Could we carve a large radio dish in the Antarctic ice? Could we carve a large radio dish in the Antarctic ice that could replace the Arecibo Observatory? Could this work, and what would be some potential limitations?

Great question!
There are many open and active antarctic permanent bases and several antarctic astronomical observatories. There are even major large projects that involve substantial drilling and removal of ice such as the IceCube Neutrino Observatory. The biggest construction project is probably the Amundsen–Scott South Pole Station.
This is certainly a valid idea to consider and should not be dismissed out of hand by "Antarctica is hard".
There's no bedrock on which to affix your dish; it will be floating in creeping ice and subject to drifting snow.
From Amundsen–Scott South Pole Station:

The new station included a modular design, to accommodate rises in population, and an adjustable elevation to prevent it from being buried in snow. Since roughly 20 centimetres (8 in) of snow accumulates every year without ever thawing the building's designers included rounded corners and edges around the structure to help reduce snow drifts.

Basically I think @PeterErwin's comment summarizes the problem nicely. For more details see their link and also Challenges That Face South Pole Architecture which mentions the following:
WIND

Constant winds result in snow pile-up on buildings.
The new station faces into the wind, and is airfoil-shaped.
The airfoil forces air into a compressed space where it accelerates.
The fast wind scours out built-up snow.
Years later, if snow still builds up, the building can be lifted two more
stories on its columns.

ICE FLOW

The station sits atop a 2-mile deep layer of ice.
Each year, the geographic South Pole is marked.
Ice (cold water), slowly drips down to the ocean with gravity.
The trail of yearly South Pole markers shows that the ice moves 33 feet per year.

ICE CREEP

The weight of the building also causes the ice to move locally.
Ice compresses and shifts away from sources of pressure.
Resulting variable rates of sinking make keeping the building level a challenge.
Architectural elements built into the design will help meet that challenge

Basically everything is built up on stilts above the surface. This lets wind blow under things and prevents snow drifting which would quickly burry a surface structure.
The South Pole Telescope consisting of a 10 meter point-able dish is also built on an elevated platform sitting on stilts to prevent drifting (see below).
You could propose heating the entire dish to de-ice and de-snow it regularly, and pumping out the melt water, but power is at a premium at the South Pole. Fuel is flown in on airplanes during a short summer season and burned in electrical generators for power. Building something that requires electrical power to constantly melt drifting snow has serious practical hurdles.
Radio telescope dishes built into fixed surface depressions
At least two very large, single dish telescopes have been built into naturally occurring depressions in local rock: Arecibo at 18.3° N and FAST at 25.7° N latitudes. Both telescope designs have large primary focus structures (small buildings where people can work) suspended high above the ground at the primary focus of the dish, and in order to view the celestial equator and to track objects as the Earth rotates, moveable cables suspended from very tall towers were designed allow these structures to move around by about in two directions by roughly +/-19° and +/-26°.
For more on that see this answer to Is there an advantage to the equatorial region of the far side of the moon for a radio telescope or would any crater on the far side work?
Arecibo succumbed to deterioration of its cables caused by environmental factors and fell down.

Is there any role today that would justify building a large single dish radio telescope to replace Arecibo?
What will succeed the Arecibo Observatory?
How could the loss of the Arecibo Observatory have been prevented?
Building a floating, ocean-going giant radio telescope?

Maintaining these kinds of necessary tall structures in the high winds and snow storms at the South Pole would be quite a challenge as well.

Sources: nsf.gov news and South Pole Station Destination Alpha

Source: nsf.gov news

Sources left and right

Source

Sources for both: Antarctic Bases and Buildings - 3 Building at The South Pole via @PeterErwin's comment.
left: "The South Pole radio telescope, the very clear and dry air at the pole make it the best place on earth to have a space telescope.
Photo Christopher Michel from San Francisco Creative Commons Attribution 2 Generic license" right: "In the 6 months of night time of the South Polar winter with the Aurora Australis overhead, July 2008 Photo: Patrick Cullis - National Science Foundation"


Is the difference in time between the sideral and synodyc month - constant? According what I read, the sideral month isn't constant (here)1, as well as the synodic month that isn't constant (here)2. Now, as far as I understood, because the sideral month isn't constant therefore the synodic month isn't constant too.

My question is: Is the difference in time between the sideral and synodyc month (around 30 degrees) - constant, or it may be various too?

The information above was taken from here:

1) synodic month. (n.d.) The Great Soviet Encyclopedia, 3rd Edition. (1970-1979).

2) sidereal month. (n.d.) The Great Soviet Encyclopedia, 3rd Edition. (1970-1979).

The time difference (synodic month) - (sidereal month) will not be constant. The reason for the variation is due to the eccentricity of the Moon's orbit: the Moon moves faster at perigee (closest to the Earth) than when at apogee (farthest from the Earth). If the Moon is near perigee around the time of New Moon, it takes less time for the Moon to "catch up" with the Sun, so the synodic month is shorter.

Also, the Moon's perigee and apogee distance changes from month to month due to the influence of the Sun. This also affects the speed of the Moon in its orbit. (The apparent speed of the Sun also changes because of the Earth's elliptical orbit around the Sun, but the variation is smaller.)

It may be helpful to take a "geocentric" view; that is, look at the motion of the Sun and Moon on the sky. In the following figure, a rare annular eclipse occurs with the Sun and Moon occulting the star Regulus ($\alpha$ Leonis). The days listed are the average sidereal and synodic periods. After 27.32 days (one sidereal month), the Moon is again aligned with Regulus. The Sun is farther east in the sky, so it takes a few more days for the Moon to catch up with the Sun and complete a synodic month (39.53 days).

Note: The distances travelled are approximate in this figure. The apparent size of the Sun and Moon are greatly exaggerated!

The two months are related by the speed of each object:

average speed of Moon = 360 degrees/27.32 days
average speed of Sun = 360 degrees/365.26 days
distance travelled by the Moon in one synodic month = T*360/27.32
distance travelled by the Sun in one synodic month = T*360/365.26

The two distances are not equal: the Moon moves an extra 360 degrees. So the length of the average synodic month, T, can be calculated by solving this equation:
$$ T*\frac{360}{27.32}-360=T*\frac{360}{365.26}$$

It should be apparent from this equation that if the speed of the Moon varies, then the duration of the synodic month T will change as well.


What does the amino acid found in the atmosphere of Venus mean? An amino acid is a part of DNA, but it has only been found on Earth. Is the amino acid maybe from one of the space vehicles that has orbited, scanned, or surveyed Venus? How can a part of DNA survive in the dense, sulfuric acid-filled atmosphere? Scientist's haven't been able to determine what organism it's from, but it hasn't been brought to Earth yet. Could this be from the water that covered most of Venus billions of years ago? Did some type of life start to evolve before all the water evaporated? I'm kinda new to this site and have always wondered about this, so thanks to anyone who has an answer!

An amino acid is any acid that has an amine in it. They are not necessarily found in proteins. They can be found in conditions similar to deep space, for example:

In a laboratory at NASA Ames Research Center in California’s Silicon Valley, the team of astrobiologists shone ultraviolet light on deep-space-like "ices," simulating conditions that are commonplace in interstellar space. Deep-space ice is common water ice laced with simple molecules. The team subsequently discovered amino acids, molecules present in, and essential for, life on Earth.

Amino acids are really simple molecules, that could arise from a number of abiotic reactions. Do not be impressed if they are found on a barren planet, for they have been found in meteorites before:

Murchison contains common amino acids such as glycine, alanine, and glutamic acid as well as unusual ones such as isovaline and pseudoleucine. A complex mixture of alkanes was isolated as well, similar to that found in the Miller–Urey experiment. Serine and threonine, usually considered to be earthly contaminants, were conspicuously absent in the samples. A specific family of amino acids called diamino acids was identified in the Murchison meteorite as well.

I'll personally only be impressed when at least a protein or sequence of DNA/RNA is found outside Earth. Then we can speculate about life out there. I do believe it exists - I just think we are getting too excited now with stuff that may not necessarily be biotic.


Average number of exoplanets in a system I was just wondering what is the normal amount of planets is in a star system, because in my mind 8 planets seems like a relatively large amount. So are we above average? Can we even see all the exoplanets in a star system? Also, do we know if belts of asteroids are common? Is it even likely for a star to have planets?

Best answer is that we don't know. The techniques that are used for finding exoplanets are incomplete - that is they are capable of finding planets around stars in some circumstances but not in others.

For example, finding stars using the transit detection technique is more sensitive to planets that are (i) big, (ii) close to their parent star and (iii) orbiting their star with an edge-on orientation from our point of view. The doppler techniques are sensitive to (i) massive exoplanets, (ii) those that are close-in to their parent star, (iii) favour detection when the line of sight lies close to the orbital plane.

Each detection method (and each observing strategy and telescope/satellite) has their own particular selection bias, but overall the picture is that the census of exoplanets is more complete for close-in massive planets, and very incomplete for planets that are either small or orbiting with periods much longer than 10 years. Then at very large separations and very long orbital periods (think 100s of years), direct imaging surveys are again capable of spotting massive exoplanets. Microlensing observations are also capable of detecting even small planets at large distances from their star.

The result of all this is that any figure for the observed fraction of stars with exoplanets or the number of exoplanets per star is a lower limit.
The exception is where researchers try to account for the various selection effects and give a figure for the fraction of stars with certain types of planet in certain types of orbit around certain types of star.

At the moment it looks like the modal number of exoplanets per star is at least 1. In other words it seems increasingly likely that most stars have exoplanets of one sort or another.

Some facts and figures. From Kepler transit data, Petigura et al. (2013) say that 22% of Sun-like stars have "Earth-sized" planets orbiting in their habitable zone (my italics), but their plots show that the overall results for planets with orbital periods in the range 5-100 days (where Kepler is most sensitive), are that nearly 60% of stars have planets from Earth- to Jupiter- sized in this range. Cassan et al. (2012) use microlensing survey data to claim that 17% of stars have "Jupiters" at 0.5-10 au from their parent star, and that Neptune-sized or "super-Earth" sized objects orbit a further (approximately) 50% and 60% respectively. The combination of these figures also suggests multiple planets are not rare and indeed the title of their paper is "One or more bound planets per Milky Way star from microlensing observations".

