How many days a month can you see a moonrise during daylight? For about 15 days before a Full Moon the moon rises during the day, and for about 15 days after a Full Moon the moon sets during the day. I think it might be possible at full moon for the moon to rise just before daylight ends, and to set just after the next daylight starts, on some occasions.
However not all of these rises/sets will be visible. Around dark moon, moonrise is lost in the rising sun, and moonset is lost in the setting sun.
Hence my question, how many can you actually see?
I know this depends on a lot of factors, so a range would be sufficient.
I think this is equivalent to asking, what elongation of the moon is needed for the moon to be visible? When both are above the horizon?

It's too hard for me to come up with reliable figures for the amount of visible moonrises/moonsets during daylight, but I can answer

what elongation of the moon is needed for the moon to be visible?

It will depend on the conditions, but according to the Wikipedia article about the Islamic calendar, the Moon may already be visible 17 hours after the conjunction with the Sun:

But, the lunar crescent becomes visible only some 17 hours after the conjunction, and only subject to the existence of a number of favourable conditions relative to weather, time, geographic location, as well as various astronomical parameters.

The synodic month is about 708.734 hours, so a period of 17 hours corresponds to an elongation of $360^{\circ} \cdot \frac{17}{708.734} = 8.6^{\circ}$.


How many constellations in the Zodiac? The astronomical zodiac contains a bunch of constellations along the ecliptic.

Some sources say there are 13 constellations in the astronomical zodiac. Other sources claim there are 12.

According to NASA (source):

  If we count all the traditional constellations, there are 13 in the
  zodiac, not 12. So here are the dates (give or take one day each year)
  when the Sun is between Earth and each of these 13 constellations.)

There seems to be some confusion out there between zodiac constellations (in astronomy) and zodiac signs (in astrology). From the research I've done so far (including the quote above), it seems as though one of them (Ophiuchus) is a "recent addition", possibly due to the Earth's precession since ancient times (source).

What (officially?) are the constellations of the astronomical zodiac?
Why do some sources list 12 and others list 13?

There are thirteen modern constellations in the Zodiac. In modern astronomy, a constellation is a specific area of the celestial sphere as defined by the International Astronomical Union. In total, there are 88 constellations.

Astronomy and Astrology are not the same thing. Astronomy is a science while Astrology is not. As such, I'll restrict myself to the historical and modern constellations of the Zodiac.

According to Encylopedia Brittanica

  Zodiac (is) a belt around the heavens extending 9° on either side of the ecliptic, the plane of the earth’s orbit and of the sun’s apparent annual path.

In historical astronomy, the zodiac is a circle of twelve 30° divisions of celestial longitude that are centered upon the ecliptic, the apparent path of the Sun across the celestial sphere over the course of the year. Historically, each of these divisions were called signs and named after a constellation: Sagittarius, Capricornus, Aquarius, Pisces, Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, and Scorpius.

In 1930, the International Astronomical Union defined the boundaries between the various constellations, under Eugène Delporte , who,

  ... drew his boundaries along vertical lines of right ascension and horizontal parallels of declination. One governing principle was that all variable stars with an established designation would remain in that constellation, as requested by the IAU’s Variable Stars committee.

"Constellations ecliptic equirectangular plot" by Cmglee, Timwi, NASA - Own work, http://svs.gsfc.nasa.gov/vis/a000000/a003500/a003572. Licensed under Public Domain via Commons.

As a result, the path of the ecliptic now officially passes through thirteen constellations: the twelve traditional 'zodiac constellations' and Ophiuchus (which was one of the 48 constellations listed by the 2nd-century astronomer Ptolemy), the bottom part of which interjects between Scorpio and Sagittarius.

Source: journeytothestars.wordpress.com

So, the 13 constellations of the Zodiac are Capricornus, Aquarius, Pisces, Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpius, Sagittarius and Ophiuchus.

As seen from the (first) figure, the ecliptic also touches the edge of the constellation Cetus, though it in not usually included in the Zodiacal constellations.

Note about Precession: Because the Earth in inclined (by $23.45^{\circ}$), it rotates like a top. This is called precession, which results in a shift in the position of the constellations relative to us on Earth. The result is that the 'Signs of the Zodiac' are off by about one month.


Why Helium-burning Stars are found in a Horizontal Branch? Why Helium-burning stars are found in a horizontal branch on the H-R diagram? Why do they have roughly the same luminosity?

This is explained in the Wikipedia article

  Stars on the horizontal branch all have very similar core masses, following the helium flash. This means that they have very similar luminosities, and on a Hertzsprung–Russell diagram plotted by visual magnitude the branch is horizontal.

To expand a little. In stars of a certain mass range, helium builds up in the core until it reaches a specific mass, at which point the "Helium flash" occurs and burning of helium to carbon and oxygen starts throughout the core. When things settle down, helium burning is going on in the core, which is more or less the same size, independently of the original mass of the star. Since this is the main power source of these stars, they all have about the same luminosity.

The variation across the branch comes from how much remaining gas there is outside the helium-burning shell -- more gas means a larger cooler star radiating the same total amount of energy


How are the elevations of the poles proportionate to distances of earth that have been traversed according to Copernicus In chapter 2 of "The Revolutions of the Heavenly Spheres" where Copernicus maintains that the Earth is spherical, the Edward Rosen translation of the book states that

  the elevations of the poles have the same ratio everywhere to the portions of the earth that have been traversed.

I'd like to know what Copernicus is referring to in the aforementioned piece of text and how it provides evidence for the spherical shape of the Earth.

Here, I believe he is referring to the altitude of the celestial poles at different latitudes on the Earth.

If the Earth is indeed a sphere then the altitude of the celestial poles above the horizon should be the same as your latitude on the Earth. That is, as you travel along a line of latitude, the altitude of the pole should change accordingly.

So Copernicus was noting that this is indeed what we observe. At the North pole the North celestial pole is directly overhead, and on the equator it is on the horizon. This observation, therefore, is direct evidence that the Earth is spherical (or at least spheroidal).

You can see the evidence of this in star trail photographs. The stars will always appear to rotate about the celestial poles. This image shows the star trails above Ecuador, at a latitude of about 1$^{\circ}$. You can clearly see that the stars are rolling directly over head; with the North and South celestial poles located on the horizon to the right and left, respectively.


How could universe inflate itself out of the very dense and curved early spacetime? Could it happen in a black hole too? Wasn't spacetime as much curved as a black hole directly after Big Bang, because mass was so densly packed? Wasn't everything like an event horizon and how could things expand across it? Could inflation happen in a black hole too so that its content does become accessible to the outside universe?

Great question! Sorry for this huge response, and it might not be a satisfying answer, but it'll address your questions.

Sadly, as with most of astronomy, the Big Bang is surrounded in mystery. It is one of the biggest uncertainties out there, and you'll come to this realization if you've researched it enough.

Most of what we can get out of the Big Bang is through extrapolation. If we observe the expansion of the Universe, we can presume that at one point, there was far less space between all the matter in the Universe. This is where things begin to break down, though.

Through General Relativity — which explains our modern understanding of gravity — and our extrapolation, we would reach the phenomena you're describing; the entire Universe would have originated as an infinitely small point (called a singularity), from which matter was incredibly hot.

However, General Relativity begins to break down at those levels, since its equations cannot explain the conditions of such a universe accurately. For one, as you hint at, an infinite density and potentially infinite mass would likely have an infinitely strong gravity.

There is no place for such infinities in a proper mathematical description of the universe, and at that point, Einstein's equations — the basis for predicting the evolution of the cosmos — collapse. These equations provide us with unreliable results and we run into some of the problems you are describing.

Quantum mechanics is more fit for such small scales, but it fails to explain how gravity acts and what its role was at the time. Thus, this tells us that QM and GR are necessarily incomplete. Their equations cannot describe the Big Bang singularity in its entirety, especially since GR (which deals with large, high-mass phenomena) and QM (which deals with small, low-mass phenomena) are incompatible and cannot describe small, high-mass phenomena.

Thus, we're left with a lot of unanswered questions: did the Big Bang create the Universe, or was it simply an event in the Universe's history? Why would the Universe be compressed to such a small scale? Would the idea of an infinitely dense universe go against the presumption of an infinite universe?

There has long been hope that these questions can be explained within a theory of quantum gravity. Quantum gravity attempts to unify general relativity and the concepts of quantum theory (QM and QFT). These hopes have recently taken on a more concrete shape, with the help of what is called loop quantum gravity, and its applications to cosmology.

Quantum gravity is still unproven and hypothetical. String theory also tries to unify GM and QFT, but is unfalsifiable. Thus, the Big Bang is a point in time that astronomers really know little about. It's unfortunate and surprising, but it's something physicist are trying to overcome.


How is the density parameter estimated via cosmic microwave background anisotropies? In 2003 and again in 2018, the density parameter was estimated based on measurements of the cosmic microwave background anisotropies. Can someone please explain how that works? Why do the anisotropies reveal something about the universe’s current density?
I don't have a background in physics or astronomy.

An appropriate explanation might begin by explaining that the cosmic microwave background (CMB) was formed in the expanding universe about 400,000 years after the big bang. An expanding gas cools, and when the gas in the universe cooled to about 3000 K, a transition tok place whereby the protons and electrons, hitherto free particles, combined to form hydrogen atoms. These atoms were almost transparent to the hot radiation field and allowed it to escape, free to travel the universe until we detect it 13.6 billion years later. In the meantime, the cosmic expansion has stretched the wavelength of the radiation from the optical/infrared to the microwaves we see now as the CMB.
At the time when the radiation was "released" there would have been instabilities in the gas - regions of slightly higher or lower density (compressions and rarefactions). For a number of reasons, these will leave a small imprint in the radiation field at that time, which is now observed in the form of tiny temperature variations in the CMB at different positions in the sky.
The "spectrum" of the temperature variations (basically, by how much the temperature varies as a function of angular separation on the sky) depends primarily on two things. (a) The scale of the compressions and rarefactions in the universe when the radiation was formed and (b) by how much and in what way the universe has expanded since then.
Factor (a) is directly dependent on how dense the universe was when the CMB was formed, which depends on how dense the universe is now and a cosmological model of how the universe has expanded. (b) Also depends on the rate of expansion, which can be expressed in terms of the Hubble constant, but also depends on the  topology of the universe - whether it is "flat" or "curved", since this can magnify or diminish the apparent size of the fluctuations.
In the end then, the appearance of the CMB now is dependent on both its formation conditions, the history of the universal expansion and the geometry of the universe. Each of these can be parameterised using standard cosmological models that include the current density parameter and Hubble parameter. These are related to conditions at any previous epoch in much the same way that the position and speed of a projectile can be used to predict is position and speed at any point in the past (or future).


