text:
Why does Jupiter have so many moons?
Jupiter has a great many moons - in the hundreds, and they're still being discovered.
What is the current theory for where all these moons came from? Are they rocks flying through space captured by Jupiter's gravity?

Mass.
The more massive a body, the larger the gap between its lowest and highest orbit; the range of speeds at which a random body entering its gravity is likely to remain as its satellite. Sun has millions of satellites if you count all the asteroids; smaller planets tend to have one or two moons at most (Pluto with five being a notable and not fully explained exception)
To a lesser degree there's a matter of shape too. A regularly round body will have more regular and stable orbit than a potato-shaped one. Jupiter, being a gas giant is perfectly round. This doesn't play that much of a role though, especially with higher orbits.
And last but not least, no destabilizing influence of other bodies. It's very hard to maintain a lunar orbit - artificial satellites around our Moon last only a couple years each, because relatively close neighborhood of Earth tends to destabilize orbit of anything orbiting the Moon. Jupiter being a single massive planet with relatively tiny (relatively to its mass) moons doesn't have them influence each other all that much.

Jupiter isn't actually so perfectly round. Its oblateness is pretty large. But yes, it is perfectly round, compared to some crazy asteroids.

@User123 Primarily it doesn't have mountains or mascons of that kind.

Bigger is better.
Most moons, especially those of gas giants, are not "formed", they are just "captured" (unlike our Moon, which could have been captured, but probably was formed in a much more exciting way).
Jupiter is the most massive planet in the solar system. It stands to reason that it has a larger region of gravitational influence (where its influence outweighs the force due to the other planets and the sun). So, it's easy for it to capture rocky masses.
If you have a look at the contours on the following image (Ignore the Lagrange points marked on it, I only want the contours)

the circular area around the Earth is more or less the area (there's a velocity dependence here which I'm not getting into) in which a moon-like body can form a reasonably stable orbit. The size of the small "well" will increase as the planet moves farther from the sun, and also when the planet is more massive.
Jupiter is both pretty far away from the Sun, and is very massive. This leads to a huge sphere of influence.
The asteroid belt may have something to do with this too, but I doubt it (it's pretty far away). However, if we assume the "half-baked planet formation" theory for the formation of the belt, Jupiter may have leeched off much of the mass that would have otherwise become part of that planet during the formative period.

You forgit Mars. Smaller is better, compared to the earth.

The number of moons also changes. Most of the moons were discovered in the late 1970s and later as a result of several explorations by automated spacecraft, including NASA’s Voyager.

Its size and mass are essential to the number of moons of Jupiter because there is a large area of gravitational stability.

Jupiter also has the strongest magnetic field of any planet, so anything passing near it, such as an asteroid, is either destroyed by gravitational tides or captured into its orbit.
Some of the orbits are nearly circular, while the moons farthest from Jupiter have more irregular orbits. The outer moons orbit in the opposite direction in which Jupiter spins, which is unusual and indicates the moons were asteroids that were sucked into Jupiter’s orbit after the initial system was formed.

Distance offset from the Sun (weaker its gravitational pull).
Several moons probably formed in tandem with Jupiter itself.
Other satellites weren't necessarily home-grown. Many of Jupiter's may have started moons drifting rock that became entrapped by the planet's gravitational pull.

Many of the moons orbit in the same direction as Jupiter. But there are those which go the opposite way. Moons that crash into one another might be destroyed in the process. Just as Jupiter acquires new moons, it's finding ways to lose some of the older ones.


text:
Is there a theoretical maximum size limit for a star?
Some stars are simply huge. Eventually, though, wouldn't there be simply too much pressure or mass for the star to sustain itself? Wouldn't it eventually collapse into a black hole?
Is there a theoretical upper limit on the size of a star, and what is it based on?

According to current knowledge, yes. If the gas cloud is too massive, the pressure of the radiation prevents the collapse and the star formation.
The article Stars Have a Size Limit by Michael Schirber, it's about 150 Solar Masses. However, there's the Pistol Star, which is speculated to be 200 SM.
In the article 'Das wechselhafte Leben der Sterne' by Ralf Launhard (Spektrum 8/2013) there's a diagram with information that when the mass is over 100 SM, the star can't form because of radiation pressure. The exact value of the limit is not speculated in the article.

@Undo Adding 2 more cents to this already superb answer: R136a1, has a mass of 265 solar masses and is currently considered to be at the limit of how big stars can become . Btw: it is assumed R136a1 once had 320 solar masses when it was born a million or so years ago.

I'm always amazed when people use Wolf-Rayet stars as limits for "how massive a star can be," since it is very most likely that Wolf-Rayet stars themselves formed as the end-result of stellar evolution of their even more massive progenitors... So clearly, the masses of WR stars that we observe are not the limit, they are just the most massive ones that we have seen because high mass stars have a short lifetime at the later stages of their nuclear evolution. To drive the point even further, there are thought to exist supermassive stars, with $m \sim10^3$ which may seed supermassive black holes.

See here... https://astronomy.stackexchange.com/questions/47550/what-if-hypothetically-ton-618-could-have-been-a-star/47555#47555

A decent part of this answer is based on the introduction to Kroupa & Weidner (2005), though I’ve obviously gone into a lot more depth on all of the references.
Our story starts, as do many concerning stellar astrophysics, with Sir Arthur Eddington. In his 1926 book, The Internal Constitution of the Stars, he derived the Eddington luminosity, the maximum luminosity $L$ a star of mass $M$ can reach (Chapter 6, pages 114-115). His derivation goes along the following lines:
I. Take the equation of hydrostatic equilibrium and the equation of radiative equilibrium:
$$\frac{\mathrm{d}P}{\mathrm{d}r}=-g\rho\tag{1a}$$
$$\frac{\mathrm{d}p_R}{\mathrm{d}r}=-\frac{k\rho H}{c}\tag{1b}$$
The relevant variables are pressure ($P$), radius ($r$), gravitational acceleration ($g$), density ($\rho$), radiation pressure ($p_R$), mass coefficient of absorption ($k$), radiative flux per time ($H$), and the speed of light ($c$). Combining $(1\text{a})$ and $(1\text{b})$ yields
$$\mathrm{d}p_R=\frac{kH}{cg}\mathrm{d}P\tag{1c}$$
II. At some radius $r$, the luminosity $L_r$ and enclosed mass $M_r$ can be related by
$$\frac{L_r}{M_r}=\frac{\eta L}{M}\tag{2a}$$
where $L$ and $M$ are the luminosity and enclosed mass at the radius of the star, and $\eta$ is some function of $r$, increasing inward from $\eta(R)=1$ at the stellar radius $R$. Given that
$$H=\frac{L_r}{4\pi r^2}\tag{2b}$$
$$g=\frac{GM_r}{r^2}\tag{2c}$$
we have
$$\frac{H}{g}=\frac{L_r}{4\pi GM_r}\tag{2d}$$
Putting this back into $(1\text{c})$, we find
$$\mathrm{d}p_R=\frac{L\eta k}{4\pi cGM}\mathrm{d}P\tag{2e}$$
III. As temperature and density increase towards the center of the star, so does the pressure due to the matter, $p_G$. Therefore, $\mathrm{d}p_G>0$. Furthermore, given that $P=p_G+p_R$, $\mathrm{d}p_R<\mathrm{d}P$. This means that $(2\text{e})$ yields
$$\frac{L\eta k}{4\pi cGM}<1\tag{3}$$
which is the criterion leading to the Eddington luminosity. There are, of course, other ways of getting this criterion, but I thought I’d give Eddington’s original one, in all its mathematical glory.
Using a suitable mass-luminosity relation for massive stars, we can then establish the mass of a star at the Eddington limit. Eddington himself took it to be in the range of 60-70 solar masses ($M_{\odot}$), though today a value of somewhere around 120 solar masses is more appropriate.
Let us take a detour to a lesser-known figure, Paul Ledoux. In 1941, Ledoux analyzed vibration modes in stars due to the usual perturbations in density, pressure, radius, temperature, etc. He came up with the stability condition of
\begin{align}
  A_k&=\!
  \begin{aligned}[t]
 \int_0^M\frac{\delta \rho_k}{\rho}\biggl[(\Gamma_3-1)\delta_k\left\{\epsilon_1+\epsilon_2-\epsilon_3-\frac{\mathrm{d}}{\mathrm{d}m}[4\pi r^2(F_1+F_3)]\right\}&\\
  -\frac{2}{3}\delta_k\left[4\pi r^2\bar{C}\frac{\mathrm{d}P}{\mathrm{d}m}+\epsilon_2+\frac{\mathrm{d}}{\mathrm{d}m}[4\pi r^2F_2]\right]\biggr] \mathrm{d}m&<1\\
  \end{aligned}\\
\end{align}
for the $k$th mode of vibration. I’m not going to explain all the variables because that’s not quite important; the important takeaway is that Ledoux took turbulent pulsations into account. His conclusion is that an exact model “probably” would lead to a limit of about 100 solar masses; using certain inexact assumptions, he found a limit of 128 solar masses.
Ledoux’s analysis laid the groundwork for the work of Schwarzschild & Härm (1958). Their stability criterion isn’t necessarily simpler, but it can be written more compactly. Specifically, the coefficient of stability, $K$, defined as
$$K=\frac{1}{2}\frac{L_P}{E_P}$$
must be negative to ensure stability against pulsations. A positive $K$ means that the pulsation amplitude is increasing; a negative $K$ means that the pulsation amplitude is decreasing.
$E_P$ is the pulsation’s energy, while $L_P$ is the rate of gain of pulsation energy and can be expanded as
$$L_P=\overbrace{L_{PN}}^{\text{nuclear}}-\overbrace{L_{PH}}^{\text{heat leakage}}-\overbrace{L_{PS}}^{\text{progressive waves}}$$
Here, $L_{PN}$ represents the rate of energy gained, while $L_{PH}$ and $L_{PS}$ represent the rate of energy lost. All of the above quantities can be calculated through some relatively simple expressions (see Equations 9-12 and 15-22). The upshot of all this is that $K$ becomes negative at birth for stars greater than 60 solar masses. This can be figured out by writing $L_P$ and $E_P$ as functions of mass, $M$, and age, $\tau$.
Now, interestingly enough, the critical age ($\tau_{cr}$) can be written as a function of mass:
$$\tau_{cr}=0.05\left(\frac{M}{M_{\odot}}-60\right)$$
where $\tau_{cr}$ is in millions of years. This means that a star of, say, 62 solar masses (to take the authors’ example) will evolve to a stable state in a quarter of a million years. We can also determine whether or not, in this time, the star’s instability will become too great and destroy it. It turns out that this is the case for stars with masses greater than 65 solar masses - putting the upper limit for the mass of a star at 65 solar masses.
Here is a graphical representation from their paper, Figure 1:

