There are two main theories for the formation of binary stars - one accepted, and one mainly deprecated.

The fission hypothesis. The fission hypothesis states that the binary system forms after the collapse of the original gas cloud into a protostar. Angular momentum is conserved, so as the extremely large cloud slowly contracts, it spins faster. After enough time, the protostar could form a dumbbell shape, one part of which eventually breaks off. There are now two blobs, and thus two protostars, which will evolve into a binary star system.
The fragmentation hypothesis. The fragmentation hypothesis states that the gas cloud fragments earlier on, due to some instability or cooling/heating effects. Each of the two fragments then evolves separately, forming a binary star system.

The fission hypothesis is no longer favored. It cannot explain the existence of systems with certain mass ratios or separations. The fragmentation hypothesis, however, can, and is accepted. It is also possible for a binary system to form through the capture of one star by another, but that requires a third body, and probably cannot explain the high fraction of binary systems in the galaxy.

Basically, then, the main theory to date is that the protostellar nebula breaks up, and then both fragments collapse independently.


Observational evidence for the existence of black holes

There are a few different features of black holes that researchers look for:

An accretion disk
Mass transfer from companion star
X-ray emission
Absorption of gas and other matter around it
Gravitational lensing

You mentioned #3 in your question; quite a few black holes, including Cygnus X-1, emit X-rays.

1 is a common feature around black holes. The material can be pulled from surrounding gases, or, in many cases, a companion star. Accretion disks generally emit electromagnetic radiation; the exact type depends on the type of material and the object it surrounds.

2 is related to 1 and 3, but it deserves its own explanation because of the effects on the donor star. Lets take mass transfer on a large scale. Take the Whirlpool galaxy. It is interacting with its neighbor, NGC 5195, via a bridge of dust and matter. What are the effects on NGC 5195? Distortion, as well as loss of structure in the area near the bridge. The same thing happens on the stellar level. A star losing mass can become distorted in the direction of its companion, potentially causing changes in brightness or other features.

4 is, I suppose related to the others, but I should note that if there were merely gas clouds surrounding the black hole, it could also swallow those up. If a black hole is not in a binary system, it may not come in contact with another star, and will instead begin swallowing other bits of matter around it.

5 is the best characteristic; a better way of detecting black holes. General Relativity predicts that massive objects bend light; using Einstein's equations it is possible to determine just how much light will be bent by such an object. This has been used in many cases, such as during the solar eclipse of 1919, to confirm Einstein's theory.

Finally, I should note two things:

These are not the only methods of detecting black holes (other effects, such as frame dragging around a Kerr black hole, offer distinctive descriptions of the black hole), but are rather straightforward
1-4 can also be signs of other compact objects such as white dwarfs and neutron stars, and are often only found in binary systems. Only by using a multitude of different observations, and careful analysis, can an object finally be designated a black hole.


What made cooler temperatures suitable for atom formation?

The temperature of a gas is a measure of the kinetic energy of the particles. For molecules you can have rotational and vibrational energy, while for single atoms you just have translational energy, or "thermal motion". At a given temperature, the particles don't have exactly the same energy, but a distribution of energies, and hence velocities.

Most (>90%) of the gas in the Universe is hydrogen. The energy needed to knock the electron off (i.e. ionize) a hydrogen atom is 13.6 eV. For a gas of $T \gtrsim 3000\,\mathrm{K}$, the fraction of particles with sufficient energy to ionize hydrogen is so high, that the majority of the atoms are ionized$\dagger$, i.e. split up in protons and electrons. This was the case in the beginning of the history of the Universe. Everytime a proton and electron met and recombined to form a neutral atom, the electron would almost immediately be knocked off again by a high-energy particle (usually an electron, but it could also be a proton or photons, since all particles were in "thermodynamic equilibrium", i.e. shared the same distribution of energies).

As the Universe expanded, the gas cooled. At some point, 380,000 years after Big Bang, the temperature had decreased enough that it was no longer possible to keep the atoms ionized, so over a rather short period of time ($\sim10^4$ years), they all recombined. This epoch is hence called the epoch of recombination.

Until this point, all photons kept scattering on the free electrons. With the electrons "trapped" in atoms, they could now stream freely, and "decouple". They have been traveling freely ever since, but since they travel through an expanding Universe, they become redshifted along the way. Since then, the Universe has expanded by a factor of ~1100, and so have the wavelength of the photons, so that today they have temperature of $3000\,\mathrm{K}/1100\simeq2.7\,\mathrm{K}$. This is what we see as the cosmic microwave background.

$\dagger$In which case they're in principle not "atoms", but a plasma. However, in astronomy it's quite normal to call it atoms anyway.


What is the difference between our time and space time?

What is the difference between time and space-time?

Space-time is time plus space.

  How does gravity affect the passage of time?

The higher the gravity of a planet or star and the closer to that body the slower the time.

  What is the speed of light and how does it relate to time?

The speed of light is 299,792.4580 km/s in vacuum, the speed at which light propagates, roughly 1.3 seconds from Earth to Moon. Velocity is distance divided by time; this applies also to the speed of light.

  How do scientists deal with timescales on the order of billions of years if time is not constant for all observers in the universe?

They treat time dependent of the observer. For different observers on Earth variations are tiny, in many cases neglectable in comparison to measurement errors, although not for precision measurements.

  How is time, or for example the age of the universe, actually measured experimentally?

There are many ways to measure or estimate ages. The age of the earth can be estimated by ratios of certain radionuclides in the oldest rocks. Estimates of the age of the universe are obtained by simulations based on observations (redshift and distance) of distant galaxies, and on observations of the cosmic microwave background. The redshift is used for a velocity estimate; together with a distance estimate (obtained by stars of known brightness) one can calculate back, when all objects of the universe should have been close together.


As matter approaches a black hole, does it speed up?

The answer is neither yes or no or possibly both.

Take a simple example. If something falls freely towards a black hole along a radial path, and is observed by someone who is far from the black hole, its velocity (according to the distant observer) is given by
$$v = -\left(1 - \frac{r_s}{r}\right)\left(\frac{r_s}{r}\right)^{1/2}c\, ,$$
(e.g. see chapter 6 of Exploring Black Holes by Taylor, Wheeler & Bertschinger - freely available) where $r_s$ is the Schwarzschild radius and the negative sign just indicates an inward velocity with $r$ decreasing.

If you plot this function (see Fig.2 in Ch.6 of Taylor et al. - freely available) you will see that initially the magnitude of the velocity increases as $r$ decreases, but as $r\rightarrow r_s$ then $v \rightarrow 0$ and the falling object appears to come to a standstill (actually, because the light from the object is gravitationally redshifted, this may not actually be observed). However, if the velocity first increases and then slows to a standstill, then it must go through a maximum!

The maximum observed speed in this scenario is achieved at $r=3r_s$ and is $0.384c$.

Of course this story is different for different observers. If you are the falling object then your speed just keeps increasing through the event horizon and towards the singularity. On the other hand, an observer who was somehow able to hover just above the event horizon would measure the falling object's speed as just below $c$ as it passed.


Derivation of the orbital parameters of a exoplanet discovered by direct imaging

All of the orbital elements can be derived by direct imaging, so long as you observe long enough to see a fair fraction of the orbit.

Kepler's third law then gives the total mass of the system and the relative orbits may give the mass ratio and hence individual masses (though this would be quite uncertain if the mass ratio was large).

Absolute masses could also come from dynamical measurements (i.e. velocities), but you can get these if the system is at a known distance. Again, because the star wouldn't move very much, this would likely lead to a highly uncertain planetary mass.

The planet mass might instead be estimated from its luminosity, an estimate of its age, and a cooling model. The radius can be estimated from its temperature and luminosity.


Both are correct, although the first can be further explained a bit. I won't give you a mathematical proof, though; Instead I'll play with characters.

Let's assume, for the sake of gedankenexperiment, that:

The speed of light is 5 characters/sec;
our universe is expanding at 1 character per 5 characters per sec.

This is our current universe, and we launch a photon from body A aiming at body E.
(Space generated each second is marked with a # symbol.)

T=0s A----B----C----D----E           Bodies
     *                               Photon     - 19 chars to E
T=1s A--#--B--#--C--#--D--#--E
      --#--*                                      17 chars to E
T=2s A--#---B-#----C#-----#D----#-E
      --#-----#-*                                 17 chars to E
T=3s A--#----B#-----#-C---#----D#-----#-E
      --#-----#-----#*                            18 chars to E
T=4s A--#-----#B----#----C#-----#---D-#-----#--E
      --#-----#-----#-----#*                      19 chars to E

I know, the graph isn't too granular, and the space generation isn't evenly distribute. I apologize for that, but it's for the sake of demonstration.

Notice that at T=2 some space is already generated between A and the photon. But that's irrelevant: E is sitting at the event horizon, and will never be reached by the photon, because the amount of space being generated between photon * and body E is equal, or superior, to the speed of light.

Given any positive expansion rate, there will be an event horizon - a point where the accumulated dilation of space is more than the amount of space a particle moving at the speed of light can travel.

A galaxy sitting initially at say, 1000 characters from A at T=0, will be at staggering 1200C at T=1 - that's 40 times our speed of light.

At T=16s, B (that was passed by the original photon at T=1) will be sitting exactly where E was relative to A, and at T=17 will fall out of our event horizon. A new photon emitted from A will never reach it.


Does the orbital variation in planetary gravity affect the Sun's corona

The gravitational effect of the planets on the Sun can be calculated very accurately. The acceleration of a planet of mass $M$ at a distance from the Sun of $r$ is
$$a = \frac{G M}{r^2}.$$
If you plug in the numbers, you'll find that the largest effect comes from Jupiter due to its enormous mass. The gravitational acceleration of Jupiter on the Sun is $2\times10^{-7}$ m/s$^2$. Even if all planets were aligned so to add up all their gravitational accelerations, the total would only be $2.8\times10^{-7}$ m/s$^2$.

