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How are the compositional components of exoplanet atmospheres differentiated?
How are exoplanetary atmosphere compositional spectra distinguished from those of the parent star(s), from the composition of the planetary surface or any other factor?  Is it actually possible to determine the atmospheric composition precisely using this method?
Are there any specific examples of such an analysis having been performed on an exoplanet?

Meta discussion on the relevance of exoplanet topics: http://meta.astronomy.stackexchange.com/questions/1/where-do-we-draw-the-line-concerning-exoplanet-questions

In theory one should just be able to determine the difference between the spectra during the star's eclipse of the exoplanet (starlight alone) and the spectra of the star and exoplanet together, but in reality equipment is not precise enough for this. To remedy this problem, the analysis is integrated over many eclipses. Some other calibrations are made as well which you can read more about in the paper A Spitzer Spectrum of the Exoplanet HD 189733b. This is called occultation spectroscopy.
There's another method used as well called transmission spectroscopy, which detects the change in light as the exoplanet passes in front of its star. You can read more about that in this paper: Infrared transmission spectroscopy of the exoplanets HD 209458b and XO-1b using the wide field Camera-3 on the Hubble Space Telescope.

Any progress in accuracy, say in masking, in the last several years?

@JerardPuckett I'm not an expert, but here is a more recent paper: http://iopscience.iop.org/article/10.3847/2041-8205/820/2/L33/meta Similar methods seem to be used.

This isn't the only technique, but combining this answer with the one in https://astronomy.stackexchange.com/questions/15037/how-do-astronomers-detect-gases-that-are-in-the-atmosphere-of-exoplanets?noredirect=1&lq=1 basically covers it.

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Why do sunspots appear dark?
Sunspots, such as this one, appear dark:

Why?

Considering how bright the sun is, the sunspots could appear relatively dark and still be quite bright.

Plotting them in dark, is just a representation, not of visible light, but of relative temperature (infra-red light).

What prior research was done here? It is easy to answer based on wikipedia and other readily available resources that are just a simple google search away. Indeed you appear to have answered your own question using just such resources. It is baffling to me that such a simple question (and your answer) have received so many upvotes.

@RobJeffries It might have received so many upvotes because it appears on the first page of Google for "sunspots dark". It's probably helping people, which isn't a bad thing. Also, this is a self-answered question, a feature of the system which users are encouraged to utilize. If you have more information, please feel free to write your own answer.

I see. However, the question could be answered by minimal research effort. Indeed the second block quotation that you use is actually from the wikipedia page for sunspots, not from the NASA resource you cite.

Typical sunspots have a dark region (umbra) surrounded by a lighter region, the penumbra. While sunspots have a temperature of about 6300 °F (3482.2 °C), the surface of the sun which surrounds it has a temperature of 10,000 °F (5537.8 °C).
From this NASA resource:

Sunspots are actually regions of the solar surface where the magnetic field of the Sun becomes concentrated over 1000-fold. Scientists do not yet know how this happens. Magnetic fields produce pressure, and this pressure can cause gas inside the sunspot to be in balance with the gas outside the sunspot...but at a lower temperature. Sunspots are actually several thousand degrees cooler than the 5,770 K (5496.8 °C) surface of the Sun, and contain gases at temperature of 3000 to 4000 K (2726.9 - 3726.8 °C). They are dark only by contrast with the much hotter solar surface. If you were to put a sunspot in the night sky, it would glow brighter than the Full Moon with a crimson-orange color!

Sunspots are areas of intense magnetic activity, as is apparent in this image:

You can see the material kind of getting stretched into strands.
As for the reason it is cooler than the rest of the surface:

Although the details of sunspot generation are still a matter of research, it appears that sunspots are the visible counterparts of magnetic flux tubes in the Sun's convective zone that get "wound up" by differential rotation. If the stress on the tubes reaches a certain limit, they curl up like a rubber band and puncture the Sun's surface. Convection is inhibited at the puncture points; the energy flux from the Sun's interior decreases; and with it surface temperature.

All in all, the sunspots appear dark because the are darker than the surrounding surface. They're darker because they are cooler, and they're cooler because of the intense magnetic fields in them.

Is there any image of the sunspot alone, so that we can see that it's actually pretty bright?

That would be a neat picture. I'll look for some, although I somewhat doubt I'll find any.

I asked because my searches failed. It would be interesting to see the structure of the area which is black on most pictures.

Sunspots are cooler because their magnetic fields inhibit the replenishing of heat from convective flows (due to the interaction between plasma and magnetic fields). This allows them to cool radiatively. The rest of the solar surface is constantly being replenished by convective cells that reheat it.
Solar plasma at the photosphere radiates roughly as a black-body, meaning that the energy (and wavelength) spectrum of radiation follows the Planck function. The Planck function scales up and down with temperature at all energies (and wavelengths). The brightness of a blackbody at a given energy (or wavelength) is determined by the value of the Planck function at that energy.
The image you showed is taken in the "photospheric continuum", which is a black-body dominated part of the radiation spectrum. So, because sunspots are cool (compared to their surroundings), this means that their Planck function is lower than their surroundings, and hence their brightness is lower, causing them to appear dark in the image.

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Why can't light escape from a black hole?
I've heard that light can't escape from a black hole. Can it? If not, why?

Related questions on Physics.SE: http://physics.stackexchange.com/questions/25369, http://physics.stackexchange.com/questions/46591, http://physics.stackexchange.com/questions/33916, http://physics.stackexchange.com/questions/8477, http://physics.stackexchange.com/questions/28297, and probably more.

A black hole has an event horizon which 'marks the point of no return'. So yes, light cannot escape from a black hole.
Why? Well, think of a 'spacetime fabric'. It's the easiest way to understand the physics at work here, in my opinion.
Usually, the fabric would look like this:

(source: whyfiles.org)
However, a black hole has so much gravity that one could say it 'rips' the spacetime fabric:

(source: ddmcdn.com)
When the light hits this area of amazingly intense gravity, it simply cannot get out - the light travels 'along' the fabric, and since there is a rip in the fabric, one could say it simply goes away - it becomes part of the singularity.
This is a simplification, of course, but it's enough to understand at least part of the physics behind this phenonenom.

Is the time dilation effect of the event horizon itself enough to prevent light from escaping? Or is that only a small component of what keeps light trapped in a black hole?

That's terrible. There is nothing there in the fabric-of-spacetime cartoon that addresses the horizon or why it's important. It's literally irrelevant--you're answering something like "what happens to stuff after it falls into a black hole?" instead.

I like to think of this in terms of escape velocity.
Escape velocity is the speed needed to escape the gravitational pull of a given object. For the Earth, that speed is 11.2 km/second (Mach 34!). When rockets blast off from Earth, they aren't trying to achieve a certain height or altitude, they're trying to reach a certain speed, the escape velocity.
Once a rocket reaches 11.2 kips*, it has attained the speed needed to leave the Earth completely. If a rocket fails to attain that speed, regardless of its height, it will fall back to the Earth. (You can imagine a magical balloon that slowly lifts you up into space, up past the ISS and most satellites, and then you let go: since you aren't going fast enough, you will fall back down, past all the satellites, and crash into the earth.)
Smaller gravitational bodies, like the moon, have smaller escape velocities. That's why the lunar landers were able to leave the moon with such a small ascent stage, compared to the massive Saturn V it took to leave Earth: they only had to go 2.4 km/second.

To escape the Sun, you'd have to go 617.5 km/second!
Fortunately for us, light goes faster than 617.5 kips, so we're able to see the light created on the Sun. However, as you increase the mass of an object, eventually the escape velocity would meet or exceed 299,792km/s, the speed of light. At that point not even light itself can go fast enough to escape the gravity well, and will always be pulled back down into the black hole.
*Short for "kilometers per second"

escape velocity for Earth is 11.2kps; however, this applies to thrown objects; you have to throw a rock at 11.2kps (ignoring atmospheric drag) in order for it to leave the Earth and not fall back; however, if your rock has an engine that can apply thrust, it can leave the Earth at a much lesser speed. The longer it is able to apply thrust, the slower it can go when leaving.

jmarina, that's interesting, I haven't heard that. Would you mind providing a link with more info or the name of the effect you're describing?

@jmarina is kinda right but the explanation is more interesting.  Escape velocity actually decreases with distance from the body you are trying to escape from.  For instance at 9,000 km up the escape velocity is about 7.1 km/s.  The reason is that if you are going that speed aimed to just miss the Earth then you would pick up extra speed from falling towards it.  And while the escape velocity from the sun  at the sun's surface is 617.5 km/s, at Earth's orbit it is only 42.1 km/s.

Ah, I see. The escape velocity is factoring in the "free" velocity you get from gravitational pull if going back down (but angled enough to miss the planet). Is that right?

@brentonstrine the free velocity you mention you get from a gravity assist: http://www2.jpl.nasa.gov/basics/grav/primer.php
the orbital velocity of the Earth is about 30km/sec, it varies a bit down to 29kps because the orbit is not an exact circle, we are actually closer to the sun in northern winter and farther by about a million km in summer.
http://nssdc.gsfc.nasa.gov/planetary/factsheet/

The argument in terms of escape velocity is deeply incorrect, since it seems to allow for escape from a black hole if a force is continuously applied - which isn't in fact possible.

@RobJeffries it's not an 'argument' just a helpful way of thinking about it for someone who is learning about black holes. I'm trying to think of another way to explain this which is both simple for a beginner to understand and doesn't have misleading implications, but I can't think of one--if you can I'm sure many people would benefit from a new answer to this question.

Don't forget that if a black hole has less than the current stable mass of a black hole (3 solar masses) then it evaporates - transforming its mass in radiation, in which case it would give off light, mostly X rays and gamma, at an increasing rate as its mass decreases, until the entire black hole is turned into a flash of hard radiation.
http://en.wikipedia.org/wiki/Hawking_radiation
However, this light is the mass of the black hole escaping in the form of the most basic form of energy.