Even later work on the same topic was done by Ziebarth (1970), among others, who extended the models to study different metallicities and compositions (Schwarzschild & Härm) focused largely on stars with compositions similar to that of the Sun). His computations found a wide range of upper mass limits - 10 solar masses for pure helium stars, and 200 solar masses for pure hydrogen stars. Most stars fall in the middle, and so will have different limits.
The actual formation of massive stars also puts constraints on the mass. Kroupa & Weidner mention Kahn (1974), who studied how radiation pressure from a protostar could drastically lower accretion rates, stopping the star from continuing to significantly grow. As applied to a young Population I star, his simplest model comes out to a limit of about 80 solar masses, although different models of the “cocoon” yield different results.
I’ll add one final note on theory. Population III stars, the hypothetical first stars in the universe, are expected to have been extremely massive; as such, they would be excellent candidates for testing the upper mass limits. According to simulations by Hosokawa et al. (2011), mechanisms similar to those discussed by Kahn would have stopped accretion at stellar masses around 43 solar masses -  a surprisingly low figure, given the expectations of how massive Population III stars should be. Additionally, as argued by Turk et al. (2009), sufficiently massive stars could fragment; in the case studied, a 50 solar mass star broke apart into two smaller core fragments.

Something I realized just now, a couple months after writing this, is that all of this assumes that the star is spherically symmetric. Most stellar models involve spherically symmetric, non-rotating stars, which allows us to make some assumptions such that the equations of stellar structure depend purely on $r$, the radial coordinate.
We have, however, seen stars - not stellar just simply stellar remnants like pulsars, but even main-sequence stars - that rotate rapidly and are thus non-spherical. Vega, for instance, has an equatorial radius 19% larger than its polar radius. If a star of mass $M$ is rotating, the equations of stellar structure should be different, and so some of the above results should also be different. I'm not sure how important this is for various theoretical limits.

The first order theoretical limit on stellar size is from the Eddington Limit. As the star collapses it is balances by radiation pressure from fusion. However, the fusion rate scales strongly with density (which is why the most massive stars have extremely short lifetime) so if the star was massive enough, the radiation pressure would probably blow it apart. In fact, this could lead to a pair-instability supernova and there wouldn't even be a black hole remnant even though the star is so massive.

Personal speculation from a non-professional: The maximum NOW may not be the same as the maximum for primordial stars because 2nd+ generation stars contain elements that require higher temperatures for fusion. Thus it may be possible for bigger stars to form now than in the beginning, although the chances of observing them would be low because they would be extremely short-lived... a sufficiently massive star composed of a high proportion of heavy elements may go supernova almost as soon as it's born.

In the modern universe normal matter is ~73.9% hydrogen, ~24% helium, with only ~2% heavier than helium, see Abundance of the chemical elements. Heavier elements don't start fusing until a substantial proportion of the hydrogen is gone. As ProfRob explains, higher atomic mass reduces the minimum star size slightly, but heavier elements are very important in star formation because they make it easier for the collapsing gas cloud to radiate away energy.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
What are the Earth-like features of Titan?
This NASA site says that hidden beneath the smog of Titan are Cirrus like clouds and This space article describes how Titan's atmosphere is similar to Earth's too.
What other features does Titan boast that make scientists so ready to boast that "Titan may well be the most Earth like planet we know of besides Earth itself"? (paraphrased)
Answers should focus on the features of Titan itself, obviously a similarity is that it orbits the same Sun, but this type of thing is not what I'm after. I am mainly searching for geological or atmospheric similarities that Titan shares all the way down to potential similarities in the composition and flow of the planetary core.

To expand a bit on Undo's answer, there are some geomorphological features on the surface of Titan that resemble similar features on Earth, suggesting similar processes.
For example, according to this NASA image comparison, some lake features have features that are similar to groundwater based salt pans on Earth. Image below:

The feature on the left is Ontario Lacus on Titan compared to the Etosha Pan of Earth on the right.  It should be noted, that the Titan feature is considerably larger than the Earth analogue.

A lot - from this space.com article:

Saturn's moon Titan may be worlds away from Earth, but the two bodies have some characteristics in common: Wind, rain, volcanoes, tectonics and other Earth-like processes all sculpt features on Titan, but act in an environment more
  frigid than Antarctica.

The article goes on to say that while Titan has these earth like features, it only receives about 1% of the sunlight that Earth receives. Because of this, the average temperature on the moon has an average surface temperature of -180 C (-292 degrees F), and water can only exist there in a super-frozen form as hard as rock.

On Titan, methane takes water's place in the hydrological cycle of evaporation and precipitation (rain or snow) and can appear as a gas, a liquid and a solid. Methane rain cuts channels and forms lakes on the surface and causes erosion, helping to erase the meteorite impact craters that pockmark most other rocky worlds, such as our own moon and the planet Mercury.

As for the planet's core, this Stanford paper has some info:

"The picture of Titan that we get has an icy, rocky core with a radius of a little over 2,000 kilometers, an ocean somewhere in the range of 225 to 300 kilometers thick and an ice layer that is 200 kilometers thick," he said.

I'm not going to quote the whole paper here, but it's a good read on the subject of Titan's core.
This Wikipedia article has some information about Titan's atmosphere:

Observations from the Voyager space probes have shown that the Titanian atmosphere is denser than Earth's, with a surface pressure about 1.45 times that of Earth's. Titan's atmosphere is about 1.19 times as massive as Earth's overall, or about 7.3 times more massive on a per surface area basis. It supports opaque haze layers that block most visible light from the Sun and other sources and renders Titan's surface features obscure. The atmosphere is so thick and the gravity so low that humans could fly through it by flapping "wings" attached to their arms. Titan's lower gravity means that its atmosphere is far more extended than Earth's; even at a distance of 975 km, the Cassini spacecraft had to make adjustments to maintain a stable orbit against atmospheric drag. The atmosphere of Titan is opaque at many wavelengths and a complete reflectance spectrum of the surface is impossible to acquire from the outside. It was not until the arrival of the Cassini–Huygens mission in 2004 that the first direct images of Titan's surface were obtained. The Huygens probe was unable to detect the direction of the Sun during its descent, and although it was able to take images from the surface, the Huygens team likened the process to "taking pictures of an asphalt parking lot at dusk".

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Why do the planets in our solar system orbit in the same plane?
(Yes I'm excluding Pluto from this the same way it was excluded for not being a planet)
Observing the planets orbit of the Sun they all seem relatively planar and roughly all orbit along the same plane.
Is this due to the way that our Solar system was formed or is this a physical phenomena observed in other systems?

To understand better, see this.

In addition to Yashbhatt's video, you may also find this one helpful.

In the protostar stage of the Sun, it was surrounded by a (spinning) gas cloud. This cloud behaved like a fluid (well, a gas is a fluid), so it flattened out into an accretion disk due to conservation of angular momentum. The planets eventually formed from the dust/gas in the disk from compression of the dust in the disk. This process won't end up moving the dust out of the plane (all the vertical force of gravity is toward the disk), so the final planet is in the plane too.
Why does an accretion disk have to be flat? Well, firstly, let's imagine the protostar and gas cloud before the accretion disk formed. Usually such a setup will have particles spinning in mostly one direction. The ones spinning in retrograde orbits will end up reversing themselves due to collisions.
In this gas sphere, there will be an equal number of particles with positive and negative vertical velocities (at a given point in time; due to rotation the velocity signs will flip). From collisions, eventually these will all become zero.
A particle revolving around a planet will always revolve such that the projection on the planet is a great circle. Thus we cannot have a particle with vertical velocity zero but with vertical position nonzero (as that would imply an orbit that isn't a great circle). So as the vertical velocity decreases, the orbit inclination decreases too. Eventually leading to an accretion disk with very little vertical spread.

By "behaved like a fluid" you meant all of that gas tended to stick together? (covalent bonding)

But to answer the part of the question about whether it's a special phenomena, this answer is implying that all solar systems are formed by spinning gas clouds. Why is that supposed be obvious? Non-spinning bodies have gravity. But they can't create solar systems because there's no counter-force to balance an orbit? Is it that as galaxies whip around, they spin things? That can't be true because systems spin in all kinds of different directions... what don't I get?

What decides the direction in which the accretion disk rotates?

@new-kid that would make an excellent new question, I'm very curious about this !

@Nico posted here

@DaveG Just by random chance, it is very unlikely that a randomly forming gas cloud will have angular moment of exactly zero.  If it did, it would probably all accrete onto the star.  In practise most clouds do have some angular moment, so they are spinning


text:
Is the moon moving further away from Earth and closer to the Sun? Why?
According to The NASA Moon Facts page:

The moon is actually moving away from earth at a rate of 1.5 inches per year.

Why is the moon moving further away from the Earth? Is this a result of the moons formation that set it in motion to spiral away from us? Or is this a resultant force of the gravity from the Sun and other large bodies?

Since it is part of your question, but not explicitly answered in any of the posts to date: no, this has nothing to do with gravitational interaction with any other bodies in the solar system.  Furthermore, moving away from the earth does not mean closer to the sun since during half of its revolution, the moon is further from the sun.

If the moon is moving away at 1.48" (which is a fairly small, insignificant dimension) per year and, I assume, this has been fairly constant over the history of the earth, that means that 10 billion years or so ago the two were touching. How does that jive with the age of the earth and moon and how would that have affected the gravitational interaction of the two bodies?

Well, neither the Earth nor the Moon (or the Sun) are 10 billion years old. The change in distance is also not constant over time.  However, this doesn't necessarily mean that the Earth and Moon weren't touching at one point in time. If this doesn't resolve your question, I would suggest you ask it as a brand new question, as is the best practice here.