Based on the smallest number of photons that can deliver information about a star, what is the largest size of the universe? I was watching a summary video about the Banach-Tarski paradox and it led me to a question.

If I (a) assume the universe is infinite and (b) assume it is possible to receive a single photon of light from any star within it that has a line-of-site to the Earth and (c) assume that the photon has travelled an infinite distance, then such a photon, if enough information could be extracted from it, would prove the universe is infinitely large.

Today we have finite observational and recording capabilities.  I believe (not being a physicist or astronomer myself) that we can't prove the universe is larger than some size, X, because we cannot detect lower than a particular number of photons, Y, to ascertain the distance they travelled.

So, assuming my assumptions aren't assinine (they very well could be), and ignoring our present observational limitations, what is the smallest number of photons from which we could discern information about the distance those photons travelled, and what distance would that represent?

So, assuming my assumptions aren't assinine (they very well could be) ...

Your words, not mine. Your "very well could be" is the case.

Assume the universe is infinite

The universe might well be infinitely large, but the observable universe is all that we can possibly see. By all signs, the universe is 13.8 billion years old. We can't see the stuff that is beyond the observable universe because not enough time has passed for light to have traveled that far.

To make matters worse, the early universe was very hot and dense. Photons traveled but a short distance before being absorbed. We can't see those very early photons because they no longer exist. This changed about 379,000 years after the universe formed, at which time the universe had sufficiently cooled so as to allow photons to roam free. The photons let loose by that clearing of the universe form the cosmic microwave background radiation.

Assume it is possible to receive a single photon of light from any star within it that has a line-of-site to the Earth

Per the big bang theory, the first stars formed about 100 to 200 million years after the formation of the universe. Stars that formed at that time beyond the observable universe: We can't see them. Due to the continuing (and possibly accelerating; google "dark energy") expansion of the universe, we will never see those stars.

There's a famous paradox related to your question: Why aren't we bathed night and day by light from an infinite number of stars if the universe is uniform and is infinitely old> This is Olber's paradox. The solution is simple: The universe is not infinitely old.

Assume that the photon has travelled an infinite distance, then such a photon, if enough information could be extracted from it, would prove the universe is infinitely large

The only information that can be extracted from a single photon is its frequency, and even that is a bit fuzzy due to quantum mechanics. A photon from a remote star is indistinguishable from a photon of the same frequency emitted a nearby lightbulb.


Options for observational astronomy For getting into a PhD program for observational astronomy,  is physics major and data science minor enough?

You can get into observational astronomy from a range of backgrounds. Some easily, some less so. Physics and data science sounds like a very fine choice as that is what can be a large part of an observational astronomer.

You probably will have to learn a bit about image processing. And possibly about the physics of the objects you are going to observe. Such adaptions are very common when one does not want to become an expert in one thing and one particular thing exclusively.

See also the answers to this question.


How long does a comet "live"? When a comet loses all his volatile material after many orbits around the Sun, it becomes a "dead" comet (like an asteroid, a centaur or whatever), see for example 2015 TB145.
Is there any way to determine the lifetime of a comet (expressed as the number of orbits before it becomes a dead comet)?

If you don't mind an oversimplified answer.   There is a relatively straight forward relationship.   The life of a comet (in orbits not years) depends on 3 factors.   It's size, it's composition and how close it passes to the sun.

Comets are often referred to as dirty snowballs but the ratio of different types of ice and how much "dirt" or silicate material varies from comet to comet.   Average density is about 0.6 g/cc.  That variation is a factor in how much light gets reflected right back off the comet, and how easily the surface melts, but to make this answer somewhat easier, I'm going to ignore composition and make a simplified assumption that all comets are made of basically ice.

With that in mind, how much a comet loses per pass around the sun can be simplified into loss of radius.   Think of a block of ice sitting in space, facing the sun.   The rate of sublimation would simply be a product of how close the ice was to the sun.  It's not far off to oversimplify melt rate in terms of loss of radius over a pass near the sun.

Halley's comet, for example has a fairly close pass to the sun, at about 0.587 astronomical units (closer than Venus).  The closer a comet passes to the sun the faster it sublimates.   Halley's is expected to last another 100-125 orits.  Source.

As mentioned above, there's some variation in terms of the density and different materials that make up a comet, but the primary factors are Perihelion and size of the comet.

Halley's has a diameter of about 8 KM. (roughly speaking, it's 16x8x8 km, ellipsoid shape).   8,000 meters over 100-125 passes suggests it loses about 60-80 meters per pass around the sun and if we estimate that the "dead comet" after 100 or so orbits retains 1-3 KM of rocky material, the ballmark melt-rate of Halley's is somewhere in the 30-60 meters per pass around the sun, or a meter every couple-few days by sublimation.   A comet with a perihelion at .8 AU or 1 AU would lose less radius per pass.

A comet that passes within 0.3 AU instead of 0.6 might lose about 4 times as much per day as it's hit by 4 times the solar energy per square meter, but it spends fewer days due to a faster orbit near the sun.   Somebody might be able to run the math based on highly elliptical orbits, perihelion and solar energy per square meter over the entire pass.   I'm afraid to try myself, but it would be something less than the square of the distance due to the faster orbital speed for smaller perihelion.

Wit this estimate in mind, Wikipedia says a sungrazing comet with a perihelion of about 0.055 AU and an initial radius of about 2-3 km could survive that pass and be about 1 km.   That would be a loss of 1,000-2,000 KM over the pass, but that's a very close pass.

I think that's the start of an answer to this question.  The closer to the sun, the more meters of radius the comet will lose per orbit and calculating the radius losing so many meters every pass gives a rough estimate of the life of the comet.   Sublimation can begin as far away as the orbit of Saturn, but at that distance the sublimation is very slow.


Does the galaxy look like a Web or a Tree? Suppose we have a distance cutoff $r_0$.
Let's say that any two stars in the galaxy with a relative distance below $r_0$ have a path between them.
Would the path-connected Milky Way look like a Web in which every star is potentially connected to all of its neighbors or would we form a type of branched Tree structure with junctions and bottlenecks?
I understand that for a high enough $r_0$, it will always look like a web. But my doubt is whether the standard-deviation of interstellar distances in the Milky way is such that there exists a value of $r_0$ for which the topology is unequivocally a Tree.

The process where you link random points within distance $r_0$ into a graph is known as continuum percolation. As $r_0$ increases from 0 at first there are just isolated clusters (binaries, randomly very close stars) that gradually merge. At a critical distance of a few parsecs these clusters mostly merge into one major galaxy-spanning cluster, and above that distance this cluster just gets more densely connected. The big network at the critical range has a scale-free structure of local clusters, bridges, and trees: neither a proper web nor a tree.
Some illustrations from my book manuscript:

It should be noted that these pictures assume a constant star density. In reality the stellar density varies, so for a given $r_0$ there is a part of the disk and bulge that are connected and parts of the outer galaxy and halo that are not connected.


Why does radio astronomy offer higher resolution images than optical? According to this lecture,"Radio astronomy has several advantages over optical astronomy...We can make the highest resolution images, and see things happening on the smallest scales."

Why is this? If resolving power, $r \propto \frac{\lambda}{B}$ then shouldn't shorter wavelength light result in smaller angular resolution i.e. better resolution?

The relation you cited holds for a single telescope. But, as also noted in the lecture you linked
"One thing that is possible in radio astronomy is to use interferometry, which combines the signals from an array of antennas as if they were all part of the same aperture. That means that the resolving power of a radio telescope is not just what it would be for each antenna alone, but is as for a single telescope the size of the separation between antennas."

This basically means you use multiple telescopes - which in case of radio astronomy are called antennas for historical reasons - and combine their measurements into a single one. To get an intuition of how big of arrays we are talking about, google images of ALMA, LOFAR (which has stations in several countries) or SKA.

In principle other wavelengths, such as optical, can also be observed using interferometry, but it is technically much more complicated to build such instruments. On top of that, optical light is the one most affected by the atmosphere and it is hard enough to take it into account for one telescope, combining such effects to create an image from multiple ones is a very daunting task.


Is there a limit to how fast spacetime can bend/warp around an accelerating object? If in the future a spaceship can accelerate to nearly the speed of light, will it encounter any kind of resistance/drag from the spacetime in front of it not being able to bend/warp fast enough around it?

It is my understanding that as an object accelerates, its mass will increase and this increase in mass should result in the object having more gravity, and this gravity will continue to increase as the object's speed increases. The creation of more and more gravity should result in an increased area of spacetime surrounding the object being bent/warped.

So, if there is a limit to how fast spacetime can bend/warp around an accelerating object, along with an increased area of spacetime being affected by an accelerating object's increased gravity, will this spaceship start to heat up and begin to melt the faster it accelerates?

Changes in the bending of spacetime propagate through spacetime at the speed of light. If I wave my hand thus the (very, very, very tiny) changes to spacetime will propagate outwards as gravitational waves at the speed of light and won't reach the Sun for 8 minutes.

No matter how massive an object or how great the acceleration, spacetime elsewhere only "feels" that acceleration after it has passed into the acceleration's lightcone.


Stellar mass of galaxies Given the magnitudes (in the i-band) of certain galaxies, I would like to calculate their stellar mass (in terms of solar masses). So far, I have calculated their absolute magnitudes and gotten to working out the mass-light ratio $M/L$ for each galaxy.

e.g. $M/L=0.563$

Values I have are the calculated $M/L$ for each galaxy, and the $i$-band apparent ($13.25$) and absolute ($-18.06$) magnitudes for the galaxy, as well as the distance ($18.44Mpc$).

