Sun's formation as "part of a star-forming-cluster..."?

Fortunately, there is an extensive and highly respected Annual Reviews of Astronomy and Astrophysics article on this very question - Adams (2010), https://arxiv.org/abs/1001.5444

At the moment it is debated whether most stars form in clusters. They may do, but there is a lot of evidence for hierarchical structure formation that results in lots of stars forming in relative isolation.

Regardless, there are a number of strands of evidence that can be used to suggest that the Sun formed in a cluster of somewhere between a thousand and ten thousand siblings - a bit like the Orion Nebula cluster.

Adams argues for this on the basis of: (1) We have a relatively well-behaved solar system inside the orbit of Neptune. This argues the Sun did not (or rather was not likely to) form in a massive and dense cluster where it suffered lots of very close encounters. On the other hand, the wacky orbits of some more distant objects like Sedna do suggest some early close encounter with another star was required. (2) If the Sun was born in a very massive cluster, it would likely have been born close to some very massive stars. The UV emission from such stars could photoevaporate a circumsolar disk before it formed planets. (3) In tension with this is the probability that the protosolar disk was enriched with radioactive supernova products. This requires the Sun to be born close to moderately massive stars (that would explode on a short timescale), which are only found in moderately massive clusters.

Adams puts these constraints together to estimate a probability that the Sun was born in a cluster of a particular size. Figure 7 of the review shows this to be reasonably sharply peaked between a thousand and ten thousand.

An order of magnitude uncertainty is at least warranted and some of the assumptions and calculations are open to criticism. But it is a starting point for any further research into the matter.


Why is WISE J0521+1025's distance so uncertain?

The distance is that reported by Bihain et al. (2013), which is based on a mean relationship between absolute magnitude and spectral type that has a lot of scatter.
i.e. in contrast to most (all?) the other objects in that list, there is no reported trigonometric parallax measurement for this very faint T7.5 brown dwarf.
In fact, if it turned out to be a binary brown dwarf, the distance could be incorrect!


What would happen if space-time began to contract?

Given that the expansion of the universe shows us that the distance between galaxies (or rather moreso between galaxy clusters) is increasing, without any change in the distances within or between atoms, or between stars within galaxies (apart from normal stellar motion), and that galaxies themselves are also not expanding, we could assume that a contraction would manifest as a decrease in the distances between galaxies/galaxy clusters.


Astronomy & machine learning. Where to start?

This recent KISSCaltech lecture by Prof. Rob Fergus at the NYU might be of interest. "An Overview of Machine Learning Approaches: Applications to Exoplanet Detection".


Demo to get Matter Power Spectrum in cosmology

This is just a Fourier transform: (let $\boldsymbol{x}=\boldsymbol{r}_2-\boldsymbol{r}_1$)
$$
\begin{aligned}
\langle \delta(\boldsymbol{k}_1)\delta(\boldsymbol{k}_2) \rangle&=\int\int d^3r_1d^3r_2\langle  \delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2) \rangle e^{-i\boldsymbol{k}_1\cdot\boldsymbol{r}_1}e^{-i\boldsymbol{k}_2\cdot\boldsymbol{r}_2}\\
&=\int d^3r_1 e^{-i\boldsymbol{k}_1\cdot\boldsymbol{r}_1}\int d^3r_2\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2)\rangle e^{-i\boldsymbol{k}_2\cdot\boldsymbol{r}_2}\\
&=\int d^3r_1 e^{-i\boldsymbol{k}_1\cdot\boldsymbol{r}_1}\int d^3x\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_1+\boldsymbol{x})\rangle e^{-i\boldsymbol{k}_2\cdot(\boldsymbol{r}_1+\boldsymbol{x})}\\
&=\int e^{-i(\boldsymbol{k}_1+\boldsymbol{k}_2)\cdot\boldsymbol{r}_1}d^3r_1\int\xi(\boldsymbol{x})e^{-i\boldsymbol{k}_2\cdot\boldsymbol{x}}d^3x\\
&=(2\pi)^3\delta_D(\boldsymbol{k}_1+\boldsymbol{k}_2)P(\boldsymbol{k}_2)
\end{aligned}
$$
Here, $\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2\rangle)$ is two-point correlation function (2pcf) in real space. If we assume our universe is statistically homogeneous, $\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2\rangle)$ should have the form $\xi(\boldsymbol{r}_1-\boldsymbol{r}_2)$. So power spectrum is the Fourier transform of 2pcf.
In addition, if we assume our universe is statistically isotropic (not true in redshift space), 2pcf can be $\xi(|\boldsymbol{r}_1-\boldsymbol{r}_2|)$ and power spectrum can be $P(k)$.


Well, yes apparently that is how it works and a Nobel prize was won in 1993 by Hulse and Taylor for demonstrating this phenomena in the object PSR B1913+16.

PSR B1916+16 is a pulsar (a rapidly rotating neutron star with a strong magnetic field that emits beamed radiation along its magnetic axis). The pulsar provides an extremely precise ticking clock with which to study the motion of stars in a binary system. The pulsar is in orbit with a companion that is also a neutron star (though not an observable pulsar). The orbit is elliptical with a period of 7.75 hours.

The period of the orbits is decreasing. That is, the two stars are gradually getting closer together. This effect is a consequence of general relativity, that predicts that such massive objects that are so close to each other will act as a source of gravitational waves. The gravitational waves have not (yet) been directly observed, but GR makes a prediction for how much energy they carry away and hence at what rate the orbital period changes.

The rate of decrease in the distance between two bodies is given by
$$\frac{dr}{dt} = -\frac{64G^3}{5c^5}\frac{(m_1 m_2)(m_1+m_2)}{r^3}$$
and leads to a merger between the two objects in a time of
$$t = \frac{5c^5}{256G^3}\frac{r^4}{(m_1 m_2)(m_1+m_2)}, $$
where $m_1$ and $m_2$ are the masses of the two orbiting bodies, $G$ is the gravitational constant and $c$ is the speed of light.

In slightly more friendly units
$$ \frac{dr}{dt} = 7.8\times10^{-19} \frac{(M_1 M_2)(M_1+M_2)}{r^3}\ au/yr,$$
where $r$ is in astronomical units and the masses are in solar masses, and
$$ t = 3.2\times10^{17} \frac{R^4}{(M_1 M_2)(M_1+M_2)}\ yr. $$

Thus you can see that the effect gets much stronger (and the merger times shorter) if you have two high mass objects (it depends on the product of the masses) with a small orbital separations.

For "normal" stars you can never get them closer than their stellar radii and generally means that the effect is too small to be important. However, compact objects (white dwarfs and neutron stars) can be brought much closer together and this is why the effect has been seen there. The Hulse-Taylor binary components
have masses of $\sim 1.4 M_{\odot}$ and are separated by an average of about 2 million km (0.013 au). Inserting this into the formula above give a merger time of 1.7 billion years. A more accurate calculation that takes account of the elliptical orbit gives 300 million years.

However, if you made the neutron stars orbit with a separation of 0.1 au (more typical for a normal stellar binary system), the merger timescale would increases to well beyond the current age of the universe.


When was it realised that most major moons orbit in the equatorial plane of their parent planets?

Since the question has sit unanswered for months, I'm trying to give an acknowledgedly incomplete answer.

In summary: By 1700 it would have been a reasonable guess. By 1800 it was becoming an observed trend.

TL;DR:

It's hard to decide when the consensus was created on the typical orbits of satellites, because there are only a few data points (planets with satellites) but we can know when those few data points where known.

Obviously, the only data point known in antiquity was the Moon, but by that time the whole idea of satellites was far from mainstream.

In 1610 Galileo discovered the four biggest moons of Jupiter and that they orbit Jupiter in roughly the same plane. However, it took until the 1660s for Cassini to discover features in Jupiter and therefore be able to realise that its moons orbited in the equatorial plane.

By 1655 Huygens had discovered Saturn's rings and Saturn's satellite Titan, both of them in the same plane. Between 1671 and 1684, Cassini discovered Tethys, Dione, Rhea and Iapetus all of them in the same plane of the rings except for Iapetus.

The rotation of Saturn is a bit harder to detect, and it wasn't measured until 1794 by Herschel.

There is an additional data point even before being able to see Saturn rotating: both Jupiter and Saturn appear elliptical in the 17th century telescopes. Although a causal relation with rotation wasn't established until Newton, as soon as Jupiter's rotation had been discovered it was known that the planet was rotating around the smallest axis of the ellipsoid.

No more data points were added for a long time, because the moons of Mars and the rotation of Uranus were discovered much later, although the first two moons of Uranus had been discovered by Herschel in 1787.

Additionally, it was known from antiquity that all planets were in similar planes, and when sunspots were observed (by Galileo) and the rotation of the Sun could be measured, the Sun's equatorial plane was close to the planes of all planets orbit. Therefore, when the idea of planets and satellites as "little solar systems" arrived, the idea of the central body rotating in the same plane of the system wouldn't have seemed strange.

Then we can summarise what an informed observer at the end of the 17th century would have known:

Jupiter's rotation, Jupiter's oblation, and Jupiter's moons were in the same plane.
Saturn's rings, Saturn's oblation and most of Saturn's moons were in the same plane.
There were only two known exceptions to the rule of "everything around the same planet in the same plane": Earth's Moon and Iapetus, which aren't in the equatorial plane - although they aren't very far away from it.

With all that at hand, I would say that guessing that most satellites orbit in the equatorial plane of its planet would have been a reasonable guess by 1700.

By 1800 the evidence would have increased with:

The rotation of Saturn had been actually observed, confirming that most satellites and the rings orbit on the equatorial plane.
Uranus had two satellites in the same plane.

With that, it seems even more reasonable to suppose that Uranus'satellites were orbiting on its (unknown) equatorial plane.


What comet's tail did Earth pass through before Halley's?

The quotation is probably referring to The Great Comet of 1861.
According to this Wikipedia article: https://en.wikipedia.org/wiki/C/1861_J1

The comet may have interacted with the Earth in an almost
unprecedented way. For two days, when the comet was at its closest,
the Earth was actually within the comet's tail, and streams of
cometary material converging towards the distant nucleus could be
seen.