The frequency of multiple planet systems is also hard to assess. All the techniques are capable of detecting certain types of multiple system, but equally all are capable of missing multiple planets even when they are there.
Kepler did detect many multiple transiting systems, so they are not rare. A study by Tremain & Dong (2012) estimated a minimum fraction of 20-30% of stars that have one transiting planet will also have others. This is on top of the 15% of Kepler planetary systems where multiple planets can be directly observed (i.e where the orbital planes are very close together).


Is there any difference among exoplanets in composition due to host star? Simply put, would, say, rocky planets orbiting red dwarfs have a noticably different chemical makeup than rocky planets orbiting B-type main sequence stars? Mutatis mutandis gas giants and the other types of host stars.

The chemical makeup of space dust -- the stuff  which agglomerates into planets -- is dependent on the history of supernovas and neutron stars and the like which created the atoms in a given region of space.  This is unlikely to bear any relationship to the size or age of the star around which said items revolve.


When is it a good aproximation to consider a star to be an ideal gas? I am currently taking a first course on stellar astrophysics, and I noticed that in some cases we use the ideal gas equation of state for stars, so we also use $\gamma =5/3$. Of course it can only be applied where there is no nuclear reaction, so there is a limit in the temperature.

Also, if the radiation pressure if relevant, you have to consider the parameter $\beta$ to calculate the total pressure and the adiabatic coefficient $\gamma$. It is also incorrect (I think) if there are degeneration or relativistic considerations.

I don't know if I have to consider other factors before I can use this equation of state and value for $\gamma$, maybe it is important if it is a convective or radiative area, or other factors I didn't consider.

My question is: can anybody tell me the limits of the ideal gas approximation? (quantitative better than qualitative but any help will be well received)

There is no hard and fast answer. To be treated as an idea gas (your title question), the particles in your gas should be point-like and they should be non-interacting if you are to use the ideal gas approximation.

This means you can take the mean separation ($\sim n^{-1/3}$) and compare that with the size of the particles - it should be much bigger. You can also compare the particle kinetic energies with the potential energy of their interactions at typical separations. eg The ratio of $kT$ to the Coulomb potential energy should be large.

The rest of your question is confused. You are asking about an adiabatic index, but that is a completely separate question from whether the gas is ideal. It depends on the equation of state, on interactions, on internal degrees of freedom, on the way energy is transported.

Relativistically degenerate gases can be ideal gases and are often assumed to be so. Do not confuse an ideal gas with a perfect gas (even though many do so).


Are there Earth rocks on Mars? Certain meteorites found on Earth have been established to come from Mars: a giant impact ejected rocks from Mars, these rocks traveled through interplanetary space, and went through the Earth's atmosphere without completely burning up.

Might it be possible to find a meteorite on Mars that comes from Earth from the same kind of mechanism?

Earth's gravity is stronger than Mars', so I guess you would need a bigger impact to send debris into interplanetary space. And the atmosphere is thicker on Earth, so the impacting body would need to be even bigger for enough of it to touch the ground. And you also need to give the rocks more energy so that they can go outwards in the Sun's gravitational well. So I'm guessing that impacts that might be violent enough are rarer for the Earth than they are for Mars.

Might it be possible for there to be Earth meteorites on Mars? And if so, is there any chance that we might stumble upon one?

Well if no one is going to answer this I will. The answer is we don't know for sure. We speculate that there should be earth rocks on mars but until we 'see' one and analyze it we will not know for sure.  The comments here all point to this answer.
Mars hit with thousands of Earth rocks possibly containing life following asteroid impacte talks about the Chicxulub impact and how it probably spread rocks to all the terrestrial planets in our solar system and the moons of all the planets.

The statement comes from Penn State University researchers who have calculated the approximate number of rocks from our planet large enough to possibly carry life that have made their way into space over the past few billion years.
Said the paper’s lead author Rachel Worth: “We find that rock capable of carrying life has likely transferred from both Earth and Mars to all of the terrestrial planets in the solar system and Jupiter. Any missions to search for life on Titan or the moons of Jupiter will have to consider whether biological material is of independent origin, or another branch in Earth’s family tree.”

The article cites its source as The BBC's Dinosaur asteroid 'sent life to Mars' which cites a 2013 paper published in Astrobiology Seeding Life on the Moons of the Outer Planets via Lithopanspermia:

The asteroid that wiped out the dinosaurs may have catapulted life to Mars and the moons of Jupiter, US researchers say.
They calculated how many Earth rocks big enough to shelter life were ejected by asteroids in the last 3.5bn years.
The Chicxulub impact was strong enough to fire chunks of debris all the way to Europa, they write in Astrobiology.
Thousands of potentially life-bearing rocks also made it to Mars, which may once have been habitable, they add.
"We find that rock capable of carrying life has likely transferred from both Earth and Mars to all of the terrestrial planets in the solar system and Jupiter," says lead author Rachel Worth, of Penn State University.


Why hasn't the Earth fallen into the sun yet? So the Earth's been orbiting the sun in a relatively stable orbit for billions of years now. during that time, it's been plowing through dust, and hit by comets and asteroids, presumably mostly from the outer reaches more than the inner parts of the solar system.

Even if the drag and increased mass should amount to a tiny, tiny fraction of Earth's mass, shouldn't that have been enough to cause our orbit to spiral inwards? Both due to slowing of the Earth's motion, and an increase in mass?

Thanks!

No not nearly enough drag.  There just isn't much dust in space, and the Earth has so much momentum that the tiny amount of interplanetary gas has a tiny effect on the Earth's orbit.

There's only about 5 particles of mostly hydrogen gas per cubic cm in the region around the Earth. It has a drag on the earth of a few thousand Newtons. But the mass of the Earth is $5\times10^{24}$ kg, so the deceleration is of the order $10^{-20}$ m/s2.

At this rate it would take longer than the age of the universe for drag to make a significant change to the orbit of the Earth.

Things were different when the planets orbited in an accretion disc. At this time, large planets could make excursions from the outer part of a stellar system to the inner, due to friction from the dust that they encounter.

The Earth does gain a little mass from dust, rocks and asteroids that continuously hit it, but it also loses a little mass to atmospheric lose. The total is slightly negative (overall Earth loses a little mass) but this doesn't affect the orbit, as orbital velocity is independent of mass.

Large bodies do change the orbit very very slightly, but comet and asteroid impacts happen from all directions, and there is little cumulative effect.

The gravity of other planets does perturb the orbit of the Earth, but not to make it spiral into the sun, but to cause the elliptical orbit to become first more, then less eccentric, and change the orientation of the ellipse.


What is the projected range of the JWST to be able to detect exoplanet atmospheres? Yesterday the K2 mission detected transit events of objects passing in front of a number of M dwarfs which could turn out to be rocky planets. If some of these planets host atmospheres, could the JWST detect IR absorption lines from molecules in these atmospheres?

What would be the expected limiting distance for these stars to allow for such measurements by the JWST?

According to this article, the TRAPPIST-1 system is supposed to be studied by JWST with the potential to detect atmospheres' composition in infrared band. So at last 40 ly (approx. 12 pc) should work. Somewhere else I read there is an assumption to study earth-like exoplanetary atmospheres at distance of 10-20 pc. But it might be better to see it from different angle: the real limit is due to the produced light energy/intensity, not the distance itself. For example, see an article on exoplanet transit spectroscopy.


Is the mass of Kuiper belt dominated by large objects? There is a problem in my example sheet trying to teach me distributions. My result is that Kuiper belt objects are dominated by large mass objects. Is this supported by observational evidence?

It depends on what you call large. There may be hundreds of thousands of icy bodies larger than 100 km (62 miles) and an estimated trillion or more comets within the Kuiper Belt. But the total mass of objects in the Kuiper belt is estimated to be less than 10% of earth's mass.  So the Kuiper belt is largly composed of small icy onjects. These objects are very hard to see with existing telescopes because the are so small and not much sunlight where they are.
References:
https://solarsystem.nasa.gov/solar-system/kuiper-belt/in-depth/
https://solarsystem.nasa.gov/news/792/10-things-to-know-about-the-kuiper-belt/
https://solarsystem.nasa.gov/solar-system/kuiper-belt/overview/


Where did the Sun get hydrogen to work with if it is in the 3rd generation of stars? As I see here, the Sun belongs to the Population I group of stars, which is the 3rd generation of the stars in our universe. 1st generation stars are Population III, 2nd generation are Population II, and 3rd generation are Population I.

When the 1st generation (Population III) of stars died, that means most of the hydrogen was burned to helium. Stars die when there is no hydrogen left. Later, the 2nd generation of stars (Population II) appeared and they fuse another portion of hydrogen into heavier elements.

If 1st and 2nd star generations burned hydrogen to helium and more heavier elements, then shouldn't like 90% of all universe hydrogen already be converted to helium and something else? If yes, then there should not be enough hydrogen to make the Sun.

UPDATE 1

Thanks for all your answers. They are very useful. Now a new subquestion appeared.
When the star dies, like our Sun, it sends out external layers and core becomes white/other dwarf. In this case, new star can be formed only from the hydrogen from the external layer. The questions what is the percentage of initial star hydrogen after burning it to helium goes from this external layer to outer space?

Most of the galaxy's gas is not incorporated into stars and remains as gas and dust. This is not really my area of expertise, but papers such as Evans et al. 2008 and Matthews et al. 2018 seem to suggest that in the Giant Molecular Clouds where most stars in the Milky Way Galaxy form, the star formation efficiency is about 3-6%. So the vast majority of the gas (94-97%) is not made into stars. In very dense environments such as globular clusters, which were formed much earlier in the Milky Way's history, the star formation efficiency get as high as approx. 30%. The canonical quoted rate for "regular" spiral galaxies like the Milky Way is about 1 solar mass of new stars are made per year, which is very low summed across the whole galaxy.

Stars also give off a fair amount of their outer, hydrogen rich outer layers during the later red giant phases when the stellar wind is stronger and the atmosphere expands a huge amount (radius of the Sun during the red giant phase will be about what the Earth's orbit is now). Also in the end state when the white dwarf is formed, it's only the core and inner layers that form the white dwarf. The typical white dwarf mass is about 0.6 times the mass of the Sun (S. Kepler et al. 2006) and so there will be a fair amount of unfused hydrogen-rich outer atmosphere left over after the star dies. For higher mass stars, even more of the mass goes into the (ejected at high speed) envelope than goes into the remaining neutron star. These high mass stars are much rarer though; most of the Milky Way's stars are faint, cool M dwarfs.