OK would whoever is flagging answers here as "not an answer" please stop? There's a description under flags explaining when they should be used, and that isn't the case for any of the answers here in this thread. If that is not clear, please consult our [help] pages or stop by in our [chat] to discuss what you think might be wrong with any of the answers provided here. Flags should be used for pressing issues with posts, not to indicate your disagreement with them. That's what voting is for! Thanks!


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How are black holes found?
Black holes have so much gravity that even light can't escape from them. If we can't see them, and the suck up all  electromagnetic radiation, then how can we find them?

NASA has announced that it has isolated the mass and location of what could be a “wandering” black hole using the Hubble Space Telescope. This is the first time in the space agency’s history it has been able to achieve this, despite there being over 100 million black holes populating our galaxy. petapixel.com/2022/06/10/… nasa.gov/feature/goddard/2022/…

To add to John Conde's answer.  According to the NASA web page "Black Holes", detection of black holes can obviously not be performed by detection any form of electromagnetic radiation coming directly from it (hence, can not be 'seen').
The black hole is inferred by observing the interaction with surrounding matter, from the webpage:

We can, however, infer the presence of black holes and study them by detecting their effect on other matter nearby.

This also includes detection of x-ray radiation that radiates from matter accelerating towards the black hole. Although this seems contradictory to my first paragraph - it needs to be noted that this is not directly from the black hole, rather from the interaction with matter accelerating towards it.

There are many, many ways of doing this.
Gravitational lensing
This is by far the most well known. It has been mentioned by the others, but I'll touch on it.

Light coming from distant bodies can be bent by gravity, creating a lens-like effect. This can lead to multiple or distorted images of the object (Multiple images give rise to Einstein rings and crosses).
So, if we observe a lensing effect in a region where there isn't any visible massive body, there's probably a black hole there. The alternative is that we are peering through the dark matter 'halo' which surrounds (and extends passed) the luminous components of every galaxy and galaxy cluster (See: Bullet Cluster). On small enough scales (i.e. - the central regions of galaxies), this is not really an issue.

(This is an artist's impression of a galaxy passing behind a BH)
Gravitational waves
Spinning black holes and other dynamical systems involving black holes emit gravitational waves. Projects like LIGO (and eventually, LISA) are able to detect these waves. One major candidate of interest for LIGO/VIRGO/LISA is the eventual collision of a binary black hole system.
Redshift
Sometimes we have a black hole in a binary system with a star. In such a case, the star will orbit the common barycenter.
If we observe the star carefully, its light will be redshifted when it is moving away from us, and blueshifted when it is coming towards us. The variation in redshift suggests rotation, and in the absence of a visible second body, we can usually conclude that there's a black hole or neutron star there.
Salpeter-Zel'dovitch / Zel'dovitch-Novikov proposals
Going in to a bit of history here, Salpeter and Zel'dovitch independently proposed that we can identify black holes from shock waves in gas clouds. If a black hole passes a gas cloud, the gases in the cloud will be forced to accelerate. This will emit radiation (X-rays, mostly), which we can measure.
An improvement on this is the Zel'dovitch-Novikov proposal, which looks at black holes in a binary system with a star. Part of the solar winds from the star will be sucked in to the black hole. This abnormal acceleration of the winds will, again, lead to X-ray shock waves.

This method (more or less) led to the discovery of Cyg X-1
Cosmic gyroscopes
Cyg A is an example of this. Spinning black holes act like cosmic gyroscopes — they do not easily change their orientation.
In the following radio image of Cyg A, we see these faint gas jets emanating from the central spot:

These jets are hundreds of thousands of light years long — yet they are very straight. Discontinuous, but straight. Whatever object lies at the center, it must be able to maintain its orientation for very long.
That object is a spinning black hole.
Quasars
Most quasars are thought to be powered by black holes. Many (if not all) of the candidate explanations for their behavior involve black holes with accretion disks, e.g. the Blandford-Znajek process.

Another thing to point out is that these observational methods typically set a density limit on the invisible object such that the only thing that would reasonably explain it would be a black hole.

I should note that this answer can be misleading, as many of these methods are only possible ways of finding them (e.g. gravitational waves have not been detected at all from anything and gravitation lensing has never found a black hole). In reality, in external galaxies we look for distinctive emission spectra or timing characteristics of the material around around the black hole (by its X-ray, UV, optical or infrared emission). Jets, in particular aren't necessarily from black holes but can be made by star formation. It's also difficult to distinguish a black hole like object from a BH.

Why don't you say which of these techniques are actually used to find black holes.

@Aaron No. Looking at binary motion tells you the mass, not the density.

@RobJeffries the comment (which was originally an answer) doesn't specifically refer to binary motion, but to the collection of methods above discussed both determine black hole mass and restrict it to some likely volume. Black holes aren't defined by their total mass, but their density. Even for the binary motion example, the closest approach of the objects can allow you to make an estimate on the likely largest size the object could be (excluding issues with projection effects if you don't have 3D orbital measurements)

A black hole can also be detected by how it bends light as various bodies move behind it. This phenomenon is called gravitational lensing, and is the most visually stunning prediction of Einstein's theory of General Relativity.
This image portrays the geometry of gravitational lensing. Light from luminous background objects are bent due to the warping of space-time in the presence of mass (here, the red dot could conceivably be the black hole in question):

Astronomers have discovered the existence of a super-massive black hole at the center of our very own Milky Way Galaxy, and has been dubbed Sagittarius A*.
Over a period of ten years, the trajectories of a small group of stars have been tracked, and the only explanation for their rapid movement is the existence of a highly compact object with the mass of about 4 million suns. Given the mass and distance scales involved, the conclusion is that it must be a black hole.

As your picture suggests, their elliptical trajectory also strongly suggests the central object is not only very massive but also extremely compact.

*No * black holes have been found using the first method outlined in this answer.

One way is by following Gamma Ray Bursts. When a black hole feeds on surrounding gas or swallows a star that got too close, they often emit gamma ray bursts which are very energetic and easy to spot (although they don't last long).
In the case of super massive blackholes, they are seemingly at the center of every medium and large galaxy. It makes where to look rather easy.

All 4 answers given prior to this one are very good and complete each other;
finding an object orbiting your target object enables you to also calculate the mass of your target object.
Matter falling into a black hole is accelerated toward light speed.
As it is accelerated, the matter breaks down into subatomic particles and hard radiation, that is, X-rays and gamma rays. A black hole itself is not visible, but the light (mostly X-rays, gamma rays) from infalling matter that is accelerated and broken up into particles is visible.
By looking toward the center of our galaxy, the Chandra X-ray space telescope has observed several black holes besides Sgr A*, indirectly, by catching the hard radiation of infalling matter flaring up as they swallow something; afterward, the black holes go dark again if there is nothing more to assimilate nearby;
http://chandra.harvard.edu/press/05_releases/press_011005.html
Here you can see some of this flaring in the swarm of black holes near the center of our galaxy.
Methods to detect black holes (which are not really holes or singularities, as they do have mass, radius, rotation, charge and hence density, which varies with radius, see http://en.wikipedia.org/wiki/Schwarzschild_radius ).

to passively detect a (stellar or supermassive) black hole, look/wait for hard radiation flares, which occur sporadically, then follow up with observations to see if you caught a grb (gamma ray burst) from an actual black hole or just a white dwarf or neutron star doing a periodic nova;
to actively detect a black hole look for gravitational lensing, which is a continuous effect, or stars orbiting at a high speed around a seemingly empty point in space, such as S2 at 5000+km/sec, around Sgr A*

http://en.wikipedia.org/wiki/S2_(star)
But there will be nothing left to see what caused it; better have some observations of that spot in the sky before it happens.

You should definitely check the time it would take for a three solar mass black hole to evaporate.

The Tolman-Oppenheimer-Volkov limit hasn't anything to do to the Hawking-radiation! It is a very hard factual mistake.

"swarm of black holes"?


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What is the current accepted theory as to why Venus has a slow retrograde rotation?
According to this NASA overview, the planet Venus is unique (amongst the major planets), Venus has a slow retrograde axial rotation, taking 243 Earth days to make one rotation (which is longer than its orbital revolution).
What is the current accepted theory as to why (and how) Venus developed this anomalous slow retrograde axial rotation?

There seem to be a few, and none are accepted by the whole scientific community. The main ones:

Venus was struck by a large body during its early formation
The spin axis flipped, as can happen with a gyroscope
The spin slowed to a standstill and then reversed, caused by the sun's gravity, the dense atmosphere and friction between core and mantle

That final one seems to be the most recent, being proposed by Alexandre Correira and Jacques Laskar in 2001. Their research seems to imply that the conditions on Venus and its distance for the sun make a retrograde spin slightly more likely than a forward one.

To a first order approximation, the third explanation has to be wrong, because (as of 2012) Venus is known to be slowing down. See Could Venus be shifting gear?. If it slowed to a standstill and then reversed, for any gravitational reason, then to first order it would now be speeding up. Of course, there might be a second order oscillation, but that's more complicated and, I assume, not what the authors (back in 2001) were proposing.