This was reported in newspapers at that time:
https://www.newspapers.com/image/424095911/?terms=comet%20tail&match=1
and also prior to the close approach of Borelli's Comet in 1903.
https://www.newspapers.com/image/85101704/?terms=%22%20borelli%20comet%22&match=1

From  this archive of C/1861 J1 (Great Comet of 1861) from http://cometography.com/lcomets/1861j1.html

Further reading:

Great Comets in History by Donald K. Yeomans (April 2007) JPL/Caltech
Comet C/1861 J1 (Great comet) in JPL Horizons database
Periodic Comets Found in Historical Records
Do Comets C/1861 G1 (THATCHER) And C/1861 J1 (Great Comet)
Have a Common Origin?
International Comet Quarterly
Forward-Scattering Enhancement of Comet Brightness. I. Background and Model (large PDF, Int Comet Quart April 2007)
Forward-Scattering Enhancement of Comet Brightness. II. The Light Curve of C/2006 P1 (McNaught)  (large PDF, Int Comet Quart October 2007) also here
Forward-Scattering Enhancement of Comet Brightness. III. Prospects for C/2010 (Elenin) as viewed from the Earth and SOHO, STEREO-A, and STEREO-B spacecraft (large PDF, Int Comet Quart August 2011?) also here


Why are the wings of many strong lines, including hydrogen lines and most metal lines, stronger in Dwarfs than in Supergiants?

Collisional broadening - which includes van der Waals and Stark broadening - is more important in the higher gravity, higher pressure/density atmospheres of dwarf stars (a factor of 100-1000 higher for dwarfs vs giants of the same photospheric temperature).
These collisional effects effectively "truncate" radiative emission and absorption processes, leading to broadening of the frequency response of a line absorption cross-section. That means strong lines have appreciable opacity away from the line centres and means the lines sample a broader range of temperatures in the photosphere.
See https://physics.stackexchange.com/a/257471/43351


Which class of objects in Solar System carries more mass?

The sun by far occupies the largest part of the solar system's mass. The mass of the sun happens to be approximately $(1.98855±0.00025)×10^{30}$ kg. Which is immensely huge. By comparison the inter-planetary medium though may occupy a large volume is nowhere dense enough to even compare to the mass of the sun which is approximately 99% of the solar system's mass.

The mass of all planets combined by comparison happens to be approximately $2.6634×10^{27}$kg.
As far as interplanetary matter is concerned, its density is to low and its volume is pretty hard to define so mass remains a highly variable quantity.

Moons and asteroids account only for a fraction of the mass of the solar system though they happen to be present in large numbers.


A semi-major axis of an ellipse defines the half diameter of the ellipse where it is longest. That's a convenient measure to compare ellipses and express its properties. Mathematically you can describe an ellipse with semi major axis $a$ (how long is it) and semi minor axis $b$ (how thick is it) as $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
Often, the eccentricity is used which is defined as $$ e = \sqrt{1 - \frac{b^2}{a^2}} $$
So for a circle with $a = b$, the eccentricity is $0$, and for the edge case, a very long ellipse with $a \rightarrow \infty$, we get a parabola which hence has an eccentricity of 1.
Now, what would a trajectory look like with an eccentricity larger than 1? We get an eccentricity larger than one, if the radix of the definition of eccentricity gets larger than 1, thus $-b^2 / a^2$ needs to be positive. We achieve that if we introduce an imaginary semi major axis (not a negative! See the linked wiki article for a brief introduction to the concept of those numbers) $a$ such that $a^2 < 0$.
With that extension we can express hyperbolic orbits with the same formulation as elliptic and parabolic ones, with a single number, the eccentricity:
$$ a \rightarrow a_r + ia_i\\ b \rightarrow b_r + ib_i $$
and thus we get a single number to describe any kind of orbit, closed and not-closed ones:
$$ e = \sqrt{1 - \frac{b^2}{a^2}} = 0 \cdots <1: \qquad\mathrm{ellipse}\qquad a = a_r; a_i = 0\\
 $$
$$ e = \sqrt{1 - \frac{b^2}{a^2}} = 1: \qquad\mathrm{parabola}\qquad a = \infty\\
 $$
$$ e = \sqrt{1 - \frac{b^2}{a^2}} > 1: \qquad\mathrm{hyperbola}\qquad a = i\cdot a_i; a_r = 0\\
 $$
Energetically, objects in elliptical orbits don't have enough energy to escape, objects on a parabolic orbit had no kinetic energy at infinity and thus just escape and it only needs a slight change in orbital energy to capture them. Objects  on hyperbolic orbits were already in motion at infinity, thus have excess orbital energy and would need to loose much more energy to capture them (and in a solar system context are very likely to be of extra-solar origin).


Before Voyager 2, what were the highest quality photos of Uranus and Neptune?

As per the books referred to in comments:

Both planets could be resolved as a disk, but no surface features could be observed on either. Spectrograms had been taken, so the general colour of the planets was known. The extreme axial tilt of Uranus was known about. But the relative warmth of Neptune was not.


What would happen if we put a small part of White Dwarf or Neutron Star on Earth?

A piece of a white dwarf would explode, because atoms would expand to their normal size. This would hold for outer layers of a neutron star, too. For inner parts of the neutron star, that's not quite clear, but that kind of matter would probably decay rapidly, also resulting in an explosion.

If the material of a neutron star could be kept stable, it would fall down to Earth's core, up to the other side of the Earth, and oscillate a while that way, before being slowed down by friction, and stopped close to Earth's center. (1 cm³ neutron star matter has a mass of about 400 million tons.)

One cm³ white dwarf matter would have a mass of about 1 ton. Assumed it could be kept stable, solid rock could withstand this load.


Table of absolute magnitudes of stars by spectral type and photometric band

There are a couple of standard papers containing the table you want.

Kenyon & Hartmann (1995) http://adsabs.harvard.edu/abs/1995ApJS..101..117K

Table A5 contains many colours for stars as a function of spectral type. You need to combine this with something that gives absolute V magnitude along the main sequence, like that of Schmidt-Kaler (1982).

An alternative, that has colours for both main sequence and pre main sequence stars (though the absolute magnitude of a PMS star is age-dependent) is found in Pecaut & Mamajek (2013).
 http://arxiv.org/abs/1307.2657

Online table at
 http://vizier.cfa.harvard.edu/viz-bin/VizieR-3?-source=J/ApJS/208/9/table5


How do we measure the brightness of the stars?

The term brightness, or apparent brightness, is used to describe how bright a star appears to us from Earth. The term luminosity is used to describe how bright the star is physically, also called intrinsic brightness.

As light spreads from a star to the viewer, it diverges inversely proportional to the square of its distance (d) from the viewer.  $\begin{equation}B \propto \frac{1}{d^2}
\end{equation}$.

The relationship between brightness (B) and luminosity (L) is described by $\begin{equation}B = \frac{L}{4\pi d^2}
\end{equation}$ (1). So we have three variables, brightness, luminosity and distance.

So out of those three, you'd need two to acquire the third. The distance is generally the hardest to acquire. The gold standard in measuring the distance to a star is stellar parallax. If you can't do parallax, then distance-independent properties of stars sometimes serve as clues to distance.

Measuring brightness is generally the easy one. CCD's are often used to measure apparent brightnesses. Brightness can be expressed in stellar magnitudes or absolute fluxes.

So to sum up, we calculate or measure distance, we measure apparent brightness and finally calculate the luminosity value from (1).


Is radiation from neutron stars delayed by time dilation?

A typical neutron star of $1.5M_{\odot}$ is thought to have a radius of around 8-10 km. This is only a factor of 2 larger than the Schwarzschild radius for a similar mass black hole.

We know that more massive neutron stars do exist. The current record holder is around $2M_{\odot}$. Most equations of state (the adopted relationship between pressure and density) for dense nuclear matter suggest that more massive neutron stars are smaller and therefore must be even closer in radius to the Schwarzschild radius.

So the premise of you question is basically correct. It is certainly true that when you deal with neutron star spectra you do have to apply significant general relativistic corrections to measured temperatures and the same corrections would need to be applied to any temporal variations.

Thus a time-variable signal from a neutron star surface will appear slower to an observer on Earth.

For the last part, I suspect that the scenario you propose is extremely unlikely. Rhoades & Ruffini (1974) first established that there must be a maximum mass for a neutron star under GR conditions, even if we allow the equation of state to harden to the point where the speed of sound is the speed of light. This maximum mass is around $3.2M_{\odot}$.  This sets an upper limit to the possible value of $GM/Rc^{2} \leq 0.405$ (see p.261 of Shapiro & Teukolsky, Black holes, white dwarfs and neutron stars). This in turn sets an upper limit the possible gravitational redshift (and time dilation factor) of 2.29.

Beyond this point the neutron star is unstable and will collapse to become a black hole. In reality the limit is probably a bit tighter than that because most proposed equations of state result in neutron stars becoming unstable at finite densities and at masses quite a bit lower than $3.2M_{\odot}$.

So I think the most time dilation you are ever going to see from a neutron star surface is a factor of $\sim 2$.


Is the mass of Kuiper belt dominated by large objects?

It depends on what you call large. There may be hundreds of thousands of icy bodies larger than 100 km (62 miles) and an estimated trillion or more comets within the Kuiper Belt. But the total mass of objects in the Kuiper belt is estimated to be less than 10% of earth's mass.  So the Kuiper belt is largly composed of small icy onjects. These objects are very hard to see with existing telescopes because the are so small and not much sunlight where they are.
References:
https://solarsystem.nasa.gov/solar-system/kuiper-belt/in-depth/
https://solarsystem.nasa.gov/news/792/10-things-to-know-about-the-kuiper-belt/
https://solarsystem.nasa.gov/solar-system/kuiper-belt/overview/


This puts us in the realm of silly numbers.

A supernova has enough energy to rip a massive star apart. Putting all that energy into one small rigid body isn't going to end well.

A hypernova gives off about $10^{46}$ Joules. We will need the formula for relatavistic kinetic energy $E=mc^2(\gamma-1)$ where $\gamma$ depends on velocity. The mass of the shuttle is about $2\times 10^6$ kg, and $c^2=9\times 10^{16}$

So $10^{46} = (2\times 10^6) \times (9\times 10^{16}) (\gamma - 1)$. Rearranging, $\gamma = 5.5\times10^{22}$.

In relativity $\gamma = 1/\sqrt{1-v^2/c^2}$ So even this massive amount of energy won't make the shuttle go faster than light. It will make it go at

99.99999999999999999999999999999999999999999998%

of the speed of light.