However, the acceleration is not what is important; rather it is the tidal forces, which are proportional to $1/r^3$ instead on $1/r^2$. The tidal acceleration felt over a distance $\Delta r$ at the surface of the Sun is
$$a_{\mathrm{tidal}} \simeq \frac{2GM}{r^3} \Delta r.$$
Now I didn't really bother to read the paper in detail, but the guy doesn't seem to write what exactly in the Sun's atmosphere should be affected by the planets. But even if we say that it's all of Sun's corona, which has a thickness of $\Delta r \sim 10^6$ km, the combined effect of all eight planets standing on a line would be roughly $\mathbf{10^{-9}}$ m/s$\mathbf{^2}$, a minor effect compared to the difference in the Sun's own gravitational field over the same distance, which is 11 orders of magnitude higher.


How did single dish (or single receiver) radio telescopes originally generate images?

They scan the object, if you point the dish a a point in the sky as the Earth rotates the dish scans across astronomical objects, then move the dish to point at a slightly different position and let it scan across the object again, and again. After a while you can re-construct an image from the scan lines in a similar way to analogue TV.


Strange bright light in the sky flying fast

This sounds kind of like a Foo Fighter.
There have been various reports of weird glowing orbs following aircraft and zig-zagging around them ever since World War two. In fact both sides assumed they were secret weapons belonging to the other side up until the end of the war.

As to what they actually are exactly, well nobody really knows. The most plausible answer is that these are examples of Ball lightning or some other electrostatic phenomenon. As far as I know, it has never been conclusively replicated in the lab.


From which country or area is the new moon visible first?

The duration of one lunation (the period between one new moon to the next one) isn't neither constant as the Moon rotates around the Earth and it around the Sun (changes between 29.272 and 29.833 days due to the perturbing effects of the Sun's gravity on the Moon's eccentric orbit), nor integer divisible by 24 hours or one Earth's rotation around its axis. So this position of the Moon on the skies where the next new moon as the first of its lunar phases will be first observable constantly changes.

     Phases of the Moon, as seen looking southward from the Northern Hemisphere. The Southern Hemisphere will see each phase     rotated through 180°. (Source: Wikipedia on Lunar phase)

Saying it differently, by the time the Moon completes one lunation (or it's synodic period) and starts the next one, it won't be positioned exactly above the same Earth's longitude (East to West) as the one it started at.


By putting a mirror in space, would we be able to see into the past?

I think the question is referring to situating a very large mirror in space facing earth.  If we were to put it several light minutes away, then events occurring opposite the mirror could be reviewed de novo with more preparation upon the warning we received upon the first light of the event arriving at earth.

For example, a supernova going off in M31 might not be under observation at the moment its light first arrives, and so the initial observations might be lost.  However, with a mirror facing M31, we would be able to observe that mirror as the event unfolded, having been warned that there was something worth watching, in advance.

Nice idea!  But it would likely be much less expensive to simply have multiple telescopes always watching "prime" starscape for unexpected events.


How is the term "solar system" defined? Could confirmation of a new planet lead to a change in this definition?

The definition of the edge of the Solar System is not one of particular scientific importance, so it is not particularly well-defined. There are a number of plausible definitions, all reasonable.  I've listed one which occur to me below in order of increasing size:

The pancake-shaped volume which contains the orbits of the major planets.
The sphere big enough to contain the major planets.
The volume inside the solar bubble.
The thickish disk which contains the Kupier Belt.
The sphere which contains the Oort Cloud.
The Sun's Hill sphere.
The irregularly-shaped volume of space where the Sun is the closest star.

As you note, the discovery of a Planet X would change some of them, but not others.

How choose to define the edge of the Solar System depends a lot on what you intend to use it for.  I very much doubt that any one will win out, at least not until the definition of the Solar System makes an important difference somewhere: government, commerce, science.

So I think there's no clear answer to your question possible.


How do I get focussed images with my T-adaptor and 6mm eyepeice?

Using the eyepiece, and no camera lens, is a very different configuration to the camera+tube+scope (case 3).  The configuration you're after is "Eyepiece projection".

I don't know exactly what you have available, but I believe you will need to fit a T-threaded extension tube to the external thread on your T-adapter. Then fit the camera onto that.

Note - before you go out and buy a new extension you can probably do a hand-held test of this by the following:

Attach the adapter to the scope, with a low-power eyepiece in the adapter (low power makes the test easier to see.)
Point the scope at anything distant (you can do this in daylight, but NOT at the sun of course!)
Now hold the camera, by hand, a few inches behind the
eyepiece+adapter. Perhaps you can use a cardboard tube to keep the stray light out.
See if you can adjust the scope's focus, and the camera position, to get a sharp image.

If this worked, then you can get a T-threaded extension tube to go on the back of your current adapter, to hold the camera in position.

See for example this link:
http://www.astronomysource.com/2011/10/13/eyepiece-projection/


Martian dust storms-what was the smallest in area, or shortest in duration storm detected?

SOLAR SYSTEM/SUN, ATMOSPHERES, EVOLUTION OF ATMOSPHERES | Planetary Atmospheres

Numerous dust storms occur each Martian year and are generally classified according to size. From the smallest to largest, they are dust devils (<10−1 km2), local storms (∼103 km2), regional storms (∼106 km2), and planet-encircling storms (>106 km2).
Local dust storms are also quite common. Based on Mars Global Surveyor (MGS) images, as many as 2000 local storms occur each Martian year. This gives a daily-averaged rate of 2–3 storms per Martian day. They have typical lifetimes of less than several days.
Regional storms have been observed at nearly all seasons, but are most frequent during southern spring and summer. Most regional storms develop within ±30° of latitude, although there is a distinct bias toward the southern hemisphere. Regional storms can last from days to weeks. These storms can drift a significant distance from their original location, and new satellite storms can develop that are quite remote from the original center.


Can weak gravitational lensing or microlensing-induced wavefront distortion limit resolution of absurdly large aperture telescopes?

As far as microlensing is concerned:

Not likely

This is a bit of a dodge on the question, but it is hard to come up with a scenario where microlensing becomes the dominant PSF term.

Microlensing occurs during conjunction with a stellar or planetary object, and results in a momentary (seconds-to-days timescale) increase in the brightness of the observed star. It is an extremely rare event, with rates around 100 in 17 million stars in the most recent surveys.

Stellar densities ($\rho$) around our solar system are on the order of 0.14 stars per cubic parsec. With such a low number density, microlensing does not pose any practical limitation on the resolution of radio telescopes. Lets do some scratch-pad estimates to see:

If we assume a synthetic aperture ($D$) of 1AU ($4.84\times10^{-6}$parsecs) , a wavelength ($\lambda$) of 1mm (similar to the view of Betelguese by the ALMA observatory) , and a desired spatial resolution ($s$) of 10,000 km at a distance $L$:

$$
\begin{align}
\frac{s}{L}  &= 1.22 \frac{\lambda }{D} \\
L &= \frac{sD}{1.22 \lambda } = 40\;\text{kilo-parsec}
\end{align}
$$

This is roughly the diameter of the milky way galaxy. So an interferometer operating in the 1mm wavelength with a 1AU baseline has a 10,000 km resolution at a distance of 40 kpc.

How many stars are between the 1AU-wide observatory and the target star? Using a cone of base diameter $D$ and height $L$ as a volume and the above stellar density, we get the average number of stars ($N$) between our aperture and the observed star:
$$ N = \rho \frac{\pi}{12}D^2 L = 3 \times 10^{-8} \; \text{stars}$$

You would have to ramp $D$ up by a factor of $10^4$ before $N \approx 1$ and 'microlensing aberrations' were more than just an extremely rare event.


Retrograde Motion: Stars in the background

Go outside.
Pick a star, any star and watch it for a while.  It moves.  As the Earth spins all the stars move across the sky, taking 23hr 56 min to move all around the sky to the same place.  This is due to the rotation of the Earth (less than 24 hours due to the orbit of the Earth). We sometimes say the "celestial sphere" appears to rotate around the Earth, due to the Earth's rotation.
The planets also move across the sky every day, taking just less than 24 hours to all around the sky.
But the planets also move relative to the celestial sphere, and since th stars don't move relative to the celestial sphere we can say that the planets move relative to the stars.  It is this motion that is of interest, since it is not due to the rotation of the Earth, but to the actual motion of the planet relative to the Earth. This motion is much slower than the apparent motion of the celestial sphere; Jupiter takes 12 years to make a full journey around the sky. But this motion is special, complex and interesting.  It is this motion that earns planets the name "wandering stars" in ancient Greek.
The stars are at a very great distance, so to all intents and purposes they are "fixed". Relative to each other, the stars hardly move at all. You would need to wait centuries before you noticed any difference. So the background of fixed stars makes a very convenient way to refer to the motion of the planets


How does inflation justify the nonexistence of magnetic monopoles?

Let me start off by stating that scientists have not conclusively proven that magnetic monopoles do or do not exist. There has never had a confirmed detection of one so the jury is still out on if a magnetic monopole can or does exist at all. That being said, if monopoles do exist, no one has a good explanation for why we haven't detected one yet. There are hypotheses, but no solid proof.

To get to your question now, I think it is an incorrect statement to say that inflation theory explains the non-existence of magnetic monopoles. What inflation theory does potentially explain, is the extremely low density of magnetic monopoles in the universe. The idea here is that magnetic monopoles, for some unknown reason, are extremely unlikely to form, even in the intensely energetic and hot crucible of the Big Bang. Some estimates say that for 1 magnetic monopole that formed, $10^{29}$ "normal" particles formed, making magnetic monopoles extremely sparse and rare.

Where inflation theory comes in is that it says the universe expanded exponentially fast. If the monopoles were already sparse before this extreme expansion, the expansion would have only served to cause the magnetic monopole density in the universe be practically zero. It is quite possible that, post-inflation, the magnetic monopole density in the universe is so low that only one exists within the observable universe. Good luck finding that single particle out of the $10^{80}$ particle haystack.

In my mind though, that doesn't really answer the question as to why we don't see magnetic monopoles. Inflation just makes them extremely sparse, but why did so few form to begin with (assuming they formed at all). As I said above, nobody knows. A potential answer (without any evidence) is that magnetic monopoles are extremely energetic particles (e.g., with energies greater than the LHC of 14 TeV). If they have such extreme energies, even the Big Bang wouldn't have formed many of them. But again, that's just a guess.

Only time will tell what the real answer is.


Has the Earth-facing side of the Moon that we see today always faced us ever since the Moon got tidally locked? Or does it precess?