There is also the theory which involves Mercury as an ex-moon of
Venus, largely based on calculations done by Van Flandern and Harrington (A Dynamic Investigation of the Conjecture that Mercury is an Escaped Satellite of Venus. Icarus 28: 435-40 (Abstract), 1976) and goes as follows (Van Flandern, Missing Planets, Dark Matter, and New Comets, 1999):

As Mercury tidally drifted outward it necessarily produced rotational
  drag on Venus, and it raised even bigger tides on the Venusian
  atmosphere causing it to circulate in retrograde direction. After
  billions of years this might impart retrograde motion on the whole
  planet.
Tides caused on Venus by Mercury while the latter was still spinning
  rapidly would have caused great interior heating and outgassing, and
  probably a great deal of surface upheaval (mountain building), too,
  causing the very dense atmosphere, the massive release of carbonate in
  the rocks as CO2 into the atmosphere, and the very high mountains.
  Mercury is massive enough to have taken much of Venus's spin in the
  1st half-billion years after formation and Venus's orbit is close
  enough to the Sun that complete escape occurs. The interchange of
  energy between Venus and Mercury would have been enormous, given
  Mercury's large mass (4 1/2 times more massive than the Moon).
Most of the iron (which eventually produces the magnetic field) in
  Venus would have been forced up into the crust by an excessively high
  spin rate, with Mercury getting most of the iron during fissioning,
  which would explain why Mercury has a stronger magnetic field than
  Venus. By contrast, the Earth's iron was not forced to the surface,
  perhaps because the Earth was not as hot and molten as Venus during
  that phase of its formation.
During its lunar phase Mercury would have acquired a prolate shape
  (somewhat elongated towards Venus) because of tidal forces.
Both planets would have been melted by tidal heating in the early
  stages following escape. If this occured before Venus differentiated,
  it might have caused Mercury's high density and stronger magnetic
  field. Subsequently, both planets would have melted from mutual tidal
  heating.
After escape, Mercury acquired greater tilt and eccentricity, and
  Venus would have lost more of its spin. Its prolate shape would have
  been reduced after escape but still maintained.
At the point of escape Mercury would have had a period of revolution
  of about 40 days, and would have retained its spin period, which would
  also be 40 days since it was locked with Venus. But tides raised by
  the Sun would slow down its spin to its present 60 days, which gives
  it a 3-2 spin-revolution ratio (3 spins per 2 revolutions, in other
  words, its rotational period is 2/3 its period of revolution, which is
  88 days), because the next stable configuration for such a body
  (Mercury mass and diameter and degree of prolateness) is this ratio,
  so it is a predicted outcome of its having been a moon of Venus.

This model, then, explains all the anomalies of both Venus and Mercury. Musser (2006) says it would require too much time for Venus to lose a moon but does not provide any reference for this, and the possibility has been corroborated by Kumar (1977) and Donnison (1978). This is the abstract from Donnison:

Kumar's (1977) suggestion that the slow rotations of Mercury and Venus are in part due to natural satellites that subsequently escaped is discussed. A more useful criterion for the escape of such satellites than that previously proposed is derived, and it is shown that this distance is sufficiently small for Mercury and Venus to make the escape of satellites a likely possibility.

And this is the abstract from Kumar:

It is suggested that the slow rotations of Mercury and Venus may be connected with the absence of natural satellites around them. If Mercury or Venus possessed a satellite at the time of formation, the tidal evolution would have caused the satellite to recede. At a sufficiently large distance from the
  planet, the sun's gravitational influence makes the satellite orbit unstable. The natural satellites of Mercury and Venus might have escaped as a consequence of this instability.

They do not, however, specifically say that Mercury was once a moon of Venus.
This is the abstract from Van Flandern and Harrington (gizidda.altervista.org):

The possibility that Mercury might once have been a satellite of Venus, suggested by a number of anomalies, is investigated by a series of numerical computer experiments. Tidal interaction between Mercury and Venus would result in the escape of Mercury into a solar orbit. Only two escape orbits are possible, one exterior and one interior to the Venus orbit. For the interior orbit, subsequent encounters are sufficiently distant to avoid recapture or large perturbations. The perihelion distance of Mercury tends to decrease, while the orientation of perihelion librates for the first few thousand revolutions. If dynamical evolution or nonconservative forces were large enough in the early solar system, the present semimajor axes could have resulted. The theoretical minimum quadrupole moment of the inclined rotating Sun would rotate the orbital planes out of coplanarity. Secular perturbations by the other planets would evolve the eccentricity and inclination of Mercury's orbit through a range of possible configurations, including the present orbit. Thus the conjecture that Mercury is an escaped satellite of Venus remains viable, and is rendered more attractive by our failure to disprove it dynamically.

That's a creative outside the box answer.    I would use the quote function, rather than just " " to indicate when you are quoting a source.  It's easier to read that way.   I would also point out that this line makes no sense to me:   "Most of the iron (which eventually produces the magnetic field) in Venus would have been forced up into the crust by an excessively high spin rate, with Mercury getting most of the iron during fissioning."    I think it's generally accepted that Tom Van Flandern was not correct in his theory.  Perhaps it was bold and innovative at the time, but less so today.

Note that Tom Van Flandern had some very non-mainstream theories...

@PM2Ring this man hated Occam's Razor-- but without men like him we'd not question ourselves nearly as much :).

The whole idea of "iron being forced up" is non-physical, equivalent to suggesting that orbital velocity depends on density. Centrifugal force in a gravitationally bound object could only weaken gravitational sorting, centrifugal sorting would require a crust with unrealistic amounts of tensile strength. And in any case, its size and density tell us that Venus has an iron core similar to Earth's.

One has to take anything written by Tom Van Flandern after the mid 1970s with a huge grain of salt, and anything written by him after the 1980s with a huge block of salt.


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Navigation using moon phases
How to navigate with possible maximal precision using moon phases?
The Moon is the brightest celestial body seen in the night sky, and it is possible to find even through moderate clouds. So it's a good natural object to use in navigation when a compass is not available.
A full Moon is quite an easy case - it's directly on the opposite site from the Sun. However, in the other phases, are there tables or mnemotechnical methods allowing to determine the north based on moon shape, assuming the time is known? Are they different depending on geographical height? Is it possible to estimate both north and the time of the night, when only the geographical height is known?

Google gives some suggestions: https://www.google.com/search?q=finding+north+using+the+moon

In other phases, the angular distance of Sun from Moon is equal to the phase of the moon, in the direction of the lit side. One of the methods could be to estimate the position of Sun based on this, and then you can, assuming time is known, estimate North.
If you don't know time, use constellations (which can form a separate question) to estimate North and the reverse of above method for time.
Now, if there are clouds and you cannot see any constellations, well, you need to be able to calculate cosines and sines and their inverse in your heads. This method, which I will not describe depends on the rate of change of moon's position, which you'll have to estimate using your hands.
I will have to search for my notes for actual calculation of time, let's see if someone else answers it. But let me tell you how to find direction just using Moon, though this is less accurate than if you already know the time.
From the center of the Moon, draw a line in the sky and follow it towards the lit side of the Moon, the spot where it cuts on the Horizon is either East or West. It is East if you are observing this in Waning Phase i.e. after a Full Moon, else it is West (if after the days of a New Moon). You can get the rest of the directions this way.
Here's how it works. The line you just drew was approximately the ecliptic, the apparent path of Sun in sky. It is at an angle of 23.5° from the equator, so that is the maximum inaccuracy that you might have for your cardinal directions. This would actually lead you off in a wrong direction initially, but the ecliptic keeps on moving, so keep on doing this for the whole night, the positions of your North will be varying but the averaged out value will be better than your initial value.


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What is the accepted theory as to why Uranus' axis is tilted so severely?
The planet Uranus is another solar system anomaly, where according to the NASA profile has an axial tilt of 97.8 degrees, also considered to be retrograde.  This NASA summary "Uranus" suggests the current theory of a large planet-sized impact earlier in its history.
Does the planet-impact theory still hold true or have new accepted theories come to light?
Most of all, are there any results from any simulations available?
A note, this is posted as a separate question to my other question "What is the current accepted theory as to why Venus has a slow retrograde rotation?" as the axial tilt is significantly different.

Well, sorry to all responses but there is no accepted theory. A theory must be founded on axioms that make predictions that are testable against evidence. To that end we don't have any theories, just specilation.

Nasa.gov speculates that
the most likely possibility is that an object with a mass close to that of Earth's collided with Uranus, causing it to rotate on its side from then on.
A recent test by Space.com suggests that "Planet Uranus Got Sideways Tilt From Multiple Impacts". These findings suggest that two or more smaller collisions probably occurred, asserting that the early solar system was probably more turbulent than most would think.

Note that all 5 of Uranus's major moons orbit "on their side" too...that is, their orbital inclination relative to Uranus's equator is very small. This is good evidence that something impacted (more accurately, flew thru) the proto-planet's accretion disk. The disk got re-oriented and the moons accreted afterwards.

There is not a single accepted theory as to what caused Uranus' tilt. Rather there are two families of explanations and we still need to collect more data to figure out which is correct.
As explained by @Timtech, Uranus' tilt might have been caused by a giant impact. This is supported by simulations showing that an impact might have been violent enough to tilt the planet and eject debris that agglomerated into moons. But this is not the only possibility.
Another explanation for Uranus' tilt is a spin-orbit resonance. As Uranus was a growing protoplanet, its disc generated a resonance, causing the disc in which it was forming to tilt.
Simulations have been done showing that the tilt might have happened during planetary migration without any giant impacts.
Three of the linked articles above are from 2020, showing that this is still a hot topic of research and there is no consensus on what happened to Uranus.

To add to the existing answers and adapting from my answer to this question about the Uranian satellites:
The single giant impact idea appears to be unlikely. In this scenario, the circumplanetary disc would re-align with the Uranian equator after the collision, but would end up with a retrograde rotation and thus produce retrograde satellites rather than the observed prograde system (Morbidelli et al. 2012).
As noted by Rogoszinski & Hamilton (2020), the timescale for tilting Uranus via a resonances between the precession of the rotation axis and the orbits of the other giant planets appears to be too long compared to the expected timescale for giant planet migration in the Nice model: they estimate ~108 years to reach a tilt of 90°. Under ideal conditions, they estimate a tilt of ~40° could be reached in ~107 years.
From the abstract:

We find that two collisions totaling to 1 M⊕ is sufficient to explain the planet's current spin state. Finally, we investigate hybrid models and show that resonances must produce a tilt of ∼40° for any noticeable improvements to the collision model.

But as noted in usernumber's answer, this is an active area of research and our knowledge of the ice giant planets is woefully incomplete.


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How to calculate the movement of the object passing near other object in space?
Assume object A is moving through the space and is passing near the other object (B). Assume the gravitational influence of other objects can be ignored. How to find the equation describing the movement of the object B?
There are 2 cases, object A is moving straightforward or it's moving on orbit (around other object).
I think the problem is quite elementary, but I couldn't find anything that could help solve that problem using the physics on the level of basic university course (I've studied computer science, so I've got only 1 semester of physics, and basic mathematical knowledge - integrals, algebra etc.).
I know the problem can be solved numerically, but I'm interested in finding the equation describing the movement.

