro black holes
  (MBHs) are predicted to be larger net emitters of radiation than
  larger black holes and should shrink and dissipate faster.

I don't think it was ever observed to actually lower mass of black holes that we observe as astronomy phenomena, the process would simply take too long for any measurable effects to manifest, but scientists did try to observe it experimentally. In September 2010, Belgiorno et al. claimed they've:

  ... produced Hawking radiation by firing an intense laser pulse through
  a so-called nonlinear material, that is one in which the light itself
  changes the refractive index of the medium.

But this is yet to be reproduced and confirmed by experiments of others.


Nomenclature of Interstellar Objects and Stars

Names are common enough for solar system bodies, but generally stars don't have names, they have identifiers from various catalogues.
In the case of Pulsars. they are named by their location in the sky.  Just like Earth has longitude and latitude, every point in the sky has a "right ascension" and "declination".  Due to the slow movement of the Earth's axis, to identify the location in the sky you also need to mention a date, though you can look up the positions of stars in 1950 (B1950) or 2000 (J2000). Also, for historic reasons, the angle of right ascension (or "longitude") is measured in hours and minutes (24hr = 360 degrees)
Now the Black Widow pulsar is located at 19hr 57 minutes of right ascension and +20 degrees of declination (in the B1950 coordinates).  So its designation is PSR B1957+20.


The three inner Galilean moons are in orbital resonances. For each orbit of Ganymede, Europa makes two orbits and Io makes four orbits.  So after 1 orbit of Ganymede, the two inner moons are in the same configuration. That is 7.154 (Earth) days.

Callisto is not in an orbital resonance, It won't be in the same position, but its orbit (of 16.689 days) is close to a ratio of 7:3 to Ganymede. So after 7 orbits of Ganymede (50.08 days) Callisto will have made just over 3 orbits (50.07 days)

If you want all known moons to be in a similar configuration you will have to wait a while, probably longer than the life of the sun (there are a lot of moons)


Are there large underground caverns on Mars?

As evidence of caverns detected on Mars, consider the following recent image taken from HiRISE instrument on the Martian Reconnaissance Orbiter :

Image source: NASA

Scientists believe that

  The hole appears to be an opening to an underground cavern, partly illuminated on the image right. Analysis of this and follow-up images revealed the opening to be about 35 meters across, while the interior shadow angle indicates that the underlying cavern is roughly 20 meters deep.

Which accompany larger openings, reported in the National Geographic article Mars Has Cave Networks, New Photos Suggest (Norris, 2007) of several larger pits detected by Odyssey, and they offer a possible origin:

  The surface of Mars is strewn with craters from meteor impacts and depressions formed by the collapse of underground chambers formed by flowing lava, the experts said.

There are many examples of lave tubes (and indeed, in other places in the Solar System), caverns etc reported in the article Mars Cave-Exploration Mission Entices Scientists (Wall, 2012), who also state, in regards to detection and exploration of these caverns that robotic means are being explored.

However, according to this NASA article, a method to find and characterise caves on Mars (or indeed Titan, and maybe other places, including the Moon) is to detect them via thermal differences caves cause in comparison to the surrounding surface.  This method is used on Earth, and

  Techniques developed
  through this research will ultimately be applied to
  locating subterranean cavities on the Martian surface.
  We anticipate one of the best techniques for systematically
  finding caves on Mars will be via remotely
  sensed thermal imagery.


What is the difference between Sphere of Influence and Hill sphere?

The two terms are used in answering different questions.

Hill Sphere: given a large mass (eg Sun) and a small mass (eg Earth), can a tiny mass (eg Moon) find a stable orbit around the small mass? (If the tiny mass goes outside the Hill Sphere of the small mass, no.)

SOI: given two large mass objects and a small object between them, (eg sending a probe from Earth to Mars), which massive object should we use as the origin of the frame of reference? (The small object is within which massive object's SOI?)

As the questions are different, different approximations are used to answer the questions.


Beneficial astronomical uses of satellite constellations like Starlink?

Partial answer for now.

tl;dr: It will be an "exercise in futility".

This is a great question! With tens of thousands of very similar satellites in LEO the number of occultations that will occur from a given observatory will be much larger and better characterizable than with the current diverse population, so "if it were useful it would have been tried already" is not a valid dismissal.

