universe - Integral calculus for the Olympiad?

This year I am going to participate at the Astronomy Olympiad, in the beta group (seniors) - 15 years old or above.
I have already participated at this Olympiad in the alpha group (juniors)... now, a year later, I am going against 12th graders that know superior mathematics, such as derivatives and integrals. From a quick scan on some problems, I've seen that these mathematical concepts do appear... Since I do not have all the adjacent theory (limits etc.), I supposed I should learn some basic properties and see how can I apply them. My goal is to know how to read a integral, or how can I put it in an easier understanding form.
So... What should I know about these derivatives and integrals? Which properties do apply in astronomy? I really appreciate all the help that I can get. And sorry if my question was a little dizzy... I hope it's alright, though.

Thank you for taking the time to read this!

Edit: For narrowing, I would love some websites that have only a few properties useful in mechanics.

speed of light measurement - Astronomy

While searching for different methods of speed of light measurement, i came across one of the method of fizeau discussed below which i cannot fully understand.

In short,in Fizeau’s apparatus, a
beam of light was shone between the teeth of
a rapidly rotating toothed wheel, so the
“lantern” was constantly being covered
and uncovered. Fizeau had a
mirror, reflecting the beam back, where
it passed a second time between the
teeth of the wheel.

I do understand the idea i.e. to measure the time during the course of light "from the wheel to mirror and then back to the mirror".But where are the mirror,lantern and wheel located? What kind of wheel is that? Does light pass through the holes (or teeth) in the wheel or light gets refracted through the wheel? Finally if a group of light particles goes through one of the hole (or teeth),then how do we know that the same group of light particles came back through which hole (or teeth) after its journey,since there are soo many light particles leaving and entering different holes (or teeth)?

What to do after first year of amateur astronomy?

I have practiced astronomy as a hobby for a while now. I have an entry-level telescope (6-inch Newtonian). In addition to observing the night sky, I have been studying physics related to optics, astronomical distance & temperature measurement, gravity, etc.

My plan is to next get a CCD camera, and do some astrophotography. It should keep me busy the next winter. However, I feel I should try to find some kind of project where I could gradually make progress. Otherwise, after learning all the basics, and seeing everything that can be seen with the smallish telescope I have, I might gradually start to lose interest.

What would you recommend as the "next stage"?

Did atoms in human body indeed come from stars?

The chemical elements in our bodies are inherited from the Earth. The Earth was formed in a disc of gas and dust swirling around the protosun 4.5 billion years ago. The material that formed the Earth was a selection of the material from that protostellar nebula that was itself once part of a larger molecular cloud.

So the atoms in our body were once part of this molecular cloud, so we need to understand how they got there.

After the first ten minutes or so, the universe contained mainly hydrogen, helium and some traces of lithium, deuterium and tritium - and that's all. No oxygen, iron, carbon etc.

Almost all of the heavier chemical elements are made inside stars. We could stop there - the atoms of carbon, oxygen, calcium etc. in our bodies must have been made in stars, and since these atoms/nuclei are stable, they must survive unchanged (you could argue about whether their electrons get swapped about in chemical reactions etc., but since electrons are indistinguishable this hardly matters).

But how do they get into a molecular cloud and what sort of stars make these elements? A couple of answers correctly identify massive stars that explode as supernovae as important. But they are by no means the only contributor, or even the most important contributor for some elements.

If we take carbon and nitrogen, these are manufactured in nuclear reactions inside stars of even a bit less than a solar mass during the horizontal branch and asymptotic giant branch stages. These stars may be less massive and produce less C and N than massive stars, but there are many more of them. The central material is mixed to the surface during thermal pulses and the outer envelope, enriched in a variety of chemical elements, is gradually lost into space via a slow wind. This is a major source of carbon, nitrogen, fluorine, lithium and a number of heavy elements - Ba, La, Zr, Sr, Pb and many others - produced in the s-process. About 50% of the elements heavier than iron are made in the s-process, which can occur in both massive stars that explode (mainly isotopes with $A<90$) and the less massive AGB stars with slow, massive winds (elements up to lead and bismuth).

Iron, nickel and many other elements such as sulphur and silicon are also produced during type Ia supernovae. This is the detonation of a white dwarf, the end stage of a low-mass star, after mass transfer or merger. Milder novae explosions caused by the ignition of material accreted onto a white dwarf also enrich the interstellar medium.

All these different processes produce distinctive patterns of element abundances.

The enriched material is swept up by neighbouring supernova explosions, by interactions with spiral arms and other molecular clouds. It cools, condenses and collapses to form a new generation of stars.