Have you considered our own continental shift movements as part of the measurement? its my understanding that the measurement is calculated by the use of a beam bouncing from the surface. If the continent on which the beam is mounted is moving also, could this affect the measurement?

Yes, the moon is moving away from Earth at around 1.48" per year. According to the BBC:

The Moon is kept in orbit by the gravitational force that the Earth exerts on it, but the Moon also exerts a gravitational force on our planet and this causes the movement of the Earth's oceans to form a tidal bulge.
Due to the rotation of the Earth, this tidal bulge actually sits slightly ahead of the Moon. Some of the energy of the spinning Earth gets transferred to the tidal bulge via friction.
This drives the bulge forward, keeping it ahead of the Moon. The tidal bulge feeds a small amount of energy into the Moon, pushing it into a higher orbit like the faster, outside lanes of a test track.

So, tidal forces are ultimately what causes this to happen.
Also, there is a Wikipedia article on tidal forces:

Tidal acceleration is an effect of the tidal forces between an orbiting natural satellite (e.g. the Moon), and the primary planet that it orbits (e.g. the Earth). The acceleration causes a gradual recession of a satellite in a prograde orbit away from the primary, and a corresponding slowdown of the primary's rotation. The process eventually leads to tidal locking of first the smaller, and later the larger body. The Earth–Moon system is the best studied case.

Thought, as the moon gets further away wouldn't the amount of force applied to and from the tide be reducing thus limiting the push of the moon away from us, effectively reducing the distance it moves away each year? Would it not reach a stable orbit where the force is too small to push it further away?

@RhysW: The tides are slowing down the Earth's rotation as well as driving the Moon farther out. Ignoring other influences, the effect would stop when the Earth and Moon are both in locked rotation, each showing the other the same face (like Pluto and Charon).

@Guillochon: That's not quite what the article says. Currently the barycenter of the Earth-Moon system is inside Earth. Eventually it will be above Earth's surface. The Earth and Moon will still orbit around their common center (as Pluto and Charon do). It's more a matter of the current definition of the word "planet"; there's no great physical significance to it. See also this question on the Physics site (posted to the old Astronomy site before it was merged into Physics).

Funny side note; due to pollution, earths gravity is increasing, counteracting the fact that the moon is moving away from us by a tiny amount and instead speeding up the moons orbit. This is because energy also causes gravity, and a warmer planet means more energy here.

@frodeborli I'm pretty sure you're confused - to the best of our knowledge, energy doesn't cause gravity. If you can show me a source that says otherwise, though, I'm always open to new ideas.

@Undo My source is Einstein. The formula E=mc^2. For any source of energy, you can calculate the gravity it causes by using E/c^2 in place of mass. Both kinetic and potential energy cause gravity.

@frodeborli - I'm extremely skeptical. How much energy are we talking about? And how much mass would that make? I'm betting it's negligible when compared to the total mass of a planet.

@frodeborli The Earth is actually losing mass-energy. See this BBC article which has some Cambridge physicists draw up an estimate of mass-energy in vs. mass-energy out.

@Donald.McLean The kinetic energy of a planet in motion quite large. The energy due to greenhouse effect in comparison is neglible, but it is there. 9 x 10^13 kilojoule will amount to about one kilo.

@called2voyage Nice. That article actually confirms my assertion; global warming contributes to a gain in mass of about 160 tonnes per year, a tiny slowdown of the loss of mass.

@frodeborli Right a tiny slowdown of the significantly larger but also ultimately (compared to the mass of the Earth) tiny loss of mass.

@frodeborli You're missing the whole point. It would be like contributing one penny a year to the US federal government and saying that you're making a dent in the US national debt.

@Donald.McLean You're missing a key word in my funny side note, quote: "a tiny amount". But being tiny does not mean that my assertion is wrong.

@frodeborli Your original assertion was "due to pollution, earths gravity is increasing". We have shown this to be false. Earth's mass is decreasing, thus its gravity is decreasing.

@called2voyage Yes, you are right. I failed to state what I assumed to be obvious in what I assumed would be taken as a fun fact - that pollution is not the only thing affecting Earths' gravity. Thank you for contributing that. Nevertheless, pollution is reducing the rate of weight loss, and reducing the rate of loss of gravity. Would it be correct to say "it is increasing, relative to the gravity it would have if we did not pollute"?

@frodeborli I believe the least easily misinterpreted way of stating it would be that it is "reducing the rate of mass loss". Using "increase" at all is misleading.

@called2voyage Yes. I wish I could take back the comment. I found the fact I intended to convey itself amusing and spontaneously wrote it as a comment.


text:
Are there ways other than the collapse of a star which have been observed to form black holes?
Every time I hear about a black hole, it's always in conjunction with the collapse of a star. Have any other processes been observed to create a black hole?

There are currently three classifications of black hole and these are stellar, super massive and miniature.
Stellar
Stellar black holes are formed by a massive star collapsing and its mass being great enough to form a black hole of moderate size
Super-massive
These black holes can have the mass equivalent of a billion suns, these most likely exist at the centre of galaxies and are probably as old as time itself.
Miniature
Miniature black holes are theoretical and are thought to have been created at the early stages of the universe. these require masses less than that of our sun and are thus very unlikely.
Other
"But Rhys" I hear you cry, "these are all from stars you fool!"
And yes, yes they are, this is because black holes are formed from mass, and lots of mass (except miniatures), coincidentally large amounts of mass tends to gravitate towards each other and form stars long before it forms black holes.
Therefore theoretically, a black hole does not even need stars, it just needs lots and lots of mass, but having that much mass without forming a star is no easy task.
Short answer, you can make a black hole from anything as long as it has enough mass, but good luck stopping this from forming a star beforehand!

Well, when two black holes collide a new black hole is created.  You could call that cheating, but this process is very different from that of a star collapsing. Also, these guys would be after your skin if you did.
When two black holes collide, there are three stages. Firstly, there's the "inspiral", which is exactly what it sounds like — the two black holes slowly spiral in to each other. Cue waltz music. Then there's the "Coalescence", where the two black holes merge. Note that prior to the merge, their absolute horizons actually grow in size, as if in anticipation of the merge. The exact behavior of the coalescence stage is not well understood yet. Finally, there's a "ringdown", where the new black hole (currently shaped like an eight) shakes down into an ellipsoid.
This has not been observed, but is something we're pretty sure occurs.

There also are primoridial black holes, from the time of the creation of the universe. They are extremely tiny (and unstable), though I doubt they can be formed at this current era.

It is theoretically possible to create black holes with radiation, it is called a  "Kugelblitz". The technology needed to create bright enougth pulses of gamma radiation is beyond our current technology, but may be possible.
The possibilities for the usage of such miniature black holes are enormous, it would solve all energy problems because they can convert between 20% to 50% of the introduced mass directly to energy. If the hawking radiation exists its possible to get 100% but enougth scifi.

Micro black holes are believed to exist, coming into existence for a very brief period of time following high energy particle collisions.  Some theorists believe that this may occur naturally in Earth's atmosphere due to cosmic rays. I read a statistic somewhere which said that at any given time there would be three black holes in our atmosphere due to this process (if it exists).


text:
What is a White Hole?
White holes are purely hypothetical at this point and are considered only because it is assumed something like it could exist as the balance to black-holes.
Reading this article seems mostly writers taking liberties with the information but poses an interesting question.
What is a White Hole?

Well according to the article "Have we seen a white hole?" (Byrd, 2011) there are some claims that scientists (specifically "The Revival of White Holes as Small Bangs" (Retter and Heller)) may have detected a white hole; however, this claim seems to be shrouded in controversy, as this observation may actually have a more stellar explanation.
However, the article from Retter and Heller offer a hypothesis as to what white holes are, from the article linked before:

We thus suggest that the emergence of a white hole, which we name a ‘Small Bang’, is spontaneous – all the matter is ejected at a single pulse. Unlike black holes, white holes cannot be continuously observed rather their effect can only be detected around the event itself.

Which, if the authors are correct, would explain why we don't see/detect them and white holes (unlike black holes) are unique because they supposedly spew matter, rather than extract.
One major caveat is that it is not clear as to where this matter that spews out comes from.

It supposedly comes from a black hole and travels there through a wormhole.

Is it a quack idea that the big bang could be the outpouring of a white hole? And also, would the existence of multiple white holes necessarily contradict conservation laws?


text:
Does the Sun rotate?
The planets rotate as an after effect of their creation, the dust clouds that compressed span as they did so and the inertia has kept it rotating ever since.
It's fairly easy to prove that planetary bodies are rotating just by watching their features move across their respective horizons.
This seems less easy to discern for amateur astronomers  in the case of the sun though.
Does the Sun also rotate as a by-product of it's creation? What evidence is there to support this? Does the sun have any discerning features that make it evident it is rotating?

Why did Galileo bother?

Yes.
It does not rotate uniformly though, different portions have a different angular velocity (as a body made of plasma, it can get away with this).
Measuring this in theory is pretty easy, we just need to track the motion of the sunspots. This isn't as simple as calculating the changes in relative positions of the sunspots, though, as the Earth is rotating and revolving, which makes the calculations harder. This measurement can be done using the celestial sphere (the field of stars that we see) as a "fixed" reference point and seeing how the Earth and the sunspots move relative to that.
Almost everything in the universe rotates/revolves, at least a little bit, because angular momentum is hard to get rid of. It can be transferred from body to body, but for a body to end up with zero angular momentum, it needs to meet another body with the exact same angular momentum and collide with it in a particular way. Given that this is pretty rare, all celestial bodies rotate.
In addition to that, a non rotating body that is revolving will eventually start spinning due to tidal forces.

So if the sun were not to rotate, it would start doing so (albeit it seems extremely slowly) thanks to its revolution around the Milky Way?