The Earth-facing side of the Moon doesn’t change currently due to any of the forces on it by other bodies (excepting librations).
From Gladman et. al.: “tidal dissipation in the satellite drives it to a state where […] the spin axis of a satellite in a generalized Cassini state will appear stationary.”  The $S_2$ Cassini state is one of the two possible eventual stable states for a general moon system.  But, according to Gladman, our Moon has "$S_2$ as the only possible Cassini state".
Before achieving a Cassini state, the tidal forces exert an average torque of $-SC\cos\theta\sin\theta$, where $\theta$ is the obliquity of the moon's orbit measured with respect to the orbit normal, $C$ is the satellite’s moment of inertia about the spin axis, and $$S=\frac{3Gm_p}{2r^3}\frac{C-(A+B)/2}{C}$$ where $m_p$ is the mass of the planet, $G$ is the gravitational constant, $r$ is the orbital radius of the satellite, and $A$ and $B$ are the other two principal moments of inertia.
Note that $S$ has $r^3$ in the denominator (also as pointed out by Stuart Robins in a comment above), which means the torque exerted by tidal forces was much higher when the moon was closer to the Earth.  In other words, since the Moon’s face towards the earth is not precessing now, it certainly wouldn’t have in the past when the potential tidal torque was higher.
I think Gladman et. al. give pretty credible evidence for their claim of tidal lock, in that they account for wobbles, highly eccentric orbits, and oblate spheroids in both the planet and the moon.  They don’t specifically account for the gravitation of other bodies like the Sun.  The above equation also gives us a comparison of relative tidal effects on the Moon by noting $m/r^3$ is 9.2e7 for the Earth/Moon system, but only 5.9e5 for the Sun/Moon system. Gravitational perturbations from the Sun and other solar system objects affect the Moon's orbital elements, but don't have a significant effect on it's rotation rate.
There is plenty of lunar geological evidence supporting Gladman et. al.’s claim that the Cassini state is stable, in that we see a fundamental asymmetry between the near and far faces of the Moon.  The near face has a thinner crust and more volcanic maria.  The far side has a thicker crust and more visible asteroid craters:

Roy et. al. claim the asymmetry between the near and far faces is due to heat from the Earth causing the early tidal locked Moon to cool more slowly on the lunar face pointing to the molten cooling Earth. If their theory is correct, then the Moon's face towards Earth hasn't significantly changed since the Earth/Moon system infancy.
The near face of the Moon doesn’t appear to be precessing currently, but could it have changed orientation from its original tidal lock configuration? There is a strong suggestion from Kadono, that an asteroid impact could have an effect on spin rate.  Unfortunately, the paper is behind a paywall, so lets do a "back of the napkin" calculation to see how much angular momentum a big asteroid impact could have on the moon.
The largest known basin on the Moon is the South Pole-Aitken basin. The size and speed of the impactor was estimated by Potter et. al. One of their models assumes a chondrite impactor with density $\rho$ of 3580kg/m3 at 100km radius $r$ at an impact velocity $v$ of 10km/s. We can calculate the mass $m=\rho 4/3\pi r^3$ or 1.5e19kg. The equation for linear momentum is $l=mv$ or 1.5e20km*kg/s.
The moment of inertia of a sphere is $I=2/5MR^2$ and we can use $M=7.342e22kg$ and $R=1737.4km$ as the mass and radius of the Moon to get $I=8.86e28$.  If we set the angular momentum of the Moon $I\omega=1.5e20$ and solve for $\omega$ giving us a resulting rotation rate change of 1.69e-9 radians per second or about 3 degrees per year. These calculations assume an impact angle parallel to the lunar surface, which is impossible, but forms an upper bound for a momentum change.  Higher angle impacts still conserve momentum, but more of the momentum goes into changing the orbital parameters rather than rotation rate.
From When did the Moon become tidally locked to Earth?, we know the tidal locking time is $$t_{\mbox{lock,Moon}}=7.12753\cdot 10^{-25}wa^6 \frac{\mbox{kg}}{\mbox{Nm}^2 \mbox{km}^3}.$$ where $w$ the spin rate in radians per second, and $a$ the semi-major axis of the Moon orbit.
I calculated the above equation, supposing the Moon's orbital semi-major axis was less than half it’s current value so long ago. If the moon got struck in the way presented above, it would tidal lock again in less than 2 years, rotating less than 6 degrees in that time.  So, no, after the Moon was first tidally locked, the same face almost certainly always pointed to the Earth.
Notes:

It's possible that the asymmetries in the Moon's composition would decrease the time to tidal lock, but it isn't needed for any of the above calculations. Even if the Moon's composition was homogeneous, tidal locking would still occur.

There are many other theories for the lunar face asymmetry, but most of them rely on a steady tidal lock through the early part of the Earth/Moon system.  An exception might be Elardo et. al. who claim the asymmetry is due to a greater abundance of radioactive isotopes on the near side of the Moon, which add heat and lower material melting temperatures.  They call the mix KREEP.

We can't completely eliminate the possibility that the Moon was hit by an even bigger asteroid then referenced above, with the evidence later being covered by volcanism.  This answer is only based on currently available evidence.


What will future populations of stars be like?

(Adams & Laughlin 1997) discuss the effect of increasing metallicity in the future. A higher metallicity increases the stellar burn rate since the density increases but
the higher opacity reduces it a bit. The total effect is nonlinear; in their model the maximal lifespan happens for $Z\approx 0.04$ and beyond that it declines. Whether there are newer and better models of this I do not know.

They also point out that the maximum stable stellar mass decreases as metallicity increases, while the mass of the minimum mass mainsequence star declines, both as a result of opacity effects.

Overall, the metallicity growth is likely to continue across the stelliferous era. While extragalactic gas may fall in, it merely adds to the hydrogen pool and continued star formation. What could stop the metallicity growth is quenching due to a galactic merger or active galactic nucleus that blows away enough gas to stop further star formation.


Is there an example of a star that is already dead, but can still be seen on Earth?

There is an accepted answer already, but there is a couple known cases of a star we know has gone supernova, and yet we can still see it.  This source describes one such unique circumstance.  The star that exploded happened to be in a galaxy that was behind another massive one from our point of view.  The alignment was just right such that the light formed an Einstein Cross.  The light from each point on the cross takes a different path to get here, and each of those paths are different lengths.  Thus, the different points on the cross show the star at 4 different times in its final years.  The Space.com article I linked was written in 2015. Scientists first noticed the supernova in 2014, and each image of the supernova arrived within a year of when it was first noticed.  If I'm understanding that article correctly, as of now (2017) all of the images of that supernova show the supernova aftermath, and not the star beforehand.  However, there was a period of time where we could see the star both pre and post-supernova at the same time.

You imply in your question that you are focusing on stars within our own galaxy.  I don't know of any such situations of Einstein Cross events closer to home that let us do this.


Critical density of the universe (no, not that one)?

You're basically asking about the Flatness Problem. As quoted from Wikipedia:

  If the initial density of the universe could take any value, it would seem extremely surprising to find it so 'finely tuned' to the critical value $\rho_c$. Indeed, a very small departure of $\Omega$ from 1 in the early universe would have been magnified during billions of years of expansion to create a current density very far from critical. In the case of an overdensity ($\rho > \rho_c$) this would lead to a universe so dense it would cease expanding and collapse into a Big Crunch (an opposite to the Big Bang in which all matter and energy falls back into an extremely dense state) in a few years or less; in the case of an underdensity ($\rho < \rho_c$) it would expand so quickly and become so sparse it would soon seem essentially empty, and gravity would not be strong enough by comparison to cause matter to collapse and form galaxies. In either case the universe would contain no complex structures such as galaxies, stars, planets and any form of life.

This is more or less what Phiteros says in his comment above. As he states, the distinction here is not that the universe wouldn't exist at all, as you state, but rather it wouldn't exist as we know it, with galaxies, stars, planets, or life of any form.

To explain a bit more, the basic idea is that the universe will have started out with a more-or-less homogeneous and isotropic distribution of matter with some density $\rho$ (this is a pretty strong assumption in its own right, and worth addressing in a completely different answer). Note that the overall curvature of the universe is directly tied to this density so you may see some sources talk about this in terms of curvature instead of density. The main point, as described in the quote above is that the evolution of the universe, i.e., the ability for galaxies, stars, planets, life, etc. to form, is dictated by a critical density, $\rho_c$. If $\rho = \rho_c$ (or as close to equal as to be insignificantly different), our universe can form as it is now. The flatness problem comes into play because the Big Bang Theory implies that if the universe did not start out with (almost) exactly $\rho = \rho_c$, it would have veered far off course from our current universe and resulted in either a Big Crunch or a Big Rip.

This idea poses a really big problem for the Big Bang Theory because it requires such a precise setup for our universe, where any betting person would've said there's no way the universe could've started out with $\rho=\rho_c$. The main resolution to this problem is by introducing inflation into the model. The main idea here is that the Universe may or may not have started out with $\rho = \rho_c$, but so long as it was close enough, the hyper inflation at the beginning of the universe would have forced the difference to be so small as to be effectively zero, negating the effects described in the quote above. Another potential, and not necessary mutually exclusive answer is found in the Anthropic Principle, which I won't delve into here.


What's the correct distance to Betelgeuse?