Are you looking for something like the speed or distance between object A and B?

Since you're interested in the equations of motion, I would solve this problem by using Lagrangian mechanics. Essentially, find the kinetic and potential energies for these two bodies, A and B.
Construct the Lagrangian:
$$L = T - V$$
where T is the kinetic energy, and V is the potential energy. Then use the Euler-Lagrange Equation to achieve the equations of motion (I would add it here, but I'm not sure of the specifics of your problem).
Two-body motion can always be constrained to a plane, so you may have to throw a constraint in there through the use of a Lagrange multiplier.

It isn't clear to me if you are merely interested in the equation so that you can plug numbers in and figure it out, or instead would like to understand how the equation is derived. Either way, is called the two-body problem and with that as a search key, it should not be hard to find online references explaining the derivation and illustrating the equation, such as http://en.wikipedia.org/wiki/Gravitational_two-body_problem and http://en.wikipedia.org/wiki/Two-body_problem
You refer to there being two cases, one where A is moving "straightforward" and one where A is orbiting B. This is the same problem, as your first one where 'A is moving straight forward' implies no influence, no gravitational effect from B at all, i.e. it isn't there. If it is there, there is a two-body gravitational effect, which makes it the second case.

Comets passing near earth can be approximated using the two-body equations. He doesn't say ignore gravity between A and B, but ignore other objects, ie. two-body. And there is no case where one object could travel in a straight line unaffected by the other object! Both would be pulled in towards each other. There is a case where one object orbits another, and one where they just pass, pulling at each other slightly, but these are actually just the same when it gets down to the nitty gritty.


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How would the pocket cellular clock work?
In a museum in Lviv, I saw a pocket cellular clock. I don't have a photo, but it was a small disc that had 2 or 3 filaments in it which were pointed at the stars (one of them was Andromeda, I think). Through analyzing the position of the stars it was possible to determine the time of the night. I'm not sure if knowing the day of the year was promised.
I couldn't find the information about the way the clock was used. Does anyone know how such device was used?

The device that you saw is called a nocturnal. It calculates the local time based on the month of the year and the position of Polaris and one or more other stars.
Since stars change position throughout the night, they can be used to determine time; but the positions of stars change throughout the year, so the input of the month of the year is needed. Polaris is used as a relatively fixed reference point.
More information:
Nocturnal (instrument) - Wikipedia


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How hot can a planet be?
Given that some exoplanets, particular "Hot Jupiters", orbit very closely to their parent star, how hot can these planets become?
What is the hottest exoplanet discovered so far?

The temperature of an exoplanet is dependent on a number of factors, including:

The distance between the exoplanet and the parent star.
The type (energy output) of the parent star.
The ability of the exoplanet's atmosphere to distribute the heat energy.

Recent discoveries, such as reported in the article "Sizzling Planet Makes Some Stars Look Cool" (Than, 2007), have found that some exoplanets are hotter than some types of stars.
For example, the exoplanet  HD 149026b  has a star-facing side of over 2000C (3700F), this planet reflects so little light that it would appear like a "glowing piece of charcoal". It is believed (from the article) that the planet is tidally locked and the 'dark side' of the planet is believed to be substantially cooler.

What is the hottest exoplanet discovered so far?

As far as I can tell, that honor goes to Kepler-70b. Estimates for its surface temperature range up to 7,143 K. The other confirmed planet in the system, Kepler-70c, may also be extremely hot. The two orbit their parent star - a hot b-type subdwarf - at distances of 0.0060 and 0.0076 AU, respectively, and thus receive quite a lot of radiation. The discovery paper (Charpinet et al. (2011)) reported that the two are likely chthonian planets, having been engulfed in the star's envelope during its red giant phase.
The Habitable Planets Catalog has lists of top exoplanet extremes. Kepler-70b and Kepler-70c top the list of hottest exoplanets, and are listed as having surface temperatures of 7,143 K and 6,351 K, respectively. I would imagine that planets close in would have much greater surface temperatures, as they are heated more; these two may also have residual heat from the engulfment.
There are several other exoplanets on the lists that top 2,000 K (there are likely more):

Kepler-13 b (3206 K)
PSR 1719-14 b (3118 K)
WASP-33 b (2505 K)
WASP-12 b (2368 K)
WASP-103 b (2342 K)
KELT-1 b (2296 K)
WASP-18 b (2258 K)
OGLE2-TR-L9 b (2213 K)
WASP-121 b (2202 K)
WASP-76 b (2039 K)

The page is a year or two old, and there are clearly other hot exoplanets that have since been discovered. However, it seems that Kepler-70b and Kepler-70c are still the hottest.


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What is the upper and lower limit of temperatures found on stars?
What are the most extreme temperatures (both hot and cold) stars have been detected  at?  Is there an upper and lower limit for the detected temperature of stars?

The answer depends on what you'd want to consider as a "star." If you're just thinking about stars on the main sequence, then you can just refer to the classical stellar type letters, "OBAFGKM" (which has relatively recently been extended to accommodate the coolest brown dwarfs with the letters "LTY"), where O-stars are the hottest stars (~30,000 K) and Y-stars are the coldest, so-called "room-temperature" stars (~300 K).
Self-gravitating, gaseous objects are incapable of fusing deuterium below about 13 Jupiter masses, and thus simply collapse and cool perpetually (as is the case for all the giant planets in our solar system). These objects can be colder than 300 K but are not technically stars as they do not undergo nuclear fusion.
For stars that leave the main sequence, two possible outcomes are a white dwarf star or a neutron star, both of which are born extremely hot: White dwarfs are born with surface temperatures of ~10^9 K, whereas neutron stars are born with surface temperatures of ~10^12 K. However, both white dwarfs and neutron stars cool as they age, with the coldest known white dwarfs being ~3,000 K, and neutron stars cooling to ~10^6 K.
So to answer the first part of your question: The coldest known stars are Y-stars (i.e. brown dwarfs) and the hottest known stars are either O-stars or young neutron stars, depending on whether you consider objects that have left the main sequence or not.
And as for strict lower and upper limits, the coldest stars possible are likely black dwarfs, which are what white dwarfs become after cooling for a very long time (>10^15 years). The hottest stars are likely the newly-born neutron stars I previously mentioned, it is very difficult to get much hotter than 10^12 K because any excess energy is carried away via neutrinos.

+1 Great answer, what are the hottest and the coldest stars ever detected. I did not know stars could be that cool, incredible!

What about http://en.wikipedia.org/wiki/Quark_star

Likely those wouldn't be any hotter than normal young neutron stars, as their surfaces would still cool via neutrino emission, which is very effective at temperatures in excess of 10^10 K.

How do you get this 10^10K limit? Theory? Could you explain exactly how you get this?

+1 But I think the hottest temperatures quoted for the NS and WD may be too high and reflect the core temperature rather than surface temperature?

This question already has a very good answer, I would just like to add a few details.
http://www.astro.ucla.edu/~wright/BBhistory.html
Says here that when the universe was 10^-33cm in diameter, its temperature was 10^32K. Therefore that should be the absolute max temperature reachable in this universe, and so the max temperature of a star should be below that; very interesting what Guillochon said above, that neutrinos carry away excess energy above 10^12K.
The color of a star gives away its temperature. It is interesting to note that the corona of a star including our Sun can be well over a million K even though the surface temperature of our star is around 6000 K.
http://en.wikipedia.org/wiki/Corona
Also, in stellar cores, hydrogen fusion into helium starts at 3 million K, while carbon fusion starts at above 500 million K, and silicon fusion starts at over 2700 million K for comparison.

Mostly irrelevant.

The hottest stars - and here, I assume that "star" excludes stellar remnants such as white dwarfs, neutron stars, and other exotic compact objects - are likely Wolf-Rayet stars, a class of hot, hydrogen-deficient stars characterized by hydrogen depletion and noticeable carbon, nitrogen and oxygen lines. The massive Population I sub-type are likely former O-type high-mass main sequence stars with exceptionally strong stellar winds.
Guillochon's answer mentions that O-type stars often have surface temperatures of about 30,000 K. Many, - if not most - Wolf-Rayet stars exceed that by drastic amounts. Some of the hottest may be the Wolf-Rayet components of the binaries AB7 and AB8, in the Small Magellanic Cloud. Both have normal O-type companions, which are also exceedingly hot. However, the maximum temperatures for the Wolf-Rayet components may be 105,000 K and 141,000 K, respectively (Wikipedia cites Shenar et al. (2016) here).
Now, here's the problem. It's notoriously difficult to determine the temperatures of Wolf-Rayet stars to the desired accuracy. Why? Well, it's largely because of their stellar winds and high mass-loss rates. Parts of the atmospheres and winds are optically thick, meaning that we can't necessarily observe where the "surface", as normally described in stellar astrophysics, lies. Therefore, let's bear in mind that the listed temperatures may be off a bit - although Wolf-Rayet stars are still quite clearly hotter than normal O-type stars.

The hottest stars that are still fusing in their cores are Wolf-Rayet stars that are on the extreme end of the WC sequence, appropriately classified as WO stars, which display prominent Oxygen emission lines. The hottest known star is WR 102, which has a spectral type of WO2 and a surface temperature of 210,000 Kelvin.
WR 102 is thought to have a mass of ~16.7 solar masses. Since this is a highly evolved Wolf-Rayet star, the majority of this mass is composed of the fusing core with a very thin radiative layer surrounding it. For reference, the threshold for being an O-type star is about 16 solar masses, with just a fraction of that mass the fusing core. This means WR 102 probably started with around 50-60 solar masses at ZAMS.
At this point it is unknown what exactly produces a WO star, whether it is an evolutionary stage following being a WC star or if it takes an extraordinary massive star that goes straight to WO after shifting through a WN stage. The number of WO stars known presently is in the single digits so there is still a lot to learn about these sort of stars.


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What is the current accepted theory for Neptune's immense wind speed?
According to "Neptune: The Other Blue Planet in Our Solar System" (Choi, 2010), Neptune, despite its great distance from the sun, is the windiest planet in the solar system, with winds reaching 2000km/h (1500 mph) as detected by the Voyager 2 probe in 1989.
Given that Neptune is 30 AU from the Sun, what are the currently accepted theories for where the planet gets energy for these immense wind speeds?