Could asteroid 'Oumuamua actually be round? News articles have mentioned that the recently discovered asteroid 'Oumuamua is theorized to be greatly elongated based on observations of brightness variation. However, this hypothesis seems to be based on the assumption that the object's albedo distribution is reasonably uniform.
What if it isn't? Could large periodic brightness variations also be explained by 'Oomuamua having two highly contrasting hemispheres like Iapetus? If the astronomers involved have already eliminated this competing explanation, how would they have done so and with what additional observational data?

The hypothesis that the object has an extreme aspect ratio comes from both the max/min ratio of the light curve and the shape of the light curve, along with some assumptions of the uniformity of the albedo as you mentioned.

If you read the BBC article Bizarre shape of interstellar asteroid and click from that page on the link that says which is published in Nature journal. it will take you to an open access draft of the Nature Letter. In it you can see the shape of the light curve. I've chopped up some sections to show below.

The "dip" is quite sharp. Remember magnitudes are a log scale, and so the dip would be even sharper on a linear scale. Since 5 magnitudes are a factor of ~100, a 2.5 magnitude dip is a factor of ~10.

This is what you would get when a high-aspect ratio objects so that is viewed end-on. A black spot would darken the curve much more gradually, being viewed for a longer fraction of the cycle.

This is not proof of the high-aspect ratio hypothesis, and the authors don't claim this. However, it's the best way to fit the curve with the fewest assumptions, as far as I can tell.

Screen shot of the on-line Accelerated Preview of the Nature Letter A brief visit from a red and extremely elongated interstellar asteroid K. Meech et al. doi:10.1038/nature25020


What was the absolute limit to the possible sizes of the first stars formed from "primordial material with no metals"? This answer to Why don't or (can't) stars be more than 325 or so times the mass of the sun? What limits their size? includes the following:

...The upper limit you refer to is for compositions similar to the Sun. For stars born in the distant past that were metal-poor, or even born from primordial material with no metals, the upper limit could be much higher.

Question: What was the absolute limit to the possible sizes of the first stars formed from "primordial material with no metals"? I'm primarily asking about the size that was possible to exist as a star. There may be other limits associated with the possibility of formation in the first place due to matter distribution and that would be interesting to know as well.
NOTE: There is also the well-received and as yet unanswered question Formation of the First Stars If this topic is better addressed as an answer there instead, then please do so and we can close this as a duplicate.

First off, let us try to clarify a few terms:

As usual in astrophysics, metal-free star means atomic number $Z \leq 3$, i.e. it only consists of the primordial elements hydrogen, helium, and lithium.
Primordial star literally means original star and refers to the first star(s) (generation) formed after the big bang. It is IMHO equivalent to metal-free, and a primordial metal-free star would be a pleonasm.
Star can be most generally be defined as massive, self-luminious gas sphere (see Lexikon der Astronomie, p. 412), which leads to the question what each of these four terms mean exactly. Massive would probably mean at least about  $0.09 M_\odot$ as 0.09 solar masses is the weight of the smallest star observed, AB Doradus, which is undergoing nuclear fusion.
A quasi-star is also worth a definition for the current question. The first paragraph of that Wiki-page summarizes it pretty neatly:

A quasi-star (also called black hole star) is a hypothetical type of extremely massive and luminous star that may have existed early in the history of the Universe. Unlike modern stars, which are powered by nuclear fusion in their cores, a quasi-star's energy would come from material falling into a black hole at its core.

ProfRob already pointed out that a key issue is the definition of star.

Or is something that starts nuclear fusion, but is always collapsing, a star?

Related is the question about the stability of a star, which is studied by Isabelle Baraffe et al. in arXiv:astro-ph/0009410 :

The stability of metal-free very massive stars ($ Z= 0; M = 120 \ldots 500 M_\odot$) is analyzed and compared with metal-enriched stars. Such zero-metal stars are unstable to nuclear-powered radial pulsations on the main sequence, but the growth time scale for these instabilities is much longer than for their metal-rich counterparts.

The metal-free stars analyzed in that manuscript would still be smaller than the quasi-stars, which require at least $1000 M_\odot$. Again a Wikipedia-quote:

Quasi stars would have had a short maximum lifespan, approximately 7 million years, during which the core black hole would have grown to about $10^3 \ldots 10^5 M_\odot$ for modern stars.

To summarize: The maximal size limit varies, depending on what we exactly regard as primordial metal-free star, and also how long such an object has to exist/ be stable. I would assume that the upper limit of primordial metal-free stars is probably larger than the upper limit of stability of $80 \ldots 100 M_\odot$ which holds for modern stars (see p. 458 in Lexikon der Astronomie).
References

Helmut Zimmermann, Alfred Weigert: Lexikon der Astronomie. Edition 8. Heidelberg/ Berlin 1999. ISBN 3-8274-0575-0. (in German)


What resources do C-Type(Carbonaceous) Asteroids contain? What resources do C-Type asteroids contain? The research I've done has come up with carbon, but what form is the carbon in? Graphite, diamonds, coal, or some other material?

Carbonaceous asteroids contain carbon in the form of chondrites (small marble-sized nodules),tiny crystals and other sorts of carbon dust. This is mixed in a sort of conglomerate with dust and nodules if silicate material, sometimes even nickel-iron nodules. What resources do they contain? They don't contain diamonds, because diamonds require enormous heat and pressure to form, so sometimes micro-diamonds are found in the craters of large meteorite impacts on Earth. Some of them probably contain valuable materials, but they would have to be very valuable indeed to be worth the trouble and expense of retrieving them from the asteroid belt. Prospectors there wouldn't restrict their attentions solely to carbonaceous types, and would probably explore large asteroids like Ceres. Asteroids and meteorites couldn't possibly contain coal, which is a biological product formed from thick layers of plant debris transformed by heat and pressure over millions of years in the bowels of the Earth.


Why does the background noise in this image of 2020 QG look like corduroy? The news item Small asteroid becomes closest ever seen passing Earth: NASA contains a "handout image" with the caption:

This NASA/JPL/ZTF/Caltech Optical Observatories handout image obtained on August 18, 2020 shows asteroid 2020 QG (the circled streak in the center), which came closer to Earth than any other nonimpacting asteroid on record NASA/JPL-CALTECH/AFP

Question: I saw what looks like a corduroy-like pattern in the background that is likely from some processing artifact. Does anybody recognize this kind of artifact? Does it look familiar?

above: NASA image from linked article. below: Corduroy source.

Fourier analysis (log power) of the image of the asteroid, the two spots demonstrating the strong linear periodic modulation of the background.
Concentric rings were seen when analyzing only the red part of the initial annotated image. Now that I average all three colors they are almost gone. The inclined line in the FT is probably that of the bright, straight linear trail of the asteroid.

import numpy as np
import matplotlib.pyplot as plt

fname = '424852de3808b4d6df1101d3b4f0790afd3b9d13.png'

img = plt.imread(fname)[..., :3].sum(axis=2) # sum 3 colors makes red circle gray to minimize impact on FFT

print(img.shape)

f = np.fft.fftshift(np.fft.fft2(img))
p = np.abs(f)**2

plt.figure()
c0, c1 = [int(s*2**-1) for s in p.shape]
hw0, hw1 = [int(s*2**-3) for s in p.shape]
plt.imshow(np.log10(p[c0-hw0:c0+hw0, c1-hw1:c1+hw1]), vmin=5, vmax=8)
plt.gca().set_aspect(16/9)
plt.colorbar()
plt.title('log10(ft power)')
plt.show()

I agree that it’s noise in a fixed pattern, but I think it’s unlikely to be related to ADC sensitivity.  Typically if you have multiple ADCs, they read out blocks of the sensor (e.g. one on each corner to read out a quadrant). And sensitivity differences across those amplifiers usually is removed pretty well by flat-fielding.
In my experience, you get things like this from time-variable electronic noise, e.g. 60 Hz noise from surrounding equipment.  (Probably the electronics are well-shielded from 60 Hz because it’s so ubiquitous, but there could be noise at other frequencies.)
Because CCD pixels are read out sequentially (e.g. the classic bucket brigade analogy), then any time-variable noise source that slightly changes a pixel value will manifest itself as a spatially varying pattern, where the exact pattern depends on the relationship of the readout frequency to the noise frequency.  You see the diagonal stripes because the noise signal wasn’t quite at the same phase when pixel 1 of row 2 was read out as it was when pixel 1 of row 1 was read out, and so on.
(Extra points for including an actual image of corduroy in your question, BTW.  )


Do celestial bodies have many asteroids orbiting them? Are there statistics on the expected number of asteroids (of various sizes) to orbit planets and moons in our (or other) solar systems?

Do we know of other natural bodies orbiting the Earth (besides the Moon) or asteroids orbiting the Moon?

If not, why not? Though the likelihood of natural captures are low, it is my understanding that there are lots of asteroids (though, of course, all orbits will eventually decay).

The short answer is not very often.   Article on that here:  can moons have moons?

Nothing can orbit the Moon long term cause the Earth affects a moon/moon-satellite orbit and it draws tidal energy from it so an object would crash into the moon.

A 2nd moon around earth might be more stable than one around the moon, but I suspect, not long term stable.   If it fell into orbital resonance with our moon, like the inner 3 moons of Jupiter, I don't know, maybe, but I suspect not, cause the Moon's tidal force on a 2nd moon would (I suspect) be too great.

From time to time, the Earth captures an asteroid, but they usually don't stay for long:  http://www.scientificamerican.com/podcast/episode/earth-usually-has-second-tiny-tempo-11-12-27/

Mars' 2 moons and many of the gas giant planet's moons are captured asteroids or Kupier belt objects and if the earth didn't have our large moon, we might have a few captured asteroids in our orbit as well, so planets absolutely can capture asteroids, but the closer to the sun (Mercury/Venus), the smaller the planet's gravitational sphere's of influence and the less stable long term.

Curiously, some asteroids even have asteroids, and Pluto in particular is a rather strange system with a comparatively enormous and very close moon, Charon and 4 outer moons that orbit around Pluto and Charon.   you can't get objects orbiting the Earth-Moon system, cause that would extend past the stable part of the Earth's Hill Sphere.   Pluto's Hill Sphere is 'about 3 times the radius of Earth's because it's so far away from any large bodies.