Analysis of "presolar grains" found inside meteorites tells us what our solar system formed from. These analyses tell us that all of the above processes were important in making the chemical elements that made up the Earth and hence those in our bodies.

[Further details on the production of elements heavier than iron (including supernovae, low-mass AGB stars, colliding neutron stars etc.) can be found in my Physics SE answer to this question. ]

orbit - Is it possible to create a traditional clock solar system?

Is it possible to create the proposed system of satellites?

Yes (in theorey). Hierarchical multiple systems (like the one proposed in the OP) tend to be stable if the periods differ by $sim 5$ or more and the orbits are near-circular. So, such a system could be stable. A slight problem may be the commensurability of the periods (the fact that their ratios are rational numbers), which implies orbital resonances. However, these resonances are rather weak (the lowest ratio is 1:24), but they may mess up the system in the very long term (a detailed analysis would be required to find out). In practice, it will be extremely hard to set-up such a system, even with advanced space engineering.

What are some values for the weights, rotational speed, and radii that would work?

Since this is a strongly hierarchical system, the masses $m$ and semi-major axes $r$ (=radius if the orbit is circular) can be worked out from Kepler's third law
$$
frac{2pi}{T} = Omega = sqrt{frac{G(m_1+m_2)}{r^3_{1-2}}}
$$
for each sub-system. Giving
begin{align}
frac{4pi^2}{1mathrm{sec}^2} &= frac{G(m_S+m_M)}{r^3_{S-M}} \
frac{4pi^2}{1mathrm{min}^2} &= frac{G(m_S+m_M+m_H)}{r^3_{SM-H}} \
frac{4pi^2}{1mathrm{hour}^2} &= frac{G(m_S+m_M+m_H+m_D)}{r^3_{SMH-D}} \
frac{4pi^2}{1mathrm{day}^2} &= frac{G(m_S+m_M+m_H+m_D+m_X)}{r^3_{SMHD-X}},
end{align}
where $r_{ABC-D}$ denotes the distance (more precisely: the orbital semi-major axis) between object $D$ and the centre of mass of objects $A$, $B$, and $C$. Since these are 4 equations for 8 unknowns (4 masses and 4 distances), there is some freedom in the design of your 'clock'.

Does the universe have a limit on how many layers of satellite systems it permits?

So far, our formula only considered gravitational forces and neglected all else. Moreover, post-Newtonian effects (general relativity) have been ignored. As long as these assumptions remain valid (orbital speeds $ll c$, orbital distances $gtrsim,$cm, solid objects with sizes $ll$ distances so that tides are neglible, no electric or magnetic fields, vacuum), there is no limit. However, in practice, there is always some elctro-magnetic field and never perfect vacuum...

planet - How precisely is planetary tilt defined (the tilt direction, not just the angle)?

A bit too simply (but not too much), it's the angle between the planet's rotational angular velocity vector and the planet's orbital angular momentum vector.

Where it gets tricky is dealing with little wobbles and such. Planets don't quite orbit in a plane because of gravitational interactions amongst the planets. This means the instantaneous (osculating) orbital angular momentum vector isn't quite constant, both in magnitude and direction. There are a number of different mean orbital elements that smooth out most of those tiny wobbles. One of those mean orbital elements sets (I'm not sure which one exactly) is used to determine the mean orbital angular momentum vector.

Planets don't quite rotate nice and smoothly because planets aren't perfect spheres and aren't rigid bodies. This gives the Moon, the Sun, and other planets a handle by which they can exert tiny little torques in the planet in question. The response of the planet to these torques is somewhat arbitrarily divided into two categories based on frequency. Very slow responses are called precession; faster responses, nutation. The non spherical nature of a planet means the planet undergoes a small torque-free nutation as well as the torque-induced precession, nutation, and polar motion. In addition to these, there are a number of terms (all small) that don't yet have a very good model behind them. This oddball terms, along with the torque-free nutation, are collectively called polar motion. The motion is fairly smooth if one ignores nutation and polar motion, and the effects are all small.

Precession, while very slow, can be quite large in magnitude. The rotational angular velocity vector that is used in determining the axial tilt incorporates precession but smooths out nutation and polar motion.

How to calculate the heliocentric velocity of an object?

The heliocentric velocity $V_mathrm{H}$ of an object is its velocity wrt. the Sun. When you measure an object's velocity, you measure it in the reference frame of Earth, which revolves around the Sun with ~30 km/s (varying a bit from aphelion to perihelion), so convert to $V_mathrm{H}$ you need to know the time of the year of the observation (unless the line of sight toward your object is exactly perpendicular to the ecliptic plane), as well as the angle between the line of sight, and the line of sight toward the Sun. This involves a number of sines and cosines that you can find in e.g. Barbieri (2006).

