text:
What scientific evidence is there to support or refute the Iron Sun Hypothesis?
Professor Oliver Manuel,  a professor of nuclear chemistry at the University of Missouri-Rolla, claims in the article Sun Is Made Of Iron, Not Hydrogen, Professor Says that

iron, not hydrogen, is the sun's most abundant element.

Specifically, he and his colleagues postulate that

"We think that the solar system came from a single star, and the sun formed on a collapsed supernova core," Manuel says. "The inner planets are made mostly of matter produced in the inner part of that star, and the outer planets of material form the outer layers of that star."

Is there evidence to support or refute Professor Manuel's theories?

Perhaps, the Earth's age is a reliable proof against their theory: 4billion years could be enough time to dissipate residual heat. We could compare this with White Dwarfs models, and account for an iron core.

That was pretty much my point regarding iron being the final product of currently understood stellar fusion processes.  If fusion had stopped, four billion years is a long time for it to be still as hot as it is.

The claims don't seem to refute the current model. It seems rather unlikely that a sun could form from the remains of a star that already died. Regardless, if he's looking for support he needs to find evidence that actually refutes the current model.

Right, I misused the word: I meant, they gave a new interpretation.

Scientists thought something similar to what is proposed here in regards to the composition of the sun, until Cecilia Payne, in what has been called the "most brilliant PhD thesis ever written in astronomy" showed that the sun is made mostly of hydrogen.

That's so removed from evidence currently available that I would not call it a "hypothesis". It has a lot of the characteristics of crackpot science, unfortunately. It's just orthogonal to everything we know about the Sun: density, spectrum, neutrino production, etc.

@FlorinAndrei, personally, I don't believe this theory either - I have had discussions elsewhere online with the Professor himself, and it still did not make sense.  By making a question like this, it allows an online searchable record of evidence to refute this theory in an objective scientific manner.

Sure, no problem. I was just saying - "refuting" it is very easy. :)

If the sun was principally composed of Iron, then this would be apparent in the spectrum of light from the sun (which is how we know what stars are made of) (but see comment by @Keith Thompson below, the spectra only tells you about the surface, rather than the core).  The fact that all solar physicists bar one seem to think it is made up of hydrogen and helium strongly suggests this is not the case.
If the sun were made of iron, then I suspect that would require a revision of our understanding of nuclear fusion, as iron is the product of the final fusion reaction that stars can support, unless of course the theory is that the sun is no longer undergoing fusion reactions and all we have been seeing for the last 4 billion years+ is the residual heat of the core.
Apparently heliosiesmology suggests there isn't that much iron in the suns core (although the physics is beyond me, so caveat lector).
As a further point, if the sun is primarily made of iron, it becomes difficult to explain why the planets (especially the gas giants) have so little of it.  It is reasonably straightforward to explain why the inner planets have lost much of their original hydrogen and helium, but rather more difficult to explain how the outer planets could have become hydrogen enriched.
Update.  Dr Manuel's website claims that "About 99% of the sun has the same composition as ordinary meteorites.".  The density of carbonaceous chondrite meteorites is about 2100-34700 kg/m3 (with porosity being an issue).  The mean density of the sun is 1410 kg/m3.  Given the temperature and pressure within the suns' core, if it had a similar composition to a meteorite its' porosity would be zero, and hence even more dense than an ordinary meteorite.  There is a problem with accounting here, if 99% of the sun had a composition similar to that of a meteorite, it would have to be full of voids to have the suns observed density.  The figures just don't add up as far as I can see (although how materials may behave at ultra high temperatures and pressures means intuition may be invalid here).
For Manuel to be right, pretty much all we know about stars must be wrong.

I'm not sure, but I can't see the problem with doing so, provided you make a link back to the question here.  It would be good for there to be a really definitive answer to this question from a real expert.

I think, these points are not valid: they proposed for a completely new model. Maybe, the astrosismology point is the only one which can really be an evidence against their theory, but I do not know anything about. @DikranMarsupial, if the answer author could still read this topic we could have a nice discussion.

@Py-ser I am no expert on this (read a couple of books on planetary science, but that is about it), my main aim in answering the question was to point to a forum more likely to have a definitive answer (while giving a couple of lines of evidence that make me highly doubtful).  However if a new model requires the overturning of what we know about nuclear fusion in stars, it becomes a very big claim, and therefore requires very big evidence.

As an aside, if you look up Prof. Manuel's articles on google scholar, they have been very sparingly cited and the person that cites them most regularly is Prof. Manuel himself.  Science is generally pretty open to people publishing unlikely ideas, but the usual consequence when the evidence provided is inadequate is that they are ignored.  That seems to be the case here.  I suspect the idea is not impossible, but it requires too many large adjustments to other areas of physics to work, and hence isn't considered plausible by the research community.

The Sun's spectrum only tells us about the composition of the visible surface. Hypothetically, if the core were mostly iron and the surface were mostly hydrogen and helium, that would be consistent with the spectrum we see. I don't take the "Iron Sun Hypothesis" at all seriously, but the spectrum isn't what refutes it.

I would have thought that an iron core would be significantly more dense than a core of metalic hydrogen.  The density of the sun can be measured by its diameter and its mass deduced from its gravitational effects.  I would have thought that an iron core would put severe constraints on the composition of the sun's outer layers.  I doubt anybody has done the calculations though as nobody seems that interested in Manuel's theory.  I'd be very interested if anyone here can comment on this (I'm a statistician not an astronomer/astrophysicist!)

@DikranMarsupial, I think you could retire your answer: people continue to upvote it, but it is wrong... 3 out of 4 of your points are not correct, and the 4th is not supported. About your last comment: why do you mention gravity effects? However, I agree on your point of not-cited = not-reliable. But that is not a scientific reason per se.

@Py-ser perhaps you would like to explain why some of the other answers are wrong, as Keith did, rather than just state it.  Or better still, write an answer of your own, that is the point of Stack Exchange sites.

As to why mention gravity, I would have thought that was fairly obvious, in estimating the density of a star, knowing its mass would be a useful piece of information, and we can estimate that from its gravitational effects (e.g. planetary orbits).  If there is a big difference in the density of iron and metalic hydrogen, then this would have big implications on the plausible composition of the Sun's outer layers if the core were primarily iron.

I also disagree with not-cited = not reliable.  Science doesn't work that way.  Being able to support an argument with citations adds to the reliability, but it certainly isn't zero in the absence of citations, and arguably citing some papers detracts from the reliability of the answer (plenty of papers get published that are simply wrong).  For reliability, you really need papers that have been accepted by the research community, for example those that are highly cited.  At the end of the day though, it is the evidence and internal consistency that matters.

Sorry if my comments seemed troll-like, I just try to keep it short in the messages. I am not able to write an answer, but this does not prevent me to say which reasons are not good to disprove the theory. As @KeithThompson already explained, many of the points in this answer are not good, like the spectrum one. I think, this is a delicate topic anyway.

@Py-ser there was no problem in the manner in which you comment, the point is that I am genuinely interested in why my answer was wrong so that I can learn (especially as one of the points I made was essentially the same as the comment you made on the original question, which suggest there is some nuance that I have missed).

Even if the answer has already been accepted, more evidences can contribute to this thought-provoking topic.
I think, as it has been mentioned, eliosismology is a good way to map the interior of the star, but I am not able to argue about it.
Another disproof, comes from neutrinos.
Interior of Sun can be investigated by neutrinos detection, which do not interact passing by the outer shells, and can directly say where they come from, and how.
Standard solar theory predict neutrinos production by p-p chain.
This has been observed, even if the neutrino flux is about $1/3$ than the expected one, because of the neutrino oscillation.
Now, this theory become solid after Bruno Pontecorvo, but I do not think we have any other evidence of observed neutrino oscillation.
Furthermore, Earth's age is helping us. We know Earth is ${\sim}4$ billion years old, that is roughly the age of the Solar System and the Sun itself.
From Niels Brendt website, we know that after such a long time, the white dwarf luminosity has become much fainter then Solar (where Solar is referred not only to the Sun, but to general G2 main sequence stars). Neutron stars cool even faster. And this is obviously not observed: if Manuel's conjecture was real, Sun should be much fainter now. To this, add that they say that their supposed SN happened something like 5 billion years ago...
As a final note, I would highlight that, not only this author has never been mentioned in literature, that is (as we know) synonym of poor quality, but he actually never published (at least since 10years or so) in any refereed paper, which is automatically translated, to me and I think to the community, as not science.

+1 for neutrino argument, I will look that up in more detail.  Dr Manuel has published in a variety of journals and conference proceedings, the lack of citations is more an indication that the research community don't find his arguments compelling (or perhaps simply wrong), rather than not being science.  Whether something is science doesn't depend on whether it is right or wrong, or even plausible.  Popperians would say it just needs to be falsifiable, which Manuel's theory clearly is.  Most scientists end up about as well cited as Manuel appears to be, it isn't unusual.

@DikranMarsupial, I mean, from a quick reasearch I didn't find any usual refereed astrophysics journals where Prof. Manuel published, like A&A or ApJ. My opinion is that, if your work does not pass through a referee critic analysis, probably it is not scientifically correct.

My point was that "not scientifically correct" is not the same thing as "not science".  I would venture that the majority of papers that do get past peer review are either wrong or misguided or not useful.  There are plenty of papers out there with several hundred citations where the method doesn't actually work on non-trivial problems (I know this as I tried replicating one this week).  Peer review is at best a basic sanity check, and it regularly fails in both directions.  Best not to be unduly harsh on the authors concerned.

@DikranMarsupial, yes, I agree that peer reviewing does not guarantee quality 100%, that is why, if you do not even pass that, it is just a game. In the end, science or not can not be a matter of opinion. But this, again, is my biased and not-much-experienced thought.