Disclaimer: Most of what is written here is based highly on theory and speculation. Voyager 2 is the one and only trip we have taken to Neptune and most of these theories are built upon the data it has sent back. Leave large room for error
Is Neptune windy?
Before we start answering why Neptune is windy we should first specify if Neptune is even windy in the first place. What proof do we have?

The Great Dark Spot

One observation was of a large abhorrence on the surface of Neptune aptly titled "The Great Dark Spot". Repeated viewings placed the movement of the GDS to be about 700mph.

Sourced from: Nine Planets: Neptune Facts

Keck Telescope

A lot of observations and images of Neptune have come from the Keck II Telescope, which is able to observe other markings moving at similar to faster speeds on Neptune.
Sourced From: KECK IMAGES OF NEPTUNE BEST EVER CAPTURED
What causes the winds on Neptune?
It's theorised that there are two potential causes to Neptune's winds. Shallow processes in the outer atmosphere or much deeper atmospheric changes that extend into the interior.
How the Mighty Winds of Uranus and Neptune Blow (paraphrased for simplicity) Claims that

The researchers found the windy layers of Neptune occupy the outermost 0.2 percent of It's mass, suggesting that shallow processes drive those winds, such as swirling caused by moisture condensing and evaporating in the atmosphere.

What is the main source of energy for these winds?
So now you are probably wondering where Neptune get's this energy from as it is so much further from the Sun than we are.
Nine Planets: Neptune Facts Claims that this is due to Neptune's own internal heat source radiating twice as much energy as it receives from the Sun.
Why does Neptune have stronger winds than Earth then?
So theoretically, if it was down to energy, it would make sense that Earth, being so much closer to the Sun, would have much more energy required to reach these speeds.
Yes and No.
The Sun is the main provider for our energy but it also creates turbulence in the atmosphere. It's cooling and heating of the layers keeps air constantly rising and falling, disturbing natural air flows and stemming the size of our storms.
On top of this Neptune's Atmosphere: Composition, Climate & Weather claims that 80% of Neptune's atmosphere is molecular Hydrogen.
Hydrogen being the smallest and lightest element requires less energy to reach high speeds than our larger, nitrogen based atmosphere. This may not be a main factor to consider but it is certainly contributing in no small amount.


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What is the naming convention for newly discovered objects?
Often, a newly discovered star is identified by a seemingly random string of letters and numbers. I'm sure that there is some order to it, though. What is the naming convention for newly discovered astronomical objects?
I'm sure NASA doesn't ask the public, as something like this would happen:

http://xkcd.com/1253/

The International Astronomical Union's Naming Astronomical Objects webpage is a comprehensive guide to how astronomical objects are named.
For example, for "Specifications concerning designations for astronomical radiation sources outside the solar system", the IAU states that the designation (or sequence of letters and numbers) follow a general form:

Acronym  ^ Sequence ^ (Specifier)

Where the Acronym

is a code (i.e., alphanumerical string of characters) that specifies the catalog or collection of sources.

The sequence

The sequence (or numbering) is an alphanumeric string of characters, normally only numerical, that uniquely determines the source within a catalog or collection. It may be a sequence number within a catalog (e. g., HD^224801), a combination of fields, or it may be based on coordinates.

and finally, the Specifier

The specifier is optional and allows one to indicate other source parameters. However, they are not required syntax and are enclosed in parentheses.

“a random string of letters and numbers” is often the identifier used by one observer or work group for their observation or work. That is, ‘string’ designators are often internal names, and never meant to be settled and permanent for wider use- even use by scientists two or three steps removed from that observer or work group.
The confusion is often due to hot-button factoids that get picked up by the lay media, due to some ‘sciency’ angle that might attract clicks. (And of course, there are the outright conspiracy pages and chatter, that pick up tidbits and lay them out for consumption, without context or deeper understanding but with an angle.) These pop sci blurbs get ahead of themselves, sometimes running the internal designation without the explanation- readers have short attention spans, and these blurbs are getting shorter and shorter. Ultimately, you get YouTube, for people who can’t even be bothered to read. Even in the case of ‘official’ releases (e. g., STSCI/JHU for Hubble blurbs) the pop sci writeups are done by journalists, not scientists who are just better at writing. There’s a general interpretation and filtering going on, which is to be expected- try reading an astronomy paper (an actual, peer-reviewed journal paper) through the eyes of a non-astronomer. It might as well be greek. Unfortunately, this interpretation and filtering is getting more and more severe as a field progresses, leaving the non-scientists further and further non-.
In the case of new objects (not ones with approved, general names), the list of observing targets hadn’t been winnowed down much, from raw detections (there are simply a staggering number of things out there, including possible duplicate detections) to confirmed celestial objects (not cosmic-ray hits or other bad pixels, ghost images from internal reflection or the Sun, etc.) to new celestial objects (not duplicates, misplacement errors, cataloguing errors, etc.), to new objects corroborated by outside review (a whole ‘nother game). The combination of faster news cycles, shorter attention spans, and presumption on the part of Google users (the “search effect”- having information but not knowledge or understanding or experience or even context) means internet users are seeing more and more temporary designations.
An observer or, nowadays, work group submits a paper with their observation/work for formal publication. Due to quality checks, literature search and prior-work reviews, outside informal consultations, and another round of checks and reviews by outside corroborators, this publication often takes over a year. Then, the paper is read and considered by the wider community- stellar astronomers for stars, cosmologists, astrophysicists, etc. for new (therefore distant) galaxies, etc. Then , the permanent naming process can begin, by the convention of that community. Stars do not get named by cosmologists, etc., and for that matter, asteroids and comets have separate naming paths despite both falling in planetary science.
The number of stars is just staggering, in our galaxy alone… and we have been resolving the exceptional stars of other galaxies for some time now. It’s pretty presumptuous to expect that the 1,000,000,000,001th star just identified by you will not get some sort of letter/number series attached to it. The closest we get is some sub-catalog, where an observing program (by plan) selects for targets that inherently fall in together. Then the nth new object is often designated as “[catalog name] n”.
The at-hand example is Luhman 16. That’s the 16th catalog entry, in the body of work compiled for a purpose (nearby dwarves), by a given org (Luhman and his collaborators). However, while K. Luhman and others were in the middle of the project, the detection went by “WISE J104915.57-531906”, the entry for yet another one of a myriad entries by the WISE observatory. It was three years between WISE detection, and formal, public announcement, and even that was given as WISE J104915.57-531906. It was still some time again before most settled on Luhman 16. And still, other catalogs have their own designations.


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What is the current estimation for how much time the Sun will function properly?
Eventually, the Sun will run out of fuel, go dark, and other bad things will happen. About how much long is it predicted that we will have a sun functioning as it is currently?

Just to add to Undo's answer, after the expansion to a red giant, the sun will become a planetary nebula, where (according to the link) the fusion reactions inside the star are 'overtaken' by the gravitational collapse, in turn causing the inner layers to condense and heat up causing the outer layers to be blown away.  After which, the hot core forms the remnant, a White Dwarf star (NASA source), which is likely to last for several more billions of years.
The image below depicts the current ideas of the expected lifecycle and timeline of the life of the sun:

Image source
How do we know what will happen to the sun?
Currently, the main method to determine the solar lifecycle is described by the article "The Future of the Sun" (Cowing, 2013) is to:

Studying stars with the same mass and composition as the Sun, the so-called "solar twins," can give us more information about our own Sun; solar twins of various ages offer snapshots of the Sun's evolution at different phases

Where the mass and chemical composition of a star provide the information needed to determine its lifecycle.

How much time until it is no longer possible to live on Earth due to these changes in the Sun ? I believe it is one Billion years, is this correct or do we have even less time than that ?

According to Wikipedia:

The Sun does not have enough mass to explode as a supernova. Instead it will exit the main sequence in approximately 5.4 billion years and start to turn into a red giant. It is calculated that the Sun will become sufficiently large to engulf the current orbits of the solar system's inner planets, possibly including Earth.

However, the Sun has more nasty tricks up its sleeves:

Even before it becomes a red giant, the luminosity of the Sun will have nearly doubled, and the Earth will be hotter than Venus is today. Once the core hydrogen is exhausted in 5.4 billion years, the Sun will expand into a subgiant phase and slowly double in size over about half a billion years. It will then expand more rapidly over about half a billion years until it is over two hundred times larger than today and a couple of thousand times more luminous. This then starts the red giant branch (RGB) phase where the Sun will spend around a billion years and lose around a third of its mass.

So, it's going to take long enough that we don't really have to worry about it.


text:
What is a parsec and how is it measured?
Astronomical units of measurement are mostly pretty straight-forward:

Astronomical Units refer to the mean Earth-Sun distance (~150 million km or 93 million miles)
Light years are the distance light travels in a year (~9.46 × 10^12 km)

Another astronomical unit of measurement is the parsec.  What is a parsec and how is it measured?

A parsec (abbreviated pc) is a unit of distance used by astronomers, cosmologists, and astrophysicists.
1 parsec is equal to $3.08567758 \times10^{16}$ meters, or $3.26163344$ light years (ly).
A few typical scales to keep in mind:
1) The disks of galaxies like the Milky Way are a few 10's of kpc (that's pronounced kiloparsecs, which are 1000's of parsecs) in size. The dark matter halos surrounding them extend out to almost an order of magnitude further.

2) Galaxy clusters come in at around 1 Mpc (that's pronounced megaparsecs, or millions of parsecs), and are the largest bound objects in the universe. Pictured below is the galaxy cluster Abell 2218.

3) The closest star to the planet Earth (not counting the sun, of course) is the star system Alpha Centauri, at a distance of just about 1.3 parsecs. While you may think that this is incredibly close (and it is by cosmological standards), it would nonetheless still take us 4.24 years to travel to it if we traveled at the speed of light.