If you further want to convert from $V_mathrm{H}$ to the reference frame in which the Milky Way's center is at rest, the Local Group is at rest, or the Cosmic Microwave Background is isotropic (the "cosmic" frame), then you use the formula you link to, adding a similar term as described above, but instead using the velocity (i.e. speed and direction) of the Sun wrt. to the given frame.

Does a black hole become a normal star again?

There's several ways this question could be answered, but they all come down to an emphatic "no" - a black hole will not return to being a main sequence star. The simplest way to see this is probably that a black hole has a much higher entropy than a star or even another type of stellar remnant of even vaguely similar mass and so there simply could not exist a spontaneous process by which a black hole develops back into a star.

A black hole once formed will stay a black hole, however it is believed that the Hawking process will lead to the black hole eventually evaporating. The time scale for evaporation though of a stellar remnant black hole is mindbogglingly long and they will not evaporate until long after stellar formation has ceased in the Universe.

eccentric orbit - How does 'centre of mass' concept work?

We know that two celestial bodies rotate around a center of mass.

"We" do not know that. Rhetorical question: Pick up an apple and drop it at high noon. Does it fall towards the solar system barycenter? The answer is of course "No." It falls away from the barycenter, towards the Earth.

I'll steal a graph from an answer I wrote to a question at the sister site, physics.stackexhange.com, Do the planets really orbit the Sun?:

Given the above graph, it appears that it's much better to say that Venus orbits the Sun (red curve) rather than the solar system barycenter (black curve). So why do people consistently use a barycentric frame rather than a heliocentric frame? The answer is simple: The barycentric frame is the frame in which the equations of motion undertake their simplest form. This is particularly so when one wants to incorporate some of general relativity into the equations of motion.

Our solar system is well-behaved. Other star systems aren't quite as well-behaved. For an example of a hypothetical system that is extremely far from well-behaved, I suggest you read Saari and Xia, "Off to infinity in finite time," Notices of the AMS.

gravity - Is the SOI a spherical region or a oblate-spheroid-shaped region?

Is the SOI a spherical region or a oblate-spheroid-shaped region?

The sphere of influence is neither a sphere nor an oblate spheroid. It is a surface with no name. An approximation of this surface is

$$left(frac r Rright)^{10}(1+3cos^2theta) = left( frac m M right)^4$$

This is neither a sphere nor an oblate spheroid, and this is but an approximation. Thefull expression is an absolute mess. Dropping the factor of $(1+3cos^2theta)^{1/10}$ (which is close to one) yields

$$r = left( frac m M right)^{2/5} R$$

Tada! The equation of a sphere!

The true surface is defined in terms of two ratios. Consider two gravitating bodies, call them body A and body B. From the perspective of an inertial frame, the acceleration of a tiny test mass toward these two bodies is given by Newton's law of gravitation. These two bodies accelerate toward one another as well, so a frame based at the origin of either body is non-inertial.

From the perspective of a frame at the center of body A, the acceleration of the test mass is the inertial frame acceleration of the test mass toward body A plus the inertial frame acceleration of the test mass toward body B less the acceleration of body A toward body B. Denote the acceleration of the test mass toward body A as the primary acceleration and the difference between the inertial frame accelerations of the test mass and body A toward body B as the disturbing acceleration. Finally, define $Q_A$ as the ratio of these two. Now do the same for a frame with origin at the center of body B. The sphere of influence is that surface where $Q_A = Q_B$.

earth - As days get longer, why isn't the added daylight split evenly between sunrise and sunset?

It's a consequence of the equation of time, which represents the time difference between apparent solar time (time as measured by a sundial) and mean solar time (a fictitious device; essentially mean solar time is time as measured by a clock). Below are two plots from the referenced website.

Figure 1: Equation of time. Currently (early January), a sundial will be slightly behind a clock and is getting slower.

Figure 2: Time derivative of the equation of time. Apparent solar time changes most rapidly from early December to early January, with the maximum change occurring on December 25.

Days are currently getting longer in the northern hemisphere, but noon is advancing by 20 to 25 seconds per day. Sunset advances by half of the increase in the length of day plus that 20 to 25 second advance in noon. Sunrise advances by half the increase in length of day less that 20 to 25 second advance in noon.