I think Mr. Marsupial's answer should get a star, but to put some of this in layman's terms, here's my answer.
The process of fusing iron into other elements, whether within a star or within a laboratory pulls more energy from the reaction than it creates.  It is the first element in star evolution that does this, and is therefore associated closely with the end-stage of a star's life.  The energy has to come from somewhere, and the models I have seen have the star eating up its outer layers very rapidly.  Perhaps within a day, which we would have noticed.  [Edit: I previously stated this as fusing elements into iron whereas it has been correctly pointed out that it is iron fusing into other elements that is the problem.]
For further evidence we can look at other stars whose makeup we know to be primarily iron through spectroscopy and compare them and their systems to ours.
In general, any useful theory needs to make definite predictions (and bonus points if they differ from the standard model) that can be either confirmed or refuted by observation and experimentation.  If I knew more about this 'Iron Sun Hypothesis', I could refute it better, but as it stands, it disagrees with enough of what we believe we know that it requires overwhelming evidence before most will pay it any attention.

I don't think that's correct. My understanding is that fusion reactions creating elements up to and including iron produce energy, and fusion reactions past that absorb energy (which is why elements past iron are produced mainly in supernova explosions). Thus a massive old star could be producing iron in its core. When the inputs to that process run out, the star collapses because fusion stops being exothermic.

I was going mostly off of 30 year old college learning, so I did some searching based on your response just to be sure.  Here is an article that talks about it.  My understanding is that the star uses up all of its energy in the production of iron, so it can't get to heavier stuff before going nova or just collapsing.  This episode of AstronomyCast is another good source of info.

I think the article is consistent with what I wrote. "At 3 billion degrees, the core can fuse silicon nuclei into iron and the entire core supply is used up in one day." So silicon-to-iron fusion produces energy, but only lasts about a day; once the silicon is depleted, the iron can't be fused any further without consuming energy. (The jump from Si, atomic number 14, to Fe, atomic number 26, does seem odd; I clearly don't entirely understand this stuff.) Elements lighter than iron can be fuel for fusion reactions; iron and heavier elements cannot.

@MBurke, Iron, not silicon, is the last element produced by esothermic reactions see here. Please, rewrite your answer.

Fixed, though I couldn't get to your link, I realized that I misspoke.

@MrBurke, perhaps you missed the point, because your edit do not regard the misconceptions exposed in your answer: fusing Silicon into Iron is exothermic reaction, so it seems to me that the base of your answer is not valid. Please, could you tell me why you can't read the book that I have linked in the comment above? Btw, you can see it almost enywhere, also here.

The base of my answer is that once you have iron in the star, nuclear fusion is no longer possible, and the star quickly dies.  That is still born out by these articles.  A quote from the article you link here, Once the core fails to produce enough energy to support the outer envelope of gases the star explodes as a supernova producing the bulk of elements beyond iron. Production of elements from iron to uranium occurs within seconds in a supernova explosion.  That is the point I had which answers the question asked and remains correct unedited.

Standard solar models reproduce the sun's mass, radius, luminosity, spectral energy distribution, line spectrum, neutrino luminosity, and vibration spectrum (helioseismological) with great accuracy.  The age inferred for the sun lines up nicely with the age of the solar system inferred from isotope ratios in meteors.  The solar models are based on the same physics that produces stellar evolution models, which fit the color-magnitude diagrams of clusters of various ages and compositions in impressive detail. In other words, the standard picture of solar and stellar structure is accepted because it explains, in detail, a stupendous amount of richly textured observational detail.  Manuel's theory, by contrast, explains nothing in particular.
As a reference, I'd point the interested reader at any of the many excellent astronomy textbooks aimed at the distribution-requirement college courses.


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Detecting Deep Impact
There have been several accounts in media articles of asteroid near misses, many describe that the asteroid was not seen before the near miss.
After reading the question and answer to How can we tell if the asteroid 2013 TV135 will hit Earth in 2032?, where it was discussed about determining the orbits of known asteroids.
The question here is what is the furthest practical distance that a previously undetected asteroid, capable of an extinction level event, can be observed to have a trajectory that will result in a collision course with Earth?

Does "previously undetected" put any limits on it (i.e. size, albedo, etc.)?

no limits, except for being large enough to cause an extinction level event, essentially furthest = best case scenario for detection

@HDE226868 and others, please note the edit I have made on the question - hopefully it is a lot clearer.

I think it was good to start with, but it's better now. I'm just trying to pull an answer together. :-)

Don't worry. I've had questions like that. View should improve.

@HDE226868 are you still putting an answer together?

Media articles are written with purpose to cause sensation.

@user2481 yes, I am aware of that, this is why I am seeking a scientific answer

@Undo if you are in agreeance, please award the bounty to the excellent answer below

Something to consider might be a large very dark (low albedo) asteroid.  I would say it is quite possible a large dark asteroid (e.g. P-type) "could" be headed for us, and we would not know it until it hit us (e.g. distance 0).  I think you are asking for the closest distance.  It appears the main method of detecting asteriods is with telescopes, but we are starting to scan the skies in infrared too.  Hopefully infrared will reveal these darker asteriods better.  Sources for further research include:
http://en.wikipedia.org/wiki/P-type_asteroid
http://www.space.com/3126-asteroids-data-sheet.html
http://www.nbcnews.com/science/space/how-we-scan-solar-system-find-dangerous-asteroids-f8C11480423
For the furthest distance, that question is trickier.  It would largely depend on what is considered an asteroid, for example, if we consider Pluto to be an asteroid, we can at least detect very large asteroids out to the distance of Pluto, and we could track its path and see if it were on a collision course with earth.
On average, it is said that we have already detected most of the globally threatening asteroids and it does not appear we are in immediate danger.  Still, I'd say we need to stay on the lookout, and especially look out in infrared to detect any dark asteroids we may have missed :-).
Summary:

Closest distance before large threatening asteriod detection: 0 (e.g. upon impact)
Furthest distance before large threatening asteroid detection: Beyond the orbit of Pluto (and most asteroids tracked are in the asteroid belt between Mars and Jupiter)

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sort: votes


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What is the reason for high lithium concentrations in Sun-like stars?
In reading the European Space Organisation paper Lithium depletion in solar-like stars: no planet connection (Baumann et al. 2010), several conclusions based on their observations include:

Lithium depletion in Sun-like stars is independent of having planets (an earlier theory), but rather is a function of the star's age.  From the article:

For solar-like stars, the lithium vs age trends for planet-hosts
  and stars where no planets have been found are statistically
  identical. Thus, the presence of a planet does not influence
  the observed surface lithium abundance.

However, they observed that

A number of solar-like stars with unusually high lithium
  abundance for their age are present in the field.

and that these stars also have an anomalously low surface gravity.
What is the current theory as to why some Sun-like stars have a relatively high lithium abundance and low surface gravity?

In the fourth point of their conclusions, they more or less (reading between the lines a bit,) that they will investigate that further (read: future publications), because current theories do not properly explain a link between higher lithium abundance and low surface gravity

@usethedeathstar this could be a start of a good answer

I could offer a couple of speculations - but there is no "theory". Incidentally, I was looking at the refereed journal paper by Baumann et al. (2010).

The atmospheric parameters of these stars are not correct. Given that the ages are derived from the gravities and temperatures of the stars and that the lithium abundances are also dependent on the gravities and particularly the temperatures, there will be correlated errors. I'm actually struggling to understand how you get stars which are "solar analogues", all of which have similar gravities, yet seem to have a spread of ages (estimated from the Teff vs log g plot) from 1-10Gyr? It is notable that there appears to be just 1/117 in Baumann's sample, compared with 10/82 in Israelian's work...
Maybe they are actually PMS stars, so the estimated ages are completely spurious since they assume that the stars are on the main sequence. I would however have expected the abundances to be 3 or higher.

Neither of these explanations seems completely satisfactory.
I mention in passing that the question of planets and lithium is far from settled. The latest salvo in the war was fired by Figueira et al. (2014) who claim there is statistical evidence for enhanced Li depletion in solar analogues with exoplanets.

Being an optimist I'm hoping that Ramirez et al. (2012) were right when they examined a larger sample than the initial study and found no lithium/planet correlation.

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Can a gas giant have an other gas planet as satellite?
Is it possible for a gas giant (A very big one) to have an other smaller gas planet as satellite?

This might interest you. . .

@HDE226868 Interessting stuff indeed, thanks, but even if it states that small gas planets exists, id doesn't state if they can be in orbit of greater gas planets? Let's say two gas planets orbits near each other, doesn't it make one  (the bigger) to drain the other's gas?

I think you misunderstood me. I wasn't saying it was a duplicate - I literally just thought you might find it interesting!

@HDE226868 I never said I thought it was duplicate, sorry if I misformuled, but indeed it interessted me! thanks, I justed pointed out how it could be linked with my question

Technically, the question should be asking about "... another smaller gas body as an exomoon."  A "planet", since the 2006 IAU resolution, must directly orbit the Sun.

It would seem so.
Think about brown dwarfs. At the lower end of the mass spectrum, they're only a couple dozen times that mass of Jupiter, and only a couple times the mass of Hot Jupiters. Some brown dwarfs have been found to have planets. Taking a look at some examples:

2M1207: Roughly 25 Jupiter masses, this brown dwarf has a planetary-mass object orbiting it, 2M1207b. This object has between 3 and 10 times that mass of Jupiter, and is fairly hot, although it orbits far from the star.
2MASS J04414489+2301513: Roughly 20 Jupiter masses, with a companion on the scale of 5 to 10 Jupiter masses that could be a planet.

Note that these brown dwarfs are very low-mass, only a bit more massive than hot Jupiters. They can have orbiting gas giants - why couldn't high-mass hot Jupiters have the same?

Video games think so: http://en.spaceengine.org/forum/10-1762-1
More generally, we know this can work for stars in a stable way: there are ternary stars where a close binary orbits a large, distant third sun. So replace the binary stars by gas giants, and there you go. And the Pluto-Charon system tells us that similarly sized bodies can orbit each other around a star.
So the physics does not preclude the configuration in and of itself.
There are the usual problems then with likelihood: gravitational capture is hard, and planet formation is generally expected to be noticeably different from star formation. The latter we can maybe get around: the gas giants are nearly to the brown dwarf stage, so that star formation mechanisms are more applicable.
According to the Selected Examples here, Jupiter has density comparable to the Sun, so I would expect this to mean we can scale the binary part of the stellar system to gas giant levels without making the system any more unstable than its stellar counterpart.
A precise knowledge of the relevant physics to assert that confidently, and to know if the stability could last for millions or billions of years, is currently beyond me.