From this I need to get the mass of the galaxy $M$ in terms of solar masses. Therefore I assume I first need to calculate the i-band luminosity for the galaxy in solar masses $L_i$. This is where I am stuck.

However, once I have $L_i$  next step would be ...
$$M_g = 0.563 * L_i$$

Ultimately, given these values, how would I go about estimating the stellar mass of a galaxy in terms of solar masses?

The relation between absolute magnitude $M$ and luminosity $L$ for stars
$$\frac{L_{Star}}{L_{Sun}}=10^{(M_{Sun}-M_{Star})/2.5}$$
should also be appliable to i-band luminosities of galaxies.

Taking the absolute magnitude 4.08 of the sun on the I-band the luminosity of a galaxy with absolute magnitude −18.06 would be
$$\frac{L_{Galaxy}}{L_{Sun}}=10^{(4.08−(-18.06))/2.5}=0.7178\cdot 10^9.$$
An order of magnitude estimate for the mass of the galaxy would be
$0.563\cdot 0.7178\cdot 10^9=0.404\cdot 10^9$ solar masses.

But $M/L$ isn't necessarily biased the same way for the sun to $M_i/L_i$ as for the galaxy.
Therefore you'll probably need to compare the spectrum of the sun with the spectrum of the galaxy to find out the ratios of the i-band fraction of the total emission. Absorption and extinction at different wavelength may be different, therefore determining kind of mean temperature of stars in the galaxy could help finding a more realistic estimate for the stellar mass.

A similiar approach has been used in this paper.


Will Earth lose the Moon before the Sun goes into supernova? Ive read on some sites and saw on youtube videos that the moon is getting away from earth by 1-3 cm a year.

Is this enough to make the Earth lose the Moon before the Sun goes into Supernova?

Im asking because I would like to do the calculations for Earths magnetic pull on this subject. It appears to me that it should be a non-linear function (there are also other factors, like if the Moon could be captured by another planet or having asteroids messing with its orbit on Earth).

Thank you.

As HDE 226868 noted in his answer, the Sun is not going to go supernova. That's something only large stars experience at the end of their main sequence life. Our Sun is a dwarf star. It's not big enough to do that. It will instead expand to be a red giant when it burns out the hydrogen at the very core of the Sun. It will continue burning hydrogen as a red giant, but in a shell around a sphere of waste helium. The Sun will start burning helium when it reaches the tip of the red giant phase. At that point it will shrink a bit; a slight reprieve. It will expand to a red giant once again on the asymptotic red giant branch when it burns all the helium at the very core. It will then burn helium in a shell surrounding a sphere of waste carbon and oxygen. Larger stars proceed beyond helium burning. Our Sun is too small. Helium burning is where things stop.

The Sun has two chances as a red giant to consume the Earth. Some scientists say the Sun will consume the Earth, others that it won't. It's all a bit academic because the Earth will be dead long, long before the Sun turns into a red giant. I'll have more to say on this in the third part of my answer.

The current lunar recession rate is 3.82 cm/year, which is outside your one to three centimeters per year window. This rate is anomalously high. In fact, it is extremely high considering that dynamics says that
$$\frac {da}{dt} = (\text{some boring constant})\frac k Q \frac 1 {a^{11/2}}$$
Here, $a$ is the semi major axis length of the Moon's orbit, $k$ is the Earth-Moon tidal Love number, and $Q$ is the tidal dissipation quality factor. Qualitatively, a higher Love number means higher tides, and a higher quality factor means less tidal friction.

That inverse $a^{5.5}$ factor indicates something seriously funky must be happening to make the tidal recession rate so very high right now, and this is exactly the case. There are two huge north-south barriers to the flow of the tides right now, the Americas in the western hemisphere and Afro-Eurasia in the eastern hemisphere. This alone increases $k/Q$ by a considerable amount. The oceans are also nicely shaped so as to cause some nice resonances that increase $k/Q$ even further.

If something even funkier happens and the Moon recedes at any average rate of four centimeters per year over the next billion years, the Moon will be at a distance of 425,000 km from the Earth (center to center). That's less than 1/3 of the Earth's Hill sphere. Nearly circular prograde orbits at 1/3 or less of the Hill sphere radius should be stable. Even with that over-the-top recession rate, the Moon will not escape in the next billion years.

What about after a billion years? I chose a billion years because that's about when the Moon's recession should more or less come to a standstill. If the Earth hasn't already died before this billion year mark, this is when the Earth dies.

Dwarf stars such as our Sun get progressively more luminous throughout their life on the main sequence. The Sun will be about 10% more luminous than it is now a billion years into the future. That should be enough to trigger a moist greenhouse, which in turn will trigger a runaway greenhouse. The Earth will become Venus II. All of the Earth's oceans will evaporate. Water vapor will reach well up into what is now the stratosphere. Ultraviolet radiation will photodissociate that water vapor into hydrogen and oxygen. The hydrogen will escape. Eventually the Earth will not only be bare of liquid water on the surface, it will be bare of water vapor in the atmosphere.

Almost all of the Moon's recession is a consequence of ocean tides. Without oceans, that tunar recession will more or less come to a standstill.


Is there a lower limit for the altitude of orbiting objects? As I understand it, one object can orbit another at a variety of altitudes, and the stability of the orbit is determined by (among other things) the speed of the orbiting object.  Go too slowly and you'll fall into the larger object, go too quickly and you'll fly off into space, but go at just the right speed and you'll orbit indefinitely.  And while an object could be in a stable orbit at Altitude X by going Speed A, it could also be in an equally stable orbit at Altitude Y by going Speed B.

If that's true (and given my limited understanding, feel free to say it simply isn't), is there a lower limit?  An atmosphere would obviously cause drag, but given a lack of atmosphere, could you theoretically have an asteroid stably orbiting an Earth-sized planet at an altitude of 50 feet, if it were going fast enough?

If not, why not?  What forces limit the closeness of an orbiting object?  Also, simple explanations would be most appreciated: I've heard of things like Lagrange points and such, but I don't have a good understanding of them.

For really low orbits over an atmosphereless body, the body needs to have a uniform density. Otherwise the gravity field is not symmetric, the orbit changes shape over time, and you end up with a crater.
NASA had trouble with this, lunar Mascons, during the Apollo era. In one case:

  "The Moon has no atmosphere to cause drag or heating on a spacecraft, so you can go really low: Lunar Prospector spent six months orbiting only 20 miles (30 km) above the surface."

In another:

  The orbit of PFS-2 rapidly changed shape and distance from the Moon. In 2-1/2 weeks the satellite was swooping to within a hair-raising 6 miles (10 km) of the lunar surface at closest approach. As the orbit kept changing, PFS-2 backed off again, until it seemed to be a safe 30 miles away. But not for long: inexorably, the subsatellite's orbit carried it back toward the Moon. And on May 29, 1972—only 35 days and 425 orbits after its release—PFS-2 crashed.

Now that the gravity field of the moon is mapped, it turns out that there are some low orbits that are reasonably stable:

  "What counts is an orbit's inclination," that is, the tilt of its plane to the Moon's equatorial plane. "There are actually a number of 'frozen orbits' where a spacecraft can stay in a low lunar orbit indefinitely. They occur at four inclinations: 27º, 50º, 76º, and 86º"—the last one being nearly over the lunar poles.


How would a small nearby black hole be detected and confirmed as such? This article suggests that there might be a tennis ball-sized black hole orbiting the Sun somewhere beyond the orbit of Neptune. According to the article, observational anomalies of trans-Neptunian objects suggest that there is an object out there about 10 times the size of Earth (I assume it to mean about 10 Earth mass). Since searches have failed to find such an object, an alternative hypothesis claims that there may be a primordial miniature black hole out there. The article goes on to state that a black hole has not be found there because searches have been for planets and the methods used to find planets are different than those used to find black holes.

My question is: how would such an object (mini black hole) be detected, to either confirm or rule out its existence?

I would say that the only way this can be done is making ever more careful observations of the TNOs that are claimed to be affected by this putative black hole.

Ultimately such observations should narrow down the location of the object. Even then, confirming it's existence will be very difficult. It will have a small lensing/magnification effect on background stars, but it would have to get very close (in terms of angular separation) to a background star for this to be observable. It's unlikely that the position could be pinpointed accurately enough to predict when such a microlensing event would take place.

Perhaps a better bet would be to send a probe, like New Horizons, to its likely location, where the gravitational effects could be studied in situ.


How do ground-based telescopes deal with earthquakes? Astronomical telescopes observe particular points in the sky for hours or even days, and they have to remain particularly steady for the duration of the observation, as far as I know. What happens when an earthquake shakes the ground? Does it ruin the observation? If so, how frequently does that happen?

Earthquakes are not uncommon in Hawaii and Chile, where several observatories are built. I only have personal experience with the observatory in La Silla, where a minor earthquake once occurred while I was observing. This was an old telescope, and the image was ruined. Modern telescopes may be able to withstand small seismic activities, but since we felt the ground shaking, I think the image would've been ruined on any telescope.

Luckily, images are not taken in one, long exposure. Rather, they're composed of multiple exposures that are added subsequently. You typically don't expose for more than 30 minutes. A faint object that requires, say, 5 hours of exposure, is then given by the median of 10 half-hour exposures. Why the median and not the mean? Although earthquakes are, after all, rare, cosmic rays quite often ruin a small region on the image, but when you take the median, such outliers are removed$^\dagger \!\!\!$.

In other words, the whole night isn't ruined, only the one exposure that was taken when the earthquake occurred.

Of greater concern are more violent earthquakes, which may in principle damage the telescope. Modern telescopes are prepared to deal with such seismic activity. For instance,

  …ESO's VLT has a complex mechanical system that is automatically activated when
  it experiences a strong earthquake of magnitude 7 or higher. There are
  a series of clamps around the edges of the mirror, which can — at a
  moment's notice — lift the entire 23-tonne mirror off the actuators
  and secure it to the telescope’s support structure.