The increase in the length of day (time from sunrise to sunset) depends on latitude. In Bogota Columbia (4°35′53″N latitude), length of day increases by only 3 to 10 seconds per day during January, growing throughout the month. That tiny change in length of day means sunrise continues to advance throughout the month of January in Bogota. On the other hand, in Reykjavík, Iceland (64°08′N latitude), length of day increases drastically from day to day, even at this time of year. Sunrise started becoming earlier in Reykjavík on Christmas day.

gravity - Why are stars so far apart?

Most stars are of a solar-mass or below. The average number of companions that each stars has (in the sense of being part of binary or higher multiple systems) systems ranges from 0.75 for stars of a solar mass to approximately 0.35 (not a well-established number) for the more numerous M-dwarfs. Let's take a compromise value, say 0.5.
The separation distribution of these multiples peaks at around 50 AU for solar-type stars, reducing to about 5 AU for low-mass M dwarfs. Again, lets take a compromise value of 20 AU. See Duchene & Kraus (2013) for all the details.

So if we take 1000 stars, then 333 of them (roughly speaking) are companions to another 333 stars, while 333 are isolated single stars. (NB This does not mean the frequency of multiple systems is 50%, because some of the companions will be in higher order multiple systems)

Thus, taking your calculation of the separation between stellar systems of 5 light years ($= 3.2times10^{5}$ AU), then the mean separation is:
$$bar{D} = 0.667times 20 + 0.333 times 3.2times10^{5} simeq 10^{5} AU, $$
but the median separation is 20 AU!

This is of course sophistry, because I'm sure your question is really, why are stellar systems so far apart?

Stars (and stellar systems) are born in much denser environments. The number density of stars in the Orion Nebula Cluster (ONC - the nearest very large stellar nursery) is about 1 per cubic light year. The equivalent number for the solar neighbourhood is 0.004 stars per cubic light year. Thus the average interstellar separation in the ONC is 1 light year, but in the solar neighbourhood it is about 6.3 light years.

The reason for this separation at birth is the Jeans length - the critical radius at which a clouds self-gravity will overcome its thermal energy and cause it to collapse. It can be expressed as
$$lambda_J = c_s left( frac{pi}{Grho } right)^{1/2},$$
where $c_s$ is the sound speed in a molecular cloud and $rho$ its density. For star forming giant molecular clouds $c_s = 0.2$ km/s and $rho=10^{-23}$ kg/m$^3$. So clouds of scale hundreds of light years could collapse. As they do, the density increases and the Jeans length becomes smaller and allows the cloud to fragment. Exactly how far the fragmentation goes and the distribution of stellar masses it produces is an area of intense research, but we know observationally that it can produce things like the ONC or sometimes even more massive and dense clusters.

From there we know that a new born cluster of stars tends not to survive very long. For various reasons - outflows, winds and ionising radiation from newborn stars are able to heat and expel the remaining gas; star formation appears to have an average efficiency of a few to perhaps 20-30%. Expelling the gas, plus the tidal field of the galaxy breaks up the cluster and disperses it into the field, which gives us the lower field star (or system) density that we see around us.

Once the stars are part of the field they essentially don't interact with each other; they are too far apart to feel the influence of individual objects and move subject to the overall gravitational potential of the Galaxy.
so your consideration of the force between stars is not really relevant.

big bang theory - Could we estimate the age of the universe based on the planar property of the Solar System?

First, the Big Bang did not happen at a point. I cannot emphasize this enough. It happened everywhere in space at the same time. You can't think of it as an explosion from one location pushing everything else away.

Furthermore, planets and stars didn't exist at the time of the Big Bang. The Solar System, for example, formed about 4.5 billion years ago, while the oldest stars - Population III stars - formed within a few million years after the Big Bang. At the beginning of the universe, everything was in an exotic soup-like state: a quark-gluon plasma.

Therefore, your hypothesis is flawed in two places.

Relation between black hole mass and radius, and our universe's

According to the standard ΛCDM cosmological model, the observable universe has a density of about $rho = 2.5!times!10^{-27};mathrm{kg/m^3}$, with a cosmological consant of about $Lambda = 1.3!times!10^{-52};mathrm{m^{-2}}$, is very close to spatially flat, and has a current proper radius of about $r = 14.3,mathrm{Gpc}$.

From this, we can conclude that the total mass of the observable universe is about
$$M = frac{4}{3}pi r^3rho sim 9.1!times!10^{53},mathrm{kg}text{.}$$
Sine the universe at large is nonrotating and uncharged, it's natural to compare this to a Schwarzschild black hole. The Schwarzschild radius of such a black hole is
$$R_s = frac{2GM}{c^2}sim 44,mathrm{Gpc}.$$
Well! Larger that the observable universe.