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VSOP for exosatellites? (Io, Titan, Ariel, etc)
VSOP87 (and the subsequent VSOP2000 and VSOP2013) provide reasonably
accurate planetary positions while still maintaining a small file size
compared to the Chebyshev polynomial files.
VSOP includes our moon, but has anyone created something similar for
other planets' moons (such as Jupiter's Io, Saturn's Titan, and
Uranus' Ariel)?
I'm thinking about trying to create something like this and wanted to
check that it hadn't already been done.

Can you try putting a little more detail in your questions.

@Juka What sort of additional details would you like?

I'm guessing you mean moons around planets in the solar system other than the Earth, rather than moons around exoplanets (https://en.wikipedia.org/wiki/Exomoon)?

Yes, there are several I know of for the Galilean moons, less so for other satellites.  One very good place to start is from the same organization that created the VSOP series, now called IMCCE.  Here is a link to their satellite ephemeris.  According to the Explanatory Supplement to the Astronomical Almanac, most of the other moons of Jupiter behave to chaotically to create a similar theory and numerical integration is required to produce accurate results.

This is really nice, thanks!

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How is the universe bordered?
This answer says that some models describes the universe as finite. How do those models describe the universe's border?
Does the border (theoretically) exist?
Is it a solid border? Do they predict what happens if matter crosses this border (maybe it simply can't)?

Can you specify some specific models you want to know more about?

Your intuition is correct.  It does not really make any sense for the universe to have a border.  Would it be an unbreakable wall? What would be on the other side?  Models with a finite universe would not have a border.  The universe would be curved into a closed loop like the surface of a sphere.  In other words, if you go far enough in any one direction you come back to the starting point.

So it's some kind of 4 dimension sphere (sphere in any direction you go in 3D) if the universe is finite? I was thinking to that a delimited border would be a non-sense

It could be a 3D space embedded in a higher dimension space, or it could be that 3D space is not Euclidean.  It might make a good question to ask if one can tell the difference.

Take a sheet of paper. Glue/tape/whatever pairs of opposing edges together without twisting. The first such gluing will give a cylinder. The second then gives you a torus (a doughnut). This is a closed 2-d surface. It has no border. Things can just loop back to where they started by heading in a straight line.
Practical note: you're likely to get a bit of a sharp bend in your torus, but that's the fault of the paper.
In fact, try drawing curves on the paper to see what happens. If it's unfolded, connect points on borders exactly opposite each other (straight up/down or left/right on the paper). Fold it into a torus, and what do you see?
When folded draw some lines on the torus that pass through the edges of the paper. As curvy as you want. Labelling each edge entry and exit will help in the next part. Now unfold it and what do you see?
Indeed, if you've ever played the old arcade game Asteroids then you are already familiar with life on a torus. The lines you're drawing will behave exactly as your ship in the game will when you think of the unfolded paper as the screen.
This is basically what is meant by a closed universe. A more precise description requires the use of topology and the concept of a (smooth) manifold. Manifolds can have boundaries (the ends of the cylinder, for example), though as far as I know all serious cosmological models are boundaryless. There would be interesting and distinctive physics on a boundary, to the point that it should have been pretty obvious if one existed in the visible universe.

Crickey, I've just been typing up the very same Asteroids and here it is. Your ninjutsu is strong. By the way, I recommend the following illustration for the torus gluing: http://mathworld.wolfram.com/images/eps-gif/TorusSquare_701.gif

The models we're talking about are in the Friedmann–Robsetson–Walker family of solutions of general relativity, which are used because they have the properties of being spatially homogeneous, i.e. they are the same at every point in space, and also isotropic, i.e. every direction is equivalent.
There are many possible finite yet boundaryless spatial geometries, and I recommend zibadawa timmy's answer for a simple illustration of a toroidal two-dimensional universe and the accompanying analogy of the classic arcade game Asteroids. It is probably the most straightforward illustration that being finite does not imply having a border while at the same time making it intuitive that we don't need to actually consider a higher-dimensional space to embed it it in, because the rules of Asteroids don't actually need an extra dimensions. Such embeddings are just convenient visualizations.
However, where the toroidal universe differs from the FRW models is that it's not isotropic. You can see this by the fact that on a torus, going in some directions will get you back where you started, whereas going in others will wind you endlessly along the torus, never quite making it exactly where you've started at. Thus, not all directions behave the same way.
There are only four kinds of spatial geometries for the three-dimensional space of our universe that are homogeneous and isotropic: the Euclidean space $\mathbf{E}^3$, the hyperbolic space $\mathbf{H}^3$, the sphere $\mathbf{S}^3$, and the projective real space $\mathbf{RP}^3$, the last of which is like a sphere but with a different global topology.

So it's some kind of 4 dimension sphere (sphere in any direction you go in 3D) if the universe is finite?

A three-dimensional sphere, actually. Probably the most important intuitive leap here is that any particular embedding, or even whether there exists any embedding, is completely irrelevant. The surface of an ordinary beach-ball is actually a two-dimensional sphere, and as a manifold it makes sense whether or not we think of it a surface of some three-dimensional object. When we're talking about the universe, an embedding is a tool for visualization, not reality.
The universe could be a $3$-sphere, which we could think of as the surface of a ball in flat four-dimensional space, but there is no need to. That four-dimensional space wouldn't be part of our universe (not in general relativity, anyway), and such embeddings aren't unique. Actually, for completely general spacetime manifolds (i.e. if we try to embed spacetime in flat higher-dimensional spacetime), they're not even guaranteed to exist, which is a good thing because we don't need them--all our measurements are within the universe.


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What is the cause of the variation from high and low mean obliquity periods of Mars?
It is reasonably well known that Mars has a greater obliquity range than Earth, due to Mars lacking a stabilising influence of a large moon.  However, the Martian obliquity seems to have gone through periods of high and low mean obliquity, as shown in the top chart below:

Image source: Mars Climate Modeling Group
What is the current accepted theory as to why there are periodic high and low mean obliquities on Mars?
(Please note, I am not asking about orbital eccentricity)

This figure comes from numerical simulation by Laskar et al. (2002), see Fig. 2. An extended figure from a longer simulation can be found in Laskar et al. (2004), Fig 10.
As you can see more clearly from the extended figure, the simulation doesn't particularly suggest that this is periodic behavior. The question is, then, what happened ~5 million years ago that changed up Mars' obliquity like that?
Laskar mentions the event in the 2002 paper, but doesn't bother to offer any particular physical cause. These are numerical orbital simulations after all, not recorded observations. As unsatisfying as it might be, I think your answer is that evidently the planets aligned, so to speak (albeit probably not in a literal line).
Fig. 8 in the 2004 paper gives a tiny bit of context in terms of uncertainty, it shows that the simulation is only precise for the last 10 million years, and then goes chaotic.
Laskar mentions that his data are free to anyone who requests it, although I can't say how accessible it would be. In theory, someone could go in there and look for correlations (e.g. maybe there was some kind of strong resonance with the Earth and/or Jupiter around that time). I looked through ADS a bit, and found a newer paper by Edvardsson and Karlsson (2008) that pretty much supports Laskar's results in over the past 8 million years, but my brief search didn't turn up anything looking into what exactly caused that anomaly.
Finally, from context it appears that Touma and Wisdom (1993) were the ones to first discuss that obliquity drop, although their paper unfortunately appears to be hiding behind a Science paywall.


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What implications does younger volcanism have on the Moon's thermal evolution?
In the paper Evidence for basaltic volcanism on the Moon within the past 100 million years (Braden et al. 2014), suggest that features such as Ina (image below) represent a far more recent age of volcanism (relatively speaking, around 100 million years).

Image source: NASA Science News, where they have a different non-volcanic theory.  However, this question is not about whether the features are volcanic or not.
From the Braden et al. article abstract:

The morphology of the features is also consistent with small basaltic eruptions that occurred significantly after the established cessation of lunar mare basaltic volcanism. Such late-stage eruptions suggest a long decline of lunar volcanism and constrain models of the Moon’s thermal evolution.

Assuming the latest models indicate volcanism as recent as within the 100 million years ago, what are the implications for lunar thermal history of younger volcanism?

The Wikipedia article cites studies suggesting volcanic activity as recently as 2 million years ago.

@JerardPuckett thank you for that, I don't use Wikipedia usually (I don't trust it) - but the Braden et al. peer reviewed paper does say within the past 100 million years, so I have fixed my statements to reflect this, but it does not change the main point of the question.

Not an answer, but have you read about Kelvin and Perry and the heat modeling of the age of the Earth?    http://www.americanscientist.org/issues/feature/kelvin-perry-and-the-age-of-the-earth/2   The Moon, smaller than the Earth, should cool faster and should have less heat of formation, but the lack of circulation would slow the cooling down pretty significantly.  The Moon also should have similar amounts of radioactive elements to the Earth, which contribute to internal heating.  Beyond that, the heat inside a large body like a moon or planet is pretty complicated.  Above my pay grade.

A younger lunar thermal history implies that the Moon was still undergoing a contraction and cooling process, which was initiated when it began accreting material together. Evidence has shown that the moon has no tectonic activity, so all geologic features are volcanic.
The mare are a more recent example of this activity, and have been shown to have fairly young ages, as you indicated. The moon still has some internal heat from its formation that may be leaking out through convection and cooling.
Even though the moon has a relatively small mass, it didn't lose its internal heat as quickly as the earth because there was no tectonic activity to facilitate it fast enough, and convection is a slow process.
As the moon cools, it contracts, creating sharp ridges and cracks in the surface (see features known as 'scarps', or 'garden'). These contractions contribute to the "deposit" of material formed in the images that paper discussed. As the cooling surface shrinks, it breaks the crust through a thrust fault, which pushes some material higher and shoves other material lower.

What may appear to be river channels on the moon are also lava flows. These were initially underground lava channels whose roofs collapsed to expose these subterranean features. More recent incidents of molten material rising to the surface and cooling can contribute to some of the younger ages recorded in the basaltic rocks.

Tldr; Evidence of younger volcanic activity confirms existing theories for the contraction and cooling process of the moon, just that it didn't end as soon as we initially thought


text:
Would Pluto keep an orbit without its moon?
I haven't studied space that much, but I know that Pluto has very little gravity, but just enough to keep a human on its surface. Would Pluto keep a regular orbit around the sun without some extra support gravity, such as its moon?