A parsec is approximately 3.26 times a light year or 206,265 Astronomical Units (AU).
According to "Cosmic Reference Guide" (Caltech), the word 'parsec' stands for "parallax of one arc second", and is (according to the website):

the length of the long leg of a right triangle, whose short leg is one astronomical unit when the angle between the Sun and the Earth, as seen from an object in space (a star for example), is one arcsecond

Illustrated (from the link above):

Also known as a measure of time for the Kessel Run.  ;)

so a parsec is about the distance between us and Alpha Centauri.

@Caters No, there are no stars exactly 1 parsec far away from us. The nearest star (Alpha Centauri) is about 1.302 parsecs.

Parsec or Parallax-arcsecond is defined as the 'distance' of something that has a parallax angle of one arc second. The International Astronomical Union (IAU) define 1 parsec(pc) as:
$$1 \text{pc} = 1/\tan(1'') \ \text{au}$$
$$\Rightarrow 1 \text{pc} = 1/\tan({1\over60}') \ \text{au}$$
$$\Rightarrow 1 \text{pc} = 1/\tan({1\over60\times60}^{\circ}) \ \text{au}$$
$$\Rightarrow 1 \text{pc} = 1/\tan({1\over60\times60} \times {\pi\over180}) \ \text{au}$$
$$\bf \Rightarrow 1 \text{pc} = 206,264.8 \ \text{au}$$
$$\Rightarrow 1 \text{pc} = (2.06265× 10^5) \times (1.496×10^{11})  \ \text{m}$$
$$\bf \Rightarrow 1 \text{pc} =  3.09×10^{16} \ \text{m}$$
$$\Rightarrow1 \text{pc} =  3.09×10^{16} \times {1\over9.46×10^{15}}\text{ly}$$
$$\bf \Rightarrow 1 \text{pc} =   3.26 \ \text{ly}$$

(image source:https://commons.wikimedia.org/wiki/File:Parsec.jpg)
Parsec or Parallax-arcsecond is the distance of something that has a parallax angle of one arc second. Surveying provides an everyday example of parallax. Go outside and look at a distant object. Hold your hand out with one finger raised and close one eye. Line up your finger so it covers the object from view. Then close that eye and open the other without moving your arm. Now, move your finger to block that eye's line of sight. The two positions your finger was at can be used to find the distance from the object. This is what a survey crew does to figure out land plot boundaries. The mathematical technique used in the calculation is trigonometry. A parsec is so large that if you could see that far that you would be looking in to the past. You are looking 3.26 years in the past because it takes 3.26 years for the light to come from a parsec away. I recommend watching the Khan Academy video Parsec definition which I found helpful.


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What proportion of planetary systems have been found with 'Hot Jupiters'
According to the NASA website "The strange attraction of Hot Jupiters", one of the main types of exoplanets that have been detected are Hot Jupiters, which are

These are behemoth worlds that orbit close to their parent stars, blocking a fraction of the star’s light when it transits in front.

Planets that are from as large as Saturn through to far larger than Jupiter, orbiting in a matter of days very close to their parent star.  Something our solar system does not have.
What proportion of star systems have Hot Jupiters within their systems?  Is it a case that Hot-Jupiter-less systems such as our own are unusual?

In 2011, about 20% of the exoplanets found were hot Jupiters. That is a lot, but it is strongly biased simply because there are the easiest planets to detect. You can detect exoplanets by transit (that means that the planet passes in front of its host star which decreases the luminosity of the observed star during the transit), hot Jupiters being closer to their host star and bigger than smaller planets will therefore be easier to detect. You can detect exoplanets by radial velocity, which means that you detect variation of the position of the host star due to planet's gravity. The closer and the heavier the planet is, the larger the variation will be, and therefore the easier to detect.
That's why it is important to say "In 2011"; as our detection techniques (and telescopes) improve, the proportion of hot Jupiters should decreases.
Edit:
To put that in perspective, you can read this article from Jason Wright in which he tries to estimate the "real" ratio of hot Jupiters orbiting around "normal" stars (F, G and K type stars). This ratio is actually much smaller than the currently observed ratio, about 1.2%.

three-quarters? that's a lot more than I expected.  Do you have any references that you can include in your answer?

Sorry, I got the number wrong (I mistook it with earlier estimates). I putted a more reliable numbers with a good reference; I also add some details about the "real" ratio one should expect.

No apologies needed - this is an excellent answer and answers both questions I had very nicely.

I completely agree with the answer from MBR. The number is actually $1.20\pm 0.38$ per cent, is published by Wright et al. (2012) and is the fraction of F, G, K stars that have a hot Jupiter defined as being larger than 0.1 Jupiter masses and having an orbital period less than 10 days. Table 2 of that paper summarises results from other workers, who obtain between 0.5 and 1.5 per cent. The paper also discusses observational biases, including metallicity.
It has long been known that close-in planet incidence is higher around more metal-rich stars. There is also a bias whereby it is easier to find planets around metal-rich stars, whereas the average star in the solar neighborhood is slightly metal-poor compared with the Sun.
A study by Gonzalez (2014) accumulates our current knowledge of exoplanetary systems and their metallicities, deriving a planetary incidence rate
$$ P_{planet} = \alpha 10^{\beta[Fe/H]},$$
with $\alpha= 0.022 \pm 0.007$ (i.e. 2.2 per cent), $\beta=3.0\pm 0.5$ and where [Fe/H] is the usual logarithmic ratio of the metallicity of the star to the metallicity of the Sun. (i.e. the Sun has [Fe/H]=0).
This calculation is done for giant planets with orbital periods less than 4 years, so not all of them would be classed as hot Jupiters. Bottom line, the number given by Wright et al. is about right on average, but it is higher for higher metallicity stars.

Just for fun I did a limited analysis of Jupiter size planets found using radial velocity or astrometry (mostly the former) techniques with the following criteria: 1) Data from the on-line sources http://exoplanet.eu/catalog/bd%2B15_2375_b/ and http://www.exoplanets.org/ giving planets discovered through Apr. 2016. 2) Msini: 0.5-10 M(J) 3) Distance to star less than 58 parsecs 4) Planets out to 5200 day periods (or 6AU) 5) Only data from K,G or F dwarfs on main sequence (V or IV/V). This was an attempt to minimize observational bias.                                                I still found a pronounced peak in Jupiters with periods less than 50 days, but less pronounced if plotting against distance (sma). However, I also found peaks in numbers of Jupiters at slightly farther out than Earth and at a distance about the same as the beginning of the asteroid belt in the solar system.The Earth peak is much wider and goes away in the period histogram. This data is dominated by planets orbiting G stars 89 of 139 (107 of 171 planets). Interestingly (or not) there is a peak in Jupiters that are in "multiple Jupiter systems" around the orbit of Venus that "washes out" the Earth peak. The percenatage of "hot Jupiters" depends on definition, but 22 of 171 have period less than 50 days and 13 have orbits less than 0.1AU.    If anyone is interested in more detail of my results please e-mail me.

@RobJeffries See above comment. Could there still be some "observational bias" that I am missing?

Yes, there are enormous biases favouring the detection of hot Jupiters and your procedure has not minimised those at all.

@RobJeffries I apologize for my ignorance. My question to you would be do you think that it is possible that the HARPS project, which has been monitoring stars less than 58pc for over 10 years now, may have missed some Jupiter size planets, with periods about that of Jupiter or less, orbiting stars they have been watching? There has been a fall off in discoveries (out of 107 "Jupiters" found in this range orbiting G stars only 18 have been found since Jan 2012/ 2 of those were in the same system and three were found in systems that already had a closer in "Jupiter" detected).

And I may have to apologise for mine. I had assumed you were dealing with a parent sample that was incomplete. If the HARPS project have surveyed every F,G,K star within 58pc with uniform sensitivity and sampling for 10 years, then yes, they should be highly complete for all Jupiters with periods of less than say 5 years, except for those around young stars. If however they have been observing selected samples, with instruments with varying (increasing) sensitivity then I suspect a bias towards hot Jupiters is still present, especially amongst the K-dwarfs.


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Does the Sun belong to a constellation?
Each new star we find is generally considered to be part of the constellation it is nearest to.
Our Sun is obviously a star, just much closer. Is our Sun part of any constellation? If so, which constellation is it a part of?

The constellations are how we see the stars from earth. The Earth revolves around the sun which is in the Milky Way Galaxy. http://www.universetoday.com/18256/where-is-the-sun/

From what conceivable distance would you have to be from our star in order to see what constellation our star is in, in relation to neighboring stars?

Constellations are human constructs to make sense of the night sky.  When you are trying to find your way around, it helps to "chunk" stars into patterns and assign those groupings names.  When I want to point out a particular object in the sky (say Polaris, the North Star), I start by pointing out a familiar constellation (say Ursa Major, the Big Dipper).  From there, I can tell my friend to follow this or that line to get them to look where I'm looking:

With the advent of computerized telescopes and large data sets, constellations are less important for professional astronomers.  However, many stellar databases use Flamsteed or Bayer designations, which assign stars to constellations.  In order to include all stars, the sky is divided into irregular regions that encompass the familiar constellations.
So, which constellations is the Sun assigned to?  Well, from the perspective of someone on the Earth, the Sun moves through the constellations throughout the course of the year.  Or rather, Sol moves through the region of the sky where some of the constellations would be seen if its light did not drown out distant stars.  Our moon and the rest of the planets move through those same constellations.  (The Greek phrase which gives us the word "planet" means "wandering star".)
The current position of the sun against the background of distant stars changes over the course of the year.  (This is important for astrology.)  It's a little easier to make sense of with a diagram:

So perhaps a better question is:

What constellation does the Sun belong to today?

Presumably an observer on an exoplanet would assign Sol to some constellation that is convenient from her perspective.  But from our perspective within the Solar system our sun, moon, and planets are not part of any constellation.

Why do you say Sun sometimes, and Sol other times? Sol is the portuguese word for Sun...

@Rodrigo: Yes. It's also the Latin word for the Sun. I hoped to convey that observers outside of our stellar system wouldn't have any particular attachment to our star. If it were close enough, they might have named it. (And while they wouldn't have named it "Sol", that's a pretty good placeholder.) My answer is trying to shift your perspective from thinking about the Sun as special (which, of course it is to us) to thinking of it as one of billions of stars.