Generally, stable orbits only exist in the inner 1/2 to 1/3rd of a body's hill sphere, so those are the 2 factors to look at, tidal effects and hill sphere (or Sphere of influence, which is related to Hill Sphere).

Sphere or influence numbers for the 8 planets and Pluto:

https://en.wikipedia.org/wiki/Sphere_of_influence_%28astrodynamics%29

Related:  What is the difference between Sphere of Influence and Hill sphere?


How to determine the mass of a runaway star? I would like to know how do astronomers estimate the mass of a star that has been ejected from its galaxy.  What kind of accuracy can be achieved?

OK, I'll assume you are talking about this star or some other high-mass "runaway star".

Most massive stars are born in rich clusters and are often part of multiple system. Is not unusual for such stars to get ejected from these regions are high velocities, either as a result of dynamical interactions in multiple systems or when they are companions to stars that explode as supernovae.

The steps to estimate the mass are: (i) Obtain very detailed optical and UV spectroscopy. From this you estimate the spectral type of the star, it's effective temperature and its (current) mass-loss rate. (ii) You combine this with a known distance (the star in the link was in the Large Magellanic Cloud, so the distance was known fairly accurately) to estimate a luminosity. (iii) You compare the luminosity, temperature and mass-loss rate with the predictions of a specialised high-mass stellar evolutionary model, which includes mass-loss and rotation. The comparison yields an estimate of the mass and age of the star.

The details are found in Evans et al. (2010). It is probably fair to say that the estimate of $90\ M_{\odot}$, is uncertain by a few tens of solar masses.


How to polar align an equatorial mounted telescope when you are close to the equator? I recently moved to southern Mexico (16 degrees N) and I wonder what would be the best way to align my EQ telescope when the polar star is not visible?

Mount a camera on the tripod and take a long exposure. You should be able to determine the concentric circles of the stars paths. The center of the circle is the North pole.


Why does the darkening of the Moon go from left-to-right during an eclipse? In anticipation of the upcoming April 2014 lunar eclipse, I decided to use the 3D virtual planetarium software Stellarium to simulate what it is supposed look like where I live, since I've never actually witnessed an actual eclipse before. I set the clock to begin at the moonrise before the beginning of the event and to tick at a rate of a few virtual-minutes per real-second, pressed play, and waited for the eclipse to begin. When it did, I saw something I did not expect. To be honest, I realized my expectations were embarrassingly naïve.

I watched the Moon rise in the east and begin its upward left-to-right path across the night sky, towards its zenith in the south. My expectation was that the moment the Moon enters Earth's shadow which marks the beginning of the eclipse would be visually apparent by the darkening of the full moon's outer limb on the right side, and the darkening would progress from right to left from there. When I saw that the darkening actually progresses in the opposite direction, I immediately realized that my understanding of the geometry and kinematics of the Earth-Moon-Sun system must be much shallower than I would have thought.

My main question is must lunar eclipses happen this way every time, and if so why? My gut is now telling me that it does, so I'm mostly interested in an answer of reasonable depth to the 'why' part of the question.

Additionally, it occurs to me that the observation of left-to-right progression of darkening of the Moon's face during a lunar eclipse is especially counter-intuitive from the point of views of ancient geocentric cosmologies based on a stationary Earth that doesn't rotate. Can this observation be taken as empirical evidence that the Earth rotates? (I could be completely off base here. If so, please correct me.)

On any given day, you see the moon rise in the east and (apparently) travel across the sky from your left to the right; therefore, you would assume (incorrectly) that the leading edge of the moon's movement must be to the right.

That is not correct.

The apparent motion you are seeing is predominantly from the Earth's ~24 hour rotation. But the orbit of the moon is ACTUALLY from west to east as it revolves around the Earth every 29.5 days. Look a the diagram below.

If you were looking down at the earth-moon-sun system far above the ecliptic over the north pole, you would see the Earth spinning in a counter-clockwise direction… and the moon is also revolving around the Earth in a counter-clockwise direction.

So forget about the Earth's rotation for a moment; it has nothing to do with the moon passing into the Earth's shadow (i.e. the eclipse.) Look at the diagram below, and place yourself on that little ball labeled "Earth" and look into the night sky. Which side of the moon is covered first as it passes behind the shadow cast by the Earth? What do you see?

Left to right.


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.


How long would it take for a black hole the mass of our sun to swallow our solar system? If a hypothetical black hole with the mass of our sun would appear in place of the sun.
How long would it take before our entire solar system (everything inside our heliosphere) to be swallowed by this black hole?

Is a common missconception to think black holes suck everything around them by definition as if the fact they are black holes gave them some kind of different gravitational influence than the one excerted by any other object with a similar mass.

The main difference between the Sun and a solar-mass black hole its the density. The Sun would contain a solar mass in a sphere with 1.3 million km in diameter while the Schwarzschild diameter of a solar-mass black hole would be below 6 km. But the mass is still the same. Take a look to Newton's Universal Law of Gravity:

$F = -G\frac{m_1m_2}{r^2}$

As you can see the gravitational force depends on the masses of the two bodies interacting, $m_1$ and $m_2$ (for example, the Sun and the Earth), but does not depend on the density of the bodies in any way, thus the interaction has the exact same behaviour in both the Sun and the solar-mass black hole scenarios. The planets would continue to circle around the barycenter of the Solar System as if the Sun was there.

Obviously, besides the gravitational influence, a lot of horrible consequences would take place. The absence of the solar light flux would freeze the entire system pretty fast. The first few days of darkness would be fine for Earth (even if air currents, migration patterns and circadian rythms would be totally disrupted). In a matter of months the Earth's surface would start to be sterilized for complex organisms unless moved to geothermal facilities. By the first year life on Earth would have been eradicated as the world oceans become a thick ice crust. After 10 years or so even Earth's atmosphere would start to collapse as air turns liquid and freeze upon arrival to the ground. With some luck the ice shell would still insulate the water below and microbial life could still be there, thriving with the geothermal energy of Earth's interior. The best option for us would be to transfer to Venus since atmospheric greenhouse gases would probably allow for larger thermal inertia and thus the surface would freeze slower than on Earth. In the end, after a few decades, even Venus would be sterilized. The entire Solar System would be a dark cold planetary ghost.

There is a mechanism to allow for the black hole to suck the planets eventually and that is gravitational radiation. We know that orbiting bodies emmit gravitational waves. These waves carry angular momentum away from the system shrinking the sizes of the orbits. We in fact have detected strong gravitational waves coming from inspiraling black holes. Therefore, with enought time, gravitational waves radiated by the planets would put them so close to the black hole as to be "swallowed", but this makes no difference with the Sun scenario we live in, orbits are decaying right now by the same process. But we need to have some orders of magnitude in mind, this process is crazily insignificant, more for small masses like the ones of planets, and even more for the huge orbital periods our planets have.

How long you would have to wait for this to happen? In the case of Earth, it would be swallowed in $10^{13}$ times the history of the universe. Yep, those are not 13 zeros of years but 13 zeros of times the universe age. Mercury, is not swallowed much earlier, in fact only half that time (which is still trillions of times the age of the universe). And for the entire Solar System to be swallowed (considering Neptune the last object of it), you would have to wait 145.000 times that. Not to mention the exponentially increasing waits for Kuiper belt objects and Oort cloud comets.

With these crazy spans of time we also have to think about other strange effects. Hawking radiation would evaporate the solar-mass black hole slowly, meaning that the central mass of the system would start to decrease. This means that planetary orbits would start to get larger (since a planet orbiting a less massive body with the same orbital speed would climb the gravitational dwell by some amount). The solar black hole would evaporate in $10^{58}$ times the age of the universe. Until the planets get free from the influence of an ancient mass that no longer exists in the Solar System. This is a way larger characteristic time period than the one required for orbital decay by gravitational radiation. So the gravitational decaying of the orbits would be the predominant effect even if the wait would be a bit larger (since the mass of the black hole decreases slowly).

So yeah, you can wait $10^{29}$ years until all the planets get swallowed by your black hole. The Sun would swallow them in the same way only that they would touch the Sun's surface even before than the event horizon of your black hole.

Not only that, the Earth is probably going to be swallowed by the Sun way before than in the solar-mass black hole scenario. We know because of stellar evolution that the Sun is going to get larger to the point of reaching our current position in just 6 billion years at max (even if there is some discussion about the solar mass loss in this era enlarging Earth's orbit and making it "safe").

If you want to keep your solar-mass black hole scenario think about those crazy trillions of times the age of the universe wait. Many things are going to happen way before that would disrupt the entire Solar System. It is expected that the orbital perturbations between the planets will eject several planets in a matter of a few billion years. That is a blink of an eye in the wait for them to decay. At these extreme timescales even close flybys of other stars will certainly dirsupt the system if planets don't do it by themselves in the first place.

So my answer is that in real life, turning the Sun into a black hole, would never end in the Solar System been swallowed by it, more like everything disspersed in interstellar space in some hundred of billion years at most.


Is lithium considered a metal in astronomy? My lecture notes are conflicting on this issue so I seek definitive clarification. In Astronomy, are metals either
A: the elements which form within stars (i.e., not Big Bang nucleosynthesis, where most of the lithium (Li) was formed as far as I know)
or
B:  everything heavier than helium (He)?