Hypothetical trans-Neptunian planets orbiting the Sun

If such a planet exists, WISE should observe it. WISE is an infrared satellite that imaged the entire sky. In particular:

it was able to detect anything with a temperature above 70-100 K, whereas the coolest known exoplanets are in the 100 K range (see histogram below, taken from the Exoplanets Catalogue);
it was able to detect objects larger than 1km up to 3 AU from the Sun, or objects of 2-3 Jupiter masses in a distance up to 7-10 light-years (so roughly up to 440,000 to 630,000 AU, see here and here);

so I would say that its existence is quite ruled out now.


Is it realistic to build a telescope powerful enough to take a photo of TRAPPIST-1's planets?

It's not out of the question, but it would be very difficult to use direct imaging for any of the exoplanets around TRAPPIST-1:

The exoplanets are small, all not much bigger than Earth. Ideally, you want a target the size of Jupiter or bigger. Likewise, they are very low-mass, and are thus small. Once more, this is a problem. Direct imaging has only had much success with large planets, of many Jupiter masses.
The exoplanets are close to the star, with the furthest one out - admittedly with a poorly-constrained orbit - having a semi-major axis of 0.063 AU (and a year of 20 days), give or take. That is tiny. Mercury's semi-major axis is more than six times that. If the exoplanets are close to the star, it makes it much harder to distinguish them from the star's glare. Granted, there are methods that do this very well, but they're not meant for planets this close.

To see what I'm talking about, look at some examples of directly-imaged exoplanets. Almost all in that list have many times Jupiter's mass and at least its radius. Their semi-major axes are many, many - sometimes thousands - times that of Earth.

I can't speculate as to what technology we'll have decades from now, but if you want to find out more about the exoplanets around TRAPPIST-1, direct imaging is not the way to go, and I suspect it will not be a very popular choice among astronomers.


The modern system of stellar classification is 2-dimensional, with one axis conveying spectral class, and the other conveying luminosity class. The spectral types are as follows:

O-type, "blue", $\geq 30000$ $\mathrm{K}$
B-type, "blue-white", $10000-30000$ $\mathrm{K}$
A-type, "white", $7500-10000$ $\mathrm{K}$
F-type, "yellow-white", $6000-7500$ $\mathrm{K}$
G-type, "yellow", $5200-6000$ $\mathrm{K}$
K-type, "orange", $3700-5200$ $\mathrm{K}$
M-type, "red", $2400-3700$ $\mathrm{K}$

There are also some extended spectral types, including D for white dwarfs, C for carbon stars, and several more for brown drawfs. The above are the predominant.

The magnitude axis consists of the following groupings:

0-type, "hypergiants"
I-type, "supergiants"
II-type, "bright giants"
III-type, "giants"
IV-type, "subgiants"
V-type, "dwarfs" (these are the main-sequence stars)
VI-type, "subdwarfs"
VII-type, "white dwarfs"

When these systems are used, a spectral class is combined with a number ranging from 0-9 which symbolizes the location between adjacent spectral classes, and the luminosity class is then appended.

For example, the Sun is a type G2V star. This means it is a G-type star on the main sequence. It could then informally also be referred to as a "yellow dwarf".

The system is more complex than this, and there are many additional suffixes and other symbols to further explain the nuances of individual stars. This does cover the basics quite well, however.


If you're near a black hole and your time is slowed down, would a supernova be observable to both you and someone outside of the blackhole's pull?

Time is relative therefore this is a bit of a nonsense question.

But yes, if the gravitational effect near the black hole is strong enough (depends on how close you are) it would affect 'when' the observer would notice it.  This is because the light must travel a 'further' distance through the spacetime to get to you since gravity is distorting it.

You may find this interesting: https://web.archive.org/web/20130424041627/http://www.thebigview.com/spacetime/spacetime.html


How is phase defined in a binary orbit with eccentricity?

Usually, how is phase=0 defined? It is relative to the line of apses perhaps with phase=0 at periastron?

The true anomaly, $\theta$, is the angle between the current location of the orbiting particle and its location in the orbit at which it is closest to the central body (called the periapsis/periastron). The word "phase" is used variously in physics and astronomy, but based on your question, you may mean $\theta$ since it is defined as being zero at periastron.

Are orbital phases of the black spot in the figure measured from some angle or time/period?

It ultimately depends on how you want to compute the trajectory/orbit of the particle (i.e., the black "spot" in the OP image) that is moving about a primary body (i.e., the star in the OP image). Alternate values can be used instead of $\theta$, for example $\displaystyle M$, the mean anomaly.


As mentioned in the comments, a ring will (ideally) only occur if you have perfect alignment of source, lens, and observer. This is true for point sources, for extended sources things get more complicated.

Double and quadruple images occur when the source is small and not in perfect alignment - typically a quasar, as in your example. For the sake of completeness, this would actually be triple or quintuple, but the third / fifth image is very close to the centre, overpowered by the lens, and highly demagnified (faint).

There are examples of rings and almost-rings, but these tend to be extended sources, i.e. galaxies. It is more likely that some part of an extended source happens to be aligned with lens and observer than it is for a point source.

(Image of the Cosmic Horseshoe)

In your example, if the host galaxy of that quasar could be seen, it might be an Einstein ring (or an oval, as the lens seems to be somewhat asymmetric).


Frankly, four bodies do not periodically line up, even if the orbits are planar, unless there exist a common multiplum. Because all your periods are integers, that holds, but be aware that your double alignments, like the double solar eclipse or double lunar eclipse is not periodic if one of the periods does not exactly follow that integer system. (so it does not hold for 31.01 days, for example. How you handle quazi-periods is talked about here).

  How often would both moons be full at the same time, overlapping each other like a bullseye?

Generally non-periodic, but can only happen at the synodic period of the two moons, that is:

.

$\frac{2418}{47}$ for the two moons. However, that must also happen when both of them is on the same line as the Sun, in other words when The relative synodic period between one of the moons and the sun, and the moons' synodic period both divides the requested period. For the inner moon and the Sun, the synodic period is $\frac{10385}{304}$. Since 47 does not have any factors in common with 304, the first common multiplum appears at.

$$\frac{10385}{304} \cdot \frac{2418}{47} = \frac{12555465}{7144}$$, or 1757.484 days. (@Jonathan gets 2418, because his model consider the Sun to stay static in the sky, but it does in fact revolve around one time a year.) The exact same period is also valid for your question 4.

  How often would lunar eclipses occur for Moon A?

Here, the answer is the synodic period of the Sun and the moon, already stated to be $\frac{10385}{304}$, or 34.161 days. That is a little bit longer than the Moon's orbital period, because the planet has moved a little in its orbit, slightly changing the Sun's position in the sky.

  How often would lunar eclipses occur for Moon B?

Synodic period here too, $\frac{26130}{257}$ = 101.673 days.


How can 'HD 140283' be older than the universe?

It's not - the current age is an estimate. The discrepancy comes from astronomer's attempts to fully and correctly measure aspects of the star, particularly its distance.  From the NASA page:

  The new Hubble age estimates reduce the range of measurement uncertainty, so that the star's age overlaps with the universe's age — as independently determined by the rate of expansion of space, an analysis of the microwave background from the big bang, and measurements of radioactive decay.

Currently, if you take the uncertainty into account, the current estimate for the star's age is 14.5 +/- 0.8 billion years  or between 15.3 and 13.7 billion years - the latter end of the estimate overlaps with the current model of the age of the Universe - so a more correct way of describing this age is that HD 140283 is at least 13.7 billion years old by this estimation.

Another paper based on this star in Nature - "Nearby star is almost as old as the Universe", a team has narrowed down the age to 13.9 +/- 0.7 billion years, stating that:

  Taking into account that experimental error, the age does not conflict with the age of the Universe, 13.77 billion years.

more concisely stating that

  The star's age is therefore at least 13.2 billion years

No doubt further work is being performed to further refine the age of the star.


Why can we see the Pillars of Creation, but not Proxima B?

First, let's talk about size. The Pillars of Creation aren't just "giant gas bubbles"; they're clouds of gas and dust four light-years long. That's enormous, about 36,000,000,000,000 kilometers. Proxima Centauri b, on the other hand, has a diameter of 0.8-1.4 Earth diameters; one Earth diameter is approximately 12,700 kilometers. That's miniscule, in comparison.

Let's look at the angular diameters of the objects. The general formula is
$$\delta=2\arctan\left(\frac{\text{diameter}}{2\cdot\text{distance}}\right)$$
where the angular diameter is in radians. The Pillars of Creation are 6,500 light-years away, conservatively, giving just them an angular length of $3.52\times10^{-2}$ degrees. Proxima Centauri b is 4 light-years away, with an angular diameter of about $2.02\times10^{-8}$ degrees. 6 orders of magnitude is nothing to sneeze at.

Additionally, the Pillars of Creation are located in an emission nebula. Light from the young stars that have just been born ionizes gas in the Eagles Nebula, which in turn emits the light. The dark gas and dust in the Pillars of Creation stands out quite sharply in all of this, making it easy to see.


Three main reasons.

Lots of things in the universe are cool, much cooler than the Sun. Therefore they emit most of their radiation in the infrared spectrum. Examples might include brown dwarfs, giant exoplanets, circumstellar discs.
Lots of things in the universe are distant and highly redshifted. Thus even if they emit their radiation at visible wavelengths, we receive it at infrared wavelengths.
Lots of things in the universe are hidden by dust, which effectively absorbs visible light. Dust is much more transparent to infrared radiation, so we can see beyond it. An example would be the centre of our Galaxy.


How do I calculate the velocity dispersion of a galaxy given an integral field spectrograph?

If the 3D data cube is sufficiently highly resolved, try to find Doppler shifts in the spectra by looking for typical absorption/emission lines. Doppler shifts correspond to radial velocities. Calculate the mean radial velocity, and e.g. the standard deviation of the radial velocities, as velocity dispersion.

The velocity dispersion of galaxies leads to a widening of spectral lines in comparison to a reference star. A couple of methods are discussed in this paper, mostly some kind of Fourier fitting, but direct fitting is used, too.

Redshifts may contain a gravitational component.

There exist data reduction software tools.

A paper describing data reduction and velocity analysis using the XOASIS tool.


To know what you would see with a telescope, you can use a field of view calculator such as this one from astronomy tools. There, you just fill in the details of your optics, and you can get an idea of what you would be able to see with your telescope.

For a Celestron Powerseeker with the focal lengths and apertures you provided, and a 3x Barlow, the Crab Nebula might look something like this

The yellow line here is the edge of the field of view. You can also use the web app to compare several different configurations. This will give you an idea of how big the object will be in your telescope.

An other factor to take into account is how bright you will be able to see the object. This depends greatly on where you will be observing (near a city or far away from any bright lights). In any case, the contrast will not be as stark as in the above image.

