tiny diamonds in meteorite "signature of a long-lost planet": why come to Earth in 2008? ref this paper in Nature re the Almahata Sitta 2008 meteorite shower.

I get that they're saying the diamonds could only have been formed within something the size of planet, which subsequently broke apart in the "early days" of the Solar System.

I get that Saturn and Jupiter are so massive relative to other Solar System bodies they've marshalled fragments into the Asteroid belt, Kuiper belt, outer disk. And that they're still disturbing those regions enough to occasionally pull things into a cometary orbit/become meteors.

This putative planet must have scattered everywhere when it first broke apart/lead to comets and meteors all over the place. But why/how did this fragment wait until 2008 to come to Earth? Where's it been for billions of years? Why didn't it either crash into something a long time ago; or form itself into a stable orbit?

Possibly related, as in: why haven't meteorites originating from early-formed planets all long gone extinct?

Many comets are such lucky ones...
That's why we see them with their icy tails evaporating: They have only recently been scattered into the inner solar system.
Just like the well-studied comet 67P, you could ask "where has 67P been for all those billion years?". For at least this comet, we know the answer: It has been orbiting in the outer solar system, until a close encounter in 1840 and a few subsequent later ones with Jupiter brought its' perihelion closer to the sun, where it started its' cometary activity.

I would place my bet on the origin of those comets being in the outer solar system, because:

The nature article you cite mentions 20 GPa pressure, for which one needs at least a Mercury or Mars type body to generate. This is not necessarily a problem, because Mars-type bodies form quick and early (we know this from theory as well as data), within the first 3-5 Million years of the solar system, when gas is still there. This is about 10 times faster than an Earth-type body, which has 10 times the mass.
It is also thought that Mars-type bodies should form the basic building blocks for larger planets - thus there should be many more of them in the early solar system.
However even if there are many, now there are essentially none in the inner solar system - the asteroid belt contains roughly 0.1% of a Mars mass. The total mass of the Kuiper belt is estimated to be far larger, by factors of 10-100.
If a giant planet instability occurred, it requires a even much more massive Kuipter belt as trigger, after the gas disappears in the early solar system.

So, Summarizing: They're probably out there in the outer solar system, merrily orbiting along.


Would it be possible to calculate the expected frequency of impact craters of all sizes on Earth Can the expected frequency of all sizes of impact craters on Earth be calculated based on the preservation of impact craters seen on the Moon? Taking into account the difference in gravity and atmosphere of the Earth and Moon, and assuming a similar age for both, it should be possible to come up with a reasonable estimate of how many meteorites, etc have hit the earth and their size distribution.

Yes. By looking at a nearby object that hasn't been subject to erosion and has been subject to the same position around the sun, you could estimate the distribution of impacts on the Earth in time and in size. The moon is excellent for this purpose.

Here's an article touching on the subject: http://adsabs.harvard.edu/full/1975LPSC....6.2597N


Are sunspots vertically displaced from the surrounding photospheric plasma? While reading a paper on the helioseismology of sunspots (Cameron, et al.), very brief reference is made to the "vertical displacement of a blob of plasma." This caused me to wonder if the emissive surface of the sunspot itself were perhaps depressed from the remainder of the photosphere. If this does occur, is there any literature on mechanisms? I would be very interested to learn more. There was one paper I came across around formation of sunspots from a toroidal field (Parker); but, it seemed to focus more on the vertical displacement of magnetic field lines rather than the plasma itself.

Thanks.