The latter.
To astronomers, a metal is any element that is not hydrogen or helium, because these elements together constitute most of the elements in the Universe, by far.
This means that, in many circumstances all other elements can be neglected, at least to first order.
By mass, H and He account for some 74% and 24% in the present-day Universe, respectively, while the next-most abundant elements are on the <1% scale. And because these elements are also heavier, if you consider the abundance by number, H and He account for roughly 99.9% of all atoms.
Lithium, on the other hand, only account for a few billionths of all atoms. Here's an overview of the abundances in the Solar System, in terms of number densities, i.e. not mass densities:

Credit: Wikimedia Commons user 28bytes, under C.C.-by-S.A.-3.0.
The reason for the zig-zag pattern is that, in general, elements with an even number of protons are more stable than those with an odd number, as discussed in this post.
Note that the $y$ axis is logarithmic, i.e. each tick mark denotes a ten-fold increase in abundance. The axis is arbitrarily scaled to $10^6$ silicon (Si) atoms, so the Si dot is at $\log A \equiv 6$. Oxygen, for instance, is at $y \simeq 7.15$, so it is $10^{7.15-6}\simeq14$ times more abundant than Si.
Are we wrong?
The terminology used by astronomers may be considered wrong by other physicists and chemists. But why do we use the term at all? To distinguish some of the elements from other elements, based on their properties, I'd say. In astronomy, it very often makes sense to consider the elements of the Universe as divided in three:

hydrogen, which is most of the mass and most of the atoms, and which provides most of the energy in stars,
helium, which adds significantly to the mass, lowering the mass fraction of hydrogen, and
all other elements, which together help cooling the gas (so it can condense), form dust grains (altering observed spectra), have a very roughly equal amount of protons and neutron (so $A\simeq2Z$), are formed in other processes than the Big Bang (except for a small part of the lithium), etc.

In physics, you may be more concerned with the material's ability to conduct electricity and hence consider sodium a metal until the pressure becomes too high, after which you call it a non-metal.
In chemistry, you may be more interested in the materials ability to form bonds, and hence do not consider arsenic and antimony metals.
So, I think you can say that there isn't a formal definition of the term "metal" — it depends on the context. From Wikipedia:

Around 95 of the 118 elements in the periodic table are metals (or are
likely to be such). The number is inexact as the boundaries between
metals, nonmetals, and metalloids fluctuate slightly due to a lack of
universally accepted definitions of the categories involved

Apart from these practical reasons, there are also historical reasons, as discussed in this answer by Rob Jeffries.


How early could we detect an asteroid the size of the one that caused the extinction of the dinosaurs? Would we know months, years ahead? How sophisticated is our technology's ability to detect it at 100% accuracy?

It could be as little notice as a few days.
There is a large variation in the amount of warning we would have before an asteroid strike similar in energy to Chicxulub.  Let’s examine the case in which we would have the least warning.
The lower bound to the estimated energy released by the Chicxulub collision is 1.3e24 joules https://en.wikipedia.org/wiki/Chicxulub_crater#Impact_specifics.
We know the object ‘Oumoamoa was moving at about 50km/s at 1 au from the sun https://en.wikipedia.org/wiki/%CA%BBOumuamua , which is the distance of the earth from the sun.  The earth itself is moving at about 30 km/s with respect to the sun, so if they struck in opposition,  the difference in velocity could be 80 km/s.
For kinetic energy impact, we know $e=1/2*mv^2$, or energy = 0.5*mass times the square of the velocity.  So $m=2e/v^2$, so m is approximately 4e14kg.  If we assume an asteroid a bit denser than iron at 1e4 kg/m^3, then the volume of the asteroid is 4e10m^3.  If we assume the asteroid is a sphere, then since the volume of a sphere is $V=4/3\pi r^3$, then the radius is $r^3 = (3/4*V/\pi)$ or $r$= 2.121km.
Let’s assume the asteroid is matte black so it hardy reflects any light, give it an albedo of 0.02.  Then the absolute magnitude of the asteroid of this size and albedo at 1 AU from an observer is 17, using the tables here https://cneos.jpl.nasa.gov/tools/ast_size_est.html.  Let’s suppose that we spot the asteroid when it is much much brighter than at one AU (or 1.5e8 km), with the same apparent magnitude as Pluto (magnitude 14).  Since magnitude is a log scale, that is $2.5^3 = 16$ times as bright.
Brightness varies as square of distance, so the asteroid will be at $1.5e8/4$, or 3.75e7 km away from earth when it’s spotted.  At 80 km/s relative velocity, the time from seeing it to impact is 4.6875e5 seconds, or about 5.4 days.
Notes:

The above outlined scenario is almost impossibly unlikely. A much more likely scenario is that an asteroid from our known asteroid belt strikes earth, in which case we would probably know it was going to collide years before the strike.

The object Oumoamoa was classified as an asteroid rather than a comet due to its lack of corona, but now it has an ‘I’ designation for the first Interstellar object which is neither a comet nor an asteroid.  One could argue that I shouldn’t use it as an example given that the question asks about asteroids, but I think it is perfectly reasonable.

We don’t ever know a collision with 100% accuracy.  There is always error in our orbit estimations.  We can’t even tell if satellites are going to collide in earth orbit, it’s just probabilities.

I don't have a good idea of how long a new object of a certain apparent magnitude will be in the night sky before it's discovered by someone.  If someone thinks we would notice this object sooner due to transits or some other reason, I may amend this answer.


What are the Similarities and Differences between ALMA and FAST? What are the Similarities and Differences between ALMA (Atacama Large Millimeter Array) and FAST (Five hundred meter aperture spherical telescope)?

What can they see, what's their resolution like? How do they compare?

This is quite a broad question, since there are far more differences than similarities between the two. This answer will focus on the first part of the question and be about the technical aspects of the two telescopes.

Basics: FAST is a "classic" filled aperture telescope. This means it consists of a single large dish that focuses incoming radio waves on a single receiver (this is simplified). In contrast, ALMA is an interferometer: Many (> 60) dishes each focus incoming radiation on their own receiver. The relative positions of the individual telescopes is known to great precsion, and both amplitude and phase of the incoming radiation measured, which allows astronomers to combine several or all of the telescopes for a single observation. The maximum achievable resolution is then not determined by the diameter of a telescope but by the distance between those furthest apart. (Again, this is greatly simplified; interferometry is complicated.) In principle, FAST could also become part of a huge interferometer called VLBA.
Wavelength: FAST is a "classic" radio telescope, observing in the centimeter to meter wavelength range. ALMA is a sub-mm telescope operating at smaller wavelengths of 0.3 mm to about 1 cm.
Resolution: since the resolution is $
\propto \lambda / d$, with $\lambda$ the wavelength and $d$ the diameter of the telescope, ALMA has a much higher resolution - smaller $\lambda$ and greater $d$. ALMAs resolution is given as 10 milliarcseconds, while in the ideal case FASTs is $1.22 \cdot 0.1 / 500 \cdot( 180 / \pi \cdot 3600) = 50$ arcseconds at 10 cm wavelength. For comparison, according to wikipedia the unaided human eye has a resolution of about 60 arcseconds.
Sensitivity:

Atmospheric transmission: FAST observes in a wavelength regime in which our atmosphere is quite transparent. In contrast, the ALMA dishes are located in one of the driest deserts of the world at an altitude of over 5000 m, in order to minimize the effects of water vapour in the atmosphere.
Collecting area: FAST has a much larger collecting area. Also (I'm uncertain in this point, please correct this if I'm wrong) you lose some sensitivity when doing interferometry compared to just pooling the collecting areas of the single telescopes in ALMA.

Point Spread Function: The PSF of a single telescope, even a radio telescope, is much simpler than that of an interferometer. So the data analysis required to turn observations into an image should be much more straightforward for FAST.

There are some differences in the required surface accuracy (imperfections significantly smaller in scale than the observed wavelength do not matter) and surface composition - the ALMA dishes are solid metal, whereas the FAST dish is perforated. This works because as long as the holes are smaller than the wavelength, the incoming radio waves see it as solid.

The receivers used to convert incoming photons to electricity for both telescopes work on the same principles. Practically, it is much more difficult for ALMA due to the high frequencies at the edge of the far-infrared spectrum. ALMAs shortest wavelengths are just a little too long for receivers used in infrared or optical telescopes to work, and almost too short for normal radio receivers to work.


How much solid matter is in the observable Universe? I want to know the volumetric ratio of empty space to planets, stars, and all other solid matter combined. Not considering any dark matter, dark energy, or black holes.
For example, you can say X percent of the observable space is just empty space. The remainings are comets, stars, planets, and so on.
Edit:

I meant solid as in all the things that are not gas.
I know that space is not exactly empty, let's consider that empty means all other things except solid.
I'm not looking for exact numbers but just an estimation.

Any non-star is negligible, their volume is practically zero.
Most matter in the Universe is gravitationally bound plasma, resulting that they do not have a well-specifiable volume, because their density continuously changes from the interstellar vacuum to many times of lead. But choosing a surface at any arbitrary point (for example, at 1 bar pressure) does not modify the result significantly (this is why we can say a radius for the Sun).
This source says that the estimated count of the stars in the visible Universe is $10^{21}$. This is obviously not very exact, there are probably better estimations which you substitute into this answer.
The majority of the stars in the Universe is a red dwarf. Their radius varies between 9% and 50% of the Sun, we get their geometric average, this is $\approx$ 21% of the Sun. Their volume is $\frac{4 \pi}{3}(0.21 \cdot 700000km)^3 \approx 1.37 \cdot 10^{16} km^3$.
Thus, the volume of all stars in the Universe is $\approx 1.37 \cdot 10^{37} km^3$.
The observable Universe is a sphere with the radius of 46billion light years.1 A light year is $\approx 9.5 \cdot 10^{12} km$. Thus, the volume of the observable Universe is $\frac{4 \pi}{3} (23 \cdot 10^{12} \cdot 9.5 \cdot 10^{12})^3 \approx 4.37 \cdot 10^{79} km^3$.
After a division, we get that
$$\frac{1.37 \cdot 10^{37}}{4.37 \cdot 10^{79}} \approx \underline{\underline{\frac{1}{3.2 \cdot 10^{42}}}}$$
part of the volume of the visible Universe is not empty space.
About 1 digit difference from the reality is possible due to the lack of not enough exact approximations.
1This is not the well-known and false 13.7 billion light years, which is the way what light could do since the Big Bang. The difference is coming from that meanwhile the Universe expanded, thus the stars we can see (as if they would be 13.7 billion ly far away) are now 46billion ly away. More details are on the wiki link.


Astrophotography - Unable to achieve focus with current setup I am new to astrophotography and am looking for guidance. I have the following equipment:

Orion StarBlast 9814 4.5" Altazimuth Reflector Tabletop telescope
1.25in Telescope Adapter Extension Tube T Ring for Canon DSLR SLR Camera DC618
17mm and 6mm lenses

When I use the lenses in the Telescope Adapter Extension, focus is not achievable. My theory is that the focal plane of most DSLR are 55mm back focus plane. Reflectors are positioned farther back than where the human eye would normally be placed, and hence it is usually difficult if not impossible to bring the camera into focus.