Distance measurements for Betelgeuse are a bit of a mess. Solutions based on parallax would be ideal, but Betelgeuse has a rather large angular diameter at most wavelengths thanks to its extended envelope; optical and infrared observations usually fall in the 40-60 mas range (see Dolan et al. 2016 for a recent review), while radio observations show a disk of emission roughly twice that size (O'Gorman et al. 2017). The parallax is expected to be much smaller that the angular diameter, on the order of ~5 mas, and so it depends strongly on the choice of the center of emission.
One of the first decent parallax results was obtained by the Hipparcos satellite in 1997, whose astrometric measurements allowed comparatively precise measurements of location, parallax and proper motion for over 100,000 stars. Hipparcos measured a parallax for Betelgeuse of $\pi=7.63\pm1.64$ mas, corresponding to a distance of $131\pm30$ pc$^{\dagger}$. This is that 427 light-year number the app cited. The Hipparcos-only result was subsequently improved upon significantly by van Leeuwen 2007, who found $\pi=6.56\pm0.83$ mas, cutting the old uncertainty in half; this would correspond to a distance of 152 parsecs. If you're going to quote a Hipparcos result, this is the one to pick.
More recent results indicate that this value is likely too low. Combining Hipparocs data with multiple multifrequency radio measurements using the Very Large Array, ALMA and e-MERLIN (Harper et al. 2008, Harper et al. 2017) gives derived values of $197\pm45$ pc and $222^{+48}_{-34}$ pc, with the former barely consistent with the purely optical results (the latter value converts to 724 light-years). These groups note that the Hipparcos stochastic astrometric solution required the addition of so-called "cosmic error" or "cosmic noise" terms to individual position measurements.
The papers above note that the photocenter at both and optical radio wavelengths does not coincide with the barycenter, and may change on timescales of months to years. Therefore, extended, long-term observations would be necessary to reduce any photospheric "jitter" or other variations which could lead to variations in emission and thus astrometric fitting. Harper et al. 2017 proposed joint ALMA and Expanded VLA/Jansky VLA observations and mm and sub-mm bands over a period several years, but also suggested that this would require "a Herculean effort" to achieve, for logistical reasons (how many telescope committees would be inclined to commit that much time up front?). Perhaps interest in Betelgeuse's recent luminosity dip could motivate this sort of observation.

$^{\dagger}$Rob Jeffries makes the point that given the pretty terrible signal to noise ratios of many of these parallaxes (Hipparcos was in particular was bad, but none of the observations are amazing), it's not clear that you can really get good, meaningful uncertainties on the distance measurement from them via $d=1/p$. I agree; it's safe to say that the jury is still out on those, and any source that claims a distance and an error should give make that quite clear.


Measuring parallax angle, same local time or the same time

The simplest way: take pictures at the same (ie 16UTC). The position of the moon in the sky will be very different for different observers, due to the cuvature of the Earth.  But this is not what you are measuring. You need to see the difference in position relative to the stars.
For more accuracy, instead of measuring the difference is position directly, you can time when the moon occludes a star. The absolute time when the moon will occlude a star will differ for different observers, and this can be converted into a difference of position of the moon, and parallax can be calculated from that.  It is easier to be accurate in timing than it is to measure position directly. With some clever maths you can just measure the length of the occulution rather than the absolute time, and this means you don't even need a accurate clock, you only need an accurate stopwatch (not so much of an issue now, but significant when these measurements were first being done)


Why are there no more images revealed by New Horizon?

http://www.planetary.org/blogs/emily-lakdawalla/2015/06240556-what-to-expect-new-horizons-pluto.html

  The transmission of the High Priority data set will be complete on July 20, and then image transmission will pause. For nearly two months, until September 14, New Horizons will switch to near-real-time downlinking of data from other, so-called "low-speed" instruments while it transmits just housekeeping information for all of the rest of the data. No new images will arrive on the ground during this time


If an event occurs at a point in space time what will an observer experience first, it's gravity or the light from it?

it's an interesting question. It could ask about any violent transient effect that produces light and gravitational waves, though "popping out of existence" is the ultimate step function.

electromagnetic (EM) radiation is slowed very slightly by the presence of interstellar medium (especially by free electrons) and it's wavelength dependent. The image below from this answer shows the lower frequency radio waves arriving later than the higher frequency ones due to such an effect.

The farther away an object is, the larger the time difference, so this technique can be used to determine a likely distance for a given object.

So if you relax the term "light" to include a wide range of wavelengths which many astronomers now do, you can see a quite a big delay.

Answers to How fast can and do gravitational waves travel? say that gravitational waves generally travel at the speed of light, and answers to Have more recent LIGO/VIRGO gravitational wave measurements narrowed down the speed of gravity further? addresses experimental confirmation. I've never heard of an "index of refraction" for gravity that deviates from unity and could slow it with respect to the speed that an EM wave would travel between the same two points, so I think it's probably safe to assume that it will always come before the EM waves.


Is it possible that a ultra-large portion of the space we live in is already inside a black hole? How could we refute this?

This isn't anything near a proof that we don't live inside a black hole, but it's a bundle of evidence that certainly goes against it, and that the Great Attractor is not, in fact, the singularity.

First off: Expansion of the universe.

As you no doubt know, the universe is expanding. In fact, the expansion is accelerating. Do black holes expand? Yes. As they suck in more matter, they can get bigger. But if that was the case, we should notice more matter coming into the universe (well, I suppose it could be from outside the visible universe, but we should still see lots of matter coming towards us). Also, the universe's expansion is driven by dark energy, and it pushes (on a large scale) everything away from each other. In an expanding black hole, there would be no reason for the matter inside the event horizon to move away from each other; only the event horizon expands.

You also made a good point in a comment below about Hawking radiation. Eventually, in the far future, when there is nothing left in the universe but black holes, black holes will evaporate via Hawking radiation (okay, they do that now, but they can still take in more matter). If our universe is a black hole, it should then contract. But we see no reason why it should. In fact, the theory that predicts the universe's eventual collapse into a singularity (i.e. the opposite of the Big Bang), the Big Crunch theory, predicts that the universe's contraction will match its expansion. Contraction due to Hawking radiation wouldn't necessarily mirror the black hole's growth. Also, the Big Crunch theory is supported only by a minority of scientists due to the evidence against it.

Second off: The motion of galaxies by the Great Attractor.

First off, see When will the Milky Way "arrive" at the Great Attractor, and what all happen then? (and not just my answer! I dearly wish @LCD3 would expand his/her comment into an answer!). The general gist of things there, as pertains to your remark about the Great Attractor, is that the galaxies aren't all moving towards it. There are doubts (see the papers I mentioned) that the galaxies previously thought to be moving towards it are, in fact, moving towards more distant objects - other superclusters. If the Great Attractor was indeed the singularity, a) all the galaxies in the universe should be accelerating towards it, which is not the case, and b) we should be moving directly towards it, and due to its gravity, not that of the superclusters beyond.

Like I said in my comments, I don't quite think your last part relates to your first part, but I'll try to address it. First, I'm not sure where you got your sources, but I can say that we don't really know what goes on inside a black hole, and I don't now how someone got that density figure (I am, of course, not the authority on black holes - see @JohnRennie for that, on Physics SE, and I could be wrong about this). The density inside the event horizon would not be very thin, however, in regions where there is lots of matter. For example, in a black hole with an accretion disk, the material that is being absorbed might not have that low a density. Also, a large volume of that density would not necessarily collapse to become a black hole, because it would not be compact enough.

Hope it helps.


Does the orbits of stars in a binary system have a common focus, which is the center of mass of the stars?

Elliptic orbits with the focus at the centre of mass (the barycentre) are the solution to the "two body problem".

The wikipedia page on the central force problem shows how the two body problem can be reduced to solving two central force problems. Therefore the elliptical solution to the central force problem also solves the two body problem, and in moreover Kepler's laws apply to such bodies.


What particles does a black hole emit when it evaporates itself?

Although black holes are widely believed to emit Hawking radiation, it should be stressed that it has not actually been observed (yet?). The Hawking radiation should consist of electromagnetic radiation/waves that have a near-perfect black body spectrum, which is at a temperature that is inversely proportional to the mass of the black hole - the smaller the black hole, the higher the temperature.

The radiation is caused because "vacuum" is not empty when considered in quantum mechanics. Particle/anti=particle pairs are created for brief moments of time, before they annihilate again. Close to the event horizon of a black hole, it is possible that one particles from a virtual particle pair travels within the event horizon of the black hole and cannot escape to annihilate with its anti-particle. In principle, these particle/anti-particle pairs could be any types of particle, but in practice they are more likely to be the lightest particles. The lightest charged particles are electron/positron pairs, since these need to "borrow" less energy from the vacuum to be created. I do not think it is essential for one of the pair to disappear, since "Unruh radiation", a close relation of Hawking radiation should also be seen whenever there is acceleration with respect to the vacuum.

To stay out of the black hole event horizon the particles must be accelerating. Accelerating charged particles locally emit electromagnetic radiation, which is then gravitationally redshifted when seen by an observer a long way from the black hole. It turns out that to be in a thermal equilibrium, the radiation must have a blackbody spectrum form.

The temperatures and amount of radiation emitted are very small for stellar-sized black holes. The temperature of a non-rotating Schwarzschild black hole is given by
$$ T = \frac{6.2\times 10^{-8}}{M}\ K,$$
where the mass $M$ is in solar maases.

The power emitted by a black hole as Hawking radiation is
$$ P = \frac{9\times 10^{-29}}{M^2}\ W$$


Zodiac signs are equal 30° divisions of the ecliptic.
In the Western tropical zodiac, these are linked to the equinoxes and solstices, and named after the nearest constellations at the time Ptolemy wrote about them.
The equinoxes precess westward 1.4° per century, so the signs currently lie 25° to 30° west of their namesake constellations.
There are also sidereal zodiacs, which account for precession and keep signs aligned with stars.

A celestial object's ecliptic longitude of date determines its zodiac sign.
If the lunar RA and Dec in question are for epoch J2000, then the Moon's epoch 2020.25 ecliptic longitude is 99.0°, in the tropical sign of Cancer (longitudes 90° to 120°).
Relative to the IAU boundaries, drawn along lines of RA and Dec for epoch B1875, this is in the constellation Gemini.

Approximating ecliptic longitude by right ascension alone can err as much as 2° in the case of the Moon, whose orbit is inclined 5° to the ecliptic.
For example, at 2020-04-06 19:40 UT, the Moon is at equatorial RA 12h4m (1° east of 12h0m) but ecliptic longitude 179° (1° west of 180°).
Since the ecliptic is oblique to the celestial equator, lines of ecliptic longitude are oblique to lines of right ascension except at the solstices.

Houses are completely different, dividing the sky in reference to the local horizon.
The precise definition depends on the particular astrological tradition; roughly, a celestial object enters the 12th house when it rises and the 6th house when it sets, decrementing house number every 2 hours or so.
The most closely related astronomical coordinate is hour angle.


How much of the Sun's disk must be covered for a visible shadow to be cast?