Yes.
The idea that sunspots are depressed slightly came as a possible explanation for the Wilson effect. The Wilson effect was discovered as the shape of sunspots as viewed from Earth changed as the Sun rotates, in a way consistent with the change in perspective looking onto a slightly depressed region. While this isn't the only explanation for the effect, it's certainly the most prevalent.
More specifically, as Solanki (2003) writes, the depressions indicate a lowering of the layer where the optical depth $\tau=1$ (keep in mind that the bottom of the photosphere is the layer where $\tau=2/3$). There are two causes mentioned: lower temperature and magnetic effects.
Sunspots are cooler than the surrounding areas, as is well known, and thus appear darker. We see the same thing apparent at the boundaries of solar granules: Cooler, darker gas sinks and lets the hotter gas (which is less dense) move upward. Additionally, the opacity $\kappa$ is temperature-dependent, which may impact how far one can see into the star.
Not only is temperature a factor, but so is the magnetic field. Sunspots are, at heart, a magnetic phenomenon, and thus the radial force equation is substantially different. Normally, in a star, the equation of hydrostatic equilibrium is
$$\frac{\mathrm{d}P}{\mathrm{d}r}=-\rho g$$
for pressure $P$, density $\rho$ and gravitational acceleration $g$. However, when the magnetic field becomes important in a sunspot on the solar surface, the force balance is
$$\frac{\mathrm{d}P}{\mathrm{d}r}=\frac{B_z}{4\pi}\left(\frac{\mathrm{d}B_r}{\mathrm{d}z}-\frac{\mathrm{d}B_z}{\mathrm{d}r}\right)$$
where $r$ and $z$ are the radial and vertical coordinates (note the change of coordinate system - $r$ is along the surface, and $z$ is perpendicular to it!). The force from the magnetic field implies a lower gas pressure and a greater depression.


How is a day measured on a gas giant? After watching this JPL video of a Cassini time lapse of Saturn, I wondered since being mostly gases, how do we know how long a day—one rotation period—is?

Is there a usable relation between how fast the upper most layer of gases is and the planet's solid core?

That the rotation period of the bulk mass of a planet is estimated through something with the magnetic field is true.

But let me maybe elaborate a bit in-depth on that.

No planetary magnetic field is a dipole. Earth has strong magnetic anomalies over the Atlantic, there may be ionic disturbances by moons like Io, or your magnetic field is just really crazy, like Uranus and Neptunes.

(A simple, low-order model of Earth's magnetic field, from GFZ Potsdam)

But no matter how it looks, it will usually be constant on lines of latitude and longitude. This in particular signifies that the field and its distortions are created somehow by the bulk mass of the planet, else the field would move relative to a surface map, in the case of Earth.
So all this talk just to establish the following: We can take the shapes of the magnetic field as if they were surface shapes, and by following them determine a rate of rotation.

Now the story doesn't end there. While this sounds nice in principle, to do this in practice, we would need to send a probe and do in-situ measurments of this changing magnetic landscape.

But what can be done from afar, and from Earth is observe ionospheric patterns. Because the ionosphere of a planet is well, ionized gas, at super-low pressures (roughly on the order of microbar to nanobar) it is frozen-in with the magnetic field that makes the particles of the plasma gyrate. This means, the ionospheric plasma follows the magnetic landscape, now translating into a plasma landscape. This plasma landscape is then usually observable from Earth by the radio signals they emit. It is also possible to see this in the ultraviolet, if there is strong plasma activity through auroral bombardment of a planet's upper atmosphere.

For your second question, that's a simple no.
First of all there is no 'uppermost' layer of atmosphere, as the atmosphere is continously decreasing in density with increasing distance from the planetary body in question.
What you're possibly thinking about are the upper cloud layers, as they visibly end at some height.
But wind speeds are much more determined by pressure gradients created by the central star heating the planet, and the atmospheres ability to cool.
There is one class of planet, however where a relation between the wind speeds at roughly 0.1 bar pressure and magnetic dynamo might exist. Those are Hot Jupiters which host atmospheres that might be sufficiently ionized for its winds to be braked down by the planetary magnetic field.
Then there would exist a limiting speed, given a certain planetary field. However this is still a field of active research, but it would give you 'a' relation between the two.


light emission of quasars I'm studying Astrophysics and I have some video's from a past course. I'm currently studying the reionization of the universe. At the current point in the class, the instructor is talking about quasars and the redshift at which the Lyman-Alpha forest disappears. This appears to occur at a redshift greater than 6. The instructor goes into finding quasars with z > 6.5. He states that quasars that far away, "They're only emitting infrared light.". I don't understand that statement. Does the instructor mean they are only visible in IR or those very old quasars only emit in IR? Why would quasars only emit in IR? I understand why we'd only see IR from them.