Is this true? Also, would a Barlow lens help achieve focus or be a waste of money? Any other suggestions?

Yes, the vast majority of small telescopes made for visual observations (like most dobsonians and other newtonian-based architectures) have very limited back focus. In other words, with the focuser retracted all the way in, the focal plane of the primary mirror is sticking out only a short distance. A large back focus is not typically seen, nor is it desirable, in a visual scope. (However, most astrographs have a large back focus.)

This is compounded by the fact that most DSLR have a flange distance (focus-flange distance) which is pretty large - over 45...50 mm in many cases.

Furthermore, an additional distance is wasted by the classic T-ring adapters, which tend to push the camera way back from the focuser's shoulder.

All these factors conspire in your case to prevent you from achieving focus.

Some ideas and solutions:

Adapters

I tell everyone who is trying to do prime focus photography with visual scopes (dobsonians, regular newtonians) to ditch the old T-ring system and just buy dedicated adapters. All you need is a piece of metal that goes in the bayonet mount on the camera on one side, and into the focuser on the other side. No need for T- components.

Some such adapters are made by telescopeadapters.com:

http://www.telescopeadapters.com/

I have two adapters like this, they save a very large amount of back focus, compared to the classic T-ring / T-adapter system. In other words, they allow the camera to get much closer to the focuser, and therefore catch the focal plane on the sensor. You just need to choose the kind of adapter that works with your camera.

Most of these adapters are 2", so in a 1.25" focuser a size adapter is needed - but that would waste some back focus distance.

Camera

Classic DSLR cameras have a very large flange distance, so a large back focus is needed to overcome it.

Other removable-lens, large-sensor cameras have a much smaller flange distance, if they are mirrorless. Examples: Micro Four Thirds cameras (Panasonic G series), Sony NEX, Samsung NX, etc. All these systems have large sensors (comparable to DSLR) which work well for astrophotography, but don't have moving mirrors, so their flange distances are tiny.

https://en.wikipedia.org/wiki/Mirrorless_interchangeable-lens_camera

I have a Panasonic G1, used with the adapter mentioned above. I only need 25 mm of back focus for this to work, which is an extremely small value.

Use a barlow

A barlow would increase the amount of back focus available. Try and see what happens. Keep in mind it cannot work miracles - if you need a very large amount of back focus, it may not work.

Use eyepiece projection

With some eyepieces, you can thread adapters on the user side, and then put the camera body (without lens) on the adapter. Move the eyepiece WAY out of focus, and it will generate a real image which can be projected on the sensor. This is not prime focus photography anymore, but it works and it's better than nothing.

However, such adapters are hard to find, and many eyepieces won't support them.

Use a dedicated USB "eyepiece camera"

You can find relatively cheap USB cameras made for imaging, some are less than $100. With one of those, you could achieve focus relatively easy. Example:

http://www.telescope.com/Astrophotography/Astrophotography-Cameras/Orion-StarShoot-USB-Eyepiece-Camera-II/pc/-1/c/4/sc/58/p/102083.uts

Would not offer the same performance like a DSLR, but it's better than nothing.

Use a telescope made for imaging

Visual scopes, like a small newtonian, are not really that great for photography, for a variety of reasons. A dedicated astrograph would always work better. There are many Cassegrain systems out there made specifically for imaging, some are relatively inexpensive (not much more expensive than a DSLR).

In many cases, you need very long exposure, and you need to track the sky while the shutter is open. Many such scopes come with a tracking mount, or are made to be installed on a tracking mount. That would make it very easy to do tracking well.

It would be very hard to do tracking with other kinds of scopes (tracking platforms for dobsonians do exist, but are not very common).

Move the primary mirror up

Think about it - if you move the primary mirror up in its cell, the focal plane will stick out. This is a way to gain back focus only if you know what you're doing. Additional spacers in the primary mirror cell could do it. Modifying the cell is also doable.

Beware, you could ruin the scope if you make a mistake. Also, if the secondary mirror is too small to begin with, you'll increase the amount of clipping.

Afocal photography

Put the eyepiece in the scope, just like for a normal visual observation. Now put the camera at the eyepiece and move it around until you capture the image. Click the shutter. This is called afocal photography.

You'll need a camera mount, that will attach to the focuser, holding the camera at the eyepiece. Something like this:

http://www.telescope.com/Astrophotography/Astrophotography-Accessories/Orion-SteadyPix-Deluxe-Camera-Mount/pc/-1/c/4/sc/61/p/5338.uts

Keep in mind that your camera is very likely too heavy for this trick to work. Also, afocal is usually the lowest quality of all techniques, and rarely works well enough.

(There are special adapters that go on the eyepiece on one end, and thread into the camera lens like a filter on the other end. These would work better, but are hard to find, only a few eyepieces support them, and your camera is too heavy anyway.)

Keep in mind that your scope has a 1.25" focuser. Even if you do get the sensor in focus, the small size of the focuser will clip the image; you will only be able to use the central part of the sensor. It is recommended to use 2" focusers for prime focus imaging with large sensors.

Also, your f/4 scope would exhibit strong aberrations at the edge anyway, which would blur the image. For prime focus imaging, longer focal ratios are usually desirable.

All in all, your scope and your camera are mismatched. You have a pretty good camera, but a very small scope that is not made for imaging. Either use a dedicated astrograph with that camera, or keep the scope and use a cheap USB "eyepiece camera".


Which came first: Galaxies  Stars  Planets? If a galaxy is defined as a collection of planetary systems (and all matter in between), and a planetary system is defined as a collection of planets circling a star (and all matter in between), and a planet (and other smaller bodies, like asteroids, moonlets, moons, etc) is basically just a very very, very large collection of minerals, stardust and gases (which make up the entire universe), which came first: The planets, planetary systems or galaxies? I've been given to understand the universe came first...but after that?

I thought I understood, at least the layman's basics, of the universe formation. However, today, I’m having trouble conceptualising the order in which they would have been formed, IF they are defined by the smaller unit.

Is this just me getting the semantics of classification and the actual chronology of each feature forming, mixed up?

VERY Basically,

The universe is made up of a complex mix of everything (matter, antimatter, darkmatter, to name just a few), and expanding 'space'.
A very, very, large area of 'space' is covered in very, very, tiny particles: made up of at first gases and later 'stardust'.
A galaxy is formed of gravity-attracted gases, and after the very first stars went supernova and exploded there was also 'stardust', which combines to create 'clumps' of interstellar clouds (cloud nebulae).
The cloud nebulae are the 'nursery' areas for star formation (both initial stars and later generation stars).
The newly created star uses remnants of 'stardust' in the cloud nebulae to form planetary systems and then, a 'short' time later the orbiting planets (and other sized objects).

So then the chronological order would be:

Universe (Largest scale) => gases (very small scale) => galaxy => interstellar clouds => stars => planets (combined to form planetary systems). And the tiny particles of gases and 'stardust', are involved in nearly all scales.

Image taken from spaceexplained

Just writing out this question has helped! I hope you don't mind me posting it anyway, as it might help others and I still need confirmation that I have, hopefully, worked myself out of my conundrum (not deeper).

The structure we see in the Universe has formed from the gravitational collapse of the matter that was once an almost smooth density field of gas ("baryons") and dark matter$^1$. The word "almost" is important here, for if it had been completely — or even non-completely but much more — smooth, then the collapse would not have had the time to happen before the expansion of space has diluted the matter enough to prevent any collapse, and we would never had come into existence.
That is, the density field was slightly clumpy, and these clumps — or overdensities — existed on all scales.
But calculating which clump sizes collapse first — stellar-sized, galaxy-sized, cluster-sized, etc. — is far from trivial. Analytical attempts has to make several approximations, but can still make quite meaningful predictions which have subsequently been backed up and refined by numerical simulations, and by now, although many loose ends still exist, we have a rather good picture of structure formation:
Gravitational collapse
An overdensity is denoted by
$\delta\equiv\rho/\langle\rho\rangle - 1$, where $\rho$ is the density of the overdensity and $\langle\rho\rangle$ is the average density.
The evolution of overdensities under the force of gravity can be calculated exactly for $\delta \ll 1$ using linear perturbation theory, but when $\delta$ becomes of order unity, the non-linear regime is entered, and severe approximations must be made, so one turns instead to numerical simulations. It turns out that if $\delta\gtrsim1.68$ (i.e. if a region in space has a density which is 2.68 times the ambient density), it will collapse. The answer to your question is then given by what size of clumps first reach $\delta\gtrsim1.68$.
First clumps
Primordial quantum fluctuations$^2$ grew in size during the (admittedly still pretty blurry) inflation, a fraction of a second after Big Bang. In the young Universe dark energy was negligible, and the dynamics of the Universe was dominated by matter. Because dark matter comprises $\sim5/6$ of the total amount, we can initially neglect the presence of gas, but when the density becomes very high, gas pressure builds up and counteracts the collapse.
The overdensities amplified as matter started to collapse. It turns out that the density fluctuations are larger for smaller scales, so the smaller the clump, the sooner it will collapse. This results in the so-called "bottom-up" formation of structure, which is in contrast to what was originally thought; namely that galaxies formed "top-down" in a monolithic collapse (Eggen, Lynden-Bell, & Sandage 1962).
However, this approach neglects both gas and the motion of dark matter particles (treating it as so-called cold dark matter). Taking into account the effects of this puts a lower threshold for the masses of the structures of $\sim10^5$-$10^6M_\odot$ (e.g. Naoz et al. 2006; Yoshida 2009). Hence the first structures that formed are believed to be minihalos roughly the mass of globular clusters.
Stars, galaxies, and clusters
Stars consist of collapsed gas and almost no dark matter, and the formation of a star thus needs the theory of hydrodynamics rather than just gravity. In order for matter to collapse to such dense structures as stars, it must get rid of much of its energy. This is not possible for the collisionless dark matter (at least in the normal sense, but this post), but gas, which can collide and cool by radiating, is able to do so$^3$.
Through radiative cooling, the minihalos thus fragmented further into first gas clouds and then the first stars when the Universe was a few 100 million years old. Subsequently conglomerations of stars merged into galaxies, which eventually formed sheets, filaments and clusters of galaxies.
Dust
The first stars were formed out of pure hydrogen and helium (and small amounts of lithium), and the material for making planets did not exist. But when these stars — which were very massive — exploded as supernovae and polluted the interstellar medium with metals$^4$ and stardust$^5$, the formation of less massive stars became possible$^6$.
Planets
Dust particles stick together to form pebbles, rocks, and planets. Planet formation is probably not possible for too massive stars$^7$, but with the formation of small stars this became possible.
Timeline
We can make the following timeline:

Minihalos
Gas clouds
Stars
Galaxies
Stardust
Planets

However, note that the merging of minihalos to larger "galaxies" (without a significant amount of stars) also may happen at earlier epochs, and in particular that star formation (and dust and planets) is a continuous process which still takes place today (although the bulk of the star formation took place when the Universe approximately 3–6 billion years old (Madau et al. 1998)).