But the Schwarzschild spacetime has zero cosmological constant, whereas ours is positive, so we should instead compare this to a Schwarzschild-de Sitter black hole. The SdS metric is related to the Schwarzchild one by
$$1-frac{R_s}{r}quadmapstoquad1 - frac{R_s}{r} - frac{1}{3}Lambda r^2,$$
and for our values we have $9Lambda(GM/c^2)^2 sim 520$. This quantity is important because the black hole event horizon and the cosmological horizon become close in $r$-coordinate when it is close to $1$, a condition that creates a maximum possible mass for an SdS black hole for a given positive cosmological constant. For our $Lambda$, that extremal limit gives $M_text{Nariai} sim 4!times!10^{52},mathrm{kg}$, smaller than the mass of the observable universe.

In conclusion, the mass of the observable universe cannot make a black hole.

---

Well, we don't fully comprehend black matter, do we? And it was just "yesterday" that we discovered the "black energy", wasn't it?

If GTR with cosmological constant is right, we don't need to "fully comprehend" it to know its gravitational effect, which is what the calculation is based on. If GTR is wrong, which is of course quite possible, then we could be living in some analogue of a black hole. But then it's rather unclear what theory of gravity you wish for us to use to try to answer the question. There's no remotely competitive theory that's even approaching general acceptance.

From the perspective of our huge ignorance, I think that 14.3Gpc and 44Gpc are not even one order of magnitude apart, which I consider a good approximation.

Actually, the point of that calculation was to show that it's at least prima facie plausible. The Schwarzschild radius calculation doesn't rule out the black hole--quite the opposite. However, it's also not appropriate for reasons I explained above. The more relevant one actually does have mass more than one order of magnitude apart, and shows inconsistency. So if GTR with Λ is correct, it's unlikely because the ΛCDM error bars aren't that bad.

However, even if we still treat it as "close enough", that does not by itself imply what you want. The question of what kind of black hole all the mass of the observable universe would make, if any, is quite different from whether or not we're living in one. The black hypothetical needs to be larger still.

The biggest point of uncertainly, though, is the cosmological constant, even if GTR is otherwise correct. If we're allowed to have very different conditions outside our hypothetical black hole, then we could still have one, but then we get into very speculative physics at best, and just complete guesswork at worst.

So treat the above answer as conditional on the mainstream physics; if that's not what you want, then there can be no general answer besides "we don't know". And that's always a possibility, although not a very interesting one.

Climate modeling of exoplanets - Astronomy

The presentation linked to seems to primarily discuss General Circulation Models, solutions to the equations of fluid dynamics which predict the behavior of a planet's atmosphere, a component of more complete Global Climate Models (confusingly, both are GCMs).

Spectral data on exoplanets is extremely limited to date and poorly constrains the composition, but a lot can be inferred from a planet's orbit and eccentricity (how far from circular the orbit is). At the simplest level, planet temperature depends on how much light it receives from its parent star, and this will vary as a planet goes around a non-circular orbit. Starting from there and making reasonable assumptions about the composition of the planet (usually starting with a solar system analogy) you can test out a GCM model in the various extreme cases we observe (hot Jupiters etc.), to be subjected to future observational constraints.

One exciting new development is looking at a planet's phase curve (the observed brightness of the planet as it orbits its parent star) and using that to infer properties of the atmosphere. One team recently found that Kepler 7b's phase curve may best be explained by the presence of clouds:

Demory, et al. 2013. “Inference of Inhomogeneous Clouds in an Exoplanet Atmosphere.” arXiv:1309.7894 [astro-Ph] (September 30). http://arxiv.org/abs/1309.7894.

Red Moon a Characteristic of all Total Lunar Eclipses?

No, not all total lunar eclipses will turn the Moon deep red. Most of them do, but not all.

If you were standing on the Moon during the eclipse, you'd see the Earth passing in front of, and obscuring, the Sun. But the Earth will never become fully dark, even when the Sun is fully covered. A bright ring will always surround the Earth. Why?

That ring is sunlight refracted by the atmosphere. It's there because the Earth has an atmosphere. You could say it's all the sunrises and sunsets of the Earth, all seen at once. It's this light that continues to illumine the Moon during the eclipse.

But why is the Moon red, instead of some other color? This is because the blue end of the spectrum is scattered more easily in all directions (same mechanism that explains why the sky is blue on Earth), whereas the red part of the spectrum is scattered less easily and moves on a straighter path along the refraction lines. The bright ring around the Earth, as seen from the Moon, is probably red, because most of the blue in it has been scattered away.