In 2006, an earthquake close to Hawaii shifted slightly the position of various things on the Subaru telescope, apparently resulting in people there working with pre- and post-2006 numbers, according to former telescope operator Dan Birchall.

Source: ESO, where you can also find more about how to prevent damage in case of earthquakes.

$^\dagger$For instance, if the values in a given pixel of 10 images are $[21,25,22,19,22,27,23,65536,18,24]$, then obviously something went wrong in the 8th exposure, and the median $(22.5)$ is more characteristic of the "true" value than the mean $(6573.7)$.


RA and Dec of Near-Earth-Objects I'm new to the forum.
This might be a silly question. I had a confusion about how RA and Dec are calculated. So, from what I understand RA is the angle made by the position vector of an object(position vector as measured with respect to ECI's origin) with vernal equinox direction. Declination is the angle made by the same position vector with the celestial equator(coincides with Earth's equator).
I wanted to make sure that the position vector of the object is considered with respect to ECI's origin. If that is in fact the case, then wouldn't the RA and Dec of a Near-Earth Object depend on the position of the Earth in its orbit ? (If it's a far away star, then the effect of orbital motion of Earth is negligible). Please correct me if I'm fundamentally mistaken in my understanding. Thanks!

If the RA and Dec in question are geocentric, then they are relative to the ECI origin.
The difference between heliocentric and geocentric positions is significant for any solar system object; on a baseline of 1.0 au, a trans-Neptunian object at 40 au has a parallax of 1.4°.
However, for near-Earth objects, the observer's location on Earth is also significant.
An asteroid passing at 0.01 au has a parallax of 0.24° on a baseline of 6370 km.
The Minor Planet Center lists observations and generates ephemerides in topocentric coordinates.


Why do stars explode? I always hear the narrator of documentaries say that a star exploded because it ran out of fuel.
Usually things explode when they have too much fuel, not when they run out of fuel.
Please explain...

Short answer:

A tiny fraction of the gravitational potential energy released by the very rapid collapse of the inert iron core gets transferred to the outer layers and this is sufficient to power the observed explosion.

In more detail:

Consider the energetics of an idealised model star. It has a "core" of mass $M$ and initial radius $R_0$ and an outer envelope of mass $m$ and radius $r$.

Now suppose the core collapses to a much smaller radius $R \ll R_0$ on such a short timescale that it decouples from the envelope. The amount of gravitational potential energy released will be $\sim GM^2/R$.

A fraction of this released energy can be transferred to the envelope in the form of outward moving shocks and radiation. If the transferred energy exceeds the gravitational binding energy of the envelope $\sim Gm^2/r$ then the envelope can be blown into space.

In an exploding star (a type II core collapse supernovae) $R_0\sim 10^4$ km, $R\sim 10$ km and $r \sim 10^8$ km. The core mass is $M \sim 1.2M_{\odot}$ and  the envelope mass is $m \sim 10M_{\odot}$. The dense core is mostly made of iron and supported by electron degeneracy pressure. The star is said to have "run out of fuel" because fusion reactions with iron nuclei do not release significant amounts of energy.

The collapse is triggered because nuclear burning continues around the core and so the core mass is gradually increased and as it does so it gradually shrinks (a peculiarity of structures supported by degeneracy pressure), the density increases and then an instability is introduced either by electron capture reactions or photodisintegration of iron nuclei. Either way, electrons (which are what is providing the support for the core) are mopped up by protons to form neutrons and the core collapses on a free fall timescale of $\sim 1$ s!

The collapse is halted by the strong nuclear force and neutron degeneracy pressure. The core bounces; a shock wave travels outwards; most of the gravitational energy is stored in neutrinos and a fraction of this is transferred to the shock before the neutrinos escape, driving away the outer envelope. An excellent descriptive account of this and the previous paragraph can be read in Woosley & Janka (2005).

Putting in some numbers.
$$GM^2/R = 4\times 10^{46}\ {\rm J}$$
$$Gm^2/r = 3\times 10^{44}\ {\rm J}$$

So one only needs to transfer of order 1% of the collapsing core's released potential energy to the envelope in order to drive the supernova explosion. This is actually not yet understood in detail, though somehow supernovae find a way to do it.

A key point is that the rapid collapse takes place only in the core of the star. If the entire star collapsed as one, then most of the gravitational potential energy would escape as radiation and neutrinos and there would be insufficient energy even to reverse the collapse. In the core collapse model, most (90%+) of the released gravitational energy is lost as neutrinos, but what remains is still easily sufficient to unbind the uncollapsed envelope. The collapsed core remains bound and becomes either a neutron star or black hole.

A second way to cause a star (a white dwarf) to explode is a thermonuclear reaction. If the carbon and oxygen can be ignited in nuclear fusion reactions then enough energy is released to exceed the gravitational binding energy of the white dwarf. These are type Ia supernovae.


Long term development of Comets When an exocomet passes by our Sun in a trajectory that leads it to establish a Solar Orbit what eventually happens to it?
In addition to its original inertia and vector being altered what are the longer term affects that occur?

Ice chunks would be melted off during closer approaches, potentially blasting non-ice bits off of the exocomet as it goes. This could affect the trajectory of the exocomet both by the inertial force of the separation and because the mass would be reduced.

Eventually it would run out of water, which means it would no longer be a comet--the streaming tail of gas wouldn't exist anymore, even on closer approaches to the sun. What would remain--assuming anything remained at all--would be a hard rocky core.

From there, the orbit would either stabilize or decay to a point that it crashed into the sun. Even a fairly stable orbit could eventually bring it close to other gravitational bodies in the solar system which could throw it into an unstable orbit, or alternatively, it could collide with a planet or asteroid.


If the earth stopped spinning what would happen to the moon A question was posed at work today, "if the earth stopped spinning what would happen to the moon"? Ignoring any effects on the Earth, what actually happens to the Moon? Does it continue to orbit the now non-rotating Earth, or does it fly off into space like a massive asteroid, or more likely some other thing that I can't conceive of?

Note, the Earth stops spinning, not orbiting the sun.

For additional curiosity, what would be the differences if this were an extremely gradual slowing vs an almost instantaneous stoppage?

Earth's current rotation is faster (~1 day) than the Moon's orbital period around Earth (~28 days). This leads to tidal acceleration -- the tidal bulges raised by the Moon rotate ahead, pulling on (and being pulled by) the Moon. Over a long period, this converts Earth's rotational momentum into Lunar orbital momentum. The Earth's rotation slows, and the Moon's orbit expands. Eventually, if nothing else interfered, Earth and Moon would become tidally locked, with the length of a day and the length of a month equal.

If Earth weren't rotating, this effect would happen in reverse. Tidal acceleration (deceleration?) would transform Lunar orbital momentum into Earth rotational momentum. The Earth would slowly start to rotate again, and the Moon's orbit would contract, again until the two were tidally locked.

I'm not in a position to run the numbers, but it should be possible to estimate the final configuration -- how close the Moon would be, and how long the day/month would be.


Question: calculating orbit using another orbit A satellite which is moving around a planet of mass $M$ in a circular orbit of radius $R$ is given a sudden thrust into the center of the planet, so that it deviates an angle $a$.
Find the latus-rectum $l$ and the eccentricity $e$ of the new orbit.

As a hint: a force directed inward does not change the angular momentum, and therefore does not change the semi-major axis nor the period. It's not clear which angle you're talking about, but knowing that the semi-major axis did not change should let you set the problem up as a simple geometry problem.


Thought experiment about the outskirts of the universe Suppose I aim at some object that is 13 billion light years away from Earth. Let's assume I have a twin that pilots a ship that travels in infinite speed, meaning that he can reach that object  and get back here in a split second. On the other hand I'm the poor brother, and I own a ship with top speed of only at 99.9% of c.

We leave earth at the same time, with the same destination. My ship accelerates at constant rate for a full day until we reach max speed, then we enjoy the travel without accelerating for a second time.

1) how long will it take me to meet my brother, and how older will he be?
2) what will be the age of the universe, according to me?
3) what will be my current speed relative to surrounding objects before decelerating? Would I be able to accelerate and reach max speed again? If yes, what's the point of accelerating later rather than sooner, and having a top speed after all?

4) And a plot twist: suppose that while in travel I build a second ship exactly the same as mine. When we're about to arrive we choose a distant point 13 billion light years away in the same direction, and I send a friend on it, who accelerates until he reaches 99.9% of c relative to me. If we repeat that for N times, will we ever reach "free space" where matter didn't reach yet, or some other cosmologically interesting place? How does it look like there? If not, a) can we tinker with any parameters in order to do it "right"? b) any other plans on getting there?

I understand relativity from what I learned in college (long time ago), is there any other theory needed to understand such traveling questions?

I will answer all of your points with zero tact and/or precision in order to fit them all in.

1) It will take you about a "split second" given you stated that it only takes your brother a split second to travel from Earth to the point 13Bly away and back again. He will be exactly said split second older and no more, because as he is traveling faster than the speed of light, the rules governing his and your aging no longer apply, meaning you have created a scenario where both his and your time's are connected and simultaneous.

2) At the point when you meet up with your brother a split second later, the universe will be a split second older... At the time you've reached your destination the universe will be [ > 13By + (the age of the universe when you left) ]. The fact that your ship takes one year to accelerate to and from the top speed is irrelevant given the duration of the travel time.

3) .999c ... Why not, assuming that you have enough "fuel" your imaginary ship should be more than capable than making a single one way trip. There given the distance and time it would take you to get there, there is close to zero reason to accelerate either sooner or later, although the whole near speed of light travel is sort of the point of your question, it would kind of be necessary.

4) The N times scenario is redundant and a little weirder than your other questions, as the speed of light is the same in all reference frames, you would never breach that boundary, however to each previous ship the rate of travel would appear exponential even though there would only be infinitesimal gains. Would you ever reach free space... no, without faster than light travel it is not possible. I would venture a guess that space would look somewhat similar to the here and now regardless where you ended up within your lifetime. a) by tinker, you could assume that you were able to travel as fast as your twin in part (1). b) Think Star Trek, impulse power was sub light speed travel, all warp speeds are some large multiple of c.