@JonEricson Just noting, the Big Dipper is an asterism, which lies in Ursa Major. They're not the same thing.

@Rodrigo Sol is also Swedish for (the) sun. In any case, every known star has a name. And while not official, as soon as we start to move outside our own little neighborhood and look at our star on a map in a wider context, "Sol" typically is what is is called because "Sun" does not as much feel like a name as a noun. And "The Sun" does not make it much prettier on the map. So "Sol" is quite simply prettier and the de facto name used when referring to our little star. For example: Sol in Elite: Dangerous.

No, it does not. The constellations are fixed (on time scales long enough for humans to consider as fixed, at least) patterns of stars which exist on the celestial sphere. This celestial sphere is a coordinate system which has the Earth at its center. From the Earth's perspective, the sun rises and sets at the same rate as the constellations, but as the Earth revolves around the sun, the constellation the sun will appear to be in will change.
However, the constellation the sun appears to be in (if projected onto the celestial sphere), is exactly how the constellations of the Zodiac are determined.

Your answer comes closest to what I think is the key point: stars that appear close together from our point of view would likely appear randomly scattered when viewed from elsewhere in the universe.

On a three-dimensional chart of the Galaxy, each of our constellations is an irregular cone with its peak at our Sun.  Thus it belongs simultaneously to all constellations, or to none.

I wonder what a chart of our galactic neighborhood (say, thousand parsecs) would look like if a line were drawn from each star to the brightest other star as seen from there.

We did an exercise on this particular subject at Secondary Modern School in 1958.
With reference to"Patrick Moor" publications a 3-dimentional model wasmade of the nearest 50 stars.
Particular reference to the "well known" ie:the plough.As the model evolved,those we are familiar with loose their familiarity because of the distance.The plough is familar if assumed all the same distance,they are not and the model highlights the lack of pattern if viewed from Orion's Belt.
Our Sun did not appear to fit into any"Constellation type pattern"from any angle and all familiar ones were lost when viewed anywhere else.
An interesting "Experiment" though.Try it.Cheers Colin

Yes, but it does not stay in the same constellation.
The Sun, Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune and all the other solar system objects move across the constellations of the Zodiac: Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpius, Sagittarius, Capricorn, Aquarius, Pices and Ophiuchus.
As of March 28, 2014, the Solar System is most likely traveling through Cancer, Virgo or Libra and will move on to Scorpius.

"... the Solar System is most likely traveling ..." -- I think you mean the Sun is most likely traveling.

I think DeathItself misunderstood the question.


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What does it mean for a star to go nova or supernova? Can I safely observe these?
What does it mean for a star when people say it goes 'nova' or super nova, what are the differences?
More importantly, can I safely observe these with an amateur telescope? I imagine they would be big enough to spot should one happen!

It's a matter of size and stellar evolution.  There are many, many types of stellar explosions.  The University of Arizona has one page that describes these types. Generally, a Nova is not what we think of (i.e. a star exploding). That's actually a Type II Supernova.
According to that site:

Novae are frequently (perhaps always) members of binary systems where the more massive star evolves more quickly and becomes a white dwarf. The less massive star loses some of its hydrogen-rich outer atmosphere to the white dwarf where it collects.  The high gravity and intense heat of the white dwarf make conditions suitable for hydrogen fusion into helium. When enough hydrogen collects, it explodes in a fusion reaction, causing a nova. The gas ejected in the explosion moves outwards at 1000-2000 km/sec.

There are (at least) two distinct types of supernovae:

The "Type I" supernovae arise when the white dwarf star member of a binary system accretes so much matter from its companion that it is tipped over the 1.4M Chandrasekar limit. The white dwarf collapses very rapidly (just a few seconds) until the infalling layers hit the very hard surface of the neutron star. The resulting shock goes back through the collapsing layers, and they explode by hydrogen fusion.

And the one we are more familiar with from popular media:

The "Type II" supernovae are the result of a massive star consuming all of its nuclear fuel and then exploding. The black line traces the evolution of a massive star schematically, until it reaches the supernova stage.

Portland Community College actually lists a lot of the variations that happen in the different types of supernovas out there.

Type I: supernovae WITHOUT hydrogen absorption lines in their spectrum
Type II: supernovae WITH hydrogen absorption lines in their spectrum.
Type II
The type II supernovae are massive stars whose iron cores collapse and then rebound, shock heating the outer layers of the star, which then explode outward.
Type I
The type I supernovae are subdivided into three subclasses, called (boringly enough), Type Ia, Type Ib, and Type Ic.
Type Ia: no hydrogen lines, no helium lines, strong silicon lines
Type Ib: no hydrogen lines, strong helium lines
Type Ic: no hydrogen lines, no helium lines, no silicon lines
Type Ib and type Ic supernovae are massive stars which lost their outer layers in a stellar wind before core collapse.
Type Ib supernovae lost their hydrogen-rich outer layer, revealing the helium-rich layer immediately below.
Type Ic supernovae suffered more mass loss as supergiants, losing both the hydrogen-rich layer and the helium-rich layer (revealing the carbon-rich layer below).
Type Ib and type Ic supernovae are essentially the same as type II supernovae.

As for observing them, since none of them are expected to be close to us (i.e. no stars near us able to go supernova), it should be safe. The problem is predicting when one will happen.  For instance, Betelgeuse is expected to go supernova "soon", but in stellar terms, that could be well over 100,000 years from now. The list of candidates in the Milky Way doesn't bode well for one in your lifetime. Although astronomers are often seeing them in other galaxies, and some are observable by amateur astronomers.  In fact, just this past August (2013) an amateur discovered a nova.

The description of the Type Ia isn't quite right.  The supernova happens just under the Chandrasekar limit. And there is little hydrogen in a white dwarf. As the white dwarf approaches this limit, its size decreases until carbon fusion begins.  And since it is formed of (electron-)degenerate carbon the temperature can rise, increasing the fusion rate without changing the pressure. So a good proportion of the star fuses at once and the whole star is disrupted. No neutron star is formed.

A supernova has brightness rivaling its galaxy. If a supernova happens in a galaxy that galaxy will suddenly have a bright spot, if that happens within our galaxy it can even rival the brightness of the moon(unless its on the other side of the galaxy which we wont be able to see it due to the black hole in the middle though if we look at a different direction we can see it due to gravity but that will appear in 2 places and it will be odd(footnote 1)) Supernova in our galaxy can be seen in broad daylight as long as they aren't too close to the sun or ecclipsed by the moon. One time a hypernova happened in our local group and it could be seen in daylight. A supernova happens every few years in the galaxy though most are close to the center where its bright anyways because of the amount of stars orbiting there, dangerous supernova are very rare(footnote 2) and happen once in about 240 million years. A dangerous hypernova is equally rare(footnote 2.1), because hypernovas have a much larger danger range, but are much less common.
Don't worry no supernova can happen so brightly that it can be harmful to our eyes, even through a telescope. the only way it can harm us is by ionizing the atmosphere causing smog to kill off all the plants and everything else. The last time something really serious happened look its in my article from footnote 2 and yeah it wasn't that nice.
However if the sun were replaced by a star going supernova, it would be 9 orders of magnitude brighter than a hydrogen bomb detonated on your eye.
Basically for lots of supernovae you can just look outside and see them every night.
Footnote one: One memorial supernovae appeared in four places:https://www.scientificamerican.com/article/scientists-see-same-star-explode-4-times/
Footnote 2: Check my other answer for 2.1 it will be in the second last paragraph for 2.2 it is at the first paragraph Link: What's the safe distance from a hypernova?


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What is a hypernova and have we observed any?
Inspired by the question "What does it mean for a star to go nova or supernova? Can I safely observe these?", I am curious about what hypernovas are?
Have we observed any occurring, or the at least the after-effects?
Is there an upper limit of how large a stellar explosion can be?

A hypernova is just a really, really big supernova.  UMass has a (tacky designed) web page that explains it.

A hypernova explosion typically has a mechanical energy output of ~ 10^53 ergs, or about a factor of 100 greater than a supernova.

Regarding aftereffect, the page says this:

The age of the hypernova remnant NGC5471B is about 30 thousand years, while the age of MF83 is about 1 million years.

As best as I am able to determine, we have not had a chance to observe any hypernova candidates in the modern era, although APOD featured the same one as mentioned by UMass:

According to New Scientist, research suggest the upper limit to a star is about 150 solar masses. These are the stars that create the infamous supernovas.his is as derived from the Eddington Limit.  Although R136a1 is an example of a star at nearly double that.  It is believed that these monsters are what are responsible for hypernovas. There is much to be learned from these stellar monsters!

Gamma Ray Burst may be caused by hypernovae. So if this is the case, we have few hundreds observations.


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Why is the Hubble Telescope in space?
Why is the Hubble Telescope in space?
Do we get enhanced clarity and range by placing it in space?
What can it achieve from space that it could not achieve from Earth?

Because the atmosphere is a jerk when you're trying to take detailed photos.

As @astromax observed, one of the primary factors that makes a space based telescope better than an equivalent telescope of the same size is scattering.
Along with scattering, there is also refraction which can be especially problematic when combined with atmospheric turbulence. In the modern era, this problem can be remedied to a certain extent using adaptive optics, but since the Hubble was designed, built, and launched before AO became practical in the 1990's, a space telescope represented the pinnacle of optical clarity for that time.
However, there is another important optical property of the atmosphere, absorption. Although the Hubble is primarily a visible light telescope, it does have instruments that cover both the near UV and near IR, both of which are absorbed by the atmosphere more than visible light.
Furthermore, there are practical advantages of space based telescopes. There is neither weather, nor light pollution in space.

There are a number of things which are important for a telescope. The first is the light collecting power of the telescope, which is simply a measure of how many photons the telescope can collect from an object. This, not surprisingly, is affected by the size of the primary mirror/lens of the telescope. The second issue is the telescope's angular resolution, which is a measure of the level of detail which can be seen within an image. Telescopes on the surface of the Earth have to worry about the atmosphere, which due to its non-uniformity causes blurring and smearing of images (this contributes significantly to what's known as the point spread function). The HST is in space to avoid atmospheric effects, and thus, it achieves higher angular resolution by doing so.