Now, if the Earth's atmosphere is full of dust particles from huge volcanic eruptions, the bright ring is a lot weaker. That makes the Moon during the eclipse a much darker shade of red. Sometimes the Moon is a very dark, dull grey during the eclipse, no red hue at all - so dark in fact that it's hard to see in the sky while the eclipse is full. This has happened some decades ago after the Pinatubo eruption.

http://en.wikipedia.org/wiki/Lunar_eclipse#Appearance

http://eclipse.gsfc.nasa.gov/LEcat5/appearance.html

EDIT:

A measure of the brightness and color of a lunar eclipse is the Danjon scale. Eclipses are rated between 0 (almost invisible, black or very dark grey) and 4 (bright orange with bluish rim).

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

http://eclipse.gsfc.nasa.gov/OH/Danjon.html

My estimate is that the last eclipse was a 3.

I can't find a list of recent eclipses rated on a Danjon scale, but here's a list of 20th century eclipses, with a measure of the magnitude of the umbra.

http://eclipse.gsfc.nasa.gov/LEcat5/LE-1999--1900.html

As you can see, the magnitude varies quite a bit.

Bottom line: each eclipse is a bit different. A majority would have some kind of orange, copper or red tint. A minority are too dark to see any color. Unusual colors (outside of the red-yellow interval) are very rarely possible. Brightness and hue varies from one instance to another, since it depends on Earth's atmosphere, which is a system that changes greatly over time.

Absolute magnitudes of stars - Astronomy

VizieR is an online source for all sorts of astronomical data published in scientific papers. As you mentioned, The HIPPARCOS catalogue contains visual magnitude data.

1. Open the query page for the main HIPPARCOS catalogue


2. Select the fields you want (defaults are ok for you)


3. Hit submit to see the results


4. You can limit the number of results and format under Preferences on the left

This table gives you the measured visual magnitude, i.e. the Apparent Magnitude ($m_V$, V column). To convert that into Absolute Magnitude ($M_V$) you need to know the distance to the star. This can be calculated using the Parallax field (Plx column).

Here's the formula for you:

$$M_V = m_V + 5 * log_{10}( Plx / 100 )$$

You can easily dump the data into Excel or something, put a formula into an extra column and calculate the Absolute Magnitude.

earth - What light source can cause a shadow of a cloud onto the next cloud above?

Scenario-
-Time & Environment-11:00pm Super Moon small town population 40,000 in Central Ca.
-Nearest body of water is 80 miles away 4 small lakes of about 1 mile in diameter.
-Pacific Ocean is 250 miles away
-Surroundings-Miles of orchards and fields for agriculture.Nearest town population of 200,000 about 10 miles.
Nearest city with population of 400,000 about 60 miles away.

-Objects in order from earth to moon-
(1) thin linear cloud across visible sky.
(2) Dark shadow from chem/con cloud.
(3) overcast cloud that filled large portion of sky with a transparency of about 20 percent based on moons visibility seen through cloud layer.
(4) Super Moon at a 12 o'clock position in sky.

comets - Do/did the asteroids contain enough water to create Earth's oceans?

This is far too soon to draw sweeping conclusions. The Philae probe has only measured deuterium for one comet. But comets are diverse, of diverse origins. We need to take a lot more samples, from many different comets, before we can conclude firmly that water has, or has not, come from comets.

The only conclusion we could draw so far from this study is that earthly water has not come exclusively from comets similar to 67P/C-G.

What you're seeing now is the usual distortion that occurs in the media after some hugely popular experiment is completed - its conclusions are blown out of proportion. But that's not how science works.

Let's write this one down, and wait for further science to be put forth. We don't know enough yet about these objects.

Theoretical limits for natural satellites having natural satellites

Wikipedia says there may be a reason (emphasis mine):

No "moons of moons" (natural satellites that orbit a natural satellite of another body) are currently known as of 2014. In most cases, the tidal effects of the primary would make such a system unstable.

This seems to be because the secondary would have to be very close to the primary - close enough that it doesn't become simply another satellite of the actual primary (i.e. the planet).

The page then goes on to talk about Rhea, a moon of Saturn that may or may not have a ring system. This gives some more information.

gravity - What would happen if Jupiter and Earth were at the same distance as the Moon is from Earth?

That's no good idea. Earth wouldn't necessarily fall into Jupiter in the short run, provided it orbits Jupiter fast enough (within about 1.7 days), and on a circular orbit, but we would risk to collide with Io, destroy it by tidal forces, or change its orbit heavily.
The other Galilean moons would get out of sync and change their orbits over time.