I would think orbit mechanics (which this question really boils down to) is on topic on Space Exploration. Should this have really been moved?

Pluto's surface gravity is about 0.06g which is about a 100 times more than just enough to keep a human on it. Martian satellite Phobos has 0.0006g which is just enough. Its other satellite Deimos has half Phobos' gravity which is low enough to escape it by jumping (or rather: by bouncing off).

The orbit of a planet does not in any way depend in its moons, mass or own gravity. The orbit is the same for a grain of dust as for a giant planet. It only depends on the mass on the Sun and the distance to the Sun. The eccentricity of the orbit is then given by the initial angular momentum of the orbiting object. This might explain it a bit better.
This was a really BIG breakthrough in physics, actually the foundation of physics, about 400 years ago. Big because it is irrefutable although it is counter-intuitive. It is really difficult to believe unless one tries to accept the actual observations and logic.

Yes, a planetoid such as Pluto will be able to orbit no matter how small its mass, so long as its angular momentum is within a range determined by its distance from the sun. In the case of Pluto and its primary moon, Charon, however, things are even more interesting.
If Charon were to magically disappear (ejection would be another story), the point on the line of Pluto's orbit at any one time would be very near the center of Pluto. There are other moons, but their gravitational tug is minimal. The orbital trace of the Pluto/Charon system, however, is not even inside Pluto.
Most planets have satellites that are very much less massive than they are, and thus the center of their systemic rotation is fairly close to the primary's center. The masses of Pluto and Charon, however, are significantly closer in size.

The Pluto–Charon system is noteworthy for being one of the Solar System's few binary systems, defined as those whose barycenter lies above the primary's surface . . . [108] This and the large size of Charon relative to Pluto has led some astronomers to call it a dwarf double planet.[109] . . . This also means that the rotation period of each is equal to the time it takes the entire system to rotate around its common center of gravity.[66]

Therefore, for practical purposes, with respect to the orbit of Pluto, it's far simpler to determine the trajectory of the Pluto/Charon barycenter -- which is completely above the surface of Pluto.

An oblique view of the Pluto–Charon system showing that Pluto orbits a point outside itself. Pluto's orbit is shown in red and Charon's orbit is shown in green.

It would is a short answer.
The long answer is it depends on how you eject the mass of the moon from Pluto's sphere of influence.
A good way to think of this is if we launch a satellite into solar orbit, it can orbit the sun in the same orbit as the Earth. Now, if we launch far to much mass from the earth then we could either move closer to, or further from the sun. This is the same way rockets propel themself.


text:
What is the process for vetting the 4,175 candidate planets Kepler has discovered?
In the National Geographic article about NASA's announcement of new planets discovered by the Kepler spacecraft, it states that there are 4,175 candidate planets. What is the process for vetting these candidate planets? How do we know they exist without knowing whether they are planets? Or are they candidates because the data hints that a planet might be there but we need more detailed data or more analysis before we can confirm it?

Planet "candidates" are Kepler Objects of Interest (KOIs)that have a transit-like light curve and have passed a number of observational tests. They are candidates, because although they do show a transit in the light curve of the star in question, there is no independent confirmation of a planetary mass.
One problem to overcome is that of "false positives". There are other astrophysical phenomena besides planets that can mimic an exoplanetary transit in a Kepler light curve. For instance a grazing incidence stellar binary can produce what looks like a transit in a light curve; so too can a chance alignment of a star with a background eclipsing binary star. There are also a number of instrumental anomalies, such that a signal in one part of a CCD or the Kepler field can produce a "ghost" at another position that might look like a transit.
If you can obtain detailed spectroscopy including accurate radial velocity measurements you can usually rule out most of these false positives, largely by getting a mass constraint for the companion.
Even where you have a very clean transit signal and can estimate a planetary radius, a further problem is that a wide range of masses produce objects with very similar radii. ie. more massive brown dwarfs have very similar radii to exoplanets. Again, only a mass estimate, either through radial velocity measurements or sometimes through "transit timing variations" if the object is in a multiple exoplanet systems.
Now the problem with the KOIs is many of these stars are way too faint ($V>14$) to do the kind of detailed spectroscopy that is easily possible on the much brighter exoplanets around stars found in say the HATNET or WASP ground-based surveys of bright stars.
So, what you can do is tackle the problem statistically, by identifying the kinds of false-positives you might have, quantifying their influence, and throwing away suspect objects (see for instance Batalha 2012). Section 4 of this paper describes in detail some of the tests that are done: e.g. comparing the depths of odd and even numbered transits to look for asymmetries that would indicate grazing incidence binaries; looking between transits for a secondary eclipse that would also indicate a stellar companion; the shape of the transit is diagnostic but cannot easily rule out a planetary candidate; searching for motion of the "photocenter" of the source - if the photocenter moves during a transit it could indicate a background, diluted, eclipsing binary lightcurve.
An early paper by Morton & Johnson (2011) claimed on the basis of a population synthesis approach, that that astrophysical false positives were limited to less than 10%. However, a recent paper by Coughlin et al. (2014) discusses how instrumental effects can be tested for by comparing the transit periods with the periods of other known objects in the Kepler field of view. They claim that around 30% of the KOIs may in fact be false positives. Either way it looks like the big majority of the KOIs are indeed exoplanets, but identifying which ones aren't will require detailed follow-up.

Just want to mention that 715 Kepler candidates were confirmed by what was coined "verification by multiplicity". This was a probability study of Kepler stars with several candidates. See this paper. http://www.astronomy.com/news/2014/02/nasas-kepler-mission-announces-715-new-worlds


text:
How to calculate distance from object to point on celestial sphere
I'd like to calculate the distance (in arcsec) from the object nearest to ra='08h55m10s' dec='-7d14m42s' to that point on the celestial sphere.
I am actually having trouble identifying the object. I have python code to run a cone search:
from astropy.coordinates import SkyCoord
from astroquery.irsa import Irsa
import astropy.units as u

loc = SkyCoord('08h55m10s',' -7d14m42s','icrs')
table = Irsa.query_region(coordinates=loc,catalog="wise_allsky_4band_p3as_psd", radius= 1 * u.arcminute)
print(table)

This prints:
designation        ra     dec   ...       angle         id
                  deg     deg   ...        deg
------------------- ------- ------- ... ------------------ ---
J085507.72-071428.4 133.782  -7.241 ... 291.84348499999999   0
J085511.22-071522.2 133.797  -7.256 ... 155.64652000000001   1
J085508.65-071354.0 133.786  -7.232 ... 337.39116000000001   2
J085509.94-071534.8 133.791  -7.260 ... 180.87171799999999   3
J085511.28-071528.1 133.797  -7.258 ... 157.48249000000001   4
J085509.47-071502.9 133.789  -7.251 ...         200.548395   5
J085513.60-071440.0 133.807  -7.244 ... 87.949470000000005   6
J085510.83-071442.5 133.795  -7.245 ... 92.439341999999996   7
J085511.39-071427.5 133.797  -7.241 ...          55.110833   8
J085513.30-071420.4 133.805  -7.239 ... 66.332006000000007   9

And this is where I am stumped. Which of these objects is nearest to the point? (I assume it must be the object in the zero index). And why? And how can I calculate the distance from the object to that point on the celestial sphere?

Re I assume it must be the object in the zero index. That's a bad assumption. The object closest to your coordinates is the eight one (index=7) on your list, J085510.83-071442.5. That tool is doing a SQL query and the results are unsorted.

Could you explain how you identified the object closest to the coordinates?

see the formula in Rob Jeffries answer. It is the correct formula.

Ok this clears my misunderstanding. Just to clarify, the smallest angular distance returned by the function is $55.110833^{\circ}=198 398.999"$, is this correct?

No! Think about it. Your query asked for objects within 1 arc minute (60 arc seconds) of your given coordinates. The search found ten objects.

You can use SkyCoord.separation to compute the separation between coordinates:
c1 = SkyCoord('08h55m10s',' -7d14m42s', frame='icrs')  # your coords
c2 = SkyCoord(133.782, -7.241, unit='deg', frame='icrs')  # first object in table
sep = c1.separation(c2)

print c1
<SkyCoord (ICRS): ra=133.791666667 deg, dec=-7.245 deg>
print c2
<SkyCoord (ICRS): ra=133.782208333 deg, dec=-7.24123611111 deg>
print sep
0d00m36.3947s

The table you printed is truncated in between. If you enlarge your terminal emulator, you will be able to see a column called dist, which is the distance between your input coordinates and the corresponding object in arcsecs:
print table['dist']
       dist
------------------
36.420772999999997
44.138542000000001
51.987606999999997
         52.838003
49.937266999999999
22.372699000000001
53.621690000000001
         12.375482
          25.27176
53.707365000000003

Then the nearest object is the one whose separation is:
print table['dist'].min()
12.375482

Or equivalently, the one whose index is:
print table['dist'].argmin()
7

The separation is calculated using the equation that @Rob Jeffries pointed out.

The shortest angular distance between two points on the celestial sphere is measured along a great circle that passes through both of them.
The equation your program is using is
$$ \cos \theta = \sin \delta_1 \sin \delta_2 + \cos \delta_1 \cos \delta_2 \cos (\alpha_1 - \alpha_2), $$
where $(\alpha_1, \delta_1)$ and $(\alpha_2, \delta_2)$ are the right ascension and declination of the two points on the sky (expressed in degrees, then converted to radians).
In terms of output from your program its harder to help, especially as you have not given the full output. The first three columns should be self explanatory - columns 2 and 3 are just the coordinates expressed in degrees. I am guessing that the third column from last, that you haven't shown, gives the separation between your object and the catalogue objects and that the penultimate column is the angle that the great circle between the points makes on the sky. That looks right because the fourth object in the list has almost the same RA as your input object but  is 67 arcseconds due south (i.e. at 180 degrees).


text:
What qualifies as a local star?
My professor used the term 'local star'. What range of distance is generally considered 'local' for a star? And what would be considered distant?