For this, you first need to estimate the magnitude of the faintest objects observable with the naked eye where you expect to be observing. If you know how to recognize a few constellations, you can look up (in a sky atlas or online) the magnitude of the faintest stars you see. For instance, if you can see M13 in Hercules, then the limiting magnitude with your naked eye is higher than 5.8. However, if you can't see the entire wings of Cygnus, then the limiting magnitude is between 3 and 4.

Once you figured this out, you can use a limiting magnitude calculator. This will give you an idea of the faintest objects you can see with the telescope.


How to get the axial tilt vector(x,y,z) relative to ecliptic

Earth is a special case since the equatorial and ecliptic coordinate systems are defined in terms of its own rotation and orbit.
Earth's north pole vector in
equatorial coordinates is
$$\vec N_{\oplus,eq} = (0, 0, 1)$$
To transform this to
ecliptic coordinates, we rotate about the $x$ axis by the obliquity $\varepsilon$ = 23.44$^\circ$ and get
$$\vec N_{\oplus,ecl} = (0, \sin \varepsilon, \cos \varepsilon) = (0, 0.3978, 0.9175)$$
In a spherical coordinate system, two angles define a unique direction.
In equatorial coordinates these are the right ascension $\alpha$ and declination $\delta$.
This IAU report, table 1,
gives $\alpha$ and $\delta$ values for each major planet's north pole as of 2000-01-01 and formulas to compute them for other years.
To convert these to rectangular form
$$(x_{eq}, y_{eq}, z_{eq}) =
(\cos \alpha \cos \delta,
\sin \alpha \cos \delta,
\sin \delta)$$
Then if you want J2000 ecliptic coordinates
$$(x_{ecl}, y_{ecl}, z_{ecl}) =
(x_{eq},
y_{eq} \cos \varepsilon + z_{eq} \sin \varepsilon,
z_{eq} \cos \varepsilon - y_{eq} \sin \varepsilon)$$
But if you want another planet's orbital plane to be the $xy$-plane,
then you also need the orbit's inclination and longitude of ascending node;
I leave that transformation as an exercise for the motivated reader.


Who gets to name Planet Nine if it is confirmed?

The cited answer is correct insofar as the IAU is not the body that makes names of celestial bodies official. However, it's authority is implicit because it is recognized by the astronomical community and the public as the foremost authority on astronomical naming, and thus the names the IAU adopts are the ones most commonly used in the scientific community (excluding objects referred to by their catalog numbers).
From one of their information pdfs:

The IAU has been the official arbiter of planetary and satellite naming since its inception in 1919. The IAU’s decisions are officially adopted by the nearly 11 000 professional astronomers who are its members, coming from more than 90 countries.
. . .
The IAU does not consider itself as having a monopoly on the naming of celestial objects — anyone can in theory adopt names the way they choose. However, given the publicity and emotional investment associated with these discoveries, worldwide recognition is important and the IAU offers its unique experience for the benefit of a successful public naming process (which must remain distinct, as in the past, from the scientific designation issues).

The IAU may reach out to the public for naming suggestions, or it may taken case-by-case suggestions by individuals. So while Batygin and Brown can call the ninth planet whatever they want - they've informally referred to it as "Phattie" - public consensus will largely rest on the results of the IAU's decision, which could, of course, end up taking a suggestion of the pair.
Anyone can come up with a suggestion, although sometimes that's a terrible idea.


In System KIC 9246715, How Far Does One Star Orbit the Other?

The AAS Nova article cites Rawls et al. 2016,
who analyze light curves and spectra of the eclipsing binary KIC 9246715 to estimate its physical properties.
Besides the stellar masses and radii in the article, their Table 2 lists orbital semimajor axis a = 211 R☉ = 147 million km = 0.98 au, and orbital eccentricity e = 0.356.
During the 171-day period, the distance between the two stars varies between 95 and 199 million km, or 0.63 and 1.33 au.


Where are we in an approximate timeline of the possibly habitable universe?

Here's a very roughly calculated partial answer.

A first generation star and solar-system would obviously not be a good candidate for life or even planets cause you can't do much with mostly hydrogen and helium.   You can't even build an ocean with mostly those 2 elements.  Stars do just fine, but planets don't.  Gas Giants and Suns only.

After the big bang: 75%-76% Hydrogen by mass, 24-25% Helium (with a teeny-tiny trace amount of Lithium).   I'm going to assume 76% / 24% for rough calculations.

Using our solar-system and the Milky-way as a model, elements by mass:

Milky-Way

73.9% Hydrogen

24.0% Helium

1.04% Oxygen

0.46% Carbon

0.13% Neon

0.11% Iron

0.10% Nitrogen

0.07% Silicon

0.06% Magnesium

0.04% Sulfur

Smaller amounts of Argon, Calcium, Nickel, Sodium, Phosphorus, Chlorine, etc.

and composition of the Sun by mass (pretty similar except for more Helium)

71.0% Hydrogen

27.1% Helium

0.97% Oxygen

0.40% Carbon

0.06% Neon

0.14% Iron

0.10% Nitrogen

0.10% Silicon

0.08% Magnesium

0.04% Sulfur

Using these numbers as a rough guideline, if 1.9% to 2.1% of the baryonic mass of the milky-way converts from Hydrogen to Carbon or heavier elements after (about) 2 generations, and we extrapolate a 1% increase in all Carbon or heavier elements to each stellar generation or every 6 billion years, then we can begin to make an estimate.

Helium percentage could stay somewhat constant at about 24% cause it's both consumed and created stars.

Your original question, how many generations, if we figure 1% increase in heavier elements (Carbon or higher) per stellar generation, 10 generations we're still looking at, very roughly, (24% helium), 66% hydrogen, 5% oxygen, 2%-3% carbon, 2%-3% other elements.  It still looks like a workable scenario to me though the stars, being made of some denser elements might have denser cores and as a result, could burn hydrogen a bit faster.    Past a certain point the ratios probably stop being ideal but I see no reason why quite a few generations couldn't still work create life-friendly solar-systems.

Our sun is expected to cast off about half it's matter into a planetary nebula before it settles down into a white dwarf and most of that cast off matter will be hydrogen and some helium.   Some time later if our sun as a white dwarf accretes enough to meet the Chandrasekhar mass and goes Type-1 Nova, it will again cast off a significant percentage of it's matter, so, to put it simply, Stars recycle a pretty good percentage of their hydrogen and helium, which is kinda cool.   If you see Rob Jeffries' brilliant answer here, stars of 7-9 solar masses will eject an even higher percentage of their elements before they go white dwarf, as much as 85% and most of what gets ejected is the lighter elements, hydrogen and helium from the outer layers.   Even smaller red-dwarf stars expel a percentage of their elements due to more active solar flares and lower gravity throughout their lives.

Add to this, the possibility of continued merger of the galaxies and dwarf galaxies in our local group, and life capable solar-systems could continue to form for quite some time, perhaps several tens of billions of years.

This answer isn't meant to be definitive, but only a rough estimate.  Corrections welcome.


Could there be a planet at the barycenter between two or more stars revolving around each other?

No.  Such an arrangement is at best "metastable".  That is, although there are periodic solutions to the three body problem (stable orbits) an infintesimal perturbation (eg the proverbial butterfly flapping its wings) will push the system off the stable orbit and into chaos.  Getting a planet to remain at the barycentre is like trying to balance a pencil on its sharpened point.
With two bodies, each orbits around the barycentre.  But with three bodies, the bodies don't orbit around the three-way barycentre. And a planet placed near the barycentre of two stars will not tend to remain in orbit around that point.
The Lagrangian point L1 is also at best metastable. Satellites that orbit the sun at the Earth-Sun Lagrangian point need to fire their engines and do regular "station keeping" to keep them from drifting away.
The L4 and L5 points can be stable. Bodies at the L4 and 5 points are called "Trojans".  However no Trojan exoplanets are known.  A Trojan planet would see the two stars separated by (a variable amount averaging at) 60 degrees


This answer in math.stackexchange provides the correct relationship between $\theta$ and $\phi$. And this answer supplies the proof.


Is it possible that Venus was once a rogue planet?

This is considered very unlikely. It is actually very difficult for things to be captured into orbits. They pick up speed as they fall in towards the larger object, and that's automatically enough speed to carry them back out. Capture either requires dust and gas to slow the body down, or very lucky gravitational interaction with a third object.

This has happened with moons, but the solar system is much more crowded than the galaxy. For something the size of Venus to pass through the solar system at all would be exceedingly rare.

Finally, it is easy to explain the retrograde rotation by other means. This is where the other question that James K mentioned in the comments comes in.


You are seeing them in color, it's an issue of the mechanics of your eyes.  Eyes are made of rods and cones, cones seeing color, rods seeing greyscale.  When you are looking through a telescope, your eye adapts to the low light conditions and the rods dominate the sensitivity.
Jupiter appears much larger than both Mars and Venus, so your eye does not adapt as much and the cones relay more information, allowing you to perceive more color for both it and Saturn (which is also larger than both Venus and Mars in our sky).
That said, Mars through a telescope, to me, appears a bright orange.  Venus, to me, is indistinguishable from white, though it should look more of a pale yellow since that is its color in visible light.
You could try looking when the sky is darker, for Mars is visible in the middle of the night.  It's possible that the blue of the sky is affecting your color perception, too.  Venus is always a morning or evening object, so you should try looking at it just a little after it has risen (looking at it just as it rises will result in atmosphere issues).


Is Mars expected to go through the tail of Comet Siding?

Short answer: Sort of

Long answer:

One week before the encounter, Wikipedia says no. The comet's nucleus will pass by Mars at a distance somewhere of about 139,000 km from the center of Mars - way too far for any predicted collision. The main tail of the comet, too, will most likely miss Mars by about "10 Mars diameters" - roughly 64,000 km. However, small quantities of dust particles could (and most likely will) stray away from the main tail and hit Mars, Deimos and Phobos. This graphic gives a good picture of the encounter:

However, Mars may pass through the comet's coma and be hit by any dust particles there.

It will certainly be observable from the Martian surface: The comet will have an apparent magnitude of -6 at its peak. It will be quite a show, but nothing extraordinarily special.

The damage to orbiting spacecraft around Mars will likely be minimal. 90 minutes after the comet passes Mars, the worst of the dust particles will hit the spacecraft. The tiny barrage will only last about 20 minutes, though, and there is a low probability of damage. In fact, the various orbiters will take pictures of the comet and try to analyze it to learn more about its properties.