Tides would be severe on Earth, not just limited to oceans, but also for "solid" ground, as long as Earth isn't tidally locked. This would result in severe earth quakes and volcanism.

Our days would be dim due to the distance to the sun. After tidal locking of Earth and ejection/destruction of Io the tides as heat source would be lost, oceans would freeze, temperature would fall to about -160°C mean temperature. During the polar night oxygen would probably condense from the atmosphere and form lakes, may be even nitrogen. By this atmospheric pressure would drop.

Since Spock is smart enough to know these consequences in advance, he wouldn't do it.

telescope - How do we convert other waves of the EMS to visible light?

It's just as simple as taking the flux at some wavelength (just a number) and using this number to represent a visible intensity.

If you only have one wavelength then you can only get a monochrome picture. However, if you have flux information at more than one wavelength, let's say three, you can use the flux at the longest wavelength to represent red (r), the middle wavelength to represent green (g) and the shortest wavelength to represent blue (b).

Put this together and you have 3 numbers representing a visible rgb signal that can be used to create a picture. Of course there may be lots of fiddling that goes on with the colour balance and the contrast to produce an effective picture, but this is the basic process.

accretion discs - When a neutron star accretes matter, will its mass increase?

A nova is generally understood as an explosion of a star. Thus comparing a nova to a stable neutron star is kind of a nonsensical thing to do.

If you're asking if a star goes nova and the core that remains is a neutron star after the resulting explosion, will it's mass increase overall? The answer is no, since it loses a lot of mass in the process of going nova. However, it is now more dense than before and the concentration of matter is in a much smaller space (the neutron star). (The explosion compresses part of the star and pushes the rest of it outward off into space).

If you're asking if a neutron star's mass will increase if you throw matter at it, the answer is yes. Just like any other form of matter. Is there a limit to how bit you can increase it? No; however, past a certain density it will compress into a black hole (which do not have a size limit as far as we know).

You may want to edit your question to clarify what you mean.

formation - Why do comets come from our local Oort cloud instead of from interstellar sources?

Since most comets are on a predictable orbit that has them circling the sun (several times), then they cannot be from outside the solar system.

Any comet that originates outside the solar system, will pick up enough velocity approaching the sun to be able to leave the solar system. Therefore, we would only see the comet once. It is unlikely that the comet approaches Jupiter (Saturn, Neptune or Uranus) where the trajectory shifts enough so that it will remain in the solar system.

Since the solar system is believe to be formed from the gravitational attraction of interstellar gas and debris, it is very unlikely that comets will be ejected unless their orbits are perturbed.

So, there is a low probability of a comet being ejected from one system, another low probability that the comet will encounter another system (after millions and billions of years) and yet another low probability that the comet gets captured in the system.

Added:
If the object was pushed out to 1 LY and it then returned as a comet, it was still under the influence the sun's gravity and would not be considered as extra-solar. The object would need to come from another system to be called extra-solar. Since the ejected objects are in a nearby orbit to the solar system, they will either return, or travel outward. Think about a satellite with a period of 12 hours. A slight push will put it in orbit 2 meters higher, and will increase the period significantly.

There is a lot of space at Neptune's orbital distance, and the intersection with Neptune's gravity is quite small (ratio of area of influence (0.1g) / surface area = ${8.6times 10^{-11}}$). The reason the Jupiter captures asteroids is because they are in the same plane and are moving in the same direction. This increases the probability of changing the orbit significantly. It could also take several hundred passes before the orbit changes significantly.

The area of influence is the cross section area where the gravity is greater than 0.1g. The surface area is the spherical surface area of the mean distance to the planet.

Jupiter - ${5.325times 10^{-8}}$ ~= 0.0000053%
Saturn - ${4.703times 10^{-9}}$ ~= 0.00000047%
Uranus - ${1.791times 10^{-10}}$ ~= 0.000000018%
Neptune - ${8.605times 10^{-11}}$ ~= 0.0000000086%

Satellite/Planetary Orbits - Astronomy

All planetary orbits contain 5 unusually stable points. These points are particularly important because they allow man-made satellites to orbit the Sun with a period equal to that of Earth’s. 3 of these points are collinear. Suppose that is the distance between the centers of mass of Earth and the Sun. Find the distance from Earth’s center of mass to either one of the other stable points in the Earth-Sun system in terms of...

(I'm not looking for a full solution; I just want to know what these points are called)

What are these points it talks about, and what's their mathematical relation to Earth's orbit?

This still isn't homework; I'm just looking for the name of these points.

weather - What makes a really good observatory site, besides altitude?