However, in LEO they are moving quite fast, thousands of arcseconds per second of time, so any measurement may be limited by shot noise for small apertures, and since the satellites themselves are of the order of a few meters, for large aperture telescopes the occultation is partial and complex; while the satellite may block light for some part of the telescopes mirror, other parts will still be receiving light.

The speed of a satellite in LEO is about 7800 m/s, so if you have an 8 meter telescope the entire event will last about 1 millisecond. There may be some really interesting way to do some kind of very complicated reconstruction for a very bright object when a spacecraft occultation is observed by a large aperture telescope, since what you have is a complex aperture that moves through a variety of states over that one millisecond.

During that time the Earth's atmosphere will be in a fairly well "frozen" state, so one solution for astronomical seeing will probably work for the entire event

But my hunch is that there will be so many small uncertainties and ambiguities that this will end up as an exercise in futility, and events will be so infrequent and non-repeating that there will be no way to check the method.

Here are some plots of apparent angular velocity of satellites in LEO from 300 to 1000 km altitude. The new, large internet-delivering constellations will be closer to 300 than 1000 km, taken from and explained in depth in this answer.

One degree per second is of course 3600 arcseconds per second of time.


There is no net internal magnetic field to perturb, so how could one induce currents?

The effect arises on Earth because some transient event (e.g., coronal mass ejection or CME) hits Earth's magnetosphere and causes a sudden $\tfrac{dB}{dt}$ which, from Faraday's law, results in induced electric fields.  These induced electric fields then drive currents in things like power grids (typically long-distance wires running east-west at high latitudes), oil pipelines, etc.

The only changing magnetic field would be that of the solar wind and the small, localized magnetic anomalies on the surface of Mars.  However, these fields are incredibly small so the $\tfrac{dB}{dt}$ would be even smaller.


What exactly is the cosmic definition of ‘now’ is?

The cosmic "now" is well-defined: It is the time for an observer that has always been at rest in the Universe's comoving coordinates, i.e. the coordinates that expand along with the Universe. Although this reference frame is no more special than any other frame, it makes sense to use this. For instance, it is the only frame in which the cosmic microwave background is (statistically) isotropic, and it is the frame in which you'll see the same number density of galaxies in all directions.

You could (rightfully) define "now" in another frame, but it wouldn't be of much use. For instance, for a neutrino released 1 s after Big Bang, the age of the Universe is some ten thousand years, but it doesn't really tell you anything about the Universe.


Why are event horizons described in Schwarzschild radii rather than diameters, and either way can they change equator to pole?

We use the Schwarzschild radius $r_s$ (rather than a diameter) because it's convenient. We're want to describe what happens in the vicinity of a black hole, so it's natural to talk about the distance from the black hole.
For example, there's the photon sphere at $3r_s/2$ (for a Schwarzschild BH), and the innermost stable circular orbit or ISCO, the smallest circular orbit in which a test particle can stably orbit a massive object, at $3r_s$

The ISCO plays an important role in black hole accretion disks since it marks the inner edge of the disk. [...] Circular orbits are still possible between the ISCO and the photon sphere, but they are unstable.
For a massless test particle like a photon, the only possible circular orbit is exactly at the photon sphere, and is unstable. Inside the photon sphere, no circular orbits exist. [...] The case for rotating black holes is somewhat more complicated.

Another phenomenon which involves the distance from a BH is gravitational lensing. The amount of deflection of a light beam passing near a BH (or indeed any source of gravity) depends on the distance $r$ of the beam's path from the centre of mass. The deflection angle is given by $\theta=2r_s/r$. I guess you could use diameter in that formula, but it's more natural to think in terms of radius. It's a bit like shooting at a target (with light beams). When you shoot at a circular target, you're concerned with how far your shots are from the centre of the target.