I hope this helps.

Other sources:

http://mars.nasa.gov/comets/sidingspring/

https://en.wikipedia.org/wiki/Mars

Update:

As of about an hour after the encounter, Mars still exists. Read this article for some not-so-juicy details.


Is the Hubble Ultra-Deep Field image representative of the entire sky?

Short answer: both.

HUDF is representative on the type of galaxies we can find at that distance (and remember that distance also means time ago).

On the other hand, we do not really know how the galaxies are grouped beyond certain level, but we think they are grouped on so-called filaments, leaving enormous empty voids.

In this sense, HUDF's density does not represent the entire sky.

http://cosmicweb.uchicago.edu/filaments.html


Reason for a correlation between Hot Jupiters and higher metallicity in Kepler data

This is a well known, well researched phenomena. Yes, there certainly is a correlation between metallicity and the likelihood of observing a hot Jupiter.

There are two classes of explanation.

(1) The correlation is real and due to the fact that it is easier to form planetary embryos from metal-rich material in the core-accretion model of giant planet formation.

(2) The observed exoplanets are those that weren't swallowed. The hosts are metal rich because they did swallow others.

Finally, it was thought possible that there could be an observational bias. Planets are easier to find and measure around high metallicity stars. This applied mainly to the Doppler technique, because high metallicity stars have stronger spectral lines. Transit detection should be much less affected.

Most think that some form of (1) is going on. (2) is argued against because the correlation still appears in stars with a wide variety of convection zone depths into which planetary material would have mixed. I'm not sure it is dead though, there have been a number of claims for abundance enhancements that look like the product of planet engulfment.

I'll add some references later. Nice work though to discover the correlation if you knew nothing of it previously.


If  you were in deep space, i,e., a spot somewhere between galaxies, then the night sky would look a lot like the sky seen by astronauts in the ISS, when they look away from both the ecliptic plane and the galactic plane, minus the individual Galactic stars.  That is, it is like our deep sky photos in those directions that are dominated by galaxies, but no atmospheric distortions, no night sky lines, and no zodiacal light.  The spectrum is dominated by different things at different wavelengths, but basically it is the sum of the light of galaxies and perhaps intergalactic stars that escaped their galaxies.  At long radio waves, it is dominated by supernova remnants and AGN as self-absorbed synchrotron radiation, peaking between 1 and 10 Mhz (Simon 1977).  The mm band is dominated by the microwave background radiation at 2.7 C. The far-ir is dominated by cold galaxy dust at 60 microns. The optical is dominated by galaxy thermal starlight at 1 micron.  The UV is dominated by hot stars at 0.2 microns (see summary).  The Xray spectrum from 3 to about 45 keV (Marshall et al 1980, Fabian 1992) is well-fitted by a ~40 keV thermal bremsstrahlung model. Above 45 kev the spectrum falls and is thought to be the sum of many AGN at all redshifts.  The spectral energy density (SED) from radio to X-ray is shown in this diagram from the paper by Hill et al. (2018) The Spectrum of the Universe:

The width of the line in the diagram shows the present uncertainties in the measurements.

Dark energy does not shine, it has an effect on the redshift as a function of time and thereby indirectly affects the shape of the background spectrum.


Once a black hole is formed, is there anything other than Hawking radiation which shortens its life?

It can merge with another black hole, if that counts. A simulation of that is illustrated here

That aside, no black hole with decay more quickly, as far as I can find out, than the "normal" decay of a non-rotating, uncharged black hole into which nothing is falling. Let's call this the "standard rate of decay".

"Feeding it" (allowing any matter or energy to fall into it), will reduce the decay below the standard rate, or even reverse it. Since the CMB pervades space and will fall into the black hole, blocking this (for instance by surrounding the hole by an opaque refrigerated shell) will be necessary to allow it to decay at the standard rate. Indeed for a black hole of mass over roughly $10^{22} kg$ the CMB would feed it faster than it decays, if not blocked.

Energy from the spin of the  black hole can be extracted using the Penrose process. Since this answer shows that rotation also reduces the emission of Hawking radiation, you can clearly speed up the decay by extracting energy from as much of the spin as possible, but the fastest decay you get is, once again, the standard rate.


You will need to find out what population II
 stars are.

When you understand what a population II star is, ask yourself two questions about each statement:

For example a) "Are population II stars out of the plane of the galaxy, and do they receive more cosmic rays", and "Would receiving more cosmic rays actually prevent life from forming". If you can answer yes to both questions, you have a potential answer.

If you judge that several answers are possible, then re-read the definition of a population II star, and ask which statement is most pertinent to the definition of such a star. Good luck.


Why do Seyfert Galaxies emit less energy than Quasars and Blazars, and how is the emitted energy measured?

The linked website seems to summarise generally accepted theory.

Seyfert galaxies appear less energetic than quasars or blazars. The energy is measured by observations of their brightness in the sky (at different wavelengths of light) combined with their distance (inferred from their red-shift). There are also Radio-galaxies

We think that active galactic nuclei don't release the same amount of energy in all directions, instead they have a "beam". They also have an accretion disc, the inner parts of which are extremely hot If we happen to be looking directly down the beam, they appear exceptionally powerful, and we call them "blazars" If we see them off centre we can see into the extremely hot central part of the accretion disk and they will be a quasar or (if less powerful) a Seyfert galaxy, and if we see them side-on they appear less powerful, but the two lobes of high-energy gas and particles emitted along the beam will be visible, and we will observe a two-lobed radio galaxy.

See these related answers: Quasars and SMBH and Difference between quasar and Active Galactic Nuclei?

http://www.astronomy.ohio-state.edu/~ryden/ast162_9/notes37.html


What would happen if a Black Hole and White Hole Collided?

From Wikipedia:

  In quantum mechanics, the black hole emits Hawking radiation and so can come to thermal equilibrium with a gas of radiation. Because a thermal-equilibrium state is time-reversal-invariant, Stephen Hawking argued that the time reverse of a black hole in thermal equilibrium is again a black hole in thermal equilibrium. This implies that black holes and white holes are the same object

The excerpt above was an argument made by Stephen Hawking, explaining that a white hole and a black hole are essentially the same object. With this in mind, we can assume that the collision will be between two black holes. The result will then be the radius of the black hole added to the radius of the white hole.


Calculating the orientation of the night sky

If you do not require arcminute precision, you can approximate Greenwich sidereal time as the Earth rotation angle $\theta(t)$.
IERS Technical Note 32 §5.4.4 gives
$$ \begin{align}
   \theta(t) &= 2 \pi (0.77905~72732~640 + 1.00273~78119~11354~48~ t) \\
      &\approx 280.46^\circ + 360.985612^\circ~ t
\end{align} $$
where $t$ is a real number of days since JD 2451545.0 (2000-01-01 12:00 TT ≈ 11:59 UTC).

For an observer at north latitude $\phi$ and east longitude $\lambda$, things should line up like this:

Local sidereal time $ = \mathrm{LST} \approx \theta(t) + \lambda$

Zenith (RA, Dec) $ = (\alpha, \delta) = (\mathrm{LST},~ \phi)$

North horizon $(\alpha, \delta) = \begin{cases}
   (\mathrm{LST + 12h},~ 90^\circ - \phi) & \mathrm{if}~\phi >= 0 \\
   (\mathrm{LST},~ 90^\circ + \phi) & \mathrm{if}~\phi < 0
\end{cases}$

East horizon $(\alpha, \delta) = (\mathrm{LST + 6h},~ 0^\circ)$

The transformation between equatorial and horizontal coordinates can be composed of two rotations, similar but not necessarily identical to those in Wikipedia: Celestial coordinate system.
For example, you could:

Start with the north horizon vector pointing at the north celestial pole, and the zenith vector pointing at (0h, 0°).
Rotate the ground counterclockwise as seen from the north horizon vector by LST.
Rotate the ground clockwise as seen from the east horizon vector by $\phi$.


How did Kepler come to the potencies in his Third Law?

If you plot the log of the period against the log of the semi-major axis then it is obvious that $P^2 \propto a^3$. Any other power law relationship simply wouldn't fit.

The following passage (from https://www.mathpages.com/rr/s8-01/8-01.htm ) seems relevant:

Is it just coincidental that John Napier's "Mirifici Logarithmorum Canonis Descripto" (published in 1614) was first seen by Kepler towards the end of the year 1616? We know that Kepler was immediately enthusiastic about logarithms, which is not surprising, considering the masses of computation involved in preparing the Rudolphine Tables. Indeed, he even wrote a book of his own on the subject in 1621. It's also interesting that Kepler initially described his "Third Law" in terms of a 1.5 ratio of proportions, exactly as it would appear in a log-log plot, rather than in the more familiar terms of squared periods and cubed distances. It seems as if a purely mathematical invention, namely logarithms, whose intent was simply to ease the burden of manual arithmetical computations, may have led directly to the discovery/formulation of an important physical law, i.e., Kepler's third law of planetary motion. (Ironically, Kepler's academic mentor, Michael Maestlin, chided him − perhaps in jest? − for even taking an interest in logarithms, remarking that "it is not seemly for a professor of mathematics to be childishly pleased about any shortening of the calculations".) By the 18th of May, 1618, Kepler had fully grasped the logarithmic pattern in the planetary orbits:


How was the core temperature of the Sun estimated?

Hydrodynamic models of the Sun allow one method of estimating its internal properties. To do this, the Mass, radius, surface temperature, and total luminosity (radiative energy emitted)/s of the Sun must be known (determined observationally). Making several assumptions, e.g., that the Sun behaves as a fluid and that local thermodynamic equilibrium applies, the stellar equations of state can be used. Numerical methods are applied to these equations to determine the internal properties of the Sun, such as its central temperature.

A great example for how to work this problem your self can be found in the undergraduate text, 'An Introduction to Modern Astrophysics' by Carroll and Ostlie (Section 10.5). The FORTRAN code to run your own stellar model is included in Appendix H.

A comprehensive review paper on how stars of different masses evolve internally (e.g., with respect to T, P, etc.) that is worth reading is:
http://adsabs.harvard.edu/abs/1967ARA%26A...5..571I

A very interesting historical overview of the development of the Standard Solar Model:
http://arxiv.org/abs/astro-ph/0209080

This (admittedly dry) paper gives you a good idea of how well the 'standard' solar models estimate the internal properties of the Sun using helioseismology and neutrino measurements to help tie down their boundary conditions:
http://adsabs.harvard.edu/abs/1997PhRvL..78..171B
The answer is that they match incredibly well (>0.2% error)

These were the least technical (but still academically published) references I could find.