• Not near an active or dormant volcano (but Mauna Kea seems to disprove this?). This is kinda a bummer because a lot of tall mountains seem to be volcanic.

There's nothing intrinsically wrong with dormant volcanos, as Mauna Kea and the observatories in the Canary Islands demonstrate. Not sure where you would have gotten that idea.

• Somewhere with clear and/or dry weather for as much of the year as possible

Clear weather is essential -- you can't observe when it's cloudy!

Dry weather is very good, especially for infrared observations (water vapor blocks a lot of infrared light).

• Not near major light pollution like cities

Yes.

• Cold weather is better than hot weather? Not sure I understand that; if it's uniformly hot or cold then I dont see how it makes a difference.

Cold climates tend to have drier air. (Antarctica is an extreme case.) Of course, being on top of a mountain means colder air, which is why Mauna Kea is good despite it's being in the tropics.

Am I missing any? Or are these wrong in some way?

You also want stable air with good "seeing", which rules out places with turbulent air and lots of wind (and is another reason mountaintops are good: they're usually above the most turbulent layers of the atmosphere).
Quoting from this page (which has a good discussion of the general topic):

Seeing ... requires at minimum a lack of extra turbulence at all atmospheric levels, and seems to be best satisfied in the convergence zones just outside the tropics, at latitudes about ± 30º. Also, minimal local turbulence is often associated with mountain peaks that reach into the otherwise undisturbed oceanic airflow, as on islands or coastal ranges (and given the direction of the planet's rotation, this generally favors western coast ranges).

the sun - Apocalypse on earth

We wouldn't feel the ejection of another planet off the solar system, since the attractive force of distant planets to Earth is very low.

Only close encounters of Earth with other planets would cause noticeable up to severe changes on Earth.
This would be caused mainly by tidal forces due to different acceleration for different parts of Earth, or by changing the orbit of the Moon. A homogeneous acceleration wouldn't cause immediate damage, but with some delay it may cause temperature changes due to the change of Earth's orbit and the changing distance to the Sun.

observation - 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?

Is there an astronomy exam I can take?

There is a Dantes Standardized Subjet Test (DSST) in astronomy. I have no idea if and how widely colleges give credit for these exams. Typically, you will start with physics and choose to specialize in astronomy. Therefore, most standardized exams are for physics which may have a handful of astronomy questions.

the sun - How likely is it that the Sun will destroy our electric society?

I should give credit here to @honeste_vivere, who pointed out to me today that there have been recent studies excited by some extremely large coronal mass ejections that were classed as "near misses" in terms of causing major disruption.

Of particular interest to you would be the event of July 2012 discussed by Baker et al. (2013). I quote from the paper

" Had the season and time of day for this CME passage been right on striking the Earth, the world would have witnessed a storm larger (possibly much larger) than the 1859 Carrington event. This most likely would have had devastating consequences for many technological systems "

and then in the discussion

"
It is the opinion of the authors that our advanced technological society was very fortunate, indeed, that the 23 July solar storm did not occur just a week or so earlier.
Had the storm occurred in mid-July 2012, the Earth would have been directly targeted by the CME and an unprecedentedly large space weather event would have resulted. In fact, there is very legitimate question of whether our society would still be “picking up the pieces” from such as severe event [see NRC, 2008].
"

It looks likely that such events are perhaps things that occur every few decades. Riley et al. (2012) suggest something as powerful as the Carrington event has a 12% chance of occurring in the next decade, with a 1% chance of something several times bigger. Even more powerful events are seen on other stars, where both flare and CME energies follow a power law relationship of the form $dN/dE propto E^{-alpha}$, where $alpha sim 2.5$ (Drake et al. 2013). Thus much larger, but rarer flares are possible and also seen on (usually) much younger and faster rotating stars. There may however be a very significant sporadic tail to the solar flare energy distribution too. Recent observations of candidate "superflares" on otherwise unremarkable solar-type stars have been reported, though the reality of these and the mechanisms are still being explored. Shibayama et al. (2013) There appear to be a group, consisting of 1% of stars that show (repeated) superflares more than 100 times the size of the Carrington event. These appear a little less likely around stars rotating as slow as the sun, but the bottom line is that, if there is nothing "special" about these stars, then these events occur every 800-5000 years (for events of energy 1e34 - 1e35 ergs) on a slow-rotating G-star. (Though you need to read carefully - it is probably a function of temperature and rotation). The authors note that superflares may be associated with the formation of very large starspots (or starspot groups). So that may be our clue to duck and cover.