The answer for your question is quite complicated. Basically, the cutoff point you are saying about depends on the amount of light, which is reflected from the certain surface experiencing the eclipse. Another factor is the sensitivity of your eye, which seems to be individual...
http://www.mkrgeo-blog.com/the-role-of-contrast-in-ability-of-human-vision/
This cutoff is nothing like the contrast difference between two separate areas, where one of them is less illuminated.
Since the albedo of the Earth is various, we shouldn't expect the fixed border between shadowed area being visible. The biggest albedo feature clouds, snow, and glacial areas, and consequently the deserts.
The satellite view is usually perfect for emphasizing the shadowed area. This is driven by the huge amount of reflected light, which comes from the outside, where Sun shines in 100%. As we are moving down to the ground, the visible cutoff changes. The phenomenon, which could explain this situation is the light transition.
The light transition defines the border between non-shadow and half-shadow surface, and next the half-shadow and full shadow surface.
This is better described here:
http://www.mkrgeo-blog.com/what-is-a-light-transition-what-examples-of-it-can-we-see/
where you have got the experiment on a white background, which reflects the light perfectly. An analog situation happens during every solar eclipse seen from the space, even the partial one. As our sight is higher from the surface, we are able to see a bigger area. Obviously our shadowed area becomes smaller, which yields eventually the regions from outside the penumbra, fully illuminated.
The details of this situation have been described here:
http://www.mkrgeo-blog.com/how-the-lunar-shadow-looks-like-during-the-solar-eclipse/

Above you can see the shadow produced by the partial phase of the solar eclipse in 2015. As you can see, it's almost the same deep as you have shown in your question. This is where I am heading to in order to make a good explanation. The image of a partial solar eclipse was taken in specific conditions, where the sun was low above the horizon. hence your penumbra remains the fuzzy oval, bringing more reflected sunlight from the outside. This is why it looks darker.
Even the ISS doesn't show the real shadow cutoff, which you are asking about. The most expressed prenumbra-affected regions are to be spotted via some satellite imagery like sat24.com or so.
See the below examples from June 21, at the eclipse beginning in Africa.

this is the most expressed penumbra, as you can see. Next, if you really wish to know roughly what minimum phase of the eclipse causes the penumbra visible, you can:
1. Draw the visible line of the penumbra cutoff, which you can spot by your eyes
2. Pick up the location just behind this line
3. Check the eclipse circumstances in this area (knowing the time provided) by some astronomical software like Stellarium, Cartes du Ciel, etc.
Remember, that this visible shadowed border will be different for the other circumstances! Imagine, that you are standing inside the umbra during the totality. In this case, your eye can even distinguish the areas brighter and darker inside the shadow! On the other hand, the prenumbral regions will appear as fully illuminated. Threading this way, the cutoff will be marked by the border of the umbra only.
In terms of the photos, as you said. It covers only 25% of the sun but the shadow is visible well. This is what I explained above. Look at the shadow itself, it is very small compared to the image frame (the area of sight). As a result, a vast majority of the surrounding area is fully illuminated, therefore reflects possibly the biggest amount of light outshining the small shadowed region effectively.


Why is this telescope so short? How hard is it to make such a fast primary?

I don't have enough room in the comments for this, so I'm writing here, although it's probably not a true answer since I know nothing about that particular telescope.

Anyway, if you look at many SCT systems, and their derivatives such as Ritchey-Chretien, Dall-Kirkham, etc, the distance between primary and secondary is often not too big.

If the secondary has negative curvature, that means one of its focal points is behind it. That's where the focal point of the primary also is. So the focal length of the primary is bigger than it seems just by looking at the picture. There's a focal point geometrically located in front of the telescope; the distance between it and the secondary or the primary depends on the design parameters of the system. It can't be too close to the secondary, or else the secondary would have to have extraordinary amounts of curvature.

Also keep in mind that there must be a gap between the edge of the primary and the inner surface of the OTA, so the primary is smaller than the visual estimate of the hole.

We also don't see the bottom of the instrument, so we don't know how far the primary mirror cell is protruding out the back of the instrument - though that's limited by the mount's big fork.

Also, look at a small SCT, such as the Celestron EdgeHD8 OTA. Visually, the tube seems a bit longer, compared to its diameter. However, the EdgeHD is a much smaller instrument. The relative amount "wasted" at the bottom by the primary cell is different. Also, the secondary is "buried" into the OTA, whereas the 0.65m scope has its secondary sticking out of the OTA.

Anyway, this is a bunch of handwaving based on visual estimates. It is possible that this is a system with a primary somewhat more strongly curved than usual.


Why can't we see distant galaxies with the naked eye?

Surely if you stared long enough, the light from them would eventually hit your eye?

Collecting light over a long span of time is how telescopes can see very dim objects. The human visual system doesn't work that way.

For one thing, even when you think you are staring at something, your eyes still dance around a bit. It's a built-in response called ocular microtremors. These microtremors appear to be an essential part of making the vision system functional.

For another, your eye does not and cannot collect light for arbitrarily long periods of time (the way a photographic telescope can). There's an immense amount of signal processing that happens in the eye and along the way to the brain. This signal processing depends on light being collected for short spans of time.

Our vision system evolved to see food, friends, and dangers under well-lit conditions. We are very good at seeing motion in broad daylight. We're not so good at seeing stationary objects, and we're not good at all at seeing barely visible sources under very dark skies.

Naked eye astronomy is limited by the nature of the human vision system. The most remote object we can see is the Triangulum Galaxy, and that's only under conditions of extremely dark and extremely clear skies.


How long does it take for a white dwarf to cool to a black dwarf?

I don't think there is an accepted definition of a "black dwarf" - it is not a term used in the scientific literature.

A popular definition that appears to circulate on the internet is that it is a white dwarf that has cooled down to the extent that it no longer emits any radiation in the visible part of the spectrum. But this is an unworkable theoretical definition. Even the coolest objects emit some radiation in the optical part of the spectrum; whether or not it is detectable will depend on the size of the object and how far away it is from us.

If a black dwarf is simply to be given the practical definition of being unobserved/unobservable at visible wavelengths then such objects exist already. Kaplan et al. (2014) have found evidence for a $\sim 1M_{\odot}$ white dwarf companion to a pulsar that they are unable to detect at visible wavelengths even with the Keck telescope down to an absolute (red) magnitude of $M_R>19.1$.

They argue that this means the white dwarf has cooled below 3000 K. This is possible within the age of the universe because more massive white dwarfs cool more rapidly because their higher interior densities lead to more rapid crystallisation and a steep decrease in their heat capacity when they enter the Debye cooling regime. A more typical old white dwarf of mass $\sim 0.6M_{\odot}$ would still be at a temperature of $\sim 4000$K and emit a reasonable amount of light at visible wavelengths.

Thus, if you are happy to accept the definition above, then black dwarfs can result from the cooling of massive white dwarfs in about 10 billion years. The plot below, taken from the paper of Kaplan et al. (2014), shows some cooling models for massive white dwarfs with hydrogen atmospheres. They easily cool below 3000 K within 10 billion years.

To address Mark Foskey's point: Once in the Debye cooling regime, the cooling rate hardly slows because the heat capacity is $\propto T^3$, whilst the luminosity is roughly $\propto T^{3.5}$ (here, $T$ is the isothermal interior temperature). Luminosity is also $\propto T_s^4$, where $T_s$ is the surface temperature.

The relevant differential equation is that the rate of change of thermal energy equals (minus) the luminosity
$$ \frac{d}{dt}\ \int T^3\ dT \propto -T^{7/2}$$

Putting this together, we find that $dT_s/dt \propto -T_s^{3/7}$ and so the cooling rate does not change much once Debye cooling starts. As a consequence the temperature plummets as $T_s \sim -t^{7/4}$ and the time taken to cool well below 1000K from 3000K would only be a few times longer (maybe a billion years) than the time to cool from 4000K to 3000K (as suggested by a simple by-eye extrapolation of the models shown in the plot).

The same treatment in the "Mestel" cooling regime, where the heat capacity is constant yields a much shallower $T_s \sim t^{-7/20}$, and accounts for the sharp "knee" at the onset of Debye cooling.


Do all stars have the potential to have life supporting planets?

Great question. The goldilocks zone is usually defined in terms of a region where the equilibrium temperature of the planet lies between some temperature limits (these temperature limits are somewhat debatable, but irrelevant for the purposes of this question - the boundary becomes fuzzy). This region can be calculated by working out how much flux is received from the star at a given radius.

In that sense - all stars have a goldilocks zone - you can always work out some distance from a star where the flux is roughly equivalent to the flux we receive from the Sun for instance. It is much closer to a dim star and much further away from a luminous star.

But, there is a difference between a habitable zone and a continuously habitable zone. If you seriously want life to evolve on a planet, then it will take time. How much time? No-one is sure but it seems to have taken a few hundred million years in our solar system. There are also good reasons to suppose that young planets are not going to be habitable - either they are still extremely hot after formation or they are being bombard by debris.

Thus you can probably exclude any star with a main sequence lifetime of less than about 100 million years (or longer if you feel conservative about how long it takes life to get going). This rules out stars of more than about 5 solar masses.

You might also rule out stars that evolve quickly. If a star changes its luminosity quickly on a short timescale then the habitable zone moves drastically too. This will happen near the ends of the lives of all main sequence stars and for all subgiant and giant stars. So for these, yes there is an instantaneous habitable zone, but no region is in a habitable zone for hundreds of millions of years or more.

You might want to consider not only the luminosity of the star, but its spectrum as well. For instance, both hot white dwarfs and low-mass M-dwarfs are reasonably stable and long-lived; they are faint, so have habitable zones close to the star (much closer than 1 au). However, for different reasons, both these types of object emit copious UV radiation. In hot white dwarfs it would be because they have hot photospheres. In M-dwarfs they can be highly magnetically active into old age, having a hot chromosphere and corona that would strongly irradiate any nearby planet. You might consider that their "habitable zones" were in fact uninhabitable. Of course an atmosphere and a strong magnetic field might counteract this.

A lot of this depends on your terms of reference and definition. The wikipedia page on habitable zones has quite a nice discussion, which emphasizes the debatable nature of this topic.