Well, it is a bit complicated to answer, because stars in galaxies do not follow expected orbital speed patterns. Mercury (near the Sun) travels at a higher speed, while Uranus (far from the Sun) travels more slowly. On the other hand, stars close to the center of a galaxy travel more slowly, compared to stars further out: https://universeadventure.org/final_frontier/dkmttr-rotnvelocities.htm . Also, the Sun is part of a local group of stars, within an arm of the Milky Way galaxy: https://www.visualcapitalist.com/cp/map-of-the-entire-known-universe/

And since the force of gravity decreases as 1/(radius squared), any tidal force on the Sun from nearby objects will be strongly dependent on the Sun's distance from them. In this case the tidal force is the difference in gravitational force between the front and back of the Sun from another object. Tidal forces are one way to affect an object's rotation on it's axis. This is the main reason we only see one side of the Moon from Earth- the moon is 'tidally locked' to the Earth because the force of gravity from the Earth is stronger on the 'front' of the Moon than the back.

All of that is to simply say that objects (e.g., stars) close to the Sun will have a far greater effect on the Sun's rotation on its axis than objects far away (e.g., the black hole at the center of the Milky Way.

Yes, the Sun rotates. This can be observed by tracking a variety of features on the Sun, such as sunspots, X-ray brightpoints, coronal holes, filaments, and small magnetic flux elements. Another way to determine the rotational speed of the Sun is to measure spectral lines at the edge of the Sun's disk and determine their redshift.
It is thought that the rotation of the Sun is due to the way the primordial gas cloud collapsed in on itself to create the Sun. Also, it is likely that the Sun originally rotated much faster when it initially formed, than it does to day. This slow-down was probably caused by 'magnetic breaking' in which strong magnetic fields threading our primordial Sun out into the solar wind resisted the rotation.
Today, the reason that the Sun rotates at a different angular speed at different solar latitudes is due to hydromagnetic effects. One cause is thought to be the nature of convection within the outer third of the Sun. Interestingly, below a the convective envelope, beneath a boundary called the tachocline, the Sun rotates as a rigid body. You can find more about this in Schou et al., 1998

+1 This answer is much more informative than the accepted answer. You pack a lot of clear explanation and physical description into ten tidy sentences!

Although its too late to reply to this interesting question but trying to add few more points.
Yes the sun rotates.
Now the question arises as to how we can check that?
We can observe this by observing sunspots. All sunspots move across the face of the Sun. This motion is part of the general rotation of the Sun on its axis. Observations also indicate that the Sun does not rotate as a solid body, but it spins differentially. That means that it rotates faster at the equator of the Sun and slower at its poles. (The gas giants Jupiter and Saturn also have differential rotation.) The movements of the sunspots indicate that the Sun rotates once every 27 days at the equator, but only once in 31 days at the poles.
You may refer this interesting article:- Does The Sun Rotate?

Here is a nice place to watch the daily movement of sunspots: http://www.spaceweather.com/

This is an old thread, but I found this interesting.   Young suns, when they just form often rotates very fast.    Not once every 31 days, but once a day or several times a day.   This is similar to the skater who spins faster when he/she pulls in their arms.    The conservation of angular momentum can make young stars spin very fast.   They slow down over time, probably due to planetary tidal drag

No they slow down due to coupling between the stellar wind and magnetic field.

Yes, Sun has differential rotation. Movement of Sun spots is one of the proofs that Sun rotates. The differential rotation causes the weird twisted magnetic fields which shows in the Sun's prominence.

Yes the sun rotates and it takes about 26.24 days.They are many methods to determine the rotation periods but most common one is by observing sunspots.
Here's a link with a detailed explanation https://en.wikipedia.org/wiki/Solar_rotation.

So, far all the answers have shed light on the experimental reasons. But you can say it with full confidence without any astronomical observations that the Sun rotates.
The logic is that any closed system have a center of mass that is unaltered unless acted upon by some external force. So, given that Earth rotates around Sun then the system forming Earth and Sun must have a center of mass which is not changing. Therefore Sun must rotate to counter the change in center of mass caused by the rotation of Earth. And so it does with any other system belonging to our solar system.

The rotation of the Sun wouldn't do anything the counteract Earth's motion. The Sun orbits the center of mass of the Solar System; that's what keeps the center of mass essentially unaltered.


text:
What are Trojan Asteroids and where are they located?
Asides from the main Asteroid Belt, there are Trojan Asteroids - what are these and where are they located?

A Trojan Asteroid is defined by the Swinburne University Astro page "Trojan Asteroids" as:

Asteroids sharing an orbit with a planet, but which are located at the leading (L4) and trailing (L5) Lagrangian points.

These are often divided into 'leading' and 'trailing', as per the diagram below (from the source linked above):

Even though the term Trojan asteroid is often associated with the 4800 or so that lead and follow Jupiter, it actually applies to any asteroid that displays this behaviour, associated with any planet.
There are numerous Trojans co-orbiting with Mars and Neptune.  Recently, single Trojans have been found co-orbiting Uranus and even Earth.

L5 is the Sun–Earth Lagrangian point located close to the Earth's orbit 60° behind Earth.
Asteroid 2010 SO16, in a horseshoe companion orbit with Earth, is currently proximal to L5 but at a high inclination.


text:
What is the current accepted hypothesis of what caused Triton's retrograde orbit around Neptune?
According to the NASA overview about Triton, this, the largest satellite of Neptune is unique in that it has a retrograde orbit around Neptune.  The page also suggests that Triton is a captured Kuiper Belt object.
What is the current accepted hypothesis as to what caused Triton's retrograde orbit?  Also, is this orbit sustainable - meaning, is it degrading?

Triton's orbit is not stable. It is degrading.

According to this source from SOA/NASA Astrophysics Data System on Tidal evolution in the Neptune-Triton system (Chyba, C. F., Jankowski, D. G., & Nicholson, P. D.),

Transcript: Many investigators have speculated on the reason for
  Triton's retrograde orbit. Lyttleton (1936) suggested that both Pluto
  and Triton originated as prograde satellites of Neptune, only to
  experience a catastrophic gravitational interaction. McCord (1966)
  showed that tidal evolution could have brought Triton inward from a
  near-parabolic orbit, lending plausibility to a capture origin for the
  satellite. Jarrington and Van Flandern (1979) suggested Triton's
  peculiar orbit and Pluto's supposed "escape" were caused by ab
  encounter with a massive "rogue" body, a conclusion contested ny
  Farinella et al. (1980). Finally, McKinnon (1984) combined momentum
  and energy constraints with an improved knowledge of Pluto's mass to
  show that all Pluto-Triton interaction scenarios for the origin of
  Triton's retrograde orbit are untenable, leaving the capture
  hypothesis as the most likely possibility.

Now, that summarizes all the theories one might have about it, at least all I could find.
As far as evolution of the system is concerned, the same document follows up to it and demonstrates how tidal effects are responsible for the evolution of the system. "Degrading" is a strange term. Let us go with evolution.
One of the possiblities stated is that Triton will reach a perfectly retrograde state and stay in it for ever! For a detailed analysis, you may read up section 6 in the link provided, "The future of Triton"


text:
How is it known that Callisto has no core?
My Astronomy book claims that scientists have discovered that Callisto, a moon of Jupiter, has no hot inner core. In fact, it says, Callisto has a core much like the nucleus of a comet.
Is this still accepted theory, and how was this information discovered?

It may have a (silicate) core, but it's very small. There also is the possibility of an internal liquid water "ocean". Currently the most probable composition is that of rock and ice, with the density of rock steadily increasing as one digs deeper (This is in stark contrast with planets like Earth, where there are discontinuities in the density between layers). Planetary differentiation has not occurred appreciably because Callisto is not tidally heated.
The composition can never be completely determined — the composition of a spherical body cannot be fully discerned from just rotation and translation1. However, we can make reasonable guesses. Basically, we can calculate its total mass and moment of inertia by observing its rotation and revolution. Then, assuming that the density ρ varies radially, we get the following two equations:
$$M = \int \rho \pi r^2dr$$
$$I = \int (\rho \pi r^2dr)r^2$$
These do not completely determine ρ, however with other reasonable assumptions on the composition and the amount of planetary differentiation it can have, we can get a good estimate for ρ.
1. It can be determined if we take a larger gravitational field and pivot it, but this ventures into the Aristotlean "Give me a place to stand on, and I will move the Earth." zone of thought experiments.


text:
What physical features determine if a planet is a major, minor or dwarf planet?
Like many, when I was growing up, we always were taught, hence always learned that there were 9 planets.  However, recently, decisions were made and all of a sudden there were 8 major planets and a series of dwarf and minor planets.
What physical features delineate a body being referred to as a major planet as opposed to dwarf planet or minor planet?

Planets
For a body to be classified as a planet it must have a few physical characteristics:
Mass
It must have enough mass to have a strong enough gravity to overcome electrostatic forces to bring it to a state of hydrostatic equilibrium.
Hydrostatic equilibrium is important because early in a planets life it is nearly entirely fluid, crust and all
Internal Differentiation
The life cycle of a planet essentially leads to denser heavier metals being at the centre of a planet, surrounded by a mantle which must at some point have been fluid (it can still be solid and called a planet as long as it used to be fluid.)
Atmosphere
This is usually driven by its mass but a planet should have an atmosphere. This means it should be massive enough to have a strong enough gravity to hold some gasses to its surface.
More massive planets are capable of keeping lighter gasses, such as hydrogen, bound to them too. I.e Jupiter.
Magnetosphere
A magnetosphere suggests that the body is still geologically active. This means they have flows of elements that conduct electricity in their interiors.
As specified by the IAU:

A celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and (c) has cleared the neighbourhood around its orbit.

Minor Planets
Minor planets  are things such as asteroids, they are usually small mass, rocky bodies. This is different from small mass bodies that might be predominantly ice or water.
Dwarf Planets
Size and Mass
The IAU have not specified an upper or lower limit for mass to be considered a Dwarf Planet, therefore it is predominantly determined by one other feature called:
Orbital Dominance
Orbital dominance is achieved when a body has cleared it's orbit of all other bodies.
For example planets are able to remove small bodies out of their area through impact, capture, or gravitational disturbance.
Any body that is incapable of doing so is therefore classified as a dwarf planet. So if Jupiter was incapable of clearing its neighbourhood of bodies then it too would be a Dwarf planet.

Where's Haumea and Makemake?

In 2006 the Internaional Astronomical Union (IAU) (that's a download link, 85KB) came to the resolution that a planet is:

a celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape, and (c) has cleared the neighbourhood around its orbit.