Look into relative simultaneity and space-time diagrams for reference on the whole twin paradox issue. Also you could read something like ... the physics of star trek for faster than light travel scenarios with actual physics explanations or problems.


Can an infinite universe explain the Fermi Paradox? Explanation of potential problem:

We're not completely sure exactly how unlikely an event Abiogenesis was, given both the extreme unlikelihood of nucleotides randomly assembling into anything making sense and the vastness of primordial ocean and extent of time it existed. And still, despite their vastness, when involved with probabilistic chain of events we're quickly getting numbers by which planetary scale pales.

It may very well be that the event was unlikely enough to be nearly impossible not only on the planetary scale, but even on the galactic scale - or even on scale of the observable universe.

My proposed solution to this problem:

Regardless, with the actual universe being infinite, it would be bound to happen somewhere. Still, occurrences of life could be separated by many times the width of observable universe, and none of them could ever contact another.

...which would offer a possible explanation to the Fermi Paradox.

Does my proposed solution address the problem I outlined above, or is there a flaw in my reasoning/understanding?

The true explanation to the Fermi paradox is not known. There is a very long list of proposed explanations, some of which are listed here:

https://en.wikipedia.org/wiki/Fermi_paradox#Hypothetical_explanations_for_the_paradox

What you're saying is literally the first one on the list, also known as the Rare Earth hypothesis.

https://en.wikipedia.org/wiki/Rare_Earth_hypothesis


Why is this moon red? I wish I could attach a good quality photograph of this red color moon I'm looking at right now.

I searched on internet but there was no lunar eclipse scheduled to occur on 23rd December 2018

in  HYDERABAD, INDIAN SUBCONTINENT around 7:20pm
Please tell me why the moon is red tonight:) ?

First of all there could not be an eclipse on December 23rd because that is one day later than the full moon.

There can be a couple of reasons for the moon looking red or orange:

It is near the horizon, and being viewed through the atmosphere over a large city.  Air pollution (for example from cars, trucks, fires etc) push small sooty particles in to the air.  These particles scatter blue light more than red, so there is less blue light coming towards you directly from the moon.
Sometimes sand from deserts can get blown up into the atmosphere, and it can get carried a long way.  The sand has a similar effect to the soot.


What is the main source of opacity in sunspots? In the first version of my answer to Are sunspots vertically displaced from the surrounding photospheric plasma?, I naively used Kramer's law to attempt to show an inverse relationship between opacity $\kappa$ and temperature $T$ due to bound-free or free-free absorption in a region with density $\rho$:
$$\kappa\propto\rho T^{-7/2}$$
Rob Jeffries pointed out that this might not be accurate in the photosphere, even though it works well deeper inside most stars (though for high-mass stars, Thomson scattering dominates and $\kappa$ is approximately constant). After some more reading, I concluded that this was right. At lower temperatures, interactions with ionized hydrogen mean that $H^-$ is the major contributor to opacity, and it follows a law
$$\kappa\propto\rho^{1/2}T^9$$
I believe the transition of dominance happens around $\sim10^4\text{ K}$, meaning that $H^-$ opacity should be the major contribution in the stellar photosphere.

At even lower temperatures, however ($\sim4000\text{ K}$), molecular scattering can be important (see Tennyson et al. (2007)). While this is not necessarily important at all points in the stellar photosphere, I think that it should be important in sunspots, which can have temperatures in this range.

What is/are the major source(s) of opacity in sunspots? Does molecular scattering dominate, or is $H^-$ still important? If both are important, to varying degrees, how can the mean opacity be calculated?

My knowledge of stellar atmospheres and atmospheric models is limited, so I apologize if this comes down to being a basic question.

In Ignjatovi´c et al. (2014), ion-atom absorption processes are discussed. They suggest that these processes are of equal, if not greater, influence on the observed sunspot opacity as absorption processes in $\mathrm{H}(1s)+\mathrm{H}^+$ and $\mathrm{H}(1s)+\mathrm{X}^+$ collisions (where $\mathrm{X}$ is one of the metal atoms $\mathrm{X} = \mathrm{Na, Ca, Mg, Si, Al}$). The main scattering processes discussed within the paper follow the photo-dissociation of $\mathrm{H}_2^+$ and $\mathrm{HX}^+$ by UV photons in the spectral region ($110$ $\mathrm{nm}$ $\lesssim \lambda \lesssim 230$ $\mathrm{nm}$).

I highly recommend reading through the paper—it is very interesting.


Would this effect occur if an ice planet was put into orbit or trajectory? Imagine an earth getting warmer. Humans decide to make their homes ready for a voluntary ice age. All the space agencies work together to get an icy planet/moon into our orbit between us and the sun.

My theory is that because the sun is heating up the ice and blowing it in the direction of earth, it'll make the Earth colder.

That final part is my question: Could ice molten in space eventually reach earth and perhaps be pushed in that direction by its heat?? Not taking into account whether or not it would benefit the earth.

I'm an amateur with no knowledge of physics, chemistry and astrology and want to learn answers to my questions.

Thanks in advance!

Welcome to Stack Exchange! I'm a little confused about your question, but I'll answer it the best to my ability.

First of all, I would like to state that such scenario is near impossible because getting the planet there would be extremely difficult.

Second of all, when the ice is being "blown", it would have to be perfectly aligned with the Earth not just in its position in orbit, but also in its angle of orbit.

Third of all, the ice would have to travel probably hundreds of kilometres, meaning that it would either 1) not make it by getting in orbit around the Sun / melting or 2) get lost in space.

The heat of the Sun can barely move things at all - I assume you're sort of thinking of a Light Sail, but that's light, and ice would be too heavy anyway.

Getting the ice off of there would also be challenging because its connected to the planet, and can't just be "lifted" by heat. That's why the Earth's surface doesn't float / scrape away by the Sun's heat.

Overall, I do not think it would be possible.

Hope that helps!


Can a moon have a moon? Is it possible for a natural moon to have a natural moon? What is the limit to when satellites on satellites becomes to unnatural for it to be possible?

Of course it is possible.

If the bigger moon (A) is far enough from a parent body so that the latter one cannot pull the smaller moon (a) away from it.
So the factors at play here are the mass of the planet, the mass of the moon (A) and the distance between them. These determine what is called sphere of influence (SOI).

So if the moon (a) orbits the moon (A) inside its SOI, it would be possible for (a) to exist there for a long time.
How long that time would be - it's a slightly more difficult question. As the tidal forces and various disturbances might eventually make the system unstable, and eventually (a) might be pulled away from (A) or even fall onto it etc.

I assume that by "unnatural" you mean "impossible"? Because usually when people talk about "non-natural" satellites, they mean "artificial", that is, "made by humans". So it has nothing to do with the type of the orbit. Natural satellites are formed, well, naturally, and artificial satellites are launched by us humans.


What are these faint streaks on the moon? I was poking around this Gigapan image of the moon's north polar region, when I noticed this area:

What is the source of these light streaks?

For position reference, here's a zoomed-out shot (I couldn't easily get coordinates from the Gigapan thing):

i would rather call the streaks you are referring to as ridges.

the surface of the moon is ofcourse not a perfect sphere and the same way the earth had ups and downs, there are ridges on the moon too.

you say that most of them are diagonal - top-left to bottom-right but on the right hand corner of the image, there are ones vertical ones. So, there is no specific orientation for these "streaks".

on the other hand, if there were streaks around a crater such that concave part of the streaks point towards the crater, they are due to shock waves travelling outwards from the point of impact.


Are the planets Mercury, Saturn, Uranus and Neptune losing mass? My understanding is that Venus, Earth, Mars, and Jupiter are all losing mass due to their gravity, temperature, and the solar wind.  But what about Mercury, Saturn, Uranus, and Neptune?  Mercury has a magnetic field and so typically would repeal the worst of the solar wind and no atmosphere to lose, only an exosphere.  I do not know the calculations to determine if Saturn, Uranus, and Neptune are losing mass though.

Anything that's working, spending energy is losing mass,  Einstein Mass-energy equivalence of SR along with radiant energy(electromagnetic and gravitational radiation) would be effecting all the planets, but at a different rate for each one.  https://en.m.wikipedia.org/wiki/Radiant_energy
  And.   https://en.m.wikipedia.org/wiki/Mass_in_special_relativity
  These should help I think


Does anything orbit the Sun faster than Mercury? Mercury's orbital period around the Sun is about 88 days.  Comets and other things have gotten closer to the Sun than Mercury does.  But has there ever been an asteroid or some other body discovered that has a shorter orbital period?  Are there theoretical constraints on if such a body can exist?
The IAU defines a planet as one that has swept out its orbit.  Does the neighborhood of Mercury extend to the Sun?

The recently discovered asteroids 2019 LF6 and 2020 AV2, each taking 151 days to orbit the Sun, have the shortest periods currently listed in the JPL Small Body Database.
Vulcanoids are difficult to detect from Earth; none are known yet.
To remain in such an orbit, Evans and Tabachnik 1999 estimate a minimum diameter of 100 m and a semimajor axis between 0.09 and 0.21 au.
There are various competing definitions for a planet's neighborhood.
Soter 2006 says:

Two bodies share an "orbital zone" if their orbits cross a common radial distance from the primary and their periods are nonresonant and differ by less than an order of magnitude.

A Sun-grazing asteroid with perihelion 0.01 au and aphelion 0.31 au (Mercury's perihelion) would orbit in 23 days, well above the 9 day minimum for that definition.


What are angular sizes of thin and thick disks in our Galaxy? What are diapasons of galactic coordinates for thin and thick disks of our Galaxy?

The sun is usually taken as the center of the galactic coordinate systems. Our solar system is well within the thin disk of our galaxy (population I stars). So taking the headline question literally, we see the thin disk as well as the surrounding thick disk in almost every direction, meaning any galactic latitude and longitude.