This is too vague a term to be useful, because the Sun does not have any kind of sphere of influence (beyond the solar system) and nor do the nearby stars affect us with their light or gravity (Oort cloud perturbations aside).
Instead it would normally refer to the stellar population in the solar vicinity. This is a mixed bag of objects, with few shared characteristics. For example, objects within 10 light years of the Sun now, will not be so in 10 million years time.
A working definition might be those stars that are close enough to the Sun that we think the census is complete to that distance. Unfortunately, incompleteness remains in all such samples at low masses.
The RECONS programme has been working hard for the last decade to complete the census of objects that are within 10pc (33 light years).
The definition of a distant star is even more vague - all stars are distant. The Sun is 5 orders of magnitude closer than the next stars, so if anything, that is the only natural division between local and distant!

When your professor refers to your "local star" he is referring to the Sun.By a "local star" he may mean any star within the "local bubble" which has a size ~100 parsecs (300 or so ly). Though he may also just be using the term loosely with no defined meaning other than nearby.


text:
What are the cloud-like blobs in the Martian southern hemisphere?
In a very recent edition of New Scientist, an article Mystery cloud-like blobs over Mars baffle astronomers (16 Feb, 2015), detailed that

on 12 March 2012, amateur astronomers around the world noticed a strange blob rising out of the planet's southern hemisphere, soaring to 250 kilometres above the surface.

According to the article, this anomaly grew to be about 1000km across and even had 'fingers' stretching to space.
The article, details several theories (some being a bit wild) including atmospheric processes, aurora and stretching reality quite a bit - aliens (from the article, not me).
An image of the plume is shown below (a still image from this YouTube clip, which states that a 2nd plume was observed some months later:

The plumes appear to have formed in a region called Terra Cimmeria.
What are scientific theories, models etc for the formation of the cloud-like blobs in the Martian southern hemisphere?

I'm not sure how I feel about questions like these. As the article says, there doesn't seem to be any consensus within the astronomical community yet on the nature of these clouds, so any answer we might give you would just be repeating speculations...

@DavidH as these observations were made in 2012, I am seeking scientific theories/research that may be available now or sometime in the near future.

This is exactly how the War of the Worlds begins! And an audio interview abut this issue Water ice at that altitude on Mars? I don't know.

From the Nature article, the authors propose two alternative explanations:

We used photometric measurements to explore two possible scenarios and investigate their nature.

For particles reflecting solar radiation, clouds of CO2-ice or H2O-ice particles with an effective radius of 0.1 micrometres are favoured over dust.

Alternatively, the plume could arise from auroral emission, of a brightness more than 1,000 times that of the Earth’s aurora, over a region with a strong magnetic anomaly where aurorae have previously been detected

Importantly, both explanations defy our current understanding of Mars’ upper atmosphere.

I think, they are the most trustworthy interpretations that one can get.
Please note that, even if the data are from 2012, the paper is published in 2015, therefore the interpretation is not outdated.

True, the interpretations are still valid - was hoping their had been peer reviewed articles about this phenomenom


text:
How many galaxies are there in the Hercules–Corona Borealis Great Wall?
The Hercules–Corona Borealis Great Wall is "to date (February 16, 2015), it is the largest and the most massive structure known in the observable universe" (source). How many galaxies is the Hercules–Corona Borealis Great Wall believed to contain?

I'm sorry for the long answer. The result is in the bottom. :)
The number of galaxies in the "wall", as in any volume of space, is the number density $n_{\mathrm{gal}}$ times its volume $V_{\mathrm{wall}}$. However, the number density of galaxies depends on their masses/sizes: Small galaxies are much more common than large galaxies. Usually we describe this by the halo mass function (HMF), which gives the (differential) number density dN/dM of dark matter halos hosting galaxies as a function of halo mass M. Depending on your favorite cosmology, it looks something like this (created with this tool):

The solid line shows the HMF today in the local Universe, i.e. at redshift zero, while the dashed and dotted line shows how it was at redshift 1.6 and 2.1, respectively, roughly 10 billion years ago, which mark the beginning and end of the wall (according to Horváth et al. 2014, A&A, 561, L12). Notice how the number of large halos grow with time, at the expense of small halos. This is due to the smaller ones accreting onto larger ones.
The HMFs are characterized by a power law at low masses with an exponential cut-off at high masses. To get the total number density we need to integrate over the entire range of masses, and here comes the problem: Whereas the upper limit is set by the maximum size a galaxy can have (roughly $3\times10^{13}M_\odot$, but the exact number isn't very important, as these galaxies are so rare), the lower limit depends on your definition of a galaxy. Large galaxies have smaller galaxies orbiting as satellites, and there isn't a well-defined threshold for when a clump of gas, stars, and dark matter is called a galaxy. These guys report on the detection of a $\sim10^3M_\odot$ clump which they call a galaxy. That's probably not what you think of as a galaxy, so for the sake of this exercise, let's use Small Magellanic Cloud-sized galaxies as our lower minimum, i.e $\sim 6.5\times10^9 M_\odot$.
The total number density of galaxies is thus found by integrating the HMF. Luckily, the online HMF tool can do that as well. The result, i.e. the number density of halos above a given mass, looks like this:

Note that the $x$ axis is given in terms of $h^{-1}M_\odot$. Since $h \simeq 0.7$, the mass of the SMC is $M_\mathrm{SMC} \simeq 5\times10^9 h^{-1}M_\odot$. From the graph, the number density of these objects is $\sim0.5\,h^3\mathrm{Mpc}^{-3} = 0.17\,\mathrm{Mpc}^{-3}$ (1 Mpc, or megaparsec, equals 3.26 million lightyears). In principle we should subtract the number density of halos above $3\times10^{13}M_\odot$, but since these are four orders of magnitude smaller, they can be neglected.
The wall starts at redshift ~1.6 and ends at redshift ~2.1. This translates into starting at a distance of 4.65 Gpc (billion parsec) and ending at 5.47 Gpc, i.e. a depth of ~800 Mpc. These numbers are in comoving coordinates, i.e. what it has expanded to today, as opposed to physical coordinates, i.e. what it was at the time it emitted the light that we see today. But the HMF is given in terms of comoving volume, so that's cool.
To get the volume, we also need the area. The finders report that it's roughly 2 Gpc times 3 Gpc, but they don't seem to state whether this is in physical or comoving coordinates (which differ by a factor (1+z) ~ 3). I suspect it's comoving, though.
Thus, the total volume is $V_{\mathrm{wall}} = 0.8 \times 2 \times 3\,\mathrm{Gpc}^3 = 4.8\,\mathrm{Gpc}^3 = 4.8\times10^9\,\mathrm{Mpc}^3$, or 166 cubic gigalightyears. Here, I have ignored the fact that the physical area spanned by the angle of observation is a bit different at the front and the back of the wall, but since these numbers are already quite uncertain, it's of lesser importance.
The considerations above are all for the average Universe. But the wall represents an overdensity in space, so the number will be larger. I can't find any estimates of the overdensity of this particular supercluster, but in general galaxies in superclusters are more numerous by a factor of $\delta \sim 5$–$10$. Let's use 5.
That is, the total number of galaxies the size of the Small Magellanic Cloud or above is
$$N_\mathrm{gal,tot} = n_\mathrm{gal} \times V_\mathrm{wall} \times \delta$$
$$= 0.17\,\mathrm{Mpc}^{-3} \times 4.8\times10^9\,\mathrm{Mpc}^3 \times 5$$
$$\sim 4 \mathrm{\,billion\,galaxies}.$$
But if you go through the same exercise for halos of masses above, say, $10^3\,M_\odot$, you'll find a number which is a million times larger.
In reality, the number density of luminous galaxies is smaller. At the high-mass end, the gas has trouble cooling sufficiently to condense and create stars, and active galactic nuclei in massive galaxies heat the gas. At the low-mass end, gas is more easily blown out of the potential well by stellar feedback from supernovae.

Oops, forgot to account for the fact that the Wall is more dense than the average Universe. I just updated the answer including this factor.

I agree regarding your measurement of the number of galaxies in the Hercules Corona Borealis Great Wall (maximum of 4 quadrillion galaxies, on average, the size of the Small Magellanic Cloud). Seems pretty reasonable. Especially by taking into account, as you do, the dark matter that anchors the galaxies. If you take the dark matter as the scaffolding of the galaxies, the number of matter dark and light, that will be involved, can be in the order of 10 to the 20th. I call this number a Ventillion, inclusive of 4 Quadrillion of luminous matter. Question of Philosophy a la Teilhand de Chardin: i

Um... okay…


text:
What is the typical resolving power of the largest optical telescopes?
I'd like to estimate the typical resolving power of the largest optical telescopes.
I've calculated the theoretical resolving power of the VLA for 21 cm light, $$R=\frac{\lambda}{B}=\frac{2.1 \times 10^{-1} m}{2.7 \times 10^4 m}=1.6 ''$$
And I know that the largest optical telescopes on Earth are GTC, Keck 1 & 2, SALT, VLT, etc.
But I don't know how to calculate a typical resolving power in practice. How can I make this estimate taking into account astronomical seeing and other effects?

The angular resolving power of a telescope is approximately $1.22 \lambda/D \times 648000/\pi$, where this is the Airy disc diameter in arcseconds, $\lambda$ is the working wavelength in metres and $D$ the diameter of the telescope primary mirror in metres. For an 8-m telescope working at 500 nm, this gives an angular resolution of 0.05 arcseconds.
There is then a separate question about whether this theoretical limit is ever achieved because of detector limitations or degradation by blurring due to atmospheric turbulence (seeing).
In practice, at visible wavelengths, seeing always trumps the intrinsic angular resolution and has a typical value of 0.5-1 arcsecond at the world's best astronomical sites. There is no set number for this. It varies with atmospheric conditions from night-to-night or even hour-to-hour. Most observatories publish statistics on this so that you can see what the mean (or median) seeing is. For instance here are some statistics for the Paranal (VLT) site, where the median is about 0.8 arcsec.
Special techniques can mitigate the seeing degradation. You may have heard of adaptive optics - you can analyse the incoming wavefront and send a correcting signal to a tip-tilt or fully deformable mirror that removes the distortion, based either on an adjacent star or a laser guide star. However, these techniques are most effective in the near infrared and are not usually attempted at visible wavelengths. See http://en.wikipedia.org/wiki/Adaptive_optics
Another method is Lucky Imaging - which can be used by amateur astronomers too. Take many pictures with short exposures. If there is a bright point source in the image somewhere, you can use this to select the best images and to shift and stack the images to produce a cleaner final product. The shorter the exposures the better this works. Very close-to the theoretical maximum resolution can be obtained in this way - but it needs a special high speed, low noise CCD and you end up throwing away more than 90% of the images.


text:
Why does radio astronomy offer higher resolution images than optical?
According to this lecture,"Radio astronomy has several advantages over optical astronomy...We can make the highest resolution images, and see things happening on the smallest scales."
Why is this? If resolving power, $r \propto \frac{\lambda}{B}$ then shouldn't shorter wavelength light result in smaller angular resolution i.e. better resolution?