$^1$And radiation, which actually dominated the energy density until the Universe was $\sim50\,000$ years old.
$^2$The term primordial quantum fluctuations might be the coolest term in physics.
$^3$There are hypotheses of so-called dark stars, but that's beyond the scope of this text (primarily because I don't understand them).
$^4$For an astronomer, the term metal means any element heavier than helium. It's easier this way.
$^5$Astronomers just call it "dust", and use only the term stardust when talking to non-astronomers, in order for them not to fall asleep.
$^6$The reason is that with the many possible electronic transitions of metals, the gas has more ways to cool, and hence may collapse at earlier times, before the proto-star clump reaches high masses.
$^7$Because the high radiation pressure of very massive stars blows away and/or destroys the dust (I think, but this is not my expertise).


SPICE Alt Az example Original question below, but my question is now about how can I supply lat/lon coordinates for other locations other than the one in the example below?

It is possible to use NASA's SPICE toolkit to produce Alt/Az coordinates?  I haven't been able to find any examples.

Update:2
The code below doesn't appear to be along the right lines.  I've posted an answer with the "most" correct code I could come up with.

Update:

I think I partly figured it out.  Apparently most people call it Elevation and Azimuth, not AltAz, which is why I couldn't find anything.  I found a SpiceyPy example which generates the Az, El for DSS-13.  But I'm at a loss as to how to specify the Lat,Lon for general locations.  Looking through the kernel file, the closest thing I see that could be a lat/lon for DSS-13 are the "TOPO_ANGLES" which are -243.2055404763616,-54.7528357325366, which is a pretty odd format if its supposed to be lat,lon.  I was unable to find the real lat/lon for DSS-13, but I was able to find it for DSS-14 which appears in the same file (FYI those coords are 116.89W, 35.426666666666N).  I verified the results with Horizons and they match exactly.

Below I provide a stripped down version of the example to only print out the el/az coordinates.  It requires a file "erotat.tm" which I also provide below.  I was able to filed all of the required kernel files by searching for the name (apparently some of them are quite old).  I was not able to find de414_2000_2020.bsp, but I was able to use de414.bsp from here.  You can install SpiceyPy (assuming you already have Python and PIP) by typing "pip install spiceypy".

import spiceypy

METAKR = 'erotat.tm'
spiceypy.furnsh( METAKR )
et = spiceypy.str2et('2007 JAN 1 00:00:00')
[topov, ltime] = spiceypy.spkpos( 'moon', et, 'DSS-14_TOPO','lt+s', 'DSS-14')
[r, lon, lat] = spiceypy.reclat( topov )

az = -lon
if  az < 0.0:
    az += spiceypy.twopi()
el = lat

print( 'Moon Az (deg):        {0:15.6f}\n'
       'Moon El (deg):        {1:15.6f}\n'.format(
           az * spiceypy.dpr(),
           el * spiceypy.dpr() )  )

spiceypy.unload( METAKR )

Contents of emrotat.tm

KPL/MK

Meta-kernel for the "Earth Rotation" task
in the Binary PCK Hands On Lesson.

The names and contents of the kernels referenced by this
meta-kernel are as follows:

File name                       Contents
------------------------------  ---------------------------------
naif0008.tls                    Generic LSK
de414_2000_2020.bsp             Solar System Ephemeris
earthstns_itrf93_050714.bsp     DSN station Ephemeris
earth_topo_050714.tf            Earth topocentric FK
pck00008.tpc                    NAIF text PCK
earth_000101_070725_070503.bpc  Earth binary PCK

\begindata

KERNELS_TO_LOAD = ( 'naif0008.tls'
                    'de414.bsp'
                    'earthstns_itrf93_050714.bsp'
                    'earth_topo_050714.tf'
                    'pck00008.tpc'
                    'earth_000101_070725_070503.bpc' )

\begintext

I kinda found a solution, unfortunately it's a few hundredths of a degree off from those specified by Horizons.  And it requires you to define a custom kernel for the observer's reference frame, which isn't bad as long as your doing a lot of observations from one location.  The code below is in Python and requires spiceypy ("pip install spiceypy").

import spiceypy
import math
#By Greg Miller (gmiller@gregmiller.net) 2019
#Released as public domain

spiceypy.furnsh('erotat.tm')

#Lat,Lon position of observer e.g. Louisville, KY
#Needs to match definition of "MYTOPO" in custom kernel mytopo.tf
lat=math.radians(38.2464000)
lon=math.radians(274.236400)

#convert lat,lon to xyz
obspos = spiceypy.georec(lon, lat, 0.0, 6378.1366, 1.0/298.25642)

#time of observation in UT
et = spiceypy.str2et('2007 JAN 1 00:00:00')

#Get the xyz coordinates of the moon relative to observer in MYTOPO frame
moon, lt = spiceypy.spkcpo('MOON', et, 'MYTOPO', 'OBSERVER', 'LT+S', obspos, 'EARTH', 'ITRF93')

#reclat function only wants a length 3 array
temp=moon[0:3]

#convert to polar coordinates
[r, lon, lat] = spiceypy.reclat( temp )

#Convert polar coords to Az, El
az = -lon

if  az < 0.0:
    az += spiceypy.twopi()

el = lat

print( 'Moon Az (deg):        {0:15.6f}\n'
       'Moon El (deg):        {1:15.6f}\n'.format(
           az * spiceypy.dpr(),
           el * spiceypy.dpr() )  )

spiceypy.unload('erotat.tm')

This is the contents of the erotate.tm kernel spec file.  You must also download all of the kernel files individually, you can generally find them by searching for their names.  Note that the last one "mytopo.tf" is one you must supply, an example is below.

\begindata

KERNELS_TO_LOAD = ( 'naif0008.tls'
                    'de414.bsp'
                    'earthstns_itrf93_050714.bsp'
                    'earth_topo_050714.tf'
                    'pck00008.tpc'
                    'earth_000101_070725_070503.bpc',
                    'mytopo.tf' )

\begintext

Below is an example mytopo.tm file.  The lon,lat go in the TKFRAME_1234567_ANGLES and is the only field you'll need to edit.  Note the odd format for lat/lon compared to how it's specified in the Python code above.  The lon is just the negative, the lat is the negative of 90-lat.

\begindata

  FRAME_MYTOPO             = 1234567
  FRAME_1234567_NAME       = 'MYTOPO'
  FRAME_1234567_CLASS      = 4
  FRAME_1234567_CLASS_ID   = 1234567
  FRAME_1234567_CENTER     = 399

  TKFRAME_1234567_SPEC     = 'ANGLES'
  TKFRAME_1234567_RELATIVE = 'IAU_EARTH'
  TKFRAME_1234567_ANGLES   = ( -274.236400, -51.7536, 180 )
  TKFRAME_1234567_AXES     = (       3,        2,   3 )
  TKFRAME_1234567_UNITS    = 'DEGREES'

\begintext


How close would the orbit of Mars have to be to that of the Earth to induce a collision? When the Earth passes by Mars they exert a slight gravitational pull on each other, but the force is so small that their orbits are not significantly disturbed.

However, if we imagined that the orbits of the Earth and Mars were coplanar and much closer to one another, then there would presumably be some critical distance at which they would move together and collide.

What is that critical distance? In other words, if we shrunk the radius of the orbit of Mars, how much would it have to shrink before a collision would occur by the natural force of gravity between the two bodies?

It's more likely that perturbations by another planet might induce a collision.

In other systems with bodies in almost the same orbit, what happens is like a game of orbital catch-up: one overtakes the other, then they swap places, so in your (rather vague) scenario, that might well be the outcome instead of an outright collision.

Some of the Saturn ring moonlets interchange positions, for example.

The solar system is stable over a timeline of millions of years (as far as we know), but there have been instabilities in the past, and due to "butterfly effects" instabilities could occur in the future.
(See also https://plus.maths.org/content/rocks-suggest-theres-chaos-solar-system)

Have a play with a simulator such as orbitsimulator.com !

http://orbitsimulator.com/gravitySimulatorCloud/simulations/1433364162018_Solar%20System.html


How to calculate a phase shift in the process of deriving Baade-Wesselinke distance? The above is the equation used in the Baade-Wesselink method to independently get distances to Cepheid variables (in the process of deriving Period-Luminosity relation).

According to the paper New Baade-Wesselink distances and radii for four metal-rich Galactic Cepheids, δθ in the equation accounts for a phase shift between the radial velocity curve and the angular diameter changes measured via surface-brightness relation. My question is how can I calculate this δθ value given that I have lists of radial velocity values and V-R values with the observed HJD dates (which is possible to be converted in terms of phase).

It is admirable that you are tackling this type of problem as a high school student, but I really think that some of these topics are out of your grasp currently, partly because of the highly technical nature, but also because of your lack of access to the necessary data.

For one thing, you say you have V-R colors as a function of time, but all these equations were calibrated to use V-K colors. From your linked paper, you're going to need equation 2 for your process to calculate $p$ and equation 3 for getting $\theta_0$. Unfortunately, these equations assume V-K colors. You can redo the analysis done in this paper to calculate $p$ and this paper to get $\theta_0$ for V-R colors, but it's not going to be easy.

For another, you're going to need measurements of the angular size as a function of time (and your cited paper suggests that only exists for a handful of Cepheids and was measured using the VLTI) as well as the RV as a function of time. The $\delta \theta$ that you reference is described exactly in the paper as "a  phase shift  between  the  radial  velocity  curve  and  the  angular  diameter changes measured either interferometrically or via the SB relation." This means if you have $RV(t)$ and $\Delta \theta(t)$, you can determine a phase shift between the two curves, presumably by fitting some equation of the form $f(t) = a\ \mathrm{sin}(bt+c)$ to both equations and saying that $\delta \theta = c_{RV} - c_{\Delta \theta}$.