The statement "They're only emitting infrared light" is wrong, or at least poorly phrased.

The Ly$\alpha$ forest

The Ly$\alpha$ forest (LAF) is caused by the spectrum of the quasar being redshifted along its way; wherever a cloud of neutral hydrogen is located, the part of the spectrum that at that at that position has been redshifted to $\lambda_0 = 1216\,\mathrm{Å}$ has a high probability of being scattered. As the probability of being scattered back into the line of sight is negligibly small, the spectrum effectively suffers absorption, but in fact the light is not absorbed.

Because the lines are caused by the Ly$\alpha$ line, it only affects the part of the spectrum blueward of the quasar's "intrinsic" Ly$\alpha$ emission (well, similar but weaker spectral featured are seen e.g. at Ly$\beta$, and for very strong absorbers such as damped Ly$\alpha$ absorbers you also see metal absorption).

The Gunn-Peterson trough

The further back in time you go — i.e. the higher the redshift you observe — the higher the fraction of the gas of the Universe was neutral, and the denser the Universe was. As you approach the Epoch of Reionization, the LAF "absorption" lines begin to overlap, until eventually, around $z\sim6$, everything blueward of the quasar's Ly$\alpha$ line is erased. This is the so-called Gunn-Peterson trough.

A quasar seen at $z \geq 6$ has its Ly$\alpha$ line redshifted to
$$
\lambda_\mathrm{obs} = (1+z) \lambda_0 \geq 8\,500\,\mathrm{Å},
$$
which is in the infrared. But it is incorrect to say that the quasar only emits in the infrared.

Reionization

The reionization started with the first light-emitting sources, most likely hard UV from the first galaxies; a non-negligible fraction may also be from quasars, but this is probably only significant at later epochs. That is, it started already at $z\sim12$ or higher, and by $z\sim6\text{–}10$ the Universe was almost fully ionized.

So if most gas is ionized at $z=6$, why is all light blueward of Ly$\alpha$ erased? This is because the Ly$\alpha$ cross section is so large: it doesn't take more than a neutral fraction of $10^{-4}\text{–}10^{-3}$ to create the GP trough.


Does Time Exist for Photons? If time slows down the closer you get to the speed of light, does it follow that time does not exist for photons. Do photons - even if they travel across billions of light years of space - arrive at the same time as they are created?
And if this is so, does this not explain the peculiarities of the double-slit experiment?

Time does not exist for photons.  Quoting Wikipedia's somewhat unnecessarily technical page:

  In relativity, proper time along a timelike world line is defined as the time as measured by a clock following that line.

The proper time between any two points along any photon's path in empty space is zero.  So the photon "experiences" no passage of time.

It's not very clear to me how this could explain the double-slit experiment. Proper Time is an entirely relativistic concept, whereas the double slit results are not explained by relativity, but are explained by Quantum Mechanics.


Do all neutron stars spin? I know that some of them are pulsars and pulsars spin very fast, but do all of the neutron stars spin? I would think that they would because of conservation of momentum, but I'm really just not sure.

I think it is absolutely safe to say that all neutron stars spin.

Conservation of angular momentum ensures that as they collapse from a massive stellar core the size (roughly) of the Earth, to something with a 10km radius, their angular velocity increases roughly as the square of the decrease in their radius (i.e. a factor of $\sim 4\times 10^5$. Thus, even if the stellar core had the slightest spin to begin with, then the neutron star will be spinning very fast indeed.

Young pulsars appears to be born with rotation periods that vary from about 0.01 s to perhaps a second or so. They then lose angular momentum as they get older.

The difficulty in answering your question without any equivocation is that once neutron stars spin down to periods of longer than 1-10 seconds or so (depending on the pulsar magnetic field strength - see picture below from Wang et al. 2011) then the pulsar mechanism switches off and the neutron star, to all intents and purposes, becomes invisible (the so-called "pulsar deathline", Zhang 2003) .