The relation you cited holds for a single telescope. But, as also noted in the lecture you linked
"One thing that is possible in radio astronomy is to use interferometry, which combines the signals from an array of antennas as if they were all part of the same aperture. That means that the resolving power of a radio telescope is not just what it would be for each antenna alone, but is as for a single telescope the size of the separation between antennas."
This basically means you use multiple telescopes - which in case of radio astronomy are called antennas for historical reasons - and combine their measurements into a single one. To get an intuition of how big of arrays we are talking about, google images of ALMA, LOFAR (which has stations in several countries) or SKA.
In principle other wavelengths, such as optical, can also be observed using interferometry, but it is technically much more complicated to build such instruments. On top of that, optical light is the one most affected by the atmosphere and it is hard enough to take it into account for one telescope, combining such effects to create an image from multiple ones is a very daunting task.

If you look at raw resolving power you get for a 10m optical telescope: $\theta_{optical}\approx \frac{5\times 10^{-7}}{10}\approx 5 \times 10^{-8}$ radian. For a synthetic aperture ground based radio telescope operating at 21cm the limiting aperture is of the order of the diameter of the Earth so: $\theta_{21cm}\approx \frac{21\times 10^{-2}}{12\times 10^{6}}\approx 1.75 \times 10^{-8}$ radian. So in this case the resolution of the radio telescope is somewhat better than that of the optical. In the absence of adaptive optics the situation is even more favorable to the radio telescope since we can achieve its theoretical resolution, but ground based optical telescopes are constrained to lower resolutions by atmospheric seeing. Also adaptive optics still do not improve the resolution of optical systems fully to the diffraction limit of the aperture.

This question may be of interest.


text:
What mechanism could have formed the Moon's Oceanus Procellarum rift-like gravity anomalies?
According to the NASA JPL web-report, Gravity Gradients Frame Oceanus Procellarum, a rift-like ring structure surrounding the Moon's Oceanus Procellarum has been detected as a gravity anomaly by the GRAIL scientific mission (as can be as the dark blue features on the image below).

Image credit: NASA/Colorado School of Mines/MIT/GSFC/Scientific Visualization Studio
What mechanism could have formed the Moon's Oceanus Procellarum rift-like gravity anomalies?

Source: JPL Article
"The gravity anomalies along the border structures are interpreted as ancient, solidified, lava-flooded rifts that are now buried beneath the surface of the dark volcanic plains, or maria, on the near side of the moon."
Because of the lava under Oceanus Procellarum, it's heavier/massive so it has a stronger gravitational pull than the area surrounding it.
Hope that helps!


text:
Are there any astronomical phenomena that could emit strong radio waves with multiples of a discrete frequency?
In the New Scientist article Is this ET? Mystery of strange radio bursts from space, it is reported that several times since 2001, astronomers have detected fast radio bursts that seem to have a frequency of multiples of 187.5.  In the article, the theory that it could be a pulsar is discounted.
Are there any astronomical phenomena that could emit strong radio waves with multiples of a discrete frequency?

The WoW! signal

lol, the what signal?

In August,1977 a researcher detected an unusual signal from the direction of Sagittarius constellation. The signal's frequency is many times stronger than background noise and it seems to be focused on the contrary the planets and stars will produce wide range of frequency. However this signal is never repeated nor explained as most sources were ruled out except of intelligence origin we may never know.

I think you've misunderstood the article - the quantity that seemed to be occurring at integer multiples of some number isn't the frequency of the radio emission but rather the dispersion measure (DM) of the source. As photons travel through the interstellar medium, interactions with free electrons mean that lower frequency photons take longer to reach the observer than higher frequency photons. The time delay of between photons with frequency $\nu_1$ and $\nu_2$ is
$$t_1-t_2=4.15\left(\frac{\text{DM}}{\text{pc cm}^{-3}}\right)\left[\left(\frac{\nu_1}{\text{GHz}}\right)^{-2}-\left(\frac{\nu_2}{\text{GHz}}\right)^{-2}\right]\;\text{ms}$$
with the dispersion measure defined as the integral of the free electron number density $n_e$ over the path between the source and the observer:
$$\text{DM}\equiv\int n_e\;d\text{l}$$
In a sense, the dispersion measure is a distance proxy. If we had some uniform $n_e$, then the DM would scale perfectly linearly with the distance to a source. That said, the Galactic electron density is not uniform in all directions; it's obviously much higher in the disk, meaning that a source inside the disk some distance from us will have a lower DM than a source in the halo the same distance away. Note that the DM isn't an intrinsic property of the source - it depends on the observer's location.
The point of the paper cited in the article (Hippke et al. 2015) is that the dispersion measures of the 11 fast radio bursts (FRBs) appeared to fall near integer multiples of 187.5 pc cm$^{-3}$. There's certainly variation, although I don't know how the deviations compare to the measurement uncertainties. The authors argue that this could be explained if FRBs are actually artificial - in other words, if there's some sort of human-made source emitting at different frequencies with each frequency emitted at a different time mimicking the effects of interstellar dispersion.
Our knowledge of FRBs has since grown as the sample size has expanded dramatically, and most known DMs vary quite a bit from the trend. I don't know if anyone has looked at this small sample again, but at this point, the extremely broad consensus is that FRBs are astrophysical (mostly extragalactic) sources. A broad range of mechanisms have been proposed over the years, with some since ruled out and others supported by further observations.
A note on pulsars: The article isn't stating the pulsars aren't FRB sources - the sample size back then was far too small to rule them out - but rather that fast radio bursts differ significantly from typical pulsar emission, which tends to be uniform. There are still viable FRB mechanisms involving pulsars, such as giant pulses of young pulsars, and neutron stars in general play a role in some currently-favored theories, such as magnetar flares.
All that said, I think it's worth answering the title question. It is indeed true that some radio sources emit at integer multiples of some frequency - namely, spectral lines from rotational transitions in molecules. In a molecule with moment of inertia $I$, the energy of the state with the angular momentum number $J$ is
$$E_{\text{rot}}=\left(\frac{\hbar^2}{2I}\right)J(J+1),\quad J=0,1,2,\cdots$$
A transition from $J$ to $J-1$ then releases an energy
$$\Delta E_{\text{rot}}=\frac{\hbar^2J}{I}$$
and leads to the emission of a photon with frequency
$$\nu=\frac{\hbar J}{2\pi I}$$
As a notable example, the $\text{CO}$ molecule$^{\dagger}$ emits a photon at $\nu\approx115\;\text{GHz}$ during the $J=1\to0$ transition, at $\nu\approx230\;\text{GHz}$ during the $J=2\to1$ transition, and so on. Keep in mind, though, that fast radio bursts are broadband sources, and while spectral radio lines will exhibit some broadening through certain mechanisms, it's simply not comparable - the emission profiles are completely different.

$^{\dagger}$Specifically, the common isotope $^{12}\text{C}^{16}\text{O}$. Other isotopes like $^{13}\text{C}^{16}\text{O}$ have corresponding lines at slightly different frequencies.

Great answer! +1 Hoping for aliens to answer mysteries like these is reminiscent of the appeal to angels  :)

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


text:
On a log-log plot of surface gravity to planet mass, what is the meaning of the y-intercept?
I am playing around with data from exoplanets.org, and am interested in the plot of surface gravity to planet mass. I reproduced this plot:

after downloading their data and performed a non-linear regression model to fit the curve. Unfortunately,the covariance matrix of my fit has infinite values, so I am now trying out a linear fit on a log-log plot, shown below:

My fit, for $y=ax+b$, is $a=0.9511$, $b=0.8631$.
I am now thinking about what I am plotting. I suspect there may not be anything interesting in $\log \frac{GM}{R^2}$ vs. $\log M$, but, regardless, I am trying to understand if there is any meaning in the y-intercept.
What are possible explanations for the value $b$?

Considering the values for your plot, I would think that the error with determining the slope and the lack of points between 0.5 and 2 Earth masses is responsible for the large deviation of an object with almost no mass having a significant surface gravity. It would be best to plot the surface gravity for objects between 0.5 and 2 Earth masses to see if there is a statistical error.

BTW, where did the data come from?

Thanks for your comments. The data are from exoplanets.org. Now that I am thinking about it, though, log(1)=0, so the y-intercept represents the surface gravity of a 1 $M_{J}$ planet, no?

You're right. I guess I should have double checked my calculation. So, the surface gravity of a planet with $log (M_O)=0$ is $log(a)=0.8631$, or $a=7.3cm/s^2$. Are you sure you have the right units? Since $9.98m/s^2=998cm/s^2$ and $log(998cm/s^2)=2.999$.

Yes this is confusing... Right now, I think that since surface gravity also depends on radius, this line is giving information about the radius. But nothing particularly useful, since $\log \frac{GM}{R^2}$ vs $\log M$ goes like $1 + \frac{k \log R}{\log M}$, after using some logarithm rules, I think. Please let me know if this does not sound correct to you, I am very new to this type of thing!

I've looked at the original data and noticed that you have only plotted gas giants (like Jupiter). It now makes more sense that a gaseous planet the mass of Earth (which would not exist due to the velocity of gases; the gases would just dissipate) would have the surface gravity about 100 times less than on Earth. Also, rocky planets increase in surface gravity as they get smaller so the value for Earth (2.998) is in the middle of the data.

Thanks for this last comment. I was trying to get rid of values that are zero, and cut out values that are less than 0.2 $M_J$, which I am now realizing is too large.

b looks like a good example of the danger of extrapolating best fit curves beyond the range of actual data. A planet of zero mass has no gravity, any other answer is illusion.