The analysis you're trying to do requires very specific data and it seems that the relevant papers on this subject have already done the analysis on all stars in which all necessary data existed. You cannot easily apply this process to any Cepheid without significant work.


The relation between the light of full moon and the distance between the sun and the earth If we want the full moon to be 2 times more bright,how we should change the distance between the sun and the earth?

Thanks for your help.

The strength of the light from the Sun scales with the inverse square of distance [note 1]. That means that we would need to have the Earth-Moon system at $\frac{1}{\sqrt{2}}$ (approx 0.7) AU for the full Moon to be twice as bright [note 2].

Note 1: The fact that the Sun is not a point source of light has only a very minor effect on the scaling.
Note 2: Technically, it will be slightly brighter because more of the absorbed radiation will be radiated away in the visible spectrum. This is however also a negligible effect.


Where are Moon's Apogee and Perigee? Do they rotate too? Where are Moon's Apogee and Perigee? Which one is inside, towards the Sun, between Sun and earth?

Do they rotate too just like the north, south node?

Does Moon always need to be on Apogee/Perigee for it to be Full/New Moon?

Where are Moon's Apogee and Perigee? Do they rotate too?

Yes they do rotate!
Apsidal precession is the rotation of the line of apses (line connecting apoapsis and periapsis and passing through the Earth). Lunar precession takes this line about nine years to rotate once around the Earth, referenced to the celestial sphere (the stars).

Source


Tunnel vision while viewing through a reflector I have a Celestron astromaster130 eq. While viewing through its eyepiece, or using a smartphone camera for photography purpose there is always a definitive no light region( it is circular in curvature). I make sure that I look directly at the eyepiece, but this tunneling makes it really difficult to view things at times. I have googled this, but am not sure of what this effect is called. Is it a general problem or is there a need to rectify some fault in the telescope itself. If not, what are the ways in which this can be effectively reduced?
Thanks.
Edits are most welcome.

This is completely normal behaviour. What you are seeing is the edge of the projection angle for the lense. This is probably sort of what you're seeing? :

Typically this is more noticable with cheaper eyepieces as the individual elements are smaller, and have a smaller field of view.


What is the difference between gas and dust in astronomy? Is there a strict difference between gas and dust?
In Earthly environment most things become gaseous if heated enough. The temperature of interstellar medium seems to range mostly between 10 and 10 000 Kelvin. Is gas/dust an analog for hot/cold, or does the phase diagram of the element in question matter too? Can metals and molecules be gas in astronomical terms?

Yes, metals and other elements and molecules can exist in gaseous form under the right conditions of temperature and pressure. A "gas" is simply one of the fundamental states of matter, as in solid, liquid, or gas (and a few other states outside the scope of this question). But as a gas, these substances exist entirely as either individual atoms, individual elemental molecules, or individual compound molecules of multiple atoms (e.g. carbon dioxide).

Dust, on the other hand, is comprised of tiny particulate matter that has undergone the stronger intermolecular bonds to create substances like ice, silicates, and carbon compounds that float around in varying densities between the stars and between the galaxies. Since these particles are still extremely small (typically a fraction of a micron across), they can appear to be a gas, but these tiny, irregularly-shaped objects still exist individually in a solid or liquid state.


Estimating Sun luminosity from first principles In a lecture, it was claimed that luminosity can be estimated by dividing how much energy there is in the Sun with the typical time it takes for a photon to get out from the centre of the Sun to its surface. I don't see how that argument holds. Why would that give me luminosity?

The thermal energy of the Sun is something like
$$E \sim  \left(\frac{3k_BT}{2}\right)(N_i + N_e)\ , $$
where $N_i$ is the number of ions and $N_e$ is the number of electrons and the interior temperature $T \sim 10^7$ K.
To first order we can consider the Sun to be made of hydrogen, so $N_e = N_i \sim M/m_p$, where $m_p$ is the mass of a proton.
Putting in the numbers gives $E \sim 5\times 10^{41}$ J.
The typical mean free path of a photon in the solar interior is $l\sim 10^{-3}$ m. The solar radius is $R\sim 7\times 10^{8}$ m. Since the photons get out by a "random-walk" diffusion process, then it takes about $(R/l)^2$ steps, each of which takes a time $l/c$.
Thus the diffusion timescale
$$ \tau \sim \left(\frac{R}{l}\right)^2 \left(\frac{l}{c}\right) = \frac{R^2}{lc} = 1.6\times 10^{12}\ {\rm s}$$.
The ratio of these two numbers gives $3\times 10^{29}\ {\rm W}$, so about 3 orders of magnitude larger than the solar luminosity.
This ratio does not give the solar luminosity.


What would we see if all stars disappeared simultaneously? If all the stars disappeared simultaneously (from the earth's reference frame), how long would it take for the night sky to go completely dark to our unaided eyes? Would we still be able to see anything after a thousand years?

The farthest object that we can see with our naked eyes as of now is the Andromeda Galaxy (2.6 million light years away), although the M83 (14.7 million light years away) has also been reported to be seen with the naked eye. (You may want to click here for more information on what the farthest objects we can see with our naked eyes, from Universe Today or see this answer from this same site.)

Assuming that just before the moment all the stars disappear, these two farthest objects are still emitting light, then we can know judging by their distance from us, that this light will only reach our eyes 2.6m years later (because 1 light year = light travelling for 1 year) from the Andromeda Galaxy, and 14.7m years from M83.

Therefore to answer your question more directly:

We would still see the stars (assuming this is what you are referring to by "anything") for up to 2.6~14.7m years, depending on what really is the farthest thing we can see in the sky.


The defintion of star/planetary/solar system In many science fiction stories, we hear the term star system that mostly refers to a star and its planets. This is often used interchangeably with the term solar system. But after some research I understand this:

Planetary system refers specifically to anything orbiting a star that is not a star itself. According to Wikipedia, this generally refers to the planets, but may include other objects like asteroids and dwarf planets. It does not refer to the star itself.

Star system refers to any system with two (binary) or more (multiple) stars, but only the stars themselves, not planets or anything else.

Solar system generally refers to our own Solar system but there are situations where "solar system" (lowercase) refers to a star and the planets or other objects orbiting it (as long as something orbits the star).

So, is it incorrect to use the term solar system/star system to refer to a star and its planets?

I think it's largely a matter of personal preference. Some people will want you to be technically correct and only use "solar system" to refer to the Sun and the objects that are under its gravitational influence. Wikipedia defines it as such:

  The Solar System comprises the Sun and the objects that orbit it, whether they orbit it directly or by orbiting other objects that orbit it directly.

NASA says that the term is more general:

  A solar system refers to a star and all the objects that travel in orbit around it. Our solar system consists of the sun - our star - eight planets and their natural satellites (such as our moon); dwarf planets; asteroids and comets.

As an astronomy enthusiast, I do differentiate between "stellar system"/"star system" and "planetary system". I'd use "planetary system" to refer solely to the planets in a system, as well as their moons, while I'd use "stellar system" to refer to the whole thing. That's just my opinion, though, and I'm unsure what professionals use. In casual conversation (especially with a non-expert), you can probably use whatever term you want. If you're using it formally, you should probably do some more research. And while you're at it, check whether or not you should capitalize "Solar" and "System" in "Solar/solar System/system". If people nag you about terminology, they'll nag you about capitalization.


Who discovered the relation between CMB fluctuations and large-scale structure formation? I haven't been able to find a reference to the first who showed that the small density fluctuations in the cosmic microwave background led to the formation of large scale structure as we know it.

I'm not an expert in this, but I'll give it a try.

I think you've got the argument the other way round here. This was never discovered. It was assumed. But why?

We know for the CMB being a redshifted blackbody-spectrum. This means the CMB consists of photons that were emitted in interaction with matter.
The CMB fluctuations inform us about the depths of gravity wells that those photons had to escape from (amongst other things). Thus we get a picture of the baryonic and dark matter density at the time of photon-matter decoupling.
This allows us to set up linear perturbation theory + simulations to look what happens to a uniform gas with the measured fluctuations on top. That's the rough idea.

You could now of course ask "who was the first to implement this?".
The earliest theoretical work that I'm aware of (Peebles 1965) linking "a possible universal blackbody radiation" with the formation of galaxies came in the year of the discovery of background radiation by Penzias & Wilson 1965.


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

I would imagine that a good quality telescope, would be about to detect and record the effects of the exoplanet transiting the parent star.

What practical considerations in terms of equipment, software etc. would be required for 'backyard' amateur observations of the effects of transiting exoplanets?

If anyone has actually tried this, what has been your experience?

This is actually quite straightforward with digital CCD's (it used to be quite tricky with film cameras as you'd have to carefully develop film that moved past the lens and assess the width of the trail)

Get yourself a good telescope - a 12" Dobsonian or above if you want to give yourself a good chance of picking out the fluctuations against the noise background. Then select a decent CCD. Five hundered pounds gets a reasonable one, but expect to pay a couple of thousand pounds for a cooled CCD, which will also help to reduce noise.

(Buying in US dollars? A reasonable CCD runs about $1000. A cooled CCD will cost you at least $1500.)

You'll want a good quality equatorial mount, with computer controlled servos to track the target smoothly over long periods of time.

Ideally you will also slave a second telescope and CCD, pointed along the same path but slightly off - this will help you cancel out cloud and other fluctuations from our own atmosphere.

Oh, and get as far away from a city as possible - up into the mountains can be a good plan :-)

Then arrange your viewing for a series of full nights. The more data points you can get, the better the noise reduction. Imagine the exoplanet is orbiting every 100 days, in order to get any useful data, you will need to track it over some multiple of 100 days. So assume you set up to track your target star for 2 years, plan for 3 or 4 star shots per night to give you a range of data points.

These 600+ days of 4 data points per night gives you a reasonable stack of data - the challenge now is to work out whether there are any cyclic variations. Various data analysis tools can do this for you. As a first step, if you find a cycle around 365 days, it probably isn't the target, so try and normalise for this (of course this will make it very difficult to discover exoplanets with a period of exactly 1 year)