It is expected that the spin down will continue, even after the pulsar has switched off, but the spin down rate itself depends on the rotation speed, so the neutron star will never come to a halt, even after tens of billions of years.


Did the night sky ever change in recorded history? I wonder whether there has ever been a major change of the firmament in recorded history, like changes in the positions of stars, changes in constellations, or stars disappearing after going supernova. There've been visible supernovae but had the star in question been visible to the naked eye before the supernova? If so, how did it change what constellation?

There are a few ways to think about this question:

Do the stars change positions in the sky ... such that the layout of say... major stars in constellations appear to move over time?

Are there events that cause sudden changes (changes you might notice overnight ... or within a few weeks or months)?

How long does it take to notice a change?

Precession
The Earth's axis mostly stays oriented in the same direction as we orbit the Sun.  This is why we can get away with saying that Polaris is the "North Star" ... because it appears to be almost directly above Earth's North Pole regardless of what time of the day or night ... or what day of the year you choose to observe (it is roughly within 2/3rds of a degree from the pole).
But it turns out the Earth's axis does "wobble" like a spinning top.  This wobble is very slow and takes thousands of years to complete a cycle (nearly 26,000 years).  At the end of this century (roughly the year 2100) Polaris will be a little closer to the pole ... only 1/2° away from the true pole (today it is about 2/3rds° away from the pole).  That is as close as it will get ... and then our pole's precession cycle will start taking it farther away from the pole.
This "precession" shift causes all stars in the sky to shift their position very slightly from year to year and it's mostly not enough to notice in a single human lifetime ... but it is easily noticed when measured across thousands of years.
There are other more subtle cycles, but Precession is noticeable across large amounts of time (thousands of years).  The position of the Sun in the sky at the equinoxes changes every year and this causes the Sun to appear to be moving across the zodiac over thousands of years.
The sky has a coordinate system used to catalog the locations of stars and other objects.  That coordinate system uses a "declination" to measure where a star is located relative to Earth's North/South ... it works much in the same way that latitude measures geographic locations on Earth.  There is also a "right ascension" value that works a bit like longitude on Earth.  Except Earth is spinning (and longitudes remain fixed to the planet) ... so we needed a way to mark a fixed position in space to have a sort of celestial longitude.  That measurement is called Right Ascension and it needs a "0" position.  So the zero position is established by the location of the Sun in the sky at the Vernal Equinox (the start of Northern Hemisphere Spring).   It even got a name... the "First point of Aries" ... named because, when it was established a few thousands years ago, it was in Aries.  Today ... it's migrated all the way over to Pisces (and is getting closer to the boundary of Aquarius).
These are changes caused by Precession and noticeable ... but mostly only when measuring over hundreds of years or thousands of years.  It is actually noticeable just from year to year ... but only with precise measurements.
Precision causes the entire sky to shift in the same direction (and by a very subtle amount).  This means that the locations of the constellations will shift ... but it wont explain changes to their shapes.   For that, we have to look at other factors.  In reality it isn't the "sky" that is shifting ... but rather the orientation of Earth's axis that changed.
Proper Motion
Stars are moving ... and they don't all move in exactly the same direction.  But the stars are so distant that the movement is mostly only noticeable with precise instruments.  Regardless, they do move.
Here for example, is the Big Dipper asterism, but I've turned on Proper Motion vectors ... the vectors (shown in a light brown color) indicates the direction of that star's motion (relative to us) and longer vectors indicate great proper motion.

Notice that many of the stars of the Big Dipper are all moving in the same general direction at about the same speed ... including Alcor/Mizar, Alioth, Megrez, Phecda, and Merak.  In this diagram, those stars are all moving up and to the left (relative to this frame).
But more importantly... notice that Alkaid and Dubhe are going in roughly the opposite direction.  These stars are moving down and to the right.  This means that over thousands of years, this asterism will become distorted and will eventually not resemble the "Dipper" shape anymore.
Incidentally, the stars in the dipper that are going in the same direction are all part of an Open Cluster called the Ursa Major Moving Group.  It isn't as obvious that it's an open cluster because it is relatively close (not all stars in Ursa Major are members of the cluster.)
Stars that physically closer to our Sun will may appear to have much faster proper motion.   It isn't so much that the stars are necessarily physically moving faster ... but since they are close, the changes are easier to spot.
Here, for example, is the proper motion of Rigel Kentaurus (aka Alpha Centauri).