The radius interval, where we find mostly stars of the thin or thick disk, varies with angles. Along the galactic plane the radius interval is across the galaxy (up to about 30 kilo parsecs, varying a bit with direction, because we are not in the center of the galaxy) for the thin disk, containing the sun as origin; perpendicular to the galactic plane it's about 1 kilo parsec to both directions for the thin disk, followed by a respective radius interval from 1 to 3 kilo parsecs for the thick disk, if I follow the Wikipedia articles referenced in the question.

The region dominated by the thick disk is the farther away from us the closer we look along the galactic plane.

There are certainly no sharp boundaries between the star populations. Actually the thick disk penetrates the thin disk, but it's much sparser populated.

The Gaia project may give a much more detailed answer about the distribution of stars within our galaxy in the next 5 to 10 years.


Will the Universe end by time stopping? I know there are several theories as to how the universe might end, if it does. Is it possible for it to end by time stopping? A friend of mine told me about this theory, but her explanation that it would be caused by heat death doesn't have me convinced.

There are several theories on how the universe might end by time stopping. The problem with cosmology and some other parts of theoretical physics is that it cannot be proven, nor reproduced or tested. This makes it often more a faith/religion than a science. So, if you notice some irony in my answer, please, it's because of this.

There are two types of theories which fantasize about the end of the universe by end of times. I freely call those type the formula type and transcendental type theories.

The first theory I will only briefly describe, because they end up transcendental anyway. There are several models on the universe that will describe what the universe (or universes) looks like and how they will evolve in time. The problem with these models is the variable time. If you have an endless time or an endless amount of universes, you end up with an endless amount of possibilities, which will result in all things will eventually happen. So, somewhere in time or a universe we will all become filthy rich and marry a gorgeous, intelligent, funny, friendly, etc., woman/man that never says no (to you). Because of this anomaly scientists have started to theorize that time must end at some point. If you believe these models to be correct, the scientists might be right. The problem in this lies that you first must believe the model is right.

The transcendental theories are a bit more difficult to explain and I will briefly leap out of astronomy into philosophy. In the 18th century a great philosopher wrote a book on the critique of pure reason (Kritik der reinen Vernunft). In this book he was looking for a transcendental doctrine of method and elements. The transcendental part of his doctrine was that it had to be build up on a priori knowledge that was not based on experience. This was his pure reason, because the elements and the methods are not tainted by (personal) experience from an observer. They are true in itself.

Why is this interesting? Because Kant (that great philosopher) said that there were two transcendental elements, space and time. I will skip space, but must say something on time, because someone thinks it might end. Kant states that time is an a priori element of knowledge in pure form. Kant uses five reasons, why time is universal and not empirical. They all can be brought back to the fact that we cannot see things/phenomena to coexist together or successively when there is no concept of time. Therefore, time is an a priori element of pure form.

Now, what has this to do with the ending of the universe in respect to time? The theory you mentioned as heat death is such a transcendental theory. The theory states that the universe expands until the universe is too big to be heated by the energy provided by the matter in the universe. This theory models that the universe follow the same rules as any thermo dynamic system and therefore at a certain point in time all energy, matter and temperature are evenly distributed in this huge universe. Because all is evenly distributed no stars will be created and all processes come to a grinding halt. Since the state before, now and after are the same, the scientists now determine this as the end of time. The formula theories also end up making some theory that time will end.

The problem with these theories is that time will never end. According to Kant time is an a priori element of pure reason and will be there despite of what we observe. Because the state of the universe doesn't change, doesn't mean that there will be no time. Time has just stopped. Theories that state that time will end, do not comply with Kant's critique of pure reason, which states that time is the formal condition of all phenomena whatsoever.

If it didn't entirely answered your question, I hope it was entertaining and educational.

Kind regards,
MacUserT


What is the duration of the era during which it will be possible to perceive a total solar eclipse? Total solar eclipses occur because the Moon and the Sun have the same apparent size in Earth's sky — the Sun is about 400 times wider than the Moon, but the Moon is about 400 times closer.
It's my understanding that the Moon is currently receding from the Earth at about the rate our fingernails grow (about 1.6 inches, or 4 centimeters, per year). Lunar retreat rates in the distant past may have been slower than this.
At some point, the Moon will no longer be able to fully block the Sun. (This question on StackExchange addresses the period from now onwards; one solution offered is that the Earth will have its last total solar eclipse in about 600 million years.)
However, what I'm interested in is determing the total duration of that period when it will ever be possible to perceive the Sun's corona (and other effects such as Baily's beads) from Earth by means of a total solar eclipse. This requires knowing both how long it is until the last eclipse will occur (~600 million years) and the time that has elapsed since the first such event happened. Note that the critical period I'm trying to discover is that window of time during which the apparent sizes of the Sun and the Moon, as seen from Earth, are approximately the same size, as they are now.
(In case anyone's interested, my reason for asking this is to try to determine the probability that we humans should have evolved at that point in time when it's possible to perceive a total solar eclipse, to further a pet theory of mine that suggests that it may not be a coincidence.)

There are a lot of competing models for how fast the Moon is retreating from the Earth, but they all agree the rate of distance increase is non-linear. That means any answer to this question is going to involve a lot of error.  I think there are two different questions here and I will address them separately:
1. How long ago could the whole circumference of the Corona be seen from a hypothetical observer on Earth?
No matter how close the orbiting moon was to the Earth, Bailey's beads and some of the corona would be visible just as the Sun became fully obscured by the Moon.
If you are wondering when the entire circumference of the corona might be visible during a solar eclipse, NASA says the corona extends to about 12 solar radii.

So the whole circumference of the corona presumably wouldn't be visible if the moon was 12 times closer to the earth.  There a few different guesses for the Earth-Moon distance in the past, and Franco and Hinnev show Webb's TD (Tidal Dissipation) model here contrasted against constant extrapolated recession models:

Here, some of the points on the plot denote geological tidal evidence for Earth/Moon proximity. If we believe their TD model, the answer is about 3.9 billion years.
1. How long ago could the angular diameter of the Sun and Moon be the same in the sky as viewed from Earth?
Remember that angular diameter of a sphere is $\delta = \arcsin(d/2D)$, where $d$ is the diameter of the sphere and $D$ is the distance from the observer to the sphere.  The angular diameter of the Sun as seen from the Earth is then $\delta_s = \arcsin(d_s/2D_s)$ and the angular diameter of the Moon is $\delta_m = \arcsin(d_m/2D_m)$, where the subscripts $s$ and $m$ denote the Sun and Moon.  Setting them equal and simplifying yields the common rule of thumb for determining if an exact full eclipse is possible: $D_m/d_m=D_s/d_s$.
Wiki gives Earth's perihelion at $D_s = 1.47095e8 km$, the Sun diameter $d_s = 695,700 km$, the Moon diameter $d_m = 1737.4 km$ so solving for $D_M$ gives us 367,346 km.  This means that the Moon would block the Sun entirely if the Earth was at perihelion and the Moon was 367,346 km from the Earth and all three bodies were colinear.
The current moon's apogee is 405,400 km, so it still would barely eclipse at 38,054 km closer.  This is about 6 Earth radii, so if the above chart is to be believed using their 'TD' model, the earliest time the sun and the moon could have had the same angular size is 1.2 billion years ago.  Of course, this guesstimate doesn't account for the fact that the sun's size isn't constant.  It also doesn't account for the fact that the Moon and Sun aren't perfectly spherical.


What study profiles could land me the job of astronomer? I realize there are very few to none universities that offer direct specialization of astronomy. If I aspired to be make astronomy my career choice, I'd likely have to pick something like physics and specialize later - or something similar - but that's just my hunch, and I'd prefer to know for sure.

What educational and professional paths optimize one's education and skill set for the occupation of astronomer? How do I become one?

What you need would be at least the following:

Physics: astronomy is physics in the end, and you'll need strong basis in physics;
Math: as for any good education in physics, you'll need a reasonable amount of math to be able to become an astronomer;
Programming: it's almost impossible nowadays to get away from it. Being an astronomer today implies lots of data processing (to reduce the data, to interpret it etc.), and it involves a lot of coding.

And, as far as I know, in lots of countries/universities, you can find at least few astronomy & astrophysics courses. In some countries, you can even find dedicated master degrees.

With all of that, you'll be ready to start a PhD in astronomy to become a professional astronomer.


What's the distinction between "planetary formation mechanisms" and "gravitational collapse"? As I was curious about the exact identical comment to my title here: https://astronomy.stackexchange.com/a/16135/10102

And as it seems no one asked it yet, I'm simply interested in the difference of the both, so:

What's the distinction between "planetary formation mechanisms" and "gravitational collapse"?

"Gravitational collapse" is not the generally used name of any planet formation model it simply refers to the collapse of a cloud of gas or collection of objects under the influence of their mutual gravity. "Planetary formation mechanisms" simply refers to mechanisms which form planets, all of which involve the mutual gravitational attraction of the planetary components at some phase.

In the context of the answer you refer to, there "gravitational collapse" refers to where a small object (like a brown dwarf) forms through the same direct collapse and fragmentation process that forms stars.

Planets on the other hand are thought to form in the disc that is inevitably created around a protostar during the star formation process. This might be through something called gas disc instability, or the more likely model at present is something known as the core accretion model.


What evidence is there of Earth-Like internal features of Europa? This question is inspired by "What are the Earth-like features of Titan?".  According to NASA's Europa overview, Europa is believed to have an iron core, rocky mantle and an ocean under the frozen surface.

Does the tidal heating of Europa's interior by Jupiter make it a smaller analogue for Earth?

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

According to this geology site Earth's core is predominantly iron and nickel.

and this space site says that Earth's mantle is predominantly silicates

So the internal similarities so far is that they both possess iron cores and silicate rock mantles. However the core of Europa is mixed with sulphur rather than Earth's nickel.

Another similarity between Europa and Earth could be the composition of the oceans. NASA speculates that Europa's oceans are also briny with salts and minerals dissolved in it. This is mostly theorised due to what we know about planet formation.

However this also comes with many differences, the Galileo spacecraft did not detect a magnetic field from Europa, which leads to speculation that, whilst Europa may have an Iron compound core, it is probable not molten any more.