A final thought, which is not in the wiki page. Multiplicity could mess things up. Planetary systems will not be stable in the habitable zone of either star in a binary system if their separation is comparable with the orbital radius of the habitable zone. If you had two stars like the Sun, their separations would have to be less than a few tenths of an au or greater than of order 10 au in order to allow something to orbit at about 1 au around one of them. Even then, there could be all sorts of dynamical instabilities which prevent long-lived planets, especially if there were other planets in the system too.


How do the sun's light rays come to the world?

The are parallel.  The picture is way off scale. Here is a scale picture of the sun and Earth

Oh dear! the Earth is too small to be seen!  Even the Earth and moon look to be in the same place.  Rays of light that are hitting your neighbourhood are coming in parallel. The light from the top of the sun and the bottom of the sun are not quite parallel (the lines in the diagram slightly converge) but light from on point on the sun will hit the all the places in your neighbourhood in exactly the same direction.
Let's zoom in

There's the Earth, and two rays from the sun.  Are they parallel? No there is an angle of 0.00052 degrees between them. The angle is only so large because the distance between the ends of the rays is a lot (about the distance from Mexico to Canada). If I'd put the ends of the rays closer together the angle would have been much less.  To all intents and purposes the rays are parallel.
What you describe is an optical illusion.  When the sun shines through gaps in the cloud you will see several parallel rays coming towards you.  Because they are coming towards you, they will appear to diverge, just like railway track appear to diverge as they come towards you.  But with railway tracks we can see that they are coming towards us, because they are on the the ground.  With the parallel rays of sunlight, there are no clues, and we see them as shining down, and so being non-parallel.  Its an optical illusion.


Could a pair of binary black holes form within a star?

I doubt two black holes could form within a single star, but they might form from a contact binary. That's a pair of stars so close that they touch and share a single atmosphere. For the stars that made the black holes for the LIGO gravitational wave discovery, just consider two very massive stars that are touching or nearly touching. They would share a single gaseous envelope for an atmosphere. Each one separately formed a black hole when it died, and then their two black holes merged.


If the moon is losing energy dragging the oceans around, why is it moving away from earth instead of falling like any other satelite would?

No. The energy budget of the Moon is actually increasing. The Earth's spin is slowing down, and that energy gets transferred to the Moon.

Why is the Moon receding from the Earth due to tides? Is this typical for other moons?

Whatever else may be happening (like the losses you mention), this transfer from Earth's spin to Moon's orbital energy is the main process responsible for raising the Moon's orbit.


Is Blue Shift just as provable as Red Shift?

The expansion of the Universe can be expressed using Hubble's law. This is that the recession velocity that a galaxy has with respect to our own is proportional to its distance from us. i.e.
$$ v = H_0 D, $$
where $v$ is the recession velocity, $D$ is the distance and H_0 is the Hubble "constant", which has a value of about 70 km/s per Megaparcsec.

Hubble's law is attributed tot he overall expansion of the universe and means that once we get further away than about 10 Mpc (or about 30 million light years), that other galaxies recede from us with velocities of 700 km/s or more. This recession produces a redshift.

However, superimposed on Hubble's law at a more local scale, individual galaxies are gravitationally interacting with each other and hence have their own peculiar motions with respect to the general "Hubble flow". In addition, don't forget our Sun rotates around the centre of our Galaxy with a velocity of some 230 km/s, so is heading in the direction of some galaxies and away from others.
These local peculiar motions means that some nearby galaxies are moving towards us (at relatively modest speeds) and do exhibit a blue shift. An example would be the Andromeda galaxy at 2 million light years, which is approaching our Galaxy at about 110 km/s (or at a relative closing speed of 300 km/s with respect to the Sun).

But there are only about 50 galaxies in our local group and a further few tens of galaxies in some other structures out to about 30 million light years, and it is only from this small sample that any can exhibit blueshifts. The vast majority of known galaxies are further way and are swept along in the general universal expansion and hence away from us.


How can the IAU 2000A vs IAU 2000B nutation comparison be reproduced?

I suspect that the problem is in your assumption that changes in $\Delta \epsilon$ or $\Delta \psi$ lead to coordinate changes of the same size. In fact, the position of the equator and celestial pole are complex functions of those angles, notably with both positive and negative terms.  If you look at Equations (8) and (9) (for the coordinates $X$ and $Y$ of the pole) in the 2008 A&A paper you linked to, you can see this dependence.  For example, in the $X$ equation, there is a $\Delta \psi \sin \epsilon_0$ term, but later there is also a $- (\psi^2_A / 2) \Delta \psi \sin \epsilon_0)$ term that could at least partially cancel it for a given value of $\Delta \psi$.  There are also cross terms between $\Delta \psi$ and $\Delta \epsilon$.

While those equations are only for an approximation to IAU 2000B, looking at Capitaine & Wallace 2006 for the exact equation (Eq. 36) shows the same behavior, i.e. that there are both positive and negative terms, as well as cross terms, involving those quantities.

So to do a comparison of accuracy, I think you would need to calculate the $X, Y$ coordinates of the pole and the position $s$ of the origin with both models, and compare those.


How to determine when Sun/Moon is aligned to two location on Earth's surface?

The Photographers Ephemeris is a very nice tool to give you the direction and time of both Moon and Sunset and -rise, made for exactly that purpose: plan and time shots to be at the right place and time for an awesome scenery.
Other programmes and tools to tell you rise and set times are

Stellarium
heavens above
time and date


Yes and no. In general, Dark Matter tends to exist in the same places as galaxies. It isn't really distributed throughout all of space uniformly. It has voids and filaments just like normal matter does. In fact, on the whole, normal matter is a pretty good tracer of where Dark Matter is. In some cases Dark Matter can separate from normal matter, and it's believed by some that Dark Matter can clump in areas where normal matter doesn't, but more or less, on a cosmological scale, it is distributed and acts like regular matter. On short timescales, Dark Matter most just sits there. There isn't mass motion or propagating waves. On long timescales and in very large views, you can see actual motion and interesting structures (e.g., Baryonic Acoustic Oscillations - technically we observe these in normal matter, but Dark Matter would be affected in the same way).

To give you an idea of this, check out the Millennium Simulation (that's a YouTube video which moves around and views the results of the simulation). This was a set of simulations of the universe as a whole, in an attempt to test out current theories about our universe and see if those theories could produce a universe that looks like ours. Suffice to say, they were pretty successful. An image of the results of the simulation is shown below. This represents the universe at a very very large scale and you can see all the dark matter and how it is distributed.

You might also want to check out the Aquarius Simulation which was very similar in nature to the Millennium Simulation. However, unlike the video for the Millennium Simulation, this video shows the evolution of the universe. This really gives you an idea of the flow and motion of dark matter over the entire history of the Universe.


Can the Gaia telescope detect small temporarily captured asteroids near its Lagrange orbit?

Yes.

Detection of inner Solar System Trojan Asteroids by Gaia (pdf):

  The Gaia satellite, planned for launch by the European Space Agency (ESA) in 2013, is the next generation astrometry mission following Hipparcos. While mapping the whole sky, the Gaia space mission is expected to discover thousands of Solar System Objects. The se will include Near-Earth Asteroids and objects at Solar elongations as low as 45 degrees, which are difficult to observe with ground-based telescopes. We present the results of simulations for the detection of Trojan asteroids in the orbits of Earth and Mars by Gaia.


Why are most discovered exoplanets heavier than Earth?

There are a number of methods of detecting exoplanets, but all of them favour detection of larger planets over smaller ones, albeit for slightly different definitions of large:

Radial velocity measurement — this detects the small movement of the star towards and away from us as the planet and the star orbit their mutual barycenter. This movement is fastest when the planet is massive (so the barycenter is further from the center of the star) and close to the star (so the orbital velocity is highest). I also needs the planet's orbit not to be "face-on" to the Earth. This method produces measurements for the $mass\times \sin(i)$ since a more massive planet in a less inclined orbit produces the same motion as a less massive planet in a more inclined orbit

Transverse displacement — this detects the small movement of the star from side to side (against the background of distant stars) as the planet and the star orbit their mutual barycenter. The displacement is largest when the planet is massive and far from the star (although distant planet require observation over a long period of time).  It works best on stars close to us.

Transit — this detects the tiny reduction in the brightness of the star when the planet moves between us. It is more likely to detect large planets, and more likely to notice if the orbital period of the planet is fairly small.


It appears to be static because it's huge beyond your imagination.

The distance to the nebula is 7,000 light years. Its apparent size is 7 arc minutes. Therefore its linear size is about 14 light years.

Think about that. The whole nebula is so big, it takes light 14 years to cross it. Any motion therein must necessarily be much, much slower. No wonder you're not seeing much change.

Data source: wikipedia

Calculation using Wolfram Alpha


How do I find the RA of sunset and sunrise in a specific location?

Use Stellarium. You can set your location and date as needed and simulate all the conditions (ignoring weather...) :)

http://www.stellarium.org/


Is a white dwarf hotter than a Red Giant?

"White" stars are typically much brighter than Red stars, as both the "color & brightness" of a star are directly proportional to the temperature. The only reason there are "bright" red stars is that their radius is incredibly large. Note that the "color" of a star is directly linked to the temperature.

The equation that best demonstrates this is the luminosity equation of a black body. Stars aren't perfect black bodies, but they are close enough that they are treated as such.

  L = 4πR²σT⁴

this equation tells us that the Luminosity (L) is proportionate to the Radius Squared (R²) and the Temperature to the Fourth power (T⁴). The bigger the brighter, or, the hotter the brighter. Meaning that for a given radius the hotter the star, the more luminous, and the same goes for stars of the same temperature, the larger the radius the more luminous.

White dwarfs on the other hand are not stars in the sense that they do not fuse anything, they simply glow due to the lingering heat that was generated during their time as stars.

As shown in the HR-Diagram, White Dwarfs are some of the hottest objects in the universe, and as stated by agtoever there has not been enough time for even the oldest white dwarf to have cooled passed something like 4800K.


Exactly how long does it take for the exposed core of a star to cool from its starting temperature (several billion K) to ~50,000 K?