I think what you have established here is just that $\rho$ tends to increase linearly with mass for giant planets, as a consequence of their radii being held almost constant by electron degeneracy pressure.
Let density depend on mass as $\rho = \rho_0 (M/M_{earth})^{\alpha}$, so that $M = (4/3)\pi R^{3} \rho_0 (M/M_{earth})^{\alpha}$
Then
$$g = \frac{GM}{R^2} = \frac{4\pi G}{3} R \rho$$
Replace $R$ with $(3M/4\pi \rho)^{1/3}$ so that
$$ g = \frac{4\pi G}{3} \left(\frac{3M}{4\pi \rho}\right)^{1/3} \rho$$
$$ g = \left(\frac{4\pi}{3}\right)^{2/3} G M^{1/3} \rho_0^{2/3} (M/M_{earth})^{2\alpha/3}$$
$$ g = \left(\frac{4\pi}{3}\right)^{2/3} GM_{earth}^{1/3} \rho_0^{2/3} (M/M_{earth})^{(2\alpha+1)/3}$$
So, bar a (highly possible) algebraic slip, if you plot $\log g$ vs $\log M$, the gradient is $(2\alpha+1)/3$, which from your plot, gives $\alpha \simeq 0.92$ - i.e. the average planet density increases almost linearly with mass.
The intercept then is
$$ b = \log \left[ \left(\frac{4\pi}{3}\right)^{2/3} GM_{earth}^{1/3} \rho_0^{2/3}\right],$$
which yields $\rho_0 \simeq 3.5$ kg/m$^3$ (NB: I subtracted 2 from your $b$ to make it SI; giving a density of around 814 kg/m$^3$ at a Jupiter mass).
The fact that density is almost proportional to mass can be found from the same dataset. e.g. see below. Of course, the value of $\rho_0$ above is nothing like the density of Earth and that is because below 0.1 Jupiter masses, the relationship breaks down, though in actual fact very few of the densities for such planets are very accurately measured (since it requires a radius from a transit), but it works well enough in the range you have plotted. The physics here is that gas giants are governed by a partially (electron) degenerate equation of state that results in them all having a similar radius from about a tenth of a Jupiter mass to about 50 Jupiter masses (albeit with considerable and largely unexplained scatter). Thus the density is proportional to mass. This relationship does not work for small rocky planets, where the radius decreases for smaller masses. Thus your nice line in log space does not continue to lower masses (see plot at bottom - NB: some of the low-mass points have huge uncertainties).

For planets of constant mean density you have:
$$
M=\rho \times 4\pi r^3
$$
and the surface value of $g$ is:
$$
g(r)=\frac{GM}{r^2}=G \times \rho\times 4 \pi \times r
$$
So for bodies of constant density the surface gravity is proportional to the radius, and the slope as $r \to 0$ tells you the density. So for bodies of equal density $\log(g(r)) \to -\infty$ as $r \to 0$
In terms of mass:
$$
g(M)={G(4\pi \rho)^{2/3}M^{1/3}}
$$
So as $M\to 0$ we have $g(M)\to 0$ and again $\log(g(M))\to -\infty$ and the intercept of the $\log-\log$ plot as $M$ goes to zero alows you to calculate the mean density.

Shouldn't that be $g = (\tfrac{2}{3}\sqrt[3]{6\pi^2})(G\sqrt[3]{\rho^2M})$? Note that the OP's question is actually mostly about $\log M\to 0$, but otherwise, I agree, the most straightforward physical thing it corresponds to is mean density, fit to the data.

Isn't the problem with this idea that $\log g = (1/3)\log M + const$, which is not what the OP gets?

@RobJeffries The OP is far-extrapolating gas giants, so there $\rho$ is also strongly dependent on $M$--but that shouldn't change the correspondence to mean density for any particular fixed $M$, which was actually OP's question as stated. Really, though, the OP's extrapolation is physically inappropriate, and for light exoplanets ($M\lesssim 4$ in units of $R_\oplus = M_\oplus = 1$), Seager et al. (2007)  $$\renewcommand{\lg}{\log_{10}}
\lg R = k_1 + \frac{1}{3}\lg{M}-k_2M^{k_3}\text{,}$$ and so $\lg g = \frac{1}{3}\lg M - 2k_1 + 4k_2M^{k_3}$.

That $4k_2$ should be $2k_2$ in my comment (typo).


text:
Is there a difference between the terms 'elliptical' and 'elongated' for galaxies?
While studying the Solar System, I found that some galaxies are either elliptical or elongated. What's the difference?

Please check a dictionary for the definitions of these words. If it is still confusing, edit the question to add the two definitions- with references- and elaborate on what about those definitions is confusing you.

tl;dr: "Elliptical" refers to a special kind of galaxies. "Elongated" is a loose term meaning that a galaxy is stretched out in one direction.
Etymologically, "elliptical" means having a shape that can be described by a certain mathematical function — the ellipse. For a three-dimensional object, the proper term is "ellipsoidal", but still it refers to being described by a certain equation. In contrast, "elongated" is a more loose term, meaning something like "significantly deviating from a spherical shape in one direction".
In the context of galaxies, the term "elliptical" refers to a certain type of galaxies that tends to be not actively star-forming, have no or very little structure (as opposed to the beautiful spiral arms of spiral galaxies), be quite massive, as well as other characteristica. Having no ongoing star formation, the massive, short-lived stars, which have blue colors, have died long ago, leaving behind the less massive, red stars. In addition, elliptical are often quite dusty, further reddening the light. Thus, ellipticals appear red in color.
The shape of an elliptical galaxy is described by the relative size of its three axes, $a$, $b$, and $c$. If two axes, say $a$ and $b$, are roughly the same size and are larger than $c$, the galaxy is said to be "oblate", whereas if they are less than $c$, it is said to be "prolate".
As in the etymological meaning, an elongated galaxy is not a specific type, but rather refers to a galaxy departing from spherical shape in one direction. This could for instance be a prolate elliptical, but it could also by any other galaxy that has been distorted by merging with another galaxy.
Usually, though (I think), it will refer to ellipticals. The "elongation" of an elliptical is defined as $10\times(1 - b/a)$, where $a$ and $b$ is now the observed axes, i.e. of the two-dimensional projection on the sky. Ellipticals are classified according to this number as E0 (being spherical, such that $a=b$) to E7 (being very elongated). In principle, you could have even higher E-numbers, but that's not observed.

Elongated means made longer. An elongated shape or object is stretched primarily in one direction.
Elliptical means having an ellipse-like shape. An ellipse is an elongated circle, where the elongation is done in a constant/linear way.
For example, a limousine often has the form of an ordinary car, elongated (an elongated car). It is not at all ellipse-like, however.

More simply, "elongated" is a general term that can be used to describe the appearance of any galaxy, and means more or less what it means in ordinary English.
"Elliptical" is a specific technical term for galaxies, and refers to galaxies which are in simplistic terms "round and fuzzy" -- made up primarily of stars on randomized three-dimensional orbits, with little or no gas and little or no current star formation, without a dominant stellar disk. They can be "elongated", but they can also be (intrinsically, or when seen at the right orientation) circular. There's a decent introductory Wikipedia article.


text:
How to run MOOG on OS X Yosemite
By what process can a user running the latest version of OS X Yosemite execute MOOG with $MOOG in bash?
Due to the removal of 32bit libraries in OS X 10.7+, the Fortran77 code for the MOOG installer cannot be compiled on OS X 10.7+. As a result, the pip installer will not work.
I have followed the instructions in the above guide, except that:

I cannot install scisoft because the link in the page does not work
I have supermongo installed independently of MOOG, so $sm works for me
I have made my computer compile Fortran code with the true version of gcc, not Apple gcc.

Unfortunately, in addition to being unable to compile the code, the pre-compiled binaries aren't working for me right now on Yosemite 10.10.4. The error I get when I $MOOG is:
open: No such file or directory
apparent state: unit 35 named /.moog/Barklem.dat
lately writing sequential formatted external IO
Abort trap: 6

But I do have .moog in my home folder, and it has the Barklem files inside, and both the directory and files have been made readable/writable with:
$sudo chmod +rw .moog and $sudo chmod +rw Bark*

Motivation: Astronomers try to solve the problem independently
Justification: AstronomySE topics include: "Setting up, using and maintaining your astronomy related equipment"

In order to execute the compiled binaries, all libraries on which the executable depends must be in the correct location. In particular, the error specified in the question is due to the fact that my .moog folder was in ~/, my user home folder, rather than /, the root of my hard drive, as explained here.
To solve the problem:
cd
mv .moog /.moog
See this link for answer on how to get MOOG to find SM libraries for plotting.


text:
Why is $g$ tied to the oscillator strength $f$ in $\log{gf}_{\odot}$
The $\log{gf}$ value of an element in a star, where $f$ is the oscillator strength, or probability of transition in an atom, is a scientifically pursued quantity with relation to the model atmosphere of the star. Why is the oscillator strength tied to $g$? And what is the motivation for including this quantity in a stellar model?

You don't make it clear, but you may be confused about what $g$ is. It is the statistical weight of an atomic energy level.
The $gf$ value, which refers to a particular transition between two energy levels in an atom/ion, is used because there is symmetry in terms of emission/absorption processes once the statistical weight is taken into account.
$$g_1 f_{12} = - g_2 f_{21}$$
If you just quoted $f$, then you would also need to know what the appropriate statistical weight was in order to calculate transition probabilities.

Ah, ok, yes I was confused. In the context of a stellar photosphere, would the statistical weight of an energy level be governed by the Boltzmann factor $\exp(-E/kT)$? So, if you know log(gf), and you know f (from physics), then could you determine an estimate of $T_{\mathrm{eff}}$?

@user44816 The statistical weight is the number of degenerate quantum states in an energy level.


text:
Why do we observe spectral lines of specific energy?
The basis of my confusion is that atomic orbitals, even when described accurately by quantum numbers, have definite energies, meaning they represent energy eigenstates of the wave function. They're stationary states, and thus separable solutions to the Schrödinger equation. The general wave function of an electron, however, is a linear combination of such states, so that general solution does not have a definite energy. Still, we observe discrete spectral lines at specific wavelengths (with some broadening).
Why do we observe spectral lines of specific energy when the wave function of the electrons in an atom are linear combinations of energy eigenstates?
It seems like the spectral lines that are observed are just the energy eigenstates, while, according to quantum mechanics, the wave representation of electrons is a linear combination of those states.