Here is a comprehensive page on the state-of-the-art in solar modelling and measuring the internal Sun using Helioseismology:
http://www.sns.ias.edu/~jnb/Papers/Preprints/solarmodels.html
(highly technical)


Will the TESS telescope be much better than the Kepler telescope?

Rather than one being better than the other, I would say that they have different targets.

Whereas Kepler focused on a specific area of the sky, TESS will cover an area 350 times bigger.

Also, most of the exoplanets discovered by Kepler are far away; TESS will focus on stars within 300 light years away.

For more info, feel free to watch this video I recently made: https://www.youtube.com/watch?list=PL3RiFKfZj3pv1ZqpFxuZinoGtUGEOankw&v=7bsxlG-rh1E


Assuming that this is a serious question... yes, Oumuamua will be tumbling, just like every other small body.

That it comes from outside the Solar System does not change that it will have three axes of rotation because it is small and thus has not accreted enough matter that it would stabilise into a single primary axis. (This doesn't preclude one of its axes being a primary spin axis.)

And just like every other small body, if we collect enough data in the form of reflected sunlight from its surface, we can construct shape and rotation models and estimate its size and composition.


How to view the surface of Mars with a 130 mm Newtonian?

Keep in mind that you have a 650mm telescope and Mars is small.  While it is currently appearing larger because it is near opposition, it is still only 22 arc-seconds wide.
Here's a simulated view of what you should see using a 650mm f/5 telescope with a 10mm Plössl type eyepiece (51° Apparent Field of View - chosen because Celestron offers that so I suspect it may be the eyepiece included with your telescope):

It will be difficult to see much detail at that size even if seeing conditions were perfect.  Add in typical seeing conditions and it becomes even more challenging.
You could use a 2x barlow or a 5mm eyepiece and that will make it a little larger ... but not much.  Using a higher power barlow or a shorter focal length eyepiece (shorter than 5mm) will likely only result in a fuzzier image that doesn't resolve much detail.  My experience is that average seeing conditions will let you increase magnification to the 1x of your telescope aperture in millimeters (e.g. 130x) and it turns out that an eyepiece with a focal length that matches the telescope's focal ratio will do that (in your case, a 5mm eyepiece ... or a 10mm eyepiece with a 2x barlow).  Going beyond that (to the theoretical 2x aperture diameter limit or 260x in your case) is exceptionally rare and requires outstanding seeing conditions.
@AaronF offered suggestions to try to avoid factors that degrade seeing conditions (e.g. if you live in an area where it is currently warm during the daytime, viewing over rooftops or asphalt, etc. where a considerable amount of heat is still radiating from those surfaces will degrade the visual quality.  But this can be improved by relocating your telescope.


Why does earth not collide with the sun?

The primary reason the Earth doesn't fall into the Sun is that it has a very large tangential velocity with which it is able to maintain an orbit. The physics is the same for describing satellites which we launch into orbit around the Earth. I've heard it described as such: "The ISS is falling towards the earth, but is moving forward so quickly that it continually misses."

In short, if you were to take a mass and insert it with zero velocity somewhere in the solar system, it would inevitably fall in towards the sun. But because almost all bodies in our solar system were formed from the same large clump of rotating gas and dust which due to conservation of angular momentum during its collapse both sped up and flattened into a disk, all of the solar system bodies (Earth, other planets, asteroids, etc), have enough forward velocity to continue orbiting the Sun without being pulled to the center.

From a physics perspective, you can see that centripetal acceleration is governed by the inward force of gravity, tangential to the direction of motion of moving objects.

In fact, the Solar Parker Probe which just launched is attempting to do exactly as you describe --- fall into the Sun (or well, get close, to take lots of measurements). To do so is actually difficult, because the probe, launching from Earth, has to lose roughly 30 km/s in velocity it already has (because that's the Earth's orbital velocity).

How it plans to do this is a cool story in and of itself, which you might be interested in, here's a good summary: https://www.theatlantic.com/science/archive/2018/08/parker-solar-probe-launch-nasa/567197/


Calculating the Solar Day of an Exoplanet?

This source doesn't give sufficient information to get the length of either a stellar or sidereal day.  The rotation rate of a body is not governed by the orbital parameters.  The Earth, for example, has a slowing spin rate as the Moon retreats due to tidal interactions.  This doesn't affect the Earth's orbital parameters.
The fields given are:
Name
Planet Status
Discovered in
Mass
Mass*sin(i)
Semi-Major Axis
Orbital Period
Eccentricity
ω
Tperi
Radius
Inclination
Update
Detection Method
Mass Detection Method
Radius Detection Method
Primary transit
Secondary transit
λ
Impact Parameter b
Time Vr=0
Velocity Semiamplitude K
Calculated temperature
Measured temperature
Hottest point longitude
Geometric albedo
Surface gravity log(g/gH)
Alternate Names


Error when calculating Alt/Az from Ra/Dec

The algorithm on that page is very simplistic and is probably the source of a good portion of that error.  It's best to test your code using a very accurate ephemeris, even if your final goal is to have a much more simple ephemeris implementation.  DE405, or VSOP87 would be good alternatives.

A small error in GMST can have a big effect on Alt/Az computations, I would expect that the simplicity of the algorithm on the page produces a good sized error in GMST.

Another common source of error is differences in time scales.  You need to make sure you propperly account for things like leap seconds, which that page specifically says it ignores.  You need to convert UTC to Terrestrial Time by subtracting 37.0 + 32.184 (where 37 is the current number of leap seconds).

Also make sure you're not comparing different coordinate systems.  Like J2000 vs "Of Date", or coordinates adjusted for atmospheric refraction.


Could the solar system survive in a void?

If you took away the Milky Way except for the Solar system (but kept the MW's satellite galaxies), the  "naked eye" night sky would feature the Andromeda Galaxy, 2.5  million light-years away, the Large Magellanic Cloud (LMC), 160,000 light-years away and the
Small Magellanic Cloud (SMC), 200,000 light years away.


Which measurements have shown that 2014 MU69 is red? (New Horizons' next target)

This was a hard one to answer, primarily because of the difficulty in tracking down information.
The Observations of MU69
The extended mission for New Horizons involved adjusting its orbit to do a close fly-by of a Kuiper Belt Object (KBO). The annoying part about this was that they needed to find one first! Ground based observations were unsuccessful due to the faint magnitudes of any potential KBOs and the crowded star field (being that they were looking towards the galactic core). Finally, HST was used and over the course of 194 orbits, between June to September 2014, the entire region of the potentially reachable sky, by New Horizon's designated $130\:\mathrm{m/s}$ delta-v, was scanned using WFC3 instrument and the F350LP filter. Image processing revealed 5 potential candidates. This is all laid out in a paper by the lead of the KBO search, John Spencer. You can choose to view any and all images taken during these observations (since it is in the public domain) by downloading them from Hubble's Legacy Archive. Of course you'll have to do some digging in the 970 total images to find the actual images which contain MU69.
Once the entire region was scanned and a set of candidates were found, follow-up observations were made in late October 2014 of each individual candidate to precisely determine orbital parameters and observational characteristics so as to narrow down which candidate was reachable and presented the most scientific advantage. In this case, the same observing group from New Horizons took the images and once again used Hubble's WFC3 and the F350LP filter. Once again, these images can all be found on Hubble's Legacy Archive. An example of a single image which from this run, containing MU39 (although you'd be hard-pressed to pick it out) can be seen here. Most of this information can be gleaned from the Hubble Legacy Archive and is also described in this paper.
The Minor Planet Catalog (props to AtmosphericPrisonEscape for pointing this out) indicates that further observations of MU69 were taken with the HST in May and July of 2015. I had a bit of trouble tracking these images down, but knowing that John Spencer was the lead in these observations, I went and looked up his 2015 HST proposal which you can read here. With the proper proposal number, I found the set of 15 images during May and July which were taken of MU69. Again, the WFC3 instrument was used with the F350LP filter.
The Color of MU69
In total, we have 38 images of MU69, all taken by John Spencer and his team using the HST's WFC3 instrument and the F350LP filter (that peaks at 5845.871 angstroms). The important point here, and the reason I detailed the observations so much in the previous section, was to determine the exact type of images taken of this KBO, so as to ascertain what science data could have been used to determine color.
Based on everything I've compiled, it appears to me that the statement of MU69 being "red" is a educated guess. Anyone who knows about telescopes will tell you that imaging an object in a single filter is not sufficient to reliably measure aspects such as color. You need at least two filters to determine that because you need to understand how the brightness of said object varies with wavelength.
However, it could be possible to make a guess at the color of this body from a single filter by making a lot of assumptions. This is purely a guess as to what they did, but you can first measure the apparent magnitude of this object in your filter, then, assuming things like size, albedo, distance, etc., you can create a model of this object's spectrum that you'd expect it to have such that it reproduces your observed image. The first paper I linked details how they did inject false KBOs into their images to ensure their KBO finding algorithms worked which indicates they had some basic ability to model KBO images in their data. If your model can reproduce the observations, you can use that model as a basis for what your KBO is likely like and from that you can say that it appears "red". Based on experience with KBOs and in partiuclar pluto, we expect they might be red and that the redness corresponds to tholin, a common, red chemical found on KBOs, so the fact that they believe it to be read is not surprising.


what is the difference between luminosity distance and distance?

The luminosity distance is the apparent distance of an object based on how bright is it and how bright it appears to be.  If you have two stars of equal luminosity, the more distant one will appear dimmer.
For local objects, the intensity of the light will obey an inverse-square law. So, for local objects (stars) the luminosity distance is equal to the actual distance.
But for very distant objects, like quasars or distant supernovae, the curvature and expansion of space-time will also affect luminosity. The luminosity distance is the distance obtained by ignoring these factors and just assuming that the intensity of the light from a distant object obeys the inverse square law.
As an example: a type 1a supernove is observed to have a magnitude 18.7, and is assumed to have an absolute magnitude of -19.3
It's luminosity distance in parsecs is $$10^{(18.7+19.3)/5+1} = 398 \text{Mpc}$$
If it has a redshift of z = 0.086 then its co-moving distance (what you would measure with a ruler) is 398/1.086 = 366 Mpc.


Why are quasars so far away that they couldn't be optically resolved in the 1950's?

Quasars have a very low density at the present epoch, but were much more common in the past. Their co-moving space density peaks at redshifts of 2-3.