The answer is of order 1 million years to cool from a standard end of He burning temperature of just over $10^8$ K to the top end of the white dwarf temperature range you give in your question. The details would depend exactly on the mass and composition of the white dwarf and there are also some theoretical uncertainties in neutrino cooling rates.
The surface of a white dwarf is not its interior temperature. If you "expose" the core to empty space then there will always be a transition (i.e. a temperature gradient) between an interior temperature and a "photospheric temperature", which is the temperature at which the plasma density falls low enough for photons to escape.
For example, white dwarfs with photospheric temperatures of 50,000 K to  6,000 K have interior temperatures of about $5\times 10^7$ K to $3\times 10^6$ K. This interior temperature applies to the bulk ($>99$%) of the white dwarf because electron degeneracy leads to high thermal conductivity and the surface layer where the temperature falls and the electron degeneracy disappears is very thin because of the high gravity.
So the cooling timescale for cores that become white dwarfs may not be as short as you think. The thin, outer, non-degenerate layer acts like an insulating jacket around a hot water tank. The cooling timescale (the time to radiate all internal heat at its current luminosity) changes from tens of millions of years at interior temperatures of $3\times 10^7$ K to billions of years at interior temperatures of $3\times 10^6$ K, thanks to the strong dependence of luminosity on temperature.
However, this story only works once the core has cooled to around $3\times 10^7$ K from its initial temperatures. Above that temperature, the main cooling mechanism is via neutrino emission and the insulating jacket does not stop them. Neutrino emission is very temperature dependent $(\sim \propto T^{10})$. So the initial cooling phase to $\sim 3\times 10^7$ K is very rapid once nuclear reactions have ceased; about a million years. After that the neutrino losses become nmuch smaller quite quickly and certainly photon emission is dominant by the time the interior temperatures are $10^7$ K and the photospheric temperature is <30,000 K.
A similar process occurs in neutron stars. Very rapid cooling by neutrinos for $\sim 10^5$ years, followed by slower photospheric cooling later. The difference is that most of the white dwarfs we see are in the photospheric cooling phase, whereas the neutron stars we can see are cooling with neutrinos. Neutron stars have such small radii that they become effectively invisible once they reach the photospheric cooling phase.


The rate of shrinkage is decreasing. The reason is that the core of Jupiter is becoming increasingly electron degenerate. Electron degeneracy pressure is almost independent of temperature. Once degeneracy sets in, an object can continue to cool with almost no change in its interior pressure, and hence no radius change. A cosmic example is a white dwarf.
This is not an on/off process, so at no point will you be able to say "Jupiter has now stopped shrinking", but Jupiter's radius will not reduce by more than about 10-20% from what it is now. The ultimate fate of Jupiter is to become a cold, ball of metallic hydrogen (with impurities).


Venus' magnitude during inferior conjunction

I found references which indicate that Venus usually goes dark.  Check out the graphs in this paper for one example.

The exceptions happen when Venus' orbit, which is slightly out of the ecliptic, brings it to inferior conjunction when projected onto our (Earth's) orbital plane, but Venus remains at a slight angle relative to our viewing lines.

From UniverseToday,

  Not all inferior conjunctions of Venus are created equal. The planet’s
  orbit is tilted 3 degrees with respect to our own and can thus pass a
  maximum of eight degrees north or south of the Sun. Venus last did
  this on inferior conjunction in 2009 and will once again pass a
  maximum distance north of the Sun in 2017. For the southern
  hemisphere, the red letter years are 2007, and next year in 2015.


Does a White hole lies behind a Black hole?

No, there is probably no such thing as a white hole in real life. A white hole is basically a time-reversed version of a black hole. Just as a black hole cannot be destroyed in classical General Relativity, a white hole can never be formed in GR. So while a white hole is technically a thing that could exist, there is no way to form one unless it was there at the beginning of the universe. We don't have any reason to think white holes existed at the Big Bang, so we think they don't exist (and never will).


How accurate are rederings of something entering Earth's atmosphere?

In the movies you reference, I doubt that it was ever the intention of a movie maker to suggest that the craft was moving at a crawl when it began to enter the atmosphere. If the craft doesn't look like it is moving very fast, the idea is probably that the point of reference (the "camera") is moving at nearly the same speed as the craft, or perhaps that the craft is so huge, that the reference point is miles away. Everything is relative.

There is essentially no realistic scenario where any object would enter the atmosphere at such a slow speed. Imagine, for example that you have an object, like Felix Baumgartner or something, which is standing approximately stationary in space above the Earth's atmosphere. Gravity will immediately begin pulling on that object and because there is no atmosphere to slow the object down yet, it just keeps accelerating as it falls. Even as it enters the edge of the atmosphere, it doesn't slow down much, because there isn't much atmosphere yet. Felix was moving more than 800 mph before he started to slow down.

The International Space Station is another object that's sort of "stationary", in a manner of speaking. It's not going anywhere because it's in a stable orbit (sort of). But in order to stay up there, it's actually moving more than 17,000 mph, relative to the atmosphere. If something is "stationary" with respect to the sun, just sitting there in Earth's orbit as Earth comes at it, the Earth's atmosphere would smack it at 67,000 mph. That's how fast Earth orbits the sun. So there's really isn't anything that would just gradually drift into the atmosphere at 1 to 10 mph.

If a craft were to use rockets to halt itself before entry, as you suggest, remember Felix again; it can become stationary with respect to the Earth, but then gravity begins its work, so if the craft wants to remain at a very slow speed through the entire decent, it has to keep firing those rockets all the way down. And certainly, nothing would burn up at that pace, other than a lot of rocket fuel. What causes the burning is not something special and hot about the atmosphere, but the fact that the object is smashing into the air (thin as it may be at that altitude) at thousands of mile per hour. It is even theoretically possible for meteors to enter at such speeds that the atoms are literally forced to hit each other's nuclei, and a part of the fire generated would actually a brief nuclear reaction.

Response to Question Edit:
(Thank you HDE. The initial question was very unclear.)
The speed at which friction completely stops is 0. Any movement at all, even in that 1 - 10 mph range, produces a small amount of heat from friction, even if the medium you are moving through is very thin air (though certainly nothing one would characterize as "burning"). You seem to be looking for the speed at which that friction would be low enough to not cause a problem, but there's no magic answer. It depends on the situation. How delecate is the craft? How much wear is acceptable? Essentially, this question becomes something very similar to asking "How fast can I drive with something sticking out the window?" You are trying to keep the entry speed low enough that the wind is not a problem for you as the density of that wind gradually increases from practically nothing to surface air density. If your craft has a heavy, durable heat sheild, properly oriented, then extraordinary speeds are still not a real problem. But if you are trying to create a sand mandala on the roof of your craft as you descend, then you are going to want to control that wind level very carefully.


Was the discovery of six exoplanets around one star as "easy" as counting six peaks in the FT?

I suspect that the record holder (as of 14/2/2017) is HD 10180 which has at least 7 planets and possible evidence for as many as 9.

Lovis et al. (2011) announced the initial discovery based on 190 radial velocity measurements taken over 6 years. The precision of the measurements was 0.3-0.9 m/s.

Section 4 of that paper describes how they go about finding the planets in the data. It is a hybrid of fourier and fitting methods. As each peak in the periodogram is found it is added to a model which is then iterated towards a best solution.

Successive planets reduce the rms scatter around the predicted radial velocity curve. Ultimately a judgement has to be made whether the improvement in the fitting statistic justifies the addition of another planet (and more free parameters) to the model. The final rms after adding 7 planets was just over 1 m/s, which is worse than the expected precision, but which Lovis et al. attribute to (probable) radial velocity jitter due to stellar activity. The orbital model is then refined to include the effects of planet-planet interactions and tidal forces.

The radial velocity amplitudes attributable to each planet range from 0.8-4.5 m/s. The most marginal detection has the smallest amplitude, but the shortest period (more cycles and hence easier to detect smaller amplitudes).

A later paper by Tuomi (2012) used a more conventional Monte Carlo Markov Chain in a Bayesian framework to fit a model of $n$ non-interacting planets to the radial velocity data. Again, there is lots of discussion (see section 3.3 and 3.4) of exactly how many planets are required to fit the data. Tuomi claims there is strong evidence for an 8th and 9th planet in their analysis.

There are a number of important assumptions made in these types of analysis. The main one is that you have to assume some sort of model for the background noise and it is often assumed to be Gaussian and non-periodic.


What would happen when the speed of the Sun's spin changes?

Does the Sun's spin have affect on the orbit of the Earth?

It has a very, very small effect. The Sun would have a larger equatorial bulge if it rotated faster, and a smaller equatorial bulge if it rotated slower. An equatorial bulge increases gravitational acceleration near the plane of the body's equator and decreases gravitational acceleration above  below the body's poles.
Since the Sun is already rotating rather slowly, it doesn't have much of an equatorial bulge. Moreover, the effects of oblateness drop off as 1/R^4, as opposed to the 1/R^2 drop-off in Newton's law of gravitation for a point mass.
The combination of small oblateness, 1/R^4 drop-off, and large distances makes the effects of the Sun's oblateness on the planet's orbits very small.
The effect is very, very small, about three orders of magnitude smaller than the general relativistic effect on Mercury's precession (which is already very small, about 43 arc seconds per century). The primary effect of the Sun's oblateness is to make Mercury precess by 0.0286 arc seconds per century. That said, the extremely precise observations of Mercury's orbit about the Sun do make that tiny precession statistically important because the uncertainty in Mercury's precession rate is down to
0.0015 arc seconds per century, according to Park, Ryan S., et al. "Precession of Mercury’s Perihelion from Ranging to the MESSENGER Spacecraft." The Astronomical Journal 153.3 (2017): 121.


Is there any real evidence to prove or disprove the existence of alien civilizations?

Is there any REAL proof that alien civilizations exist in outer space?

No.

  What if someone would say that Earth is the only inhabited planet in the whole Universe, how would you respond?

Depends on what their goal is. If their goal is to convince me of that, I'd just try to avoid the conversation. If their goal is to come to an understanding of the current body of scientific knowledge on the matter, I'd talk about how we don't know that that statement is true.

  I need to write a 5 paragraph argumentative essay to prove that alien civilizations exist. I need to write an introduction, 3 paragraphs supporting my statement that aliens exist, and a conclusion. In each of the three "body" paragraphs I need to describe in detail a piece of evidence supporting my claim.