Conclusion

Interior Similarities:

Iron compound core
Silicate rock mantle
Salty, briny, Oceans

Differences:

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


What does the velocity dispersion of a galaxy mean? Also, what is the advantage of getting information about the velocity dispersion of a galaxy?

Definition of the velocity dispersion

From the title of your question, I'm unsure whether you actually know what "dispersion" means: The dispersion of some numbers is the spread around their mean, usually taken to mean their standard deviation. If you measure the velocity of several light sources (from the Doppler shift of their spectral lines) that are gravitationally bound, they will have some average velocity$^\dagger$. But since the light sources — which could be stars or gas clouds — move around each other, some move a little faster, and some move a little slower. This will cause their combined emission line to be broadened, instead of being a very narrow line. The central area of this line that emcompasses roughly 68% of the area of the line has a certain width. If the distribution is Gaussian (which is typically the case), then this width, divided by two, is the standard deviation. This is the usual definition, I think, but note that other definitions could be used, e.g. the full width at half maximum which differs from the standard deviation by a constant factor of $2\sqrt{2\ln2}$ for a Gaussian.

A relation between the mass, size, and velocity dispersion

The velocity dispersion $\sigma_V$ of gravitationally interacting "particles" such as stars in a galaxy, or galaxies in a cluster, provides a measure of the total mass and its distribution. The virial theorem, which describes the total energy of a bound system, can be expressed to show that the components of a galaxy of mass $M$ and radius $R$ will have characteristic velocities of the order of the velocity dispersion
$$
\sigma_V = \sqrt{\frac{GM}{CR}},
$$
where $G$ is the gravitational constant, and the observationally determined factor $C$ depends on the geometry and the actual mass distribution, as well as on whether the system is rotation-dominated (such as disk galaxies) or dispersion-dominated (such as elliptical galaxies) (Binney & Tremaine 2008). For dispersion-dominated galaxies, $M$ refers to the total, dynamical mass of the system, and $C \simeq 6.7$ for various galactic mass distributions (Förster Schreiber et al. 2009). For rotation-dominated galaxies, the appropriate mass is the mass enclosed within $R$, and $C \simeq 2.25$, again averaged over various galactic mass distribution models (Epinat et al. 2009).

The velocity dispersion is rather easy to determine with spectroscopy, since it broadens the spectral lines (other processes also broaden the lines, so one must be careful to disentangle different effects). With photometry, the size of the galaxy can be determined (note though that the size can differ when observed in different bands). Thus, even if you don't know whether a galaxy is rotation- or dispersion-dominated, a moderately precise measure of its mass — one of the main characteristics of a galaxy — can be obtained (e.g. setting $C=5$ won't be too far off, and in astronomy we're usually happy with such small errors).

Other relations

Because of various scaling relations, $\sigma_V$ can also be used to probe other quantities, or other quantities can be used to probe $\sigma_V$. For instance, the M-sigma relation relates the mass $M_\mathrm{BH}$ of a galaxy's central, supermassive black hole to $\sigma_V$ (through $M_\mathrm{BH} \propto \sigma_V^4$, roughly), and the Faber–Jackson relation relates the luminosity $L$ of a galaxy to $\sigma_V$ (through $L\propto\sigma_V^4$, more roughly).

$^\dagger$In astronomy, this average velocity is called "the systemic velocity"), and is given by the expansion rate of the Universe (the Hubble law), plus in general a small velocity in a random direction (the "peculiar velocity" due to local dynamics).


What measurements helped us in understanding 'Oumuamua's location and speed? What measurements helped scientists understand 'Oumuamua's location and speed? I wanted to know how it was achieved and with what accuracy if possible.

Like most asteroids and comets, 1I/'Oumuamua's trajectory was determined entirely by measuring its position in optical images over several days.
The earliest data came from automated, ground-based surveys such as Pan-STARRS and the Catalina Sky Survey,
then targeted follow-ups from various other asteroid observers.
The object's evident extrasolar origin motivated additional observations with more sensitive instruments including the Hubble Space Telescope.
'Oumuamua did not pass near enough to the Earth for a radar observation.
The Minor Planet Center collects those observations here.
In the table of orbital elements, the 0.4 arcsecond residual is the root-mean-square difference between positions computed using those elements and positions actually observed.
In the observation table, the magnitude column indicates which photometric filter was used, e.g. G or R for greenish or reddish bands of visible light.
Pan-STARRS filter w passes a wide band including most of the visible band and some near infrared (Tonry et al. 2012 table 4).
JPL's estimate of 'Oumuamua's orbital elements uses the same observations and includes uncertainties.
They estimate about 1000 km error in perihelion distance (q), 23 seconds error in perihelion time (tp), and 1 arcsecond error in inclination (i) and ascending node longitude (node).
If you follow the Ephemeris link from there, and specify
Ephemeris Type: VECTORS
Table Settings: quantities code=2x
then HORIZONS gives Cartesian position and velocity uncertainties based on dynamical simulation.
These are larger now than when it was observed, and still larger a few months before discovery.


Where will I be able to see the Lyrid meteor shower? So supposedly the Lyrids meteor shower will take place between early morning April 16th to 22nd - correct me if I am wrong. I am aged 15 and am interested in skywatching, I just wondered will it be visible from the northern Hemisphere? Or will it be like the lunar eclipse and only be visible in the Western Hemisphere? Thanks.

You should definitely be able to see this meteor shower. It will be in the north east from your viewpoint near the star vega.

http://www.astronomy.co.uk/skymap

http://earthsky.org/astronomy-essentials/earthskys-meteor-shower-guide

You may find these websites useful. If you're interested in amateur astronomy, there is free software for interactive star maps. I use one called stellarium. I'm also an amateur astronomer and I've learned so much with it. You may enjoy it as well.


What is the connection between moons, rings, and gaps? I was watching this video and they said that the rock probably has a moon. Is the reason they assume it has a moon is because there is a gap between the rings created by the moon's orbit around the rock? Or the rings created by the moon breaking down in its orbit and leaving debris behind in its orbit?

It's mainly by the clear definition of the rings, and their mere existence.

Without a moon the rings would be short-lived, hence unlikely to be detected just in time shortly after they've formed. And they would tend to wash out to broader rings. The space between the rings seems to be empty. That's easier to explain by one or more moons keeping the gap clean, and keeping the rings on their track (shepherd moons).

A way of thinking of the shepherd moons:
Our Earth's moon slows down Earth's rotation by tidal forces and friction.
Think of a ring a little inside the orbit of a small moon as an analogon of Earth inside the orbit of Moon: It's slowed down by tidal forces and friction (non-elastic collisions between ring particles). Slowing down ring particles results in an orbit closer to the planet/asteroid. This way a small moon can act repulsive to a nearby ring, confining it.

More details about the dynamics of planetary rings, including confinement by shepherd moons, see this paper, p. 491 ff.


Reasoning questions on eclipses why doesn't a solar eclipse occur in every new moon?

I know one reason is the inclination of Earth's orbital plane. But is there another?

Let me put it this way.  If the Moon's orbital plane where exactly aligned with the Earth's orbital plane (which, by the way, includes the sun, not surprisingly) then there would be a solar eclipse on Earth during each New Moon.  But, in fact, the Moon's orbital plane moves about by many degrees in a very complicated fashion.  Figuring out why this was and how the plane moves kept Isaac Newton busy for a few years.

Both the Moon and the Sun subtend about 1/2 of a degree on the sky.  The distance to the Moon also changes considerably, so its angular extent varies around this 1/2 degree. So, to get a total solar eclipse, they need to be aligned to a very high degree and the Moon needs to not be near the far edge of its distance range.


If you put all planets between earth and moon what would happen? It's often discussed that you can fit all the planets in the space between the Earth and the Moon. But if they actually did teleport there instantly what would happen next?

Assuming their initial direction and velocity was the least problematic, the most likely to "work". Would they all smash together or fling apart? Or do we have any idea?

https://futurism.com/you-can-fit-all-of-the-planets-between-earth-and-the-moon

Talking about instantly teleporting planets isn't very meaningful if you want to get scientific answers, as PM 2Ring pointed out. But is is almost certainly not physically impossible for them to have trajectories that let them all pass between the earth and moon at the same time. But the earth and moon would have to start out farther apart than they are now because all the planets passing in between would pull them closer together, and we want them to be at the current distance just at the moment the planets pass between.

The biggest issue here is tidal forces. Planets pull harder on the near side of other objects than on the far side, and the effect is much stronger the closer you are. Whichever planet was almost touching Jupiter would be torn apart by this difference in force - Jupiter's gravity would be pulling on the near side faster than the far side could catch up. The only way to avoid this is to have the planet passing by very fast, so it would only be in the area of maximal tidal forces for a very short time.

What exactly happened would depend on the exact trajectories. Obviously there are an infinite number of ways a bunch of things can fly past each other. The reason everything doesn't fall into Jupiter in this scenario is that they are moving fast enough and in the right directions. They will all definitely curve towards it, though. Its gravity would dominate the situation, but the gravity of all the planets would have to be considered if you were a super-powerful civilization trying to make this happen.


Can a Ritchey–Chrétien telescope be collimated using a standard Newtonian laser collimator? Is it possible to use a standard Newtonian laser collimator, such as this one to collimate an RCT?

If so, is the procedure any different than collimating a Newtonian (i.e. make the laser go into the hole by adjusting the mirrors)?

No, the fundamental designs of the two types of telescope are in this respect quite different.

The Newtonian laser collimator works by shining a laser through the eyepiece holder to the secondary mirror which reflects the beam onto the primary. The primary, usually a parabolic mirror but always concave, is adjusted so that the beam is reflected back to the secondary mirror such that the reflected beam returns to its point of origin.