The evolving space density of bright quasars. From https://ned.ipac.caltech.edu/level5/Madau/Madau7.html.
The fact that the quasar population density evolves was one of the key arguments against steady state cosmology.


Looking for a Solar System Simulator including the smaller celestial body, for macOS

Celestia can run on MacOS. With is, you can travel in the solar system and in time. It can give you a good idea of how far things are from each other.


Presumably you're supposed to start by calculating the minimum orbital period a comet can have for a 10,000 AU aphelion distance.

Then divide that period (number of years) by five, assuming we see on average one of these every five years.

That will give you an estimate for how many there are out there.

(I would also tell the person setting the question that there are some pretty wild assumptions implied in the question!)


What is the current accepted theory as to why Mercury, despite its size, has a similar density to Earth?

The most-widely accepted hypothesis at the moment is that Mercury was struck by a large impactor that removed a significant fraction of its mantle (I believe this theory was originally proposed by Cameron & Benz in 1987, and the qualitative theory hasn't changed very much). For planets that are close to their parent stars (such as Mercury), the collision with the secondary body likely occurs at high velocity due to the rapid orbit speeds there (> 40 km/s). As this is much larger than a typical rocky planet's escape velocity (~10 km/s), material is lost from the planet in these collisions. This collision is also more likely to be "grazing," as "direct hits" are rare, and thus the outer parts of the planet are preferentially removed.

Because the mantles of rocky planets are primarily composed of silicates, which have a lower density than iron that resides in their cores, the collision results in a surviving planet with a higher average density.


What is the exact time Earth takes to revolve around its axis?

Our day is 23 hours and 56 minutes long, and slowing by an infinitesimal (but measurable) amount each year due to tidal losses.

Our day has a connection with the weather, in that the sun drives all our weather systems, so heating over each part of the globe happens every day, but aside from that, your question doesn't make much sense.

Weather changes may come at the same time where you live (on a 24 hour cycle) but here in Scotland, we still have low reliability on even a 3 day weather forecast, because the weather systems that impact the UK are so complex as heating from the sun drives various air flows.

After your update, I still cannot understand what you mean. There is slippage in accuracy, but this is counteracted by leap seconds and leap years. It has nothing to do with weather. The shortest and longest days happen when they happen and are measurable. They help us define the year.


How is Comet SWAN C/2020 F8 doing? Will it become easily visible by eye?

The SWAN C/2020 F8 comet is passing near Earth, but its magnitude will be around 5. This means that you need to have a rather dark sky and good enough eyes to be able to see it.

From the comet observer's database

In addition to this, the comet will be very low on the horizon during the night. Here is its track from Wikipedia

Early June (when it will be rather bright), SWAN will pass near Capella which is roughly 15-20° above the North-West horizon at 22:30. This means that you need to have a pretty clear horizon to see it.

TL;DR : It will be visible by the naked eye if you have a clear horizon and a dark sky. So it won't become easily visible by eye at any point in time during this pass near the Sun.


Is gravitational lensing a good way to search for negative mass?

One reference I found on this subject was Izumi et al. (2013), who note that the images produced by lensing around a negative mass are radially distorted while lensing by a positive mass are tangentially distorted, and that this effect can be distinguished without prior knowledge of the lens position by measuring the alignment of lens images.

I am not an expert in general relativity so I'm not able to evaluate their results, but assuming the paper is correct it does look like negative mass could indeed be detected by observations of gravitational lensing. The question of whether negative mass exists is another matter, so far it doesn't look like a likely prospect.


How to orient ESO's Milky Way Panorama in a 3D model

Here is a star map (from http://www.datasync.com/~rsf1/fun/sm-newd.gif ) given in Galactic coordinates. It appears to be presented in a similar (but not the same, since it is rectangular) projection to the one you show above.

The Galactic centre is in the middle and the Large and Small Magellanic clouds are to the lower right and oriented in the same way.

As you can see from this map, your assumption is completely correct. Ursa Major and Polaris should be situated towards the upper left hand side of the picture. Conveniently it should be on the opposite side and roughly equi-distant from the Galactic plane as M31 (which is clearly visible in the image) and at a similar (slightly lower) Galactic longitude.

I have added a copy of your image at the bottom and marked where I think Polaris is. I suggest you get hold of the higher resolution image available from Serge Brunier if you want to pick out individual stars. The way you transfrom this into equatorial coordinates for different times of the year is not as simple as you suggest.


Can we use Breakthrough Starshot spacecraft to detect any unknown massive object in our solar system?

This is an interesting question but the answer is that this is not feasible nor efficient at all.

You can regard each photon of light coming from outside the Solar System acting as a probe. We can measure the deflection angle of them by analysing the gravitational microlensing effect a planet would produce on them. For that you need two things; a planet massive enought to be spotted by this effect (predicted mass for Planet 9 is in place) and enough photons coming from different directions to have a chance of some of them passing close to Planet 9. And here's the thing: the net flux of incoming photons from all the stars and galaxies from all directions into the Solar System is still not enough to have even the slightest chance of some particular ray to be deflected or blocked on its way to an observer on Earth that would be just looking in that precise moment in that precise direction (as Planet 9 misaligns with that particular star from which the photons are coming after a few minutes). This could be changed by continuous observation of the entire sky, so maybe when LSST opens in 2022 we might have a chance to apply this method of detection.

So, if all the photons coming to Earth from outside the Solar System could be used as natural probes that might show Planet 9 by microlensing or occultation, immagine how puny would humanity's effort be in comparison if we tried to launch even just a fraction of a millionth  of the amount of photons currently traveling towards us from all directions.

There's also the inverse method (which is more similar to the 'launching probes to the outside' idea): photons coming from the Sun (which are a lot more than those coming from the stars and are been "launched" in all directions with higher coverage of each square degree of sky) could hit Planet 9. We need the information of those "hitting probes" to get to Earth and this is difficult, many of our solar photon probes are going to be reflected in directions unaccessible to us and many more are going to be absorbed by the atmosphere and surface materials of Planet 9. But since the Sun is emitting so many photons a few percent might actually come back to the inner solar system in just a few days. Those reflected photons are our best chance and indeed they are the ones we are searching for when we try to find direct evidence of Planet 9.

No human attempt to launch trillions of probes to the region of the sky we expect to find Planet 9 (even if we manage to reduce it by many orders of magnitude) will be as productive as the Sun launching so many photons continuosly and Planet 9 reflecting them back to Earth. To compete with that we would probably need more mass than the mass of the entire Solar System to make the tiny probes and no gain at all compared to the reflected photons that are in place right now.

I guess there is a minor detail we could talk about: reflected photons have to actually hit the target and our tiny probes could just pass close to it and signals would reveal, by some Doppler shift, a change in speed and thus the gravitational influence of the body. So actually we have to talk about the cross-section of the planet in a different way in both situations. For the photons coming from the Sun to be reflected back the considered cross-section would be the cross-area of the planet which for a planet the size of Neptune is $1.9 \cdot 10^9 \; km^2$. In the case of the nano-probes the cross-section can be expanded to the region where the gravitational influence of the planet ceases to be sufficiently large to perform a detectable change in speed for the tiny probe. I'm not going to calculate this but let me make a liberal guesstimate by using the expected Hill sphere for Planet 9, which is $15 \; AU$ in radius and thus would make for a cross-sectional area of $1.6 \cdot 10^{19} \; km^2$. That might be great for having some chances but the probes would have to be moving extremely slow for the gravitational influence to be able to operate for enough time as to make the deflection be apparent, and this is a constraint that makes the project again unfeasible since our probes would be moving at some percents of the speed of light (if not then we would have to wait centuries or even millennia for the experiment to be completed).


Have any more "white dwarf pulsars" been discovered or searched for?

Edson et al. (2017) list two other candidates that may be white dwarf pulsars.

AE Aquarii has been described as pulsar-like (e.g. Ikhsanov 1998) and was the best case for a white dwarf pulsar before AR Scorpii came along. From what I understand, the case for AE Aqr containing a white dwarf pulsar is not watertight. Blinova et al. (2018) explain the high spin-down luminosity as being the result of interactions with an accretion disc and the white dwarf having a magnetic field typical of that in intermediate polars, rather than a pulsar-like mechanism.

Zhang & Gil (2005) proposed that the radio source GCRT J1745-3009 is caused by a white dwarf pulsar. As far as I can tell, the case for this one isn't particularly secure either because there are other possible explanations. Follow-up observations have not yet confirmed the nature of the source.

I'm not aware of any specific surveys for such objects but so far it seems that AR Scorpii is the best example of a white dwarf pulsar.


Can close binaries have a very eccentric orbit?

There are surprisingly many examples of short period, eccentric binry systems.

A very incomplete list would include

KIC4544587 P=2.19 d  e=0.288 +/- 0.026 Hambleton et al. (2013)

HD174884 P=3.66 d  e=0.2939 +/- 0.0005 Maceroni et al. (2010)

CoRoT 102918586 P=4.39 d  e=0.249 +/- 0.005 Maceroni et al. (2013)

HD313926 P=2.27d  e=0.209 +/-0.001 Rucinski et al. (2007)

In each of these cases, the stars are relatively warm, with thin (if any) subphotospheric convection zones. Tidal circularisation is much more effective in stars with thick convection zones (giants, or cooler main sequence stars). Therefore it is possible that all of these objects (and, as I say, there are many other examples) are too young to have circularised.

Some theory is presented by Zahn et al. (2005). Indeed, circularisation times are very short - basically you expect everything with a solar mass or below to be circularised if its period is less than 6 days, but the process is less efficient for higher mass stars. Circularisation will essentilly take place during pre-main-sequence evolution. Since a circular orbit represents the minimum energy configuration, the only way you are going to get non-circular orbits in solar-type and lower mass stars is if they are very young (less than some tens of millions of years), the binary system formed after their pre-main-sequence evolution or if there is an injection of energy via a dynamical interaction or an eccentric third body in a much wider orbit.


A comet orbits the Sun in the same plane as Earth's orbit

I'll just give an outline about how I might try to answer these questions, rather than answering them outright.
Answer the first question using the circular form of the Vis-Viva equation: $v=\sqrt{\frac{\mu}{r}}$, where $\mu$ is the standard gravitational parameter ($4\pi^2 AU^3/yr^2$) and $r$ is 1 AU.  If your answer isn't $2\pi$ AU per year, you probably did something wrong with the units.
For the second question, try to determine the shape of the comet's orbit using the inequality for escape velocity: $v_e \ge \sqrt{\frac{2\mu}{r}}$.  Once you have the orbit shape, the rest of the problem falls out pretty easily using known equations for that orbit.