That will be difficult, given that there is no such proof. Best you could do is argue that alien civilizations likely exist, given a certain set of assumptions.

  I don't even know where to start. I mean, I want to find out the opinion of some scientists who think that we are alone, and then disprove their claim in my essay.

That is not a good goal for an essay. Don't start with the goal of disproof. Start with the goal of understanding.

  I need to know the latest scientific evidence on the existence of alien civilizations. I don't mean like alien abduction stories or any of that science fiction stuff. I need some real proof that will be convincing and not undermine my argument.

Look at the Drake Equation and recent research that concludes that there are many planets in the galaxy.

  How can you know for sure if there are alien civilizations in outer space?

You would need observational evidence.

  What do they look like? On which planets do they live? What kind of technology do they use?

All unknown.

  Why are we not alone in the Universe?

This question presumes the conclusion that we are not alone. We might be alone.

  What evidence disproves the claims of Earth chauvinists?

None.


where is the boundary between dark night and grey night

What a nightmare - there is no accepted definition.

A common method is to use the fraction of the night (between the times of astronomical twilight) that the moon spends above the horizon. The problem is that this does not correlate perfectly with fractional lunar illumination or sky brightness. It does provide a good method of deciding when a night is "dark", but you get a  bit of a mixed bag of "grey" nights.

It is found that the sky brightness is pretty much determined by fractional lunar illumination (FLI - the fraction of the moon's visible hemisphere illuminated by the Sun) once it exceeds about 0.15; so this is the definition that is currently used at the Isaac Newton Group of Telescopes.

Dark $0\leq$ FLI $<0.25$; Grey $0.25\leq$ FLI $<0.65$; Bright $0.65 \leq$ FLI $\leq 1$, where the FLI is calculated at 0h UT (close to midnight at the ING).

See  http://www.ing.iac.es/PR/newsletter/news6/tel1.html for further discussion (in which I only just noticed I got a mention in the acknowledgements ;) ).


If an orbit is shifting due to orbital precession, is it still a Keplerian orbit?

No orbits are strictly Keplerian, unless you really monkey about with your coordinate systems. However, a precessing orbit is pretty darned Keplerian.

The relationship between the various orbital elements can be a little complicated, and precession is traditionally thought of as the rotation of the periapsis of the orbit around the more massive body.  If we look at the orbital elements, we can think of it instead as a changing term to the longitude of the ascending node.  So we could capture the longitude of the ascending node, $\Omega_0$ at some initial time $t_0$.  Then, at any time t, the longitude of the ascending node would be $\Omega=\Omega_0+(t-t_0)\omega_p$, where $\omega_p$ is the rate of precession.  The rest of the orbital elements can be treated as constant.


What percentage of a spiral galaxy is the center/bulge?

The term you're looking for is called the bulge-to-disk (size) ratio. Sort of by definition, the answer depends on the morphology of the galaxy, i.e. how "late-type" spiral it is. "Sa" spirals are the ones that resemble ellipticals the most, and hence have large size ratios (of order, but below, unity), whereas bulges in Sc galaxies are (less than) one-tenth the size of the disk.

(EDIT: In fact, the way you phrase your question, you're looking for bulge-to-total, rather than bulge-to-disk, but there's a 1:1 correspondence between. Also, in retrospect I think you're interested in the mass ratio, but I interpreted "how big" as the size ratio. Peter Erwin's answer discusses the masses.)

The ratio is obtained by fitting the luminosity distribution of the bulge and the disk separately. Typically both are fitted as exponentials, with scale lengths $R_\mathrm{b}$ and $R_\mathrm{d}$, respectively, but other forms (e.g. Sérsic profiles) are also used. In this case, the effective radius $R_\mathrm{eff}$ is used, i.e. the radius inside which half the light is emitted. Furthermore, the answer will depend the band in which you observe the galaxy (i.e. IR, optical, UV, …)

The following figure (from Möllenhoff 2004), shows the ratio $R_\mathrm{eff,b}/R_\mathrm{d}$  as a function of "Hubble type" going from 1 (Sa), to 3 (Sb), to 5 (Sc). I've annotated examples of galaxies of Hubble type 1, 3, and 5.

You see that the ratio goes from $\sim0.5\text{–}1$ for early-type spirals, to $\sim0.01\text{–}0.2$ for late-type spirals.

The different symbols correspond to different filters going from the I band (infrared) denoted by squares, to the U band (ultraviolet) denoted by circles; they have been offset slightly along the $x$ axis for visualization purposes — you see that the ratio decreases slightly for shorter wavelengths. In other words, the bulge is less prominent, the bluer the light you consider.


Could the James Webb Space Telescope detect biosignals on exoplanets?

From what I understand, James Webb, if used in conjunction with a successful starshade (being developed at MIT), should be able to detect close in planets orbiting nearby stars. However, getting good atmospheric spectra of these planets directly (from planet's blackbody IR emission) is unlikely. What we must hope for is that TESS, which should be going up in 2017, will find a few nearby stars with transiting planets. Then, James Webb will be able to look for atmospheric absorption lines from stellar light passing through a planet's atmosphere during transit. This method may still be limited to large (Jupiter size) planets.
In an ideal situation (say looking at absorption lines of a super-earth) there are many "bio-signatures", but one of the easiest to detect would be an ozone line in the infrared. By itself, this would not be proof, even though there would need to be a constant replenishing source of O2 in the atmosphere to maintain the O3. If methane could also be found we could rightly get VERY excited since methane and oxygen don't co-exist very well.


Basically, go back to the pre-planet particle clouds, and as explained in

geek.com,

  To answer the question we have to go back to the beginning, to the
  formation of a solar system like ours. Before a star and its planets
  exist, there’s just a cloud of disorganized gas and small molecules.
  This is often called a molecular cloud or “stellar nursery.” The Eagle
  and Orion Nebulae are some of the more famous stellar nurseries we
  have been able to observe with the Hubble Space Telescope. These
  clouds are composed mostly of molecular hydrogen, which wouldn’t be
  able to congregate outside the dense molecular clouds. These clouds
  can be of any size — not just massive structures like the Orion
  Nebula. Over time the energy of the molecules in the cloud pushing
  outward can be overcome by slower molecules collapsing together
  farther in. As long as there is sufficient mass in the molecular
  cloud, it continues collapsing in toward the center until it reaches a
  high enough mass to fuse hydrogen and become a new star. The spin that
  we see quite clearly now is related to the process of the molecular
  cloud collapsing. The original cloud was very, very large and made up
  of many individual molecules and small clumps of matter. On that
  scale, there is some small amount of rotation within the cloud. It
  could be caused by the gravity of nearby stellar objects, local
  differences in mass as the cloud churns, or even the impact of a
  distant supernova. The point is, most molecular clouds have at least a
  little rotation.

  As the cloud collapses to form a star, it has what physicists call
  angular momentum. This is the movement an object has as it rotates
  around a central point. In a large system like a molecular cloud, each
  particle has some angular momentum, and it all adds together across a
  very wide area. That’s a lot of momentum, and it is conserved as the
  cloud continues to collapse. But how does that get us to objects that
  spin and orbit?

  Imagine a figure skater spinning around with arms outstretched. That
  is a model of angular momentum just like a collapsing cloud of gas.
  When the arms are drawn inward, the rotational velocity goes up
  because the total angular momentum is conserved unless there is some
  external force acting on it. There are such forces acting on the
  figure skater, but less so on a collapsing molecular cloud. So if a
  molecular cloud was maybe a light year across, then collapsed down to
  be just a fraction of that, it would be a huge change in size. Just
  like the figure skater pulling her arms in, the velocity must increase
  to conserve angular momentum and therefore form a spinning
  protostellar disc. It is from this orbiting matter that all the
  planets form, and of course, they are also spinning and orbiting in
  the same direction because of the conservation of angular momentum.

Quoting from wikipedia as to the mystery of  the misbehaving planets:

  All eight planets in the Solar System orbit the Sun in the direction
  that the Sun is rotating, which is counterclockwise when viewed from
  above the Sun's north pole. Six of the planets also rotate about their
  axis in this same direction. The exceptions—the planets with
  retrograde rotation—are Venus and Uranus. Venus's axial tilt is 177
  degrees, which means it is spinning almost exactly in the opposite
  direction to its orbit. Uranus has an axial tilt of 97.77 degrees, so
  its axis of rotation is approximately parallel with the plane of the
  Solar System. The reason for Uranus's unusual axial tilt is not known
  with certainty, but the usual speculation is that during the formation
  of the Solar System, an Earth-sized protoplanet collided with Uranus,
  causing the skewed orientation.[6]

  It is unlikely that Venus was formed with its present slow retrograde
  rotation, which takes 243 days. Venus probably began with a fast
  prograde rotation with a period of several hours much like most of the
  planets in the solar system. Venus is close enough to the Sun to
  experience significant gravitational tidal dissipation, and also has a
  thick enough atmosphere to create thermally driven atmospheric tides
  that create a retrograde torque.


You are right that it's surprising that Titan, being just a moon, has a thick atmosphere. Usually, the answer includes magnetism: Earth has an atmosphere because the liquid magma inside the planet produces a magnetic field. This magnetic field changes the paths of the particles in the solar wind, thus preserving the volatile gases intact. Mars did use to have an atmosphere, just like Earth, but it being farther away from the Sun, the magma froze and lost its magnetic properties.
Titan itself doesn't have a magnetic field, but Saturn does. Saturn's magnetosphere is produced by the motion of the super-compressed hydrogen gas inside the planet (metallic fluid). The field is so strong it encompasses the satellites, including Titan.
Saturn doesn't have the same magma as Earth. Earth is a rocky planet. This means that it was formed so close to the sun that light gases (such as hydrogen) could not coalesce due to the high temperature and the solar wind. Therefore, the inner planets (Mercury to Mars) are composed mostly of rock and metal. It's the liquid iron in the outer core that makes Earth magnetic.
Gaseous planets instead have a small solid metal/rocky core (therefore, no magnetic field from there), and a huge layer of light gases (hydrogen and helium). The gases are not usually magnetic, but under such an enormous pressure they take a "metallic" structure, which means they can conduct electricity like a metal. This same property allows them to generate a magnetic field.