A Dwarf Planet is:

a celestial body that (a) is in orbit around the Sun, (b) has sufficient mass for its self-gravity to overcome rigid body forces so that it assumes a hydrostatic equilibrium (nearly round) shape1, (c) has not cleared the neighbourhood around its orbit, and (d) is not a satellite.

The same resolution also said that anything else, except satellites, is classified as a "Small Solar-System Body".


text:
What is the difference between asteroids, comets and meteors?
Some celestial objects seem to be referred to as asteroids, some as comets, and some as meteors or meteorites.
What is the distinction between all of these different objects? Are any of them the same?

The objects you are refering to are actually two different objects: asteroids and comets. Meteor and meteorite are other names for an asteroid, at a given time of its interaction with our planet. We'll get to that.
So first, what is the difference between an asteroid and a comet?
A comet is a small solar system body that display a "coma" (an atmosphere of a sort) and sometimes a tail passing close to the Sun. They are mostly made of ice and dust, as well as some small rocky particles. We distinguish two kind of comets, with short or long orbital period. The short orbital period ones originated from the Kuiper Belt, a region composed of small bodies beyond the orbit of Neptune. The long orbital period ones originated from the Oort cloud, a scattered disk of icy planetesimals and small bodies laying around our solar system.
An asteroid is a small body, composed mostly of rocks and metals. In our solar system, they can be originated from the asteroid belt, laying between Mars and Jupiter, or from the orbit of Jupiter (the Jupiter Trojans), or from actually almost everywhere in the Solar System.
A small asteroid ('meteoroid') that enters the Earth's atmosphere becomes a meteor, what we also call a "shooting star".
Eventually, a meteor that was massive enough not to be completely distroyed entering the Earth's atmosphere and hitting the ground is a meteorite.

A "shooting star" can also be a comet, right?

@SarahofGaia No, as noted in the answer a "shooting star" is a meteor.

"A small asteroid ('meteoroid') that enters the Earth's atmosphere becomes a meteor, what we also call a 'shooting star'."              ____________________________                                                                                But comets much more often than asteroids provide the material for meteor showers. Comet 1P/Halley is identified as the source of the Eta Aquarids and the Orionids; Comet 109P/Swift-Tuttle is the source of the Perseids; and Comet Thatcher is the source of the Lyrids. I suppose that comets more often than asteroids produce meteors.

Asteroids are what are called minor planets - small inner solar system rocky bodies. Especially between Mars and Jupiter.
Comets typically have highly elliptical orbits but they don't have to. The key defining feature of a comet is that it has a gaseous atmosphere and a tail when it passes close to the Sun. Comets are generally loose aggregations of ice, dust, rocks etc.
Meteoroids are very small rocky or metal bodies, from a single particle up to about 1m diameter. If they hit the atmosphere, the visible glowing trail of particles is called a meteor. And if it hits the ground it is called a meteorite.

Asteroids and Meteoroids are very different in size. Asteroids are huge usually about few kilometers big.
Whereas Meteoroids can be so small from centimeters to about few meters wide.
Both are very different in size. Sometimes, when two asteroids collide they might throw away few percentage of their rock which falls into planets as meteorites.


text:
What is the relevance of the Tropics and the Polar Circles? Do they Exist on other planets?

The middle line is the equator, the second and fourth line are the tropics, the first and fifth are the polar circles.
What is the relevance of these bands on Earth? Do these exist on other planets? What purpose do they serve?

Good question, I wonder if these have analogues for Mars, which shares a similar axial tilt as Earth.

The Tropics (Capricorn in the South and Cancer in the North) are the furthest point from the equator where the sun reaches a point directly overhead at any time during the year. The Arctic and Antarctic circles are the points at which the sun remains above the horizon for a full rotation of the Earth (i.e. 24 hours), commonly called "Midnight Sun". Any planet that has axial tilt while have similar lines.

This page has a solar climate zone map for mars with the equator, the Tropic of Pisces, the Tropic of Virgo, and the Polar Circles depicted.

@called2voyage good find.


text:
Do other planets in our solar system experience eclipses or is this unique to Earth?

Here on Earth we get to experience lunar and solar eclipses. Is this pure coincidence that everything happens to line up?
Do other planets in the solar system experience this phenomenon or is this something unique on Earth?

Pretty much every planet with a moon can have eclipses - you'll have seen photographs of Saturn with shadows cast from its moons, like this one from wodumedia.com:

If you were in a balloon where that shadow is, you would see a solar eclipse. It wouldn't be as exciting as from Earth, as the Sun would appear so much smaller from this distance.
If you want the planet to eclipse the moons, this happens far more often than on Earth, as the moons are so much smaller than Saturn. They often are completely shadowed.

Eclipses are common, but solar eclipses like the ones our moon causes would be unusual since it takes a combination of factors, the angular size of the moon and sun are nearly the same from the viewpoint of Earth so that the moon (depending on where it is in its slightly ellipitcal path around the Earth) can cover the sun nearly exactly.
NASA has a nice picture of an eclipse as seen from the Mars Curiosity rover showing an eclipse by a moon with an angular size less than that of the Sun.
Phobos eclipsing the Sun

One also does not need to be on a planetary body to experience an eclipse. NASA also has an absolutely gorgeous picture of Saturn eclipsing the sun as viewed by the orbiting Cassini spacecraft: http://apod.nasa.gov/apod/ap110904.html

One of my favourite photographs that, @aleppke. Dr Porco has stated in interviews since that they asked people to wave at Cassini about an hour an a half before the photo was taken, so that when the light hit the probe it would include people waving from Earth. :)


text:
Why do stars appear to twinkle?
Sometimes at night you will look up to the stars and they will appear to twinkle, getting brighter and darker in bursts.

Why does this happen? Is this because of our atmosphere? Would they twinkle if you were just in space? How about looking from another planet in our solar system?

Stars tend to twinkle for two main reasons: first, stars are very far away (the closest star is about 4 light-year from the Sun) and are therefore seen as point sources. Second, Earth has an atmosphere. Earth's atmosphere is turbulent, and therefore all images view through it tends to "swim". Therefore, sometimes a single point in "object space" is mapped to several points in "image space", and sometimes it is not mapped at all. Since stars are seen as single points, they sometimes seems brighter, sometimes even seems to disapear.
If you look at it in another planet of our solar system, it will depends on the planet's own atmosphere. If you look at stars on Mars, the atmosphere being very thin, the stars won't twinkle that much. Same for Mercury. On Venus, the atmosphere is so thick than you won't see anything apart from the atmosphere itself (if you are not crunched by the atmosphere pressure, by the way...).

Well atmospheric pressure doesn't affect us all that much, not nearly as much as the gravity from the planet itself. Venus and Earth have similar gravity so the pressure is what makes us feel like we weigh more than we actually do in a thick atmosphere. Air pressure itself varies day by day, hour by hour, even second by second which is why a barograph(like a barometer but it graphs the pressure instead of just measuring it) has significant and tiny variations in air pressure. And we have been able to survive high and low air pressure for thousands of years.

You couldn't survive the pressure on Venus (or the Temperature).   The atmospheric pressure on the surface of Venus is equivalent to nearly 3,000 feet deep in the ocean.   That would kill a human pretty quick, though there are animals that can survive those pressures.   Elephant Seals and Sperm Whales for example.

Twinkling is caused by turbulence AND refraction in the atmosphere. The refraction bends the light from the star and the random motion from the turbulence causes the bending to change direction, making the star appear to twinkle.
This will happen with any object, but it is most noticeable in stars because they are so far away that they appear like a point of light. The effect is less obvious in planets since we can resolve their size, though on particularly turbulent times, planets may also appear to be twinkling.

@Aaron-Could the smallest star on that picture be closer than the larger flickering stars? And does size determine distance or does the mass determine the size?


text:
If the moon wasn't receding from Earth, what would be the impact on the weather and tides?
Moon is slowly receding from Earth, which means that after its formation it was much closer to the Earth than now. How would weather look like if it wasn't receding at it would be now as close as at the beginning?
Surely, the tides would be much higher, but how big would be the difference? I've also watched document stating that the hurricanes would be much stronger. Was there any simulations, how much stronger they would be?

An interesting BBC snippet about this http://www.youtube.com/watch?v=XR0N4XKwgHM

Tidal amplitudes roughly scale inversely with distance to the fourth power (this comes from the tidal force scaling as $r^{-3}$ and the Earth's potential scaling as $r^{-1}$). At its formation, the Moon was approximately seven Earth radii away from Earth, or about ten times closer than it is today. Thus, the amplitude of the tide on Earth would have been approximately $10^4$ times larger than it is today. As the typical ocean tide is 1 m, this means ocean tides would have been 10 km high on Earth! Of course, the collision that formed the moon turned the surface of the Earth molten, so really we're talking about a 10 km tall molten tide, which is far less scary...(kidding).
Imagine all the noise the ocean makes when it crashes on a beach. That noise requires energy to produce, and is partially driven by the action of raising/lowering the ocean via tides. Now imagine how loud a 10 km wave crashing would be. This is why the Moon receded quite quickly from the Earth shortly after forming, the tides enable the dissipation of a lot of orbital energy.
As for the hypothetical effect of this supertide on the present-day Earth's weather, the Earth's troposphere (which contains pretty much all of the Earth's weather phenomena) extends 10–20 km from the ground. Considering the height of the ground is changing by a similar amount due to the tide, I would imagine the effects on the weather would be quite dramatic.

10km tide? It's hard to imagine, wouldn't such tide be drawn by moon gravity deep into land?

The energy of the sound of waves crashing comes from the moon. I am going to mention this to my daughters next time we are at the beach with the moon visible. Thank you for this wonderful insight.

@dotancohen, most of the wave energy comes from the Sun: heat from the Sun warms air, which causes winds, which drives the waves across the ocean.  It's the energy of the tides which mostly comes from the Moon, and isn't as dramatic these days.  (Well, outside of places like the Bay of Fundy.)


text:
Why does Mars appear to retreat across the sky?
If you plot the movement of Mars across the Earth's sky it appears to move back on itself during its orbit.
Why does it appear to have a loop shape from Earth?