As I said in the comments, our usual notions of geometry are not very useful in the highly curved spacetime near a BH. And if you want to get close to a BH, you need to be in an orbit travelling near the speed of light, and / or your spacecraft needs to be capable of the extremely high acceleration required to hover near a BH (and you & your ship need impossible strength to cope with the insanely huge g force resulting from such acceleration). Travelling at high speed adds additional relativistic effects on top of those GR effects due to the BH's gravity.
To describe the locations and times near the black hole, you need to choose a coordinate system. Schwarzschild coordinates are often used to describe a non-spinning black hole, but they have a coordinate singularity at the EH (event horizon), so they're annoying to work with when you want to talk about objects crossing the EH. It's a lot like how the latitude & longitude on Earth break down near the poles. You can't travel further south than the South Pole at 90°S, and the pole itself doesn't have a well-defined longitude. If I tell you that I'm 1 km north of the South Pole, I could be anywhere on a circle of 1 km radius.
Fortunately (as I mentioned in the comments), GR is very flexible regarding coordinates, and there are quite a few standard coordinate systems used in various circumstances, you can see a list of some of them on Wikipedia's page Category:Coordinate charts in general relativity. However, that flexibility makes things complicated, and difficult to describe properly without using advanced mathematics. So popular treatments of GR gloss over those difficulties (or avoid them completely), and that has led to a popular understanding of GR elements that is somewhat distorted.


If I look beyond (about same region, higher redshift) a galaxy, will I see its progenitors?

I think I understand what you mean and the answer is: specifically no; but generally yes.

If we look out into space, we can divide up what we can see into redshift slices of arbitrary thickness. The light we see from galaxies in each slice has been travelling for a certain amount of time, so we are seeing that "piece of space" at a particular cosmic epoch corresponding to its redshift.

If we look past that galaxy to a higher redshift slice, then what we are seeing there is a different patch of space at a younger cosmic epoch. i.e You cannot see the same galaxy as it was at a different time. We see each and every galaxy only once.

But in general you will be seeing younger galaxies at higher redshift. In an LCDM model then smaller galaxies build up into larger ones, so there will be a higher density of galaxies at higher redshifts (even after accounting for cosmic expansion), but the galaxies they will produce by merging etc. are not yet visible to us. It is possible that we will see them in the future (depending on the exact values of the cosmological parameters and what redshift we are talking about) as that distant patch of space recedes from us and becomes older (in cosmic terms).


What are the main differences between solar flares and coronal mass ejections?

The following definitions are according to NASA.

"Solar Flares" are gigantic explosions from the surface of the Sun, occurring near sunspots along the dividing/neutral line between opposite magnetic field areas.

"Coronal Mass Ejections" (CME's) are:

  huge bubbles of gas threaded with magnetic field lines that are ejected from the Sun over the course of several hours.

Although the Earth's magnetic field protects the biosphere from the worst of these forms of 'space weather', according to the website "Space Weather: What impact do solar flares have on human activities?", the effects that do affect us are:

Disturbance of the ionosphere, disrupting radio communications.
heating, hence expansion of the atmosphere would cause drag on satellites, degrading their orbits, also affecting the accuracy of GPS.

From the Solar Weather site, an important distinction between CME's and solar flares in how they affect the Earth:

  Coronal mass ejections are more likely to have a significant effect on our activities than flares because they carry more material into a larger volume of interplanetary space, increasing the likelihood that they will interact with the Earth. While a flare alone produces high-energy particles near the Sun, some of which escape into interplanetary space, a CME drives a shock wave which can continuously produce energetic particles as it propagates through interplanetary space. When a CME reaches the Earth, its impact disturbs the Earth's magnetosphere, setting off a geomagnetic storm.

Very severe CME's can also potentially disrupt power grids and communications.


How to use spectral flux density data while using Planck's law giving spectral radiance?

As I understand it, spectral flux is defined as
$ F_{\nu} = \int_\Omega I_{\nu}(\theta, \phi)\cos{\theta} d\Omega $
where $d\Omega$ is the solid angle element over which the integral is performed. Here $I_{\nu}$ is the specific intensity. The $\nu$ subscript denotes frequency dependence.
According to A. Choudhuri "Astrophysics for Physicists" p. 25, blackbody sources are isotropic emitters therefore we can drop the angular dependence in the spectral intensity and $I_{\nu}(\theta, \phi)$ becomes your Planck formula $B_{\nu}(T)$.
Hence
$ F_{\nu} = \pi B_{\nu}(T)$