 Escape velocity at 0.5 AU is
 $$v_e=\sqrt{\frac{2*4\pi^2}{0.5}}=\sqrt{16\pi^2}=4\pi$$

 which is the exact speed given in the question. That means the orbit
 is parabolic.  From  Bogan's table pulled from
 Bate/Mueller/White's Fundamentals of Astrodynamics, for a parabolic
 orbit, $$r=\frac{2q}{1+\cos{\theta}}$$ where $r$ is the distance from
 the central body, $q$ is the closest distance of the orbit, and
 $\theta$ is the angle from periapsis or true anomaly. Solving gives us
 $\theta=\pi/2$.

 The same table gives us the angle of velocity relative to the
 perpendicular to the radial direction: $$\phi = \theta/2 = \pi/4$$

 So the orientation of the comet's velocity is at
 $\pi/4$ or 45 degrees off a tangent line to a circle around the Sun
 with radius 1 AU as the comet passes through it.


If we had the right technology could we see a distant star in detail?

The only limitations would be related to building an instrument large enough.

There's a limit to the size of the finest detail a telescope can see. "Size" here is angular size, the angle that the detail is covering in the visual field. For a given wavelength of light, the smallest angular size depends on the aperture (diameter) of the telescope in a linear fashion: double the aperture, and you cut the angular size in half.

Rule of thumb: 100 mm aperture has a resolving power of 1 arcsecond, for visible light (not exactly, close enough). Therefore 200 mm aperture has 0.5 arcsec resolving power, etc.

By building larger and larger telescopes, the angular size of the smallest details becomes smaller and smaller. But are there any limits here?

There aren't any cosmologic mechanisms that I'm aware of that would prevent telescopes from operating at arbitrarily large resolving powers (arbitrarily small angular sizes). But we can't build arbitrarily large instruments.

The biggest monolithic (single mirror) telescope has an aperture of 8.4 meters. Using segmented mirrors, the aperture aim of the current largest project is 30 meters - the Thirty Meter Telescope, still in a very early stage, completion date unknown.

Beyond that size, large scale interferometers are the only option, with synthetic aperture varying from dozens of meters to kilometers, and larger projects are being discussed.

In theory, very large scale interferometers could be built in space. But there might be inherent limits for the aperture of such systems. Anyway, this is hypothetical.

EDIT: Giant telescopes in general do not require a continuous surface of glass. Interferometers almost by definition do not use a single reflector. You could build a gigantic telescope or interferometer where the active reflective surfaces are just these tiny chunks at the periphery of a huge perimeter. Most of the area in between is empty / unused.

EDIT2: The performance of the "tiny chunks" reflector is not the same as the performance of the full size mirror, as mentioned in the comments.


Moons are not always tidally locked.

They are formed with a certain spin angular momentum that will be dragged by asymmetries on the hosting planetary body. Those asymmetries then backreact onto the moon and drag it's spin to become synchronized with the orbit angular momentum.
How rapidly this process happens is roughly a function of the tidal forces
$$F_{Tid} = G \frac{m_{Moon} \cdot m_{Planet}}{r^3} d$$
exerted by those asymmetries. Here $d$ is the moon size, $r$ the distance between planet and moon and the other quantities are masses and the gravitational constant.

The time $t_{lock}$ this locking takes can be estimated by comparing the excess angular momentum $L_{ex} = r^2m_{Moon}(\omega_{actual} - \omega_{orbit})$ with those acting forces:
$$t_{lock} \approx L_{ex} / \left(r F_{Tid} \right) \approx \frac{r^5}{ d \cdot m_{Planet}}$$

The latter equation makes it fairly obvious that this is not a process that happens at a certain magical orbit or at a certain time. It is an asymptotic process, and the only coincidal thing is that we observe any given moon at a time where it may not have lost enough spin angular momentum yet.
This should hopefully also clarify some confusion from the comments:

Yes, the tidal locking time is also dependent on the mass of the host-planet, but much more sensitive to the planet-moon-distance. Note that this derivation is independent of the moon's mass, and only dependent on it's size. The literature knows much more sophisticated variants of this, that also incorporate a (realistic) weak dependency on the moon's mass.
For Hot Jupiter exoplanets it is thought they'd be in tidal lock when the asteroseismologic age of the system is bigger than the derived tidal lock time.

Concluding
1.) Any moon would be tidally locked if we could wait long enough.
2.) There are much more effects / torques to consider for a detailed understanding than I did here.


Yes, comets spin although measuring it can be tricky due to the coma and outgassing from the nucleus. It's easiest to measure the rotation period when the comet is inactive near aphelion although this is more difficult as the comet is fainter, necessitating a large amount of large (>4m) telescope time which is difficult to obtain.
Searching the JPL Small Body Database (select Object Kind=Comets, set  Limit by object characteristics using rot_per (h) greater than 0, add rotation period (h) to the Output Fields) should produce a list of 21 comets with measured rotation periods:

The initial rotation is likely primordial since the pre-solar nebula the comets formed from would have some initial rotation from the galactic rotation which can be increased as the proto-comets contracted during formation. The spin rate of the nucleus can be changed by the torques caused by the activity-driven outgassing from the nucleus; this was observed for Comet 67P from in-situ measurements during the ESA Rosetta mission (Keller et al. 2015) and also have been observed to change in rotation from remote observations (abstract from Bodewits et al 2018 for Comet 41P)


One example is that of the core of a Thorne–Żytkow object, or TŻO. It consists of a high-metallicity M class red giant or supergiant with a neutron star at its core. Generally, TŻOs form when a neutron star collides with a suitable red giant/supergiant. The neutron star settles into the center of the other star.

Eventually, the main source of energy for the TŻO comes from gravitational contraction of the envelope as gas from inside the star falls toward the surface of the neutron star. In this case, the change in mass $\dot{M}$ of the neutron star is (see Thorne & Żytkow (1977)):
$$\dot{M}=4\pi r^2\rho v_{\text{in}}\mathscr{R}\tag{1}$$
The difference between $(1)$ and the underlying equation behind Bondi's formula is the factor of $\mathscr{R}$, a correction factor for relativistic redshift, which is given by
$$\mathscr{R}=\sqrt{1-\frac{2GM_{tc}}{c^2r}}\tag{2}$$
where $M_{tc}$ is the total core mass. This formula may look familiar; it comes from the Schwarzschild metric. You need $\mathscr{R}$ to properly describe accretion onto any compact object. This is something needed not just in the case of TŻOs. However, Bondi's formula can be used nonetheless, if you multiply by $\mathscr{R}$.

That said, Bondi accretion can be used without relativistic compensations in a number of ways:

Mass transfer in certain binary systems, where an envelope moves matter from one star to another.
Protostellar clusters, where many young protostars move about, accreting gas.
Galaxy clusters, where galaxies move through the intergalactic medium (IGM).


Help understanding this unsettling image of Titan, Epimetheus, and Saturn's rings?

This NASA page says this photo was taken on April 28 2006.

Using Celestia, I managed to find the picture from Cassini that best lines up with the photo. It doesn't match up precisely, but that's to be expected as the calculated orbital elements of all these moons (and cassini) in the software won't necessarily match up to reality precisely.

Below is the zoomed out version of this shot. You can see Titan in the centre and Epimetheus as a dot on top.

And here is the top down shot from Cassini to the moons. Circled are Epimetheus and Titan.

So to answer your question: Titan is really big compared to epimetheus (about 50x), Titan has an atmosphere and so appears fuzzy (It is actually in focus, everything in space is very far away and so is effectively at infinity for focusing purposes), and the rings are highly oblique, so you only see a small slice of them.


The idea of magnification is not relevant to astro-photography, what is relevant is the image scale.

The image scale depends on the focal length of the objective and the size of the sensor. The Barlow effectively modifies the focal length, so if you have a $B\times$ Barlow and an objective of focal length $f_O$ the effective focal length is $f_e=B\times f_O$.

If your sensor has side of length $l$ then the field of view that corresponds to that side is (in radian) $\theta_{rad}=l/f_e$ or in degrees $\theta_{deg}=(l/f_e)\times(180/\pi)$

So if your sensor has a short side length of $\approx 0.015 \mbox{m}$ and a the telescope has a focal length of $1.2 \mbox{m}$ and you are using a $\times2$ Barlow you have the short side of the image corresponds to a field of view of $\approx 21$ minutes of arc.


How to obtain distance modulus and Total extinction from Isochrone fitting to CMD?

The isochrones are not straight lines in the CMD. Extinction moves the stars both redward (right) as well as fainter (down) in the CMD. The vertical shift (the extinction $A$) is calculated from the horizontal shift (the reddening) using an extinction law. (e.g. $A_V = 3.1 E(B-V)$). The effects of extinction and distance (which only moves the stars vertically) can therefore be separated.
In the case of the paper referred to, the globular clusters have a horizontal branch and a giant branch. The horizontal branch is almost ...horizontal and the giant branch is almost vertical. This enables the authors to be independently sensitive to a vertical and horizontal shift of the model isochrones, as described above.
+3. There are simple conversion factors published in several papers. Extinction is wavelength dependent, so the spectrum of a star is attenuated in a wavelength dependent way  The reddening in any colour (e.g. E(B-V)) is just the difference in extinction at the two wavelengths (well it needs to be integrated over the photometric bands) i.e. $A_B - A_V$. These differences in extinction coefficients can be calculated for any pair of bands. They depend on the wavelength dependence of extinction, aka the extinction law, but also on the intrinsic spectrum of the star (which is frequently ignored). Extinction laws  can be empirically determined by observing stars with a known intrinsic spectrum.

'4. Does it matter? Have you compared a parsec with an astronomical unit?


How many tsunamis have been caused by meteorites falling?

None in historical times.
A tsunami is caused by a displacement of water, so for a meteorite to cause a measurable tsunami it would have to be very large.  The largest known impactor in historic times, in Tunguska, was about 50m diameter could have created waves, but probably not a tsunami.  There is evidence that a much larger impact, he Eltanin impact, 2.1 million years ago in the pacific ocean off Chile (a 1.5km diameter impactor) also did not create a tsunami.  The waves generated were short wavelength turbulent waves and the energy was lost in friction. Tsunami need to be very long wavelength to cross oceans and cause damage remotely.
Much larger impacts would create tsunami, and the size of the tsunami may exceed that created by earthquakes. See my source Are ocean impacts a serious threat."