@Ken G This simple java animation illustrates the principle very well. The smaller animation to the right shows how it looks from Earth (green). Moving the cursor over the animation shifts frame of reference, but it still all looks the same as seen from Earth. (To the detriment of pre-Kepler theoreticians!)

The effect is called apparent retrograde motion.
What happens is that Mars has a 'direction opposite to that of other bodies within its system as observed from a particular vantage point' when this loop occurs.
That's a bunch of words that don't mean a lot to me. A picture is worth a thousand clearer words:

(Imagine this turned sideways and you get the effect in your image)
Basically, because Earth and Mars are orbiting the sun at different rates, our vantage point of Mars changes for each combination of points in the orbit of each planet.
On this scale, the background of stars is pretty much stationary - any apparent movement of the stars due to this effect is going to be negligible. Thus, the stars are our point of reference.
As our vantage point of Mars changes, it appears to shift directions on the stellar background, creating the effect you describe.

And note you'll see the same things with cars on a highway.  If your car, in the left lane, is moving faster than a car in the right lane, and both of you are approaching a long arching curve to the left, then look at the other car against the reference of a line of distant trees as you both enter that gradual curve.  When the other car is well ahead of you, the fact that it is rounding a corner will make it appear to advance against the distant trees.   But as you catch up and pass that car, still in the turn, you will see that car back up against the trees.  Then as you get well ahead, again the other car will appear to advance against the trees, if you are still in the curve by then.   This is called "parallax."


text:
What is the origin of the OBAFGKM classification system?
Once a body has been identified as a star it is classified into a subsection of either O,B,A,F,G,K or M (I think there are also 3 more letters recently added)
What is the origin of this classification system? Do each of the letters stand for anything?

Just an interesting note : The mnemonic used to remember the sequence is "O Be A Fine Girl Kiss Me"

"O Be A Fine Girl Kiss Me" - to which my ninth grade Earth Sciences teacher added "Right Now" punctuated with a final "Smack."

The Wikipedia article on stellar classification has the full list of letters:

Class
Surface Temperature
Notes

O
> 33,000 K

B
10,000 - 33,000 K

A
7500 - 10,000 K

F
6000 - 7500 K

G
5200 - 6000 K

K
3700 - 5200 K

M
2000 - 3700 K

L
1300 - 2000 K
L T and Y are those new letters

T
700 - 1300 K

Y
< 700 K

History
This Cornell doc explains much of the history behind the system:

In the 1890s, many scientists were interested in developing a classification scheme for the stars. Edward C. Pickering at Harvard University, together with his assistant Williamina P. Fleming, assigned stars a letter according to how much Hydrogen could be observed in their spectra: stars labeled A had the most Hydrogen, B the next most, and so on through the alphabet. There were 22 types in all: this scheme was rather cumbersome, and it wasn't clear what its physical significance was.

In 1901, another of Pickering's assistants, Annie Jump Cannon, also begun to work on the classification sequence. Her meticulous observations led her to simplify the 22-type scheme into a simple sequence of temperature: OBAFGKM. Not only was the scheme much simpler than the previous, but it related the amount of hydrogen observed to a physical property of the stars for the first time.

She rearranged the previous scale, and skipped or merged some entries?


text:
Have we ever observed a large meteor hit the Moon?
Since the moon has no atmosphere, its surface has many more meteorite impacts than Earth's does.

Image source
Have we ever observed a meteor strike the surface of the Moon?

While it's true that the moon's negligible atmosphere does increase its vulnerability to impacts, I suspect (having not done the research) that a better explanation for its more heavily cratered surface is its lack of plate tectonics. When we look at the moon, we are seeing billions of years of accumulated impacts. Surface features on the Earth simply don't last that long because the crust is constantly being deformed and recycled.

Yes, we have, and the impacts occur very frequently according to "Bright explosion on the moon" (NASA):

For the past 8 years, NASA astronomers have been monitoring the Moon for signs of explosions caused by meteoroids hitting the lunar surface. "Lunar meteor showers" have turned out to be more common than anyone expected, with hundreds of detectable impacts occurring every year.

On March 17, 2013 a significant impact occurred that could have been observed from the Earth.
Updated information about lunar impacts is found at the Marshall Space Flight Centre's Lunar Impacts page.  Below is a map of the impacts that occurred on the lunar surface between 2005-2008, all of which, according to the article "Amateur Astronomers see Perseids hit the Moon, could have been seen from the Earth (with a telescope):

There have been instances throughout our history when, so is believed, people on the ground have witnessed lunar impacts. One of the more famous ones was in 1178 AD, when monks witnessed fire coming from the moon; However, according to the article "Historic lunar impact questioned", the accuracy of this historical account is questionable.


text:
Does the Moon have any oxygen in its atmosphere?
Since the moon has gravity, it's almost impossible that there aren't some gasses trapped on the surface by the moon's gravity. Has any free-floating oxygen been found on the Moon? If so, in what concentration?

Although according to Wikipedia about composition of the moon soil includes 45% Oxygen. https://en.m.wikipedia.org/wiki/Geology_of_the_Moon

The Moon's atmosphere is very thin compared to the Earth, so thin that it is usually said to have no atmosphere.  The Moon's gravity is not strong enough to retain lighter elements, so they escape into space.
Apollo 17 carried an instrument called the Lunar Atmosphere Composition Experiment (LACE).  Oxygen is not listed as one of the elements it found on this NASA web page.  The principle gases found were neon, helium and hydrogen.  Others included methane, carbon dioxide, ammonia, and water.  Additionally, ground instruments have detected sodium and potassium in the lunar atmosphere.
The recently launched Lunar Atmosphere and Dust Environment Experiment (LADEE - pronounced laddie, not lady) should be able to tell us more soon.
Update:
According to this article at Spaceflight Now, LADEE did detect oxygen in the Moon's atmosphere.

How silly!  It should be pronounced as in Ladurée

Just to add to GreenMatt's answer, according to the article "The Lunar Atmosphere: History, Status, Current Problems and Context" (Stern, 1999), the lunar atmosphere is in fact a tenuous exosphere, which the authors describe as being composed of

"independent atmospheres" occupying the same space.

This is further elaborated in "The Lunar Dusty Exosphere: The Extreme Case of an Inner Planetary Atmosphere" (NASA), that

in direct response to these intense and variable environmental drivers, the Moon releases a low density neutral gas forming a collisionless atmosphere. This ~100 tons of gas about the Moon is commonly called the lunar surface-bounded exosphere

There is also an ionosphere, due to (from the ANSA article):

Ions are also created directly either by surface sputtering or subsequent neutral photoionization, forming a tenuous exo-ionosphere about the Moon.

The authors also suggest that ionic oxygen may be present due to surface sputtering.
Due to solar radiation and the solar wind, dust particles become charged as well, and can be subsequently listed from the lunar surface.


text:
What is the current accepted theory as to why Mercury, despite its size, has a similar density to Earth?
According to the NASA web page overview about Mercury, despite the planet being just a bit larger than our moon, it's density is about 98.4% of Earth's.  This high density suggests a comparatively large iron core (compared to the rest of the planet).
Recent studies reported in "NASA Antenna Cuts Mercury to Core, Solves 30 Year Mystery" (NASA, 2007) have confirmed the presence of a molten core, surrounded by a silicate mantle (depicted in the illustration below (from the article):

The question is then, what is the current accepted theory as to Mercury has a still-molten core and a density like that of Earth?

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


text:
What is the current accepted theory for the fate of hot Jupiters?
It is well established that one main feature of many hot jupiters is their close proximity to their parent star, usually the equivalent of being within the orbit of Mercury.  So, these planets are gas giants and are very hot (hence their category).
However, a few discoveries have lead to questioning about what is the fate of these planets.
example 1: HD 209458b  a.k.a. "Osiris"
According to the NASA page "Dying Planet Leaks Carbon-Oxygen", Osiris is doing more than 'evaporating', it is leaking carbon, oxygen alongside hydrogen in an envelope behind the planet that has been detected from Earth.  The significance of carbon and oxygen is stated from the article:

Although carbon and oxygen have been observed on Jupiter and Saturn, it is always in combined form as methane and water deep in the atmosphere. In HD 209458b the chemicals are broken down into the basic elements. But on Jupiter or Saturn, even as elements, they would still remain invisible low in the atmosphere. The fact that they are visible in the upper atmosphere of HD 209458b confirms that atmospheric 'blow off' is occurring.

It is stated in the article that Osiris is likely to become a hypothesised class of exoplanet known as a Chthonian, which is defined in "Evaporation rate of hot Jupiters and formation of Chthonian planets" (Hebard et al. 2003) as

new class of planets made of the residual central core of former hot
  Jupiters

These would be similar in size to Earth, but considerably denser.
example 2: CoRoT-7b
According to the NASA article "Most Earthlike Exoplanet Started out as Gas Giant", CoRoT-7b is an Earth-sized planet where a hot Jupiter usually is found, they describe it as

is almost 60 times closer to its star than Earth, so the star appears almost 360 times larger than the sun does in our sky," Jackson said. As a consequence, the planet's surface experiences extreme heating that may reach 3,600 degrees Fahrenheit on the daylight side. CoRoT-7b's size (70 percent larger than Earth) and mass (4.8 times Earth's) indicate that the world is probably made of rocky materials.

The high day time temperature means that the star-facing side of the planet is likely to be molten, any tenuous atmosphere is also blasted away.  Scientists estimate that many Earth-masses may have been boiled off.  It also seems that the decreasing mass is causing the planet to be drawn closer to the star  -cause more material to be boiled off, hence the mass to decrease.
To summarise on of the scientists in the article:

You could say that, one way or the other, this planet is disappearing before our eyes,"

The question
As these are only 2 examples of a possible process, the question is, what is the current accepted theory as to the fate of hot Jupiter exoplanets?
Could this also be the reason a hot Jupiter does not exist in our solar system?

I'd just like to add Kepler-70b (https://en.wikipedia.org/wiki/Kepler-70b) and Kepler-70c (https://en.wikipedia.org/wiki/Kepler-70c) as two more examples of hot Jupiters that have met a similar fate and ended up as Chthonian planets.