A RCT, or other Cassegrain-like telescopes have a convex, possibly hyperbolic, mirror as the secondary. While it is possible to shine the laser through the eyepiece holder to the secondary mirror, and the mirror can be adjusted to reflect the laser back to the point of origin, it cannot be reflected onto the primary and then back to the secondary*. Any adjustment of the secondary to reflect the laser back to the origin only ensures that the incident surface of the secondary mirror is perpendicular to the line-of-sight through the eyepiece holder. It does not ensure that the secondary mirror is aligned correctly with respect to line-of-sight through the eyepiece, and neither with respect to the primary mirror.

* Notwithstanding that the two mirrors could be misaligned so badly that this could occur at some particular point.


Globular cluster star density as a function of distance from the center Context: I want to simulate globular clusters in a simple way, just to display the positions of stars.

Assuming isotropy, what would be a reasonable model of the stellar number density as a function or r, the distance from the center of the cluster?

I'm assuming that such a model would have some free parameters as well.

The usual thing is a King model.

There are indeed free parameters. These are the central density, the "core radius" and a tidal truncation radius.

The background and rationale for these models is given in the link. They provide a pretty good representation of the surface density of globular clusters (or indeed open clusters). They require a numerical scheme to "deproject" from the plane of the sky to 3D.

If you find the nitty-gritty of Abel integrals too tricky in the deprojection, then you could always approximate with a Plummer model. This is analytically and computationally easier to deal with, but lacks a bit of physical realism. Central density and a characteristic radius are free parameters here.


Solar Elevation Angles -- Anomaly? "The solar elevation angle is the altitude of the sun, the angle between the horizon and the centre of the sun's disc." - Wikipedia

The meridian is when the sun is directly facing earth and is at exactly 90° overhead at some point along the longitudinal line at a given location.

My question is, if the sun is 93 million miles away and travels overhead at 90° between the Tropic of Capricorn and the Tropic of Cancer, how is it that there are Meridian solar elevation angles reported from relatively nearby cities on Earth of less than 89.99°?

Let's take for instance Calgary, Canada as of today May, 04 2016 -- estimated ground distance from Calgary to where the Sun is at 90° on Calgary's longitudinal line is ~3000 miles. The solar elevation angle (aka "altitude") is reported by the timeanddate website to be 55°! -
TimeandDate

Reversing the angle, using right triangle math:

The base angle, our solar elevation angle, of a right angle triangle is:
    "solar elevation angle"= arctangent (h/a)
    89.998151749049 = arctangent (93000000/3000)

So using right triangle math alone with the presumption of the Sun being 93 million miles away and base distance of 3000 miles we get an expected solar elevation angle of 89.998151749049°.

That makes sense, since if something is 93 million miles away from us and you think of it as a triangle, then the distance between any place on earth and the location on earth where the sun is at exactly 90° overhead -- you should expect the other angle of the triangle to be at near 90°, connected by a relatively tiny sliver of distance in a much elongated triangle to the center point of the distant sun.

If we go the other direction and leave out the presumption that the sun is 93 million miles away. We take h as unknown, we know the reported solar elevation angle of 55°, we can again use right triangle math:

    Let a=3000, solar elevation angle=55°
     h = a * tangent ("solar elevation angle")
     4,284.4440202263 = 3000 * tangent(55°)

So according to my calculations, based on the advertised solar elevation angle of 55° at Calgary, Canada on May, 04 2016 at Meridian, at a ground distance of 3000 miles from where the Sun is at exactly 90° overhead --  the Sun is actually only 4,284.4440202263 miles high!

** Update **: The math would work out for a sphere Earth with a radius of 4000 miles, using the exact solar position and adding  corrective angles due to a presumption of a spherical curvature:
    Calgary, Canada: 51°03′N 114°04′W
    Location of Sun at Calgary Meridian: 16°23′N 114°04′W
    Distance = 2395.386 miles
    Radius of Earth=4000 miles
    Arc distance = 4790.772 miles
    Width of arc = 4509.52059 miles
    Angle to center of the Earth=68.622754110849°
    68.622754110849°/2=34.31137705542450°

This math utilized calculations that may have had a bit less acurracy but to make it work you need to add a bandaid padding of approximately 34.31137705542450° to Calgary,Canada's advertised angles. If you do you arrive at 55°+34.31137705542450° = 89.31137705542450° which is close to what I would expect to see. If, we are truly living on a Flat Earth, then the calculations would be telling us that the Sun height is only 4,284.4440202263 miles high and we can use simple trignometry.
Arc Calculator
Coordinate Distance Calculator

If the math is difficult for you, you can check with an online triangle calculator such as: Ke!san

Your confusion may stem from the way we sometimes draw the Sun (not to
scale) as though it's close to the Earth.

In reality, as you note, the lines of vision to the Sun are nearly parallel:

Note that the angles (measured from the red line, the horizon, to the
arrow) are still different, but for entirely different reasons.

I've actually stolen @Andy's answer (and @Tanenthor's comment), but
I'm hoping the diagram makes it easier to understand.

Re the parallel lines, the 11810x60 image below may help explain: even
though the lines eventually converge (on an object about 10 times the
distance to the moon, still nowhere near the distance to the Sun),
they are nearly parallel when leaving Earth.


What does oxygen on comet 67P/Churyumov-Gerasimenko mean? Recently, the Rosetta mission found oxygen on the comet 67P/Churyumov-Gerasimenko. According to theories, oxygen should not be in the state in which it is on the comet. What are the previous theories that didn't support this, and what is the explanation for the oxygen-rich environment? How will this change our understanding of the formation of the Solar System?

It's not that oxygen should not be in this state, but rather we had not discovered it (e.g. on other comets). One of the reasons for this is that interstellar O2 is very reactive and quickly reacts to form e.g. H20 and O3.

Current models also favoured O2-less comets.

Interestingly, the paper you're referring to also mentions the O2 abundance was rather constant over the half year of observation, which hints to the O2 actually being primordial, i.e. from when the solar system condensed.

To reconcile these  observations with models, a couple of possible solutions are offered: a slightly warmer primordial cloud (from which the solar system formed), or reactions with ices.


Can hot Jupiters cause solar flares? I'm very new at Astronomy, and my knowledge is sparse. I've tried to be conscientious about my Wikipedia research but there's going to be a lot of things I don't know. Thanks for your patience.

BACKGROUND

I had read that hot Jupiters can cause superflares due to magnetic recombination. But checking Wikipedia I came across the following:

  Not all planetary transits can be detected by Kepler, since the planetary > orbit may be out of the line of sight to Earth. However, the hot Jupiters > orbit so close to the primary that the chance of a transit is about 10%. > If superflares were caused by close planets the 279 flare stars
  discovered should have about 28 transiting companions; none of them
  actually showed evidence of transits, effectively excluding this
  explanation.

- Wikipedia

QUESTION

Given this, and among modern astronomers in general, is the theory that hot Jupiters can cause superflares on the outs? Or is the article just saying that hot Jupiters can't explain these particular stars, while the model remains valid elsewhere? Do we have any solid evidence of a gas giant causing solar flares, super or otherwise?

Just going off the Wikipedia article you posted, it says the hot jupiter superflare theory was abandoned.

  The flares were initially explained by postulating giant planets in
  very close orbits, such that the magnetic fields of the star and
  planet were linked. The orbit of the planet would warp the field lines
  until the instability released magnetic field energy as a flare.
  However, no such planet has showed up as a Kepler transit and this
  theory has been abandoned.


Could Dark Energy be a "Cosmic Gravity Background" Could dark energy (the accelerating expansion of the universe) be the result of gravity from the dense past universe "pulling" outward from all directions? The universe theoretically started out extremely dense and hot, and we can see the light reaching us as the cosmic microwave background. I wonder if that same region and denser portions beyond are exerting a gravitational effect that is stretching out everything within toward it, effectively being a cosmic gravity background. If so, is this the explanation for dark energy?

There is in fact a cosmic gravitational wave background. These waves are expected to be stochastic, having originated in the early universe (much earlier than the cosmic microwave background). Random fluctuations were subsequently stretched during inflation, making them observable over many wavelengths. A good and reasonably up-to-date introduction I read is Lasky et al. (2016), which gives some of the most recent constraints on the parameters of the frequency spectrum $\Omega_{\text{GW}}(f)$. The great thing about the wide spectrum is that a number of different tools (pulsar timing arrays, interferometers like LIGO, etc.) can be used to constrain different parts of it, both directly and indirectly.

That said, dark energy is almost certainly not a result of the gravitational wave background. First, there's no theoretical mechanism to explain how such an effect would arise. Second, the background is probably not strong enough to have an influence on the current expansion of the universe (see the graph here to see how weak it is). Finally, dark energy is a relatively recent phenomenon, arising several billion years ago. The background was around long before then, and yet had no effects.


Journal standard for plotting section of skymap? I'm a physics graduate student hoping to plot ~10°x10° sections of the sky in right ascension and declination (eventually I will want to plot points on this section at appropriate locations given RA and Dec, preferably in Python). I've seen square plots on the order of arcseconds, where curvature has a small effect, but that won't apply to these larger plots. I'm not familiar with astronomy journals or a module which aids in this sort of plotting.

Is there a commonly accepted method for displaying intermediate-sized sections of a sky-map? How is it implemented?

I don't think there's a journal standard for this.  An easy and commonly used method is to plot latitude $\delta$ versus longitude $\alpha$, where the aspect ratio is scaled by $\mbox{cos}(\delta_0)$ so the projection is orthonormal at the image center $(\alpha_0, \delta_0)$.  Depending on your region size and on $|\delta_0|$, this may or may not come out looking distorted.   You might be able to pick a coordinate system (i.e. equatorial or galactic) where the latitude is never large enough to give major distortion.

Another common thing to do, arguably better when you're imaging extended objects, is to use a tangent-plane projection, also called a gnomonic projection.  This projection depends on the chosen center, which is usually the image center.  It is included in some plotting and analysis packages, but none that I use regularly.  I recently coded this up for the Astropy package but haven't really tested it yet.

Projections like Aitoff, Hammer, and Mollweide have been included in many plotting packages, but usually only for whole-sky projections.  You could try to write your own versions but I'm not sure it would be worth the trouble.

(Whatever you do, make sure longitude increases to the left since you're looking at the sky not the ground...)