When an interaction occurs that results in the absorption or emission of a photon, we are not dealing with stationary states. Stationary states are solutions of the time-independent Schrödinger equation, but the absorption and emission of photons is a time-dependent phenomenon.
Spectral lines in fact do not have a definite energy. They have a natural width associated with the time taken for the transition to occur.
The probability that a transition might occur resulting in the absorption or emission of a photon of a given frequency can only be tackled by considering a time-dependent Hamiltonian. What one finds is that transition probabilities become large for frequencies close to the those corresponding to energy differences between stationary states.

This is very helpful, where can I find equations showing that the "transition probabilities becomes large for frequencies close to those corresponding to energy differences between stationary states"?

@user44816 I would just point you to any textbook on quantum mechanics which deals with radiative transitions.


text:
What does Stephen Hawking mean by 'an infinite universe'?
In the recent $100m search for extra terrestrial life project project, Stephen Hawking is quoted in the following way:

"We believe that life arose spontaneously on Earth," Hawking said at Monday's news conference, "So in an infinite universe, there must be other occurrences of life."

My understanding is that scientists believe that the universe isn't infinite - there there is a finite number of stars, it's continuously expanding but has a finite size at a given time.
What's Stephen Hawking refering to here?

Due to the finite speed of light, and the finite age of the Universe, only a portion of it is observable. When people talk about "the size of the Universe", "the number of stars in the Universe", etc., they usually refer to the observable Universe, i.e. the sphere in which we are centered, and which has a radius given by the distance light has been able to travel in the 13.8 billion years since the Big Bang. Note that since the Universe is expanding, this radius is more than 13.8 billion light-years. In fact it's 46.3 billion light-years.
Observations indicate that, on large scales (i.e. above roughly half a billion light-years), the Universe is homogeneous (the same everywhere) and isotropic (the same in all directions). Assuming that this is indeed true is known as the cosmological principle. If the rest of the Universe follows this principle, then there are three possible overall "versions" of universes that we can live in. We call these versions "flat", "closed", and "open". Whereas a globally closed universe would have a finite extent, globally flat or open universes must be infinitely large.
The observable Universe is, within measuring uncertainties but too a very high precision, flat (e.g. Planck Collaboration et al. 2016). Hence, we might think that the whole Universe is, in fact, infinite. But sort of like standing in a large forest with limited visibility doesn't tell you whether the forest is just larger than you can see, or if it's infinitely large, we can't with our current theories and observations know whether the Universe is finite or infinite.

Well, no. All the observations indicate is that the chunk we see is flat and homogeneous. There is no possible observation that could positively conclude that the universe is infinite, for obvious reasons. There's always the question of what may be far beyond the current limits of observation.

Yes, you're right, of course. I do like to think that the rest of the Universe is similar to what we see, though. $\Omega_\mathrm{tot}$ is so close to unity, and in the past is has been tremendously more close to unity. But you're right, the only thing we can say with some certainty is that the Universe is much larger than what we can see.

I believe his meaning is being misinterpreted as a claim that the universe actually is infinite.  If someone says "in an infinite universe", they are not asserting that the universe is indeed infinite (as no scientist could ever claim to know), they are simply saying "if the universe is infinite."  It is merely a case from which to begin the discussion.   For example, if you were having a discussion about whether to pass a seatbelt law, you might start with "in a world where people always make the safer choice in everything they do, we would not need seatbelt laws."  That would not be a claim we live in such a world, it is merely setting a kind of baseline for going on to talk about what is the actual situation.

A relativistic universe which is expanding faster than light (like ours) is effectively infinite for all practical purposes.
Also due to its relativistic nature and faster than light expansion, even if you assume it's not infinite at some given moment, it still doesn't have any edges or borders - for you. You're in some random place in it, the maximum speed of any possible interaction is speed of light, and the universe is expanding faster than light - then anything that happens at some (real or imaginary) "edge" is outside your realm of existence. It is effectively infinite - for you.
Give it enough time and it would grow as big as you want. Give it asymptotically infinite time and it will grow asymptotically infinite.
As to what the "edge" might be, see Eternal Inflation. This is a model in modern cosmology where local bubbles like ours have stopped inflating (they still expand, but not at the tremendous rate of the initial inflationary phase); however, inflation continues forever outside the bubbles and at the edges. Therefore the bubbles keep growing indefinitely at faster than light speeds. Because speed of light is a limit for any interaction inside the bubbles, for any internal observer each bubble is effectively infinite for any practical purpose.
Be aware that there is no proof that this is actually the case, but this is a model that fits well what we now observe.

EDIT: TBH, I'm not even sure this is a question for StackExchange. It's very open-ended, and we don't really have the conclusive answers. All we can say for now is that the universe appears to be infinite for all practical purposes, but we can't know for sure. So often scientists like Hawking just simplify the language and refer to the universe as "infinite" without any of the qualifiers that would be required in a strict context.
I don't think there's any single, final answer here.

But isn't the fundamental assumption behind Hawking's statement that there is an infinite number of planets? How does "effectively infinite" address that?

It doesn't. There's a fundamental difference between "absurdly humongous beyond imagination" and "infinite". But I think most astronomer/physicists think that the Universe is in fact infinite, lest it would conflict with the cosmological principle, which we wouldn't like.

Also, +1, but note that the term "expanding faster than the speed of light" necessitates a a chosen distance scale, since expansion rate is proportional to distance. The part of the Universe that is closer to us than roughly $10^{10}$ lightyears recede at speeds lower than that of light.

@PeteBecker In a relativistic universe (like ours), there is an upper limit to the speed of any interaction. Nothing can reach arbitrarily far away, arbitrarily quickly. That being the case, there is no practical difference between an universe infinite-right-now, and an universe which is expanding faster than light over very large scales. In the Eternal Inflation model, the bubble doesn't even have an edge - the "edge" is the place/time where/when the universe is young, perpetually created, eternally inflating. An observer there would simply see a younger universe, that's the only difference.


text:
How to specify SM libraries for MOOG
I am running MOOG on OS X Yosemite and with $MOOG running the abfind driver I can force fit elemental abundances with an appropriate input line list and model atmosphere. MOOG crashes, however, when I try to use it to generate plots, even though I know I have SuperMONGO (sm) installed on my computer because $sm works fine. In particular, MOOG will crash with the following errors:
*******************************************************************************
                  MOOG IS CONTROLLED BY DRIVER abfind
*******************************************************************************

wav. correl.:  slope =  -7.791E-07  intercept =   7.505  corr. coeff. =  -0.122
Fonts file not found
File graphcap is not defined
Can't get graphcap entry for nodevice
No such device nodevice
File graphcap is not defined
Can't get graphcap entry for x11
No such device x11 -bg black -title MOOGplot -geom 700x800+650+000

How can I get MOOG to find these libraries?

Please consider this meta post with an answer (5 net upvotes) justifying a question like this : http://meta.astronomy.stackexchange.com/questions/299/are-questions-about-installing-astronomy-related-python-packages-on-topic

MOOG will search for the libraries through a .sm file. To solve this error, download an example sm file like this one, and specify where your libraries are in this file. Then use mv .sm ~/.sm to place this in your home folder, then MOOG should be able to create plots with SM.


text:
What is the accuracy of an artificial satellite at a certain magnitude?
When we say that 'accuracy of gaia satellite is 10 $\mu$as at $V = 10$ magnitude in position and annual proper motion'. What does this mean? Does it change over time?

What you are referring to is the precision with which the astrometry (position) of stars will be measured with the Gaia satellite.
The sentences you refer to in Perryman et al. (2014)  are "Gaia will achieve accuracies of some 10μas (microarcsec) in positions and annual
proper motions for bright stars (V ∼ 10), degrading to around 25μas at V = 15, and to around 0.3 mas (300μas) at V = 20 (Lindegren et al. 2008)."
What this means is that after the 5-year Gaia mission is complete, the angular position (or coordinates) of 10th magnitude stars on the sky will be determined with a relative precision of 10 millionths of an arcsecond.  This will also be the accuracy with which the annual tangential motion (the so-called proper motion) of stars can be measured. i.e. 10th magnitude stars will have an uncertainty in their proper motion of $\pm 10 \mu$as/year.

Could you give the author name (or a link) please.

http://iopscience.iop.org/0004-637X/789/2/166/

Thanks for the reply. However, I still didn't understand the sentence 'will be determined with a relative precision of 10  millionths of an arc second'. Lets say there are 2 stars A & B with equatorial coord. (a1, b1) and (a2, b2) then how does the above sentence fits in this context.

@raj ?  It means that $a_1$ is determined to somewhat better than 10 microarcseconds, as are $b_1$, $a_2$ and $b_2$. All with respect to a coordinate frame which I believe is based on distant quasars.


text:
Help plotting a trip to the Moon!
Take me to the moon!
I know this is literally rocket science but, I would like any sources or formulas that would help me plot a trip to the Moon. I'm not sending anything, nor do I ever plan to, I just want to do the next best thing by plotting the trip. So if anyone could give me any info as to how I can plot the trip, that'd be great! :)
P.S. I'm perfectly aware that the moon isn't always the same distance away, that it's like hoping onto a taxi in New York City, and how it's got a lot of work (I wouldn't mind doing it really), I'd just like to plot the trip. And please don't hate on me for asking, let's just be civil adults here. :)

This is a bit hypothetical for Astronomy, but probably more so even for physics and I'm not sure if it would fit space exploration either? Anyway, what kind of information are you looking for exactly? How to calculate the length of the trip and the future location of the Moon at that time?  How to get there quickly? Delivery of a specific payload? Using an existing ship or some hypothetical ship? Minimizing costs? Maximizing stay period?

I would like info as to/or like how NASA did it in Apollo 11 with the Saturn V

the Dover book by R.R, Bate at al: Fundamentals of Astrodynamics should do it.

I'm voting to close this question as off-topic because this is a question about space exploration http://space.stackexchange.com/

This website has a mathematical in depth analysis of the Apollo 11 translunar trajectory. It looks to be a reasonably reliable resource but I've only had a quick skim through so you should check on the sources cited.
There should be enough information there to make a computer program to calculate the orbits.