This is a fairly loaded question in that it depends heavily on what a "hot Jupiter" actually is defined to be. What is "hot"? What is a "Jupiter"? In reality, there's a continuum of planetary masses and distances from their parent star, and in the literature you'll commonly see references to "hot Neptunes", "hot Saturns," etc.
The predominant theory as to how giant planets form is that they first coalesce from rock and ice beyond the ice line, the distance from the parent star at which water becomes solid. This distance is approximately where Mars lies today in our solar system. What's surprising about "hot gas planets" is that they are found within this ice line, significantly within. This implies that after they formed their cores, they migrated closer to their host stars via some currently undetermined process (for which there are several good candidates, but for now let's assume that the existence of hot planets shows that at least one of these processes operates quite regularly).
And what about the word "hot"? Well, for the planets that are closest to their parent stars, there is known to be a radius anomaly: The radii of these planets are significantly larger than models of giant planet structure irradiated by their host stars would predict. So I would define "hot" planets as gas giants whose radii are larger than what would be predicted by the standard models.
Now that we got some of the definitions out of the way, there's the question of survival. When giant planets are close to their parent stars, they become tidally locked. As a consequence, there is very little energy tidally dissipated on the surface of the giant planet, the shape of the planet is fixed and there are little internal motions. However, the giant planet also raises a tide on its host star as well, and because it takes a lot of angular momentum to change the spin of an object with 1,000 times more mass, the host stars are almost never going to be tidally locked to their closest planet.
The rate at which energy is dissipated within the star is highly uncertain, and this uncertainty is typically swept into a fudge parameter "Q," the quality factor, with lower quality factors reflecting more dissipation. "Q" is measured for certain bodies in our own solar system (i.e. Earth and Jupiter) and in some stellar binaries, but is highly variable from body to body, ranging from about 10 for the Earth to 10^8 for some stars.
Whether a planet survives to be observed today depends on how long the orbital decay time, which is determined by Q, compares to the age of the system. For some systems, such as WASP-12b and WASP-19b, which feature highly inflated hot Jupiters, Q is estimated to be small enough to cause them to fall into their host stars in a surprisingly short time (< 10^7 years).
One other possibility is that the gas surrounding the rock/ice core is blasted away by the tremendous amount of heat deposited into the planet. This leaves you with a relatively low-density planet that's somewhat devoid of iron, as the cores of giant planets form further from their host stars than the rocky planets. There are a few candidate close-in, Neptune-mass objects that may have been produced as a result of them losing the bulk of their atmospheres in this way (Example: GJ3470b).
As for our own solar system, the formation of a hot Jupiter would have likely destroyed the inner solar system as it migrated close to the Sun, owing to the fact that it would violently perturb the inner planets' orbits. Additionally, the Sun would likely be enhanced in metals owing to the accretion of metal-rich material from this giant planet. While it's potentially possible that there was a hot Jupiter in our solar system before the other planets formed, it currently seems unlikely.

Thank you for your answer, do you have further references for your answer?  Specifically for the potential life cycle of the hot Jupiters (et al), and for our solar system?


text:
What is the predominant element in the dust of the Sombrero Galaxy?

The Sombrero Galaxy contains a rather peculiar ring of dust orbiting it (seen as the dark ring on the outer edge). What is the prevalent element in this dust? Carbon?

Follow-up question: http://astronomy.stackexchange.com/questions/421/how-far-apart-is-the-dust-in-the-dust-lane-of-the-sombrero-galaxy

I think that first, we have to properly appreciate the size of the Sombrero Galaxy. It is roughly 50,000 light years (15 kilo parsecs) in diameter. That might be only half of the diameter of our own Milky Way galaxy, but still makes each and every pixel on the photograph you're attaching in your question stretch more than 100 light years in distance.
That's a huge distance, so we can probably agree that clouds made of nearly anything and at least as dense as what you'd on average expect the interstellar medium to be comprised of, would obscure even the strongests light sources within that galaxy. This is important to realize, so we somewhat lower your expectations; These areas aren't tightly packed with anything, we're just talking of such vast distances that the light has to travel through, that even almost nothing will be enough to stop it from shining through.
Now to your question what it is comprised of. It is in essence matter from the interstellar medium that stops this light from reaching us, so nothing in particular that you wouldn't expect also in similar areas between stars within our Milky Way, or indeed any other galaxy that is roughly in the same age period of its development. Since the Sombrero Galaxy isn't even all that far away in cosmological sense, merely around 29 million light years, it has an apparent magnitude of ~ 9 and is a rather good target for observations of even hobbyist astronomers with enthusiast telescopes. More importantly though, you aren't looking that far into the past. Again, in the cosmological sense. For comparison, our Sun is roughly 4.5 billion years old, and we're observing the Sombrero Galaxy as it was 29 million years ago.
So what is this interstellar medium made of? On average, it is comprised of mostly gas clouds of hydrogen and helium forming during the primordial nucleosynthesis (read: Big Bang) and at most a few percent of heavier trace elements, some also in gas form and an itsy-bitsy percentage of these would combine into even heavier dust particles, mostly made of carbon, silicon, and oxygen (source for the interplanetary dust composition: NASA APOD). Remember, when I say clouds and dust, they are incredibly thin. So thin, you'd have to be traveling at a significant portion of C to even notice it much, unless you're looking for it with extremely sensitive equipment, like Voyager 1 and 2 do, for example. But even with such low particle density, some regions would be slightly denser, and some slightly thinner. And that is what creates that majestic dust lane of the Sombrero Galaxy.
They are interstellar medium, mostly made of primordial hydrogen and helium, and dust made of carbon, silicon, oxygen and other trace elements, and all of it hundreds of light years in length and obscuring light from the galaxy's stars to penetrate through at varying levels, depending on background light strength, proximity, and local interstellar medium cloud density. The color of the cloud says nothing of its composition. It is mostly black with parts of it shifting towards red hues due to molecules of gas and dust in the interstellar medium scattering blue light from the source of light more than they scatter red light. I.e. the same reason why the Sun or the Moon appear more red when they're low on the horizon.


text:
What practical considerations are there for amateur observations of transiting exoplanets?
Obviously, I am not referring to actual viewing of the exoplanets themselves, but detecting their effects on the brightness of the light emitted from the parent star (as in the diagram below from The Institute of Astronomy, University of Hawaii).

I would imagine that a good quality telescope, would be about to detect and record the effects of the exoplanet transiting the parent star.
What practical considerations in terms of equipment, software etc. would be required for 'backyard' amateur observations of the effects of transiting exoplanets?
If anyone has actually tried this, what has been your experience?

This is actually quite straightforward with digital CCD's (it used to be quite tricky with film cameras as you'd have to carefully develop film that moved past the lens and assess the width of the trail)
Get yourself a good telescope - a 12" Dobsonian or above if you want to give yourself a good chance of picking out the fluctuations against the noise background. Then select a decent CCD. Five hundered pounds gets a reasonable one, but expect to pay a couple of thousand pounds for a cooled CCD, which will also help to reduce noise.
(Buying in US dollars? A reasonable CCD runs about $1000. A cooled CCD will cost you at least $1500.)
You'll want a good quality equatorial mount, with computer controlled servos to track the target smoothly over long periods of time.
Ideally you will also slave a second telescope and CCD, pointed along the same path but slightly off - this will help you cancel out cloud and other fluctuations from our own atmosphere.
Oh, and get as far away from a city as possible - up into the mountains can be a good plan :-)
Then arrange your viewing for a series of full nights. The more data points you can get, the better the noise reduction. Imagine the exoplanet is orbiting every 100 days, in order to get any useful data, you will need to track it over some multiple of 100 days. So assume you set up to track your target star for 2 years, plan for 3 or 4 star shots per night to give you a range of data points.
These 600+ days of 4 data points per night gives you a reasonable stack of data - the challenge now is to work out whether there are any cyclic variations. Various data analysis tools can do this for you. As a first step, if you find a cycle around 365 days, it probably isn't the target, so try and normalise for this (of course this will make it very difficult to discover exoplanets with a period of exactly 1 year)

"quite straightforward"... and then you tell us it takes a couple of years, a mountain location and a few thousand quid of equipment. +1 for the wry British humour..

I don't understand this answer. Are you suggesting trying to discover transits in this way?! 4 data points per night is completely insufficient. My reading of the question is the desire was to observe known transiting planets, which can be done in a few hours.

IEEE Spectrum recently ran a piece on detecting exoplanets:
DIY Exoplanet Detector -
You don’t need a high-powered telescope to spot the signature of an alien world

Canon EOS Rebel XS DSLR. With old manual-focus lenses now useless to most photographers, I was able to acquire a 300-millimeter Nikon telephoto lens...
   An Arduino-controlled star tracker...

More info here: Detect known exoplanet with DSLR/telephote lens
The star in question, HD 189733A, has a visual magnitude of 7.6. Dimmer targets would naturally require more light gathering power than an old telephoto lens can supply.

I'd question the statistics of that detection, but that's some feat. I had no idea you'd be able to get anything at all with consumer-level hardware!

@Warrick Yes, a bigger objective lens'd give you less noise, but cameras have gotten unspeakably sensitive this past decade.

If you stick to observing "hot Jupiters", then this is very much in reach of "amateur" technology.
I would agree that a 10-inch+ telescope is probably needed, along with a cooled CCD.
Hot Jupiters (giant planets orbiting close to their parent stars) produce transit signals of about 0.01-0.02 mag amplitude. The transits last of order a couple of hours, occur every 1-10 days and the times of transit are well predicted. In principle you could collect all the data you need in about 6 hours of observing. But, the amplitude of the transit dip is small, so you need to get very precise differential photometry. Your best bet is to observe targets which have plenty of other stars in the same CCD field that can act as comparisons - this may mean you need a CCD with a wide field of view. On the other hand, you must make sure that the seeing disk of the star is well-sampled by the CCD pixels (e.g. in your typical seeing is 2 arcseconds, each CCD pixel must not image more than 1 arcsec on the sky, and preferably less).
Other strategies for success involve observing at low airmasses, which should improve the quality of the differential photometry, and don't bother with anything observed even through the thinnest cirrus. Observing multiple transits will allow you to improve your data by "phase-folding" on the known planet orbital period.


text:
Why is the Moon receding from the Earth due to tides? Is this typical for other moons?
After reading the Q&A Is the moon moving further away from Earth and closer to the Sun? Why? about the tides transferring energy to the Moon and pushing it from Earth, I have a question:
How is that energy actually being transferred to the Moon? The creation of tides requires energy, so I would expect that this should take energy from the Moon, slow it and cause it to eventually fall back to the Earth. Why isn't that happening?
Finally, if this is the general mechanism, would other moons that orbit around planets with a liquid surface and causing tides, be receding from their parent planets?