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What is the current theory for the formation of the Earth's Moon?
Given that the Earth and Moon are often said in the literature to have very similar geochemistry, what is the current theory as to how the Earth's Moon formed?

The current accepted theory is known as the Giant Impact Hypothesis, where according to this NASA webpage "Origin of the Earth and Moon" (Taylor) a Mars sized object collided into the early Earth.
This theory allows explanations of (from the link above):
The chemical makeup:

The giant impact hypothesis is consistent with our ideas for how planets were assembled and explains some important features of the Earth-Moon system, such as why the Moon has only a tiny metallic core.

and in terms of orbits:

To account for the amount of angular momentum in the Earth-Moon system, Cameron estimated that the object would need to be about 10% the mass of Earth, about the size of Mars. (Angular momentum is the measure of motion of objects in curved paths, including both rotation and orbital motion. For the Earth and Moon this means the spin of each planet plus the orbital motion of the Moon around the Earth.)

Image source
However, recently, there has been some revisions to this theory.
According to NASA's page "NASA Lunar Scientists Develop New Theory on Earth and Moon Formation", based on concerns that the Mars-sized would likely to have had a different composition to the Earth (inconsistent with the current similarities in geochemistry).
The new hypothesis is (from the link above):

After colliding, the two similar-sized bodies then re-collided, forming an early Earth surrounded by a disk of material that combined to form the moon. The re-collision and subsequent merger left the two bodies with the similar chemical compositions seen today.

This is also discussed in "Huge Moon-Forming Collision Theory Gets New Spin" (Wall, 2012), basing the revised hypothesis on the rotation rate, which is theorised to have been very fast, from the article:

Earth's day had been just two to three hours long at the time of the impact, Cuk and Stewart calculate, the planet could well have thrown off enough material to form the moon (which is 1.2 percent as massive as Earth).

Further discussion, based on the geochemical makeup is discussed in "Geochemical Constraints on the
Origin of the Earth and Moon" (Jones and Palme) conclude that:

Although none of these observations actually disproves
  the giant impact hypothesis, we find it disquieting that the
  obvious consequences expected of a giant impact are not
  observed to be fulfilled.

So, there is still some question as to precisely how the Moon was formed.

A recent study reported in the Phys.org article "Study crashes main Moon-formation theory" , based on the Nature Geoscience paper "A multiple-impact origin for the Moon" performed numerical modelling that suggests that

the Moon could instead be the product of a succession of a variety of smaller collisions.

The specific strength of such a theory is that "Numerous "impactors" would have excavated more Earth material than a single one" according to the authors, allowing for a stronger composition-based correlation with the Earth.
The authors also contend that

sub-lunar moonlets are a common result of impacts expected onto the proto-Earth in the early Solar System and find that the planetary rotation is limited by impact angular momentum drain.

The process is simplified as follows (from the article):

An interesting consequence of this model, according to the authors is that:

"Building the Moon in this way takes many millions of years, implying that the Moon's formation overlapped with a considerable portion of Earth's growth,"


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Why do some planets have rings?
Some planets, specifically Jupiter, Saturn, Uranus, and Neptune in our solar system, have planetary rings.

Why do some planets have rings? How are they made and from what? Most importantly, will I be able to observe the rings on any planet with an amateur telescope?

Rings are made up of tiny (and not so tiny) pieces of rock and ice that are in some way the bits "left over" from the formation of the planet. The theory involves the Roche limit - and is that particles that are already within this limit can't accrete into a larger body because of the tidal forces involved.
Another theory is that they are formed when a moon comes closer to a planet than the Roche limit, the tidal forces cause it to break up and form a ring. Though the presence of "shepherd" moons in the rings of Saturn does hint that this may not be major source of material.
Both explanations, to me, imply that you'd only get major ring systems around larger (gas giant) planets, though it doesn't preclude rings around smaller (rocky) planets. This seems to be borne out by our solar system where the gas giants have rings whereas the rocky planets don't.
Source
As to being able to see them, you should be able to see the rings of Saturn with an amateur telescope which has a 50 - 100 powers of magnification. With binoculars you'll probably see a misshapen blob.
Source

Your best bet is a telescope at least 50x magnification. Over 100x you will get decent visibility of the rings.

I think being distant from the sun helps too.   All the rocky planets in our solar system are close to the sun and the large ones further out.    I'm not sure a gas giant could keep a ring system if it was as close to the sun as earth or mars.

In an estimated 50 millions years, Phobos will also get crushed by Mars' gravity. Will it form a ring ? Who knows... But I'm pretty sure this is possible :p Though I agree with you on the fact that gas giants have rings more often due to their huge gravity, I wouldn't consider rocky planets having rings impossible. Universe is big.


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What evidence is there of Earth-Like internal features of Europa?
This question is inspired by "What are the Earth-like features of Titan?".  According to NASA's Europa overview, Europa is believed to have an iron core, rocky mantle and an ocean under the frozen surface.
Does the tidal heating of Europa's interior by Jupiter make it a smaller analogue for Earth?

This Image, Originally sourced from NASA shows the theorized interior of Europa:

According to this geology site Earth's core is predominantly iron and nickel.
and this space site says that Earth's mantle is predominantly silicates
So the internal similarities so far is that they both possess iron cores and silicate rock mantles. However the core of Europa is mixed with sulphur rather than Earth's nickel.
Another similarity between Europa and Earth could be the composition of the oceans. NASA speculates that Europa's oceans are also briny with salts and minerals dissolved in it. This is mostly theorised due to what we know about planet formation.
However this also comes with many differences, the Galileo spacecraft did not detect a magnetic field from Europa, which leads to speculation that, whilst Europa may have an Iron compound core, it is probable not molten any more.
Conclusion
Interior Similarities:

Iron compound core
Silicate rock mantle
Salty, briny, Oceans

Differences:

Iron core mixed with sulphur not nickel
Core likely to be solid, not molten
Lack of internally generated magnetic field


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What is the evidence that galaxies rotate?
Do all galaxies rotate?  If so, do they rotate in the same direction?
What observations have been made of galaxy rotation?

Rotation:
One of the easiest way to show that galaxies rotates are Doppler observations of atomic or molecular lines. For example, if the observed galaxy is seen edge-on and rotates clockwise, the left side of the galaxy is going away from you, and is then Doppler-shifted in the red. The right side, going toward you, is Doppler-shifted in the blue.
This kind of observations was realized already in the '60, in radio.
The spiral patern observed in spiral galaxies is also a clear signature of their rotation. Note that spiral are linked to density waves in the galactic disk (see for example Lin & Shu 1964).
Direction:
As for their directions, you can convinced yourself that there is no tendency, just looking at this Hubble telescope gallery. You can also think that galaxies probably rotate by angular momentum conservation during their formation history, and that it highly dependent on initial conditions, that are clearly random.

The spiral pattern itself does not at all let us know it is spinning.  Really, the spiral pattern is a bit of a mystery, knowledge of it is shambolic.  It seems to be an overdensity that moves through the galaxy and to us appears like a kind of twirly pattern.  It says nothing about whether it's actually spinning.

What you are referring to are density waves, and as I was pointing it out with Lin & Shu (1964), these waves are linked to the galaxy rotation. I don't think there is any other way of forming this waves and pattern than with differential rotation.

The angular momentum can't be random because it is conserved.. so galaxies must balance each other to zero.

Regarding spirals, I managed to get spirals by ejecting matter at the two ends of a bar and letting them attracted back by the bar via inverse square gravity. If the process happens at different times more arms can be generated this way and the number is always even and related to the same ends. This suggest the flow in the spirals is outward then back circulation.

We have not discovered every galaxy in existence, nor have we watched each to see how it is spinning. Most of this answer will heavily be speculation. You have been warned!
We have three classifications of galaxy:
Spiral, Elliptical and Irregular.
Irregular galaxies are thought to be quite young, and predominantly gasses that are yet to combine.
Elliptical galaxies are thought to be like a disc, or elongated disk in shape.
Spiral galaxies are theorised to be the older of the three, and it is heavily theorised that they are spiral in shape due to a super massive black hole at their center.
We can see that it is thought elliptical galaxies will eventually become spiral galaxies

E classifications are elliptical. S classifications are spiral.
Therefore if we extrapolate the pattern and theories it could potentially be said that in the course of a galaxies life it will eventually become a spiral, orbiting a super massive black hole as more and more of its mass is drawn to the centre.
Ok so if it can be theorised that most galaxies will eventually spiral a black hole it leads us to believe that the rotation of a galaxy will then be because of the rotational direction of a black hole, right?
This NASA theorist says there may be such a thing as a backwards black hole, one that spins in the opposite direction to its accretion disks, which in this case I assume would be the galaxy.
Looking at observations of galaxies leads to most of them, spinning with the arms trailing, but there are cases where the arms are facing the direction of the spin!
Conclusion
So Yes, we have observed the spin of galaxies, no, not every galaxy rotates, irregular galaxies have no central point of mass of which to rotate about. As for if they all spin in the same direction, that is extremely hard to observe without more data points and further study.

I strongly disagree with some statements that are absolutly wrong. For example, the spiral shape of the galaxies is not due to the presence of a black hole. Spiral arms are features linked to density waves in the galactic disk.

@MBR The most helpful thing to do, for everyone, is to post your own answer with links, or edit his answer where you think it's incorrect.

Effectively, this answer is more than speculative. As underlined by MBR, many (most) of these statements are simply wrong. Maybe answers here should be based on something else than speculation.

This post contains several errors: 1: Although the picture of the Hubble sequence shown was originally constructed by Hubble to illustrate an evolution from ellipticals to spiral, we now know that this is not how things work. Rather, ellipticals are probably formed by mergers of similar-mass galaxies. 2: As stated by @MBR, spirals do not owe their shape to BHs, but to enhanced star formation in the arms caused by density waves from stellar feedback. 3: The accretions disks around central BHs are small (~0.01 lightyears) disks of hot gas, not the entire galaxy.