In this frame (current time) the star is in the upper-left section of this chart.  It is in Centaurus (hence its name) but near Circinus.  But notice it has a VERY long proper motion vector which is headed toward the right side of this chart in the direction of Hydra.
Its proper motion vector is much longer than the other stars ... but Rigel Kentaurus is just a little over 4 light years away (one of our closest neighbors) -- so its relative motion is more noticeable.
While Rigel Kentaurus makes the top-20 list for stars with the highest proper motion, it's not the winner ... that spot goes to Barnard's Star (in Ophiucus).  But Barnard's Star isn't very bright... at magnitude 9.5 it isn't noticeable to ordinary human vision and requires a telescope to see it.  But it is moving just a little under 3x faster than Alpha Centauri A & B.
If you're interested, here's a top-20 list:  https://www.cosmos.esa.int/web/hipparcos/high-proper-motion
Transient Events
Certainly events such as bright comets would change the sky -- but those aren't stars.  Super-Novae events, on the other hand, are stars ... and these are easily noticeable if they happen in our home galaxy.  They don't happen very often.  The estimate is that, on average, they happen in a galaxy our size about once every hundred years.  But's not like a clock where they happen with periodic regularity ... huge amounts of time may pass without any of them ... and then several could occur timed much closer together.
Since Super-Novae events can be quite bright, they can even be visible in the daytime sky.  They don't last particularly long ... they quickly brighten up ... then begin to dim over weeks and ... after a few months they may no longer be noticeable (but they will leave behind a Super Nova Remnant -- deep sky nebulae such as the Crab Nebula and Veil Nebula are examples of Super Nova Remnants.)


How can cosmic inflation make an infinite universe homogeneous? As it is explained in this video, one of cosmic inflation's observable effects is the homogeneity of our universe. Inflation allows two points on the different sides of the observable universe to be causally connected at some point in time, so they may exchange their mass densities and temperatures, which are then going to be about the same for both of them.

While I do understand, that this effect doesn't have to end on the edge of the observable universe and might go on for whatever distance the rate of inflation will make it to, I do not understand how this could be valid for the entire Universe if it is infinite. From my guessing, if the Universe is infinite, there will allways be two points, that was never causally connected, no matter how fast it expanded at the inflation period. That will mean, that, while the Universe should be homogeneous on the sufficiently large scales, it doesn't have to on the scales even larger.

I guess I either don't fully understand the idea of cosmic inflation or don't get the particular way the Universe is infinite in.

Inflation is used to explain why the observable universe is extremely homogeneous.

Without inflation, we can do the following crude calculation. The cosmic microwave background was formed about 300,000 years after the big bang, at a redshift of about 1100. Thus causally connected regions at the epoch of CMB formation would have a radius of $\sim 300,000$ light years, which has now expanded by a factor of 1100 to be $3.3\times 10^{8}$ light years in radius.

This can be compared with the radius of the observable universe, which is currently around 46 billion light years. This means that causally connected regions should only be $\sim 4 \times 10^{-7}$ of the observable universe, or equivalently, patches of CMB of $\sim 2$ degrees radius on the sky are causally connected. This is clearly not the case as the variations in the CMB are no more than about 1 part in $10^{5}$ across the whole sky.

Inflation solves this by allowing previously causally connected regions to inflate to become larger than the entire observable universe.

You appear to understand this quite well, so I am not entirely clear what your question is. We cannot know whether the entire universe is homogeneous, since we cannot measure it. The cosmological principle is an assumption that appears to hold approximately true in the observable universe, but need not apply to the universe as a whole. Indeed it is not absolutely true in the observable universe otherwise it would be quite uninteresting, containing no galaxies, clusters, or other structure. I think the only requirement on inflation is that it blows up a patch of causally connected universe so that it becomes much bigger than the observable universe at the current epoch.


Algorithm to stack astronomical images I'm looking for a simple algorithm to compare astronomical images (of the same sky region) against each other, compute their movement and rotation, to finally stack them.

At the moment I'm already having a more or less working algorithm. First I extract all the stars out of an image (including information like brightness and FWHM), and then I walk through all the resulting "points" and create triangles out of the current point and those two other stars that have the shortest distance to this star.

This list of triangles is created for every image. After this I take one image as a reference and then I walk through the list of triangles in the reference image to find a triangle in the other image with the same length of each side of the triangle (I also "allow" some tolerance due minimal relative differences of the star positions in each image). For this matches I calculate movement and rotation relative to the reference image. The last step is to find the matched triangles that have to same relative movement and rotation like the other matches. This is done by calculating the standard deviation, sorting out triangles that are not within 1 or 2 sigma and repeat this process until I have a very small standard deviation.

The last part, finding "valid" triangles with the same movement/rotation, is working fine. The problem is that sometimes I have only like 2 or 3 "valid" triangles out of 300 initial triangles. All other triangles have side lengths different to those of the reference image.

So I assume it's the way I generate my initial triangles which causes the problem. Sortings stars by their brightness and using this data to generate the triangles also doesn't work. So is there a better way to create the initial triangles in all the images?

This page about a commercial product goes into some detail about their algorithm.  It does the triangle matching you describe, with something like simulated annealing to get a more optimal solution.

The accepted answer to this closely related question recommends Hugin panorama software; it's open source, so you should be able to glean the algorithms used.


Is any 3 body system known? https://en.m.wikipedia.org/wiki/Three-body_problem

Have three or more celestial bodies rotating each other in a stable manner ever been observed, or it is only a theoretical problem?

Answering in the spirit of the question, I think he's asking if there is ever a chaotic 3 body system that's long term stable, or, to put it another way, a 3 body system without a standard hierarchy where it's stable.

The answer is no.   3 body systems without Heirarchy are never stable for very long.   They can certainly exist for a while, but they aren't long term stable.

What do we mean by Hierarchy

The Sun-Earth-Moon is an example of a long-term-stable 3 body system in heirarchy.  It will remain stable for billions of years.    The Sun is pretty much in the center.   The Earth-Moon system orbits the sun and the Moon also orbits the Earth.   In a sense, the Moon orbits both the Sun and the Earth.    The Moon can orbit the Earth because it's inside the stable part of Earth's Hill Sphere.

Hierarchies beyond 3 might exist, such as Sun, Planet, Moon, Moon's satellite but there are no natural satellites of Moons in our solar system.

Multiple planets

Obviously, multiple planets around a comparatively massive Central star can be long term stable, and just for clarity, lets call a system stable if it lasts a billion years.    Mercury, Venus, Earth, Mars, Juptier, Saturn, Uranus, Neptune are obviously stable and mathematical models say that they should remain stable for at least the next 3.5 billion years, after which, there's a chance Mercury could become chaotic and fly out of it's orbit, flying into the sun, crashing into another planet or perhaps leaving the solar system.

Any of these orbital systems, for any kind of long term prediction of where the planets will be in a billion years, requires n-body calculations which are very complex and inexact, but they can be calculated within a range of error, which makes it possible to say that a system is stable, for a couple billion years anyway.   You can't, for example, ignore the other planets and calculate where the Earth will be in 50 million years.  The planets affect Earth's orbit but they don't destabilize it.   This is called perturbation theory.

And the final example, binary star system with a planet orbiting the two stars.

This XKCD discusses the binary star system, and from same:

  A planet orbiting two stars can't get too close to them or its orbit
  becomes unstable. If it gets too close, the irregular tugging from the
  gravity of the two stars as they orbit will eventually cause the
  planet to crash into one of them or get flung out of the system.

  For a system with two similar-sized stars, this "critical radius" is
  around six times the distance between the two stars.

What this means is that any 3 body system has to follow the rules if it's going to be stable.   You can have several planets orbiting a central star, and those planets will perturb each other, sometimes casting a planet out but you can also have stability.

You can have Sun, Planet, Moon, but the Moon must be inside the stable region of the Hill Sphere.

And you can have one object orbiting two central objects, like Tatooine, or, like the 4 satellites that orbit Pluto-Charon.

But anything outside those rules is an unstable 3 body or n-body system and it may take some time, but it's ultimately unstable and unlikely to last anywhere close to the billion(s) of years or so that stable orbits tend to last.

So a star with two planets who's orbits cross each other (note, Pluto and Neptune don't actually cross, they just appear to cross when drawn in 2 dimensions).  Their orbits don't come close to crossing.

Horeshoe orbits are 3 body orbits that might last a fairly long time, but I don't believe they are truly long-term stable (if anyone can verify that, please do), but intuitively, I don't see how horeshoe orbits could last very long.

When you have a 3 body system where there's no structured orbit and all 3 bodies kind of dance around each other.   That system isn't stable.

If properly balanced, a 3 body system outside the rules of stability might last for hundreds or thousands or orbits, but in celestial mechanical time that's not stable.

Mathematically some 3 body orbits have been worked out, but they are stable in the same way that a pencil is stable if you balance it perfectly on it's point in a room with no wind.   While this article says that they found 13 solutions, none of these solutions would actually work because space isn't perfectly smooth.  It's full of objects with gravity.   So, like the balanced pencil that gets knocked over by a slight gust of wind, a passing rogue planet would throw any of these "solutions" out of whack and they would progressively destabilize.  Where as a stable system (sun/earth/moon) can hande some perterbations and remain stable.

This diagram below is an example of a perfectly stable 3 body orbit in a computer program, but like the pencil, a small push on any one of the 3 objects would send it towards instability.

Hope that's not too long.  I can clean up if needed.


Logarithmic scale for large distances? I understand this is an icky subject, but I recently got interested in units for large distances for applications in cosmology and what not (after hearing about the redefinition of the kilo and kelvin), and I was surprised to see many very weird units which seem difficult to use.

So I was thinking in astronomy distances only get larger and larger, so I looked to see if I could find any logarithmic scale for distances and I couldn't find any. Then again, I'm not an astronomer and not suited to argue anything.

Are there any logarithmic scales for larger distances, and if so why aren't they being used? And if non exist, is there a reason why(is it simply a bad idea)? Or is it simply a matter of keeping historical background?

Having some experience with the only logarithmic unit in regular use (the dB, for sound intensity), I have to say: please, no.

A logarithmic representation can be handy in some cases. This chart is an excellent example of what can be done if you choose the scale on your image correctly.

Note that this has been done without introducing a new unit for distance, it just plots log(x) instead of x on the vertical scale.

But if you were to use a logarithm in the fundamental unit for distance, you'd introduce some problems:

cosmological distances no longer match other distance units and any conversion between them involves using a log function.
calculations involving cosmological distances get more complicated because you've now got to use log() everywhere.
you lose the linear relationship between numbers: object X is twice as far away as object Y, but the difference in your unit becomes log(2) making distances far less intuitive.


Are gravitational waves part of dark energy? Do the current models of the universe consider the energy release of gravitational waves?

E.g. iirc, the last black hole merger detected had a gravitational wave energy equal to ~3 solar masses (hope I'm not mistaken in this fact).

Same for other mergers, and supernovae: the energy related to their gravitational waves should sum up to something.

Gravitational waves cannot be part of dark energy, because -- just like electromagnetic waves -- their energy density dilutes as the universe expands: both due to the simple expansion of space and due to the cosmological redshifting that the expansion produces.

Dark energy, on the other hand, has an energy density that remains constant as the universe expands (truly constant if it's the cosmological constant, very close to constant if it's something stranger like quintessence), so it can't be made up of GWs.

GWs probably don't contribute much to the energy density we attribute to dark matter, either, since dark matter is diluted by the cosmological expansion but not by redshift. So if the energy density attributed to DM ten billion years ago were due to GWs, then it would be significantly lower than the DM energy density we measure today. (Also, current cosmological observations