Note the recession only happens when the primary is spinning faster than the satellite.  When the satellite is moving faster than the primary (like Phobos and Mars), then it spirals in, not out.

@РСТȢѸФХѾЦЧШЩЪЫЬ:   Um, "closer to the Sun"?  It also gets further from the Sun, half the time...

"Finally, if this is the general mechanism, would other moons that orbit around planets with a liquid surface and causing tides, be receding from their parent planets?" Although implicitly covered by the answers, I thought I would add here that yes, if a moon has come to be a satellite of a planet where the orbit is not geostationary, then the same effect will occur to a varying degree. I suppose a particularly dense gas atmosphere might produce the same effect, though the arrangement for something relatively exotic like that to happen is probably unlikely.

It's pretty simple, actually.

The Moon creates tides. Due to tides, the water bulges out towards the Moon (and also on the opposite side).
But the Earth also rotates pretty fast (once a day), faster than the Moon orbits the Earth (once a month). There's friction between the rotating Earth, and the watery bulge created by tides. The rotation of the Earth "wants" to rotate the bulge faster.
In effect, the rotation of the Earth drags the tidal bulge forward - the bulge is always a bit ahead of the Moon. When the Moon is at meridian, the tide is already decreasing.
So there's a bit of extra watery mass on Earth, a little bit ahead of the Moon. This watery bulge interacts gravitationally with the Moon.
This has two effects:

it slows down the rotation of the Earth, gradually sucking energy from it (the Moon pulls the bulge, and therefore the Earth, "back")
that energy is dumped into the Moon's orbital motion, effectively "pulling" it forward

When you dump energy of motion into an orbiting body, it settles into a higher orbit - higher orbit means more energy. Therefore, the transfer of energy from Earth' spin to Moon's orbit gradually makes Moon's orbit larger and larger.
This only happens because the Earth is spinning faster than the Moon orbits it. If the Earth was tidally locked to the Moon (spinning exactly as fast as the Moon orbits it), then no transfer would happen. If the Earth was spinning slower than the Moon's orbit, then the transfer would be opposite (from Moon's orbital motion to Earth's spin).

Note: Counterintuitively, a satellite with more energy actually moves slower, but in a higher orbit. The extra energy goes into raising the orbit, not into making its speed faster. Why this happens exactly is a whole 'nother discussion.

Is there a tidal wave in the Earth's molten mantle too?

The whole Earth, including the "solid" crust and its soft interior, experiences a tide due to the Moon; on a planetary scale there are no true solids. It's called Earth tide. The amplitude is on the order of dozens of centimeters. http://web.ics.purdue.edu/~ecalais/teaching/eas450/Gravity3.pdf

@user104372 Energy does not just exist in the form of kinetic energy. In this case, the total energy (kinetic plus potential) of a wider orbit is larger. This is really basic physics that you are arguing about.

Re It's pretty simple, actually. It's not that simple.  The true picture is much more complex than is this simple picture. The tidal bulge as depicted in the image does not exist. If it did exist, high tide would occur shortly after lunar culmination (and then 12 hours and 25 minutes after that). This is very rarely observed. In fact, that tidal bulge cannot exist. To get the correct picture, one would have to integrate the effects of the oceans on the Moon over a long period of time (preferably 18 years or more). Our models aren't there yet.

You have correctly identified that the tidal forces are transferring energy from the Earth to the Moon. This energy causes the Moon's orbit to get larger thus slowing it down.
It's a bit counter intuitive, but if you think about it the Earth spins a rate of 1 spin per day The Moon is orbiting the Earth with a period of approximately 27.3 days. If it were to speed up it's orbit would actually decrease bringing it closer to the Earth.
To answer your final point all other moons cause tides on their parent planets and are moving away from them, but the effects are much smaller due to the larger difference in sizes. The Earth/Moon system is unique in the Solar System as the ratio of the sizes of the bodies are relatively close to each other.

The other planet/moon systems do also have this property though. The moons slowly spin outwards.

I still don't get how speeding up the Moon would decrease its orbit, from what I remember, the faster the body moves, the more distant is the orbit...

@ŁukaszLech That's a misconception, the square of the orbital period is equal to the cube of its average distance (Kepler's 3rd law), but as the size of the orbit only scales linearly with distance, the velocity scales as r^(-1/2), i.e. decreasing with distance.

Examples: At 150 miles up, orbital speed is 17000mph. At 22000 miles up it is only 7000mph.

Interesting. But when the body is slowing, the centrifugal force reduces, therefore is no longer able to match gravity, and the body falls? This is what happens to low orbit satelites?

@ŁukaszL.There is no such thing like centrifugal force. There is only one force involved: gravity. It is the force that curves the otherwise rectilinear movement into an orbit. The lower the gravity, the more straight the orbit, that is, the bigger.

@RoryAlsop It is true that a circular high orbit has a lower velocity but orbits are not perfectly circular. A non-circular orbit constantly trades kenetic energy for gravitational potential energy. If you start in a circular orbit and apply a small prograde acceleration then initially your speed will increase but you also make the orbit non-circular. This non-circular orbit converts kinetic energy into potential energy reducing your speed.


text:
How far apart is the dust in the Sombrero Galaxy's dust lane?
In his answer to this question, TildalWave made this remark:

I think that first, we have to properly appreciate the size of the Sombrero Galaxy. It is roughly 50,000 light years (15 kilo parsecs) in diameter. That might be only half of the diameter of our own Milky Way galaxy, but still makes each and every pixel on the photograph you're attaching in your question stretch more than 100 light years in distance.

Here's the photograph:

Since it almost feels like I can see individual 'pieces' of dust in the lane, it would follow that the particles must be very far apart. Just how far apart are they? Are they grouped in clumps, or relatively evenly space?

I think your assumption is incorrect. You can't see individual pieces. You are seeing the variation in particle density, and looking at the photo - there does look to be clumping, and there are bound to be tidal effects.

Note the 'feels like' - I know it's incorrect, but the dust certainly doesn't look very dense.

The "structure" you are seeing has nothing to do with the size of dust particles. You are seeing the structure of clouds of absorbing dust, which indeed can be hundred of light years across. The interstellar medium is "clumpy" on these sorts of scales.

The sizes of dust particles varies enormously, and thus the density of dust particles depend upon which sizes we are talking about. The dust size distribution can be described by a power law with a slope of roughly $-3.5$; that is, the density of 0.1 µm particles of is $10^{-3.5}$ times that of 0.01 µm particles, or roughly 3000 times smaller.
Moreover, the discussion is complicated by the fact that we don't really have a formal definition of what is dust. But one "definition" takes the minumum size as a conglomeration of a few molecules, and the maximum size as "what has time to grow in a formation process" (these dust particles can later "stick together" and grow to form pebbles, rocks, asteroids, and planets. That's why any definition will be arbitrary).
In a dust-dense region of the interstellar medium, this results in a typical mean density of roughly one dust particle per cubic centimeter.

Based on 1995A&A...303..673E,
How far apart they are?
No idea. Somebody else might help to provide a density distribution.
Are they grouped in clumps or relatively even spaced?
This is what I really wanted to answer. They are in a concentric ring-like structure around the galaxy. The maximum optical depth is 0.8 with a mean value of 0.3. By the way, optical depth gives you a nice idea of how far apart they might be, not a good density distribution but you can get the physical picture in mind.

this doesn't answer the question

It does answer part of the question. Partial answers are acceptible.


text:
What implications do younger Earth and Moon have on Late Heavy Bombardment genesis hypotheses?
The Late Heavy Bombardment (LHB) or the Lunar Cataclysm is when the inner solar system, including the Earth and the Moon, underwent multiple and sustained heavy impacts early in their history (around 3.7-4 billion years ago). This event formed many of the major impact scars we see on the Moon, (presumably as well as much of the impacts observed on Mars and Mercury). Two main theories are based on perturbations in either the Kuiper Belt, or in a primordial larger asteroid belt.
What implications would a younger Earth and Moon, discussed in "Moon and Earth may be younger than originally thought" (Lars Borg et al., 2011) have on hypotheses on what caused the LHB?

From the article: "In fact, its age may be 4.36 billion years old." This is still older than the estimated age of the Late Heavy Bombardment, so it wouldn't affect that hypothesis much. In fact, it may actually help the hypothesis, as it means the Moon would have more of its interior molten (though probably not by much). It's suggested that the magma from underneath the crust filled the lunar mare after the giant impacts through fissures in the crust. (Wikipedia)

authors: ديناصور الأمة


text:
What observational constraints are there in detecting the presence of volcanism on exoplanets?
This question is somewhat related to my earlier question How are the compositional components of exoplanet atmospheres differentiated?, but this about a specific surface-atmospheric phenomena - volcanism.
Using our solar system as a rough analogue (where asides from Earth, Venus, Io and Triton have active volcanism, probably more), volcanism should be not uncommon amongst exoplanets given the right conditions.  Obviously, we could probably only detect either the massive shield, plateau volcanism, or prolonged volcanism from multiple vents.
What observational constraints are there in detecting the presence of volcanism on exoplanets?

Exoplanets are too far away to send satellites or to image them directly. So there is no way to go there and say: there it is a volcano. My guess is that we have to guess from what we know working in the solar system. And we might just get a statistical probability that the planet is active.
I would say that there are two cases:

If the planet is rocky and near enough to the star or a giant planet, you might expect volcanism caused by tidal effects, similarly to what happen on Jupiter's Io moon. In few words: the gravitational force of the nearby big planet/star deform the planet and this deformations are converted to heat due to friction. If the gravitational forces are large enough, you can end up having volcanism
If you are able to measure the spectral lines in the planet atmosphere you could try to identify signatures of gas or dust typically associated with volcanic activity. I don't know what you would want to look for, but my guess is that you want to search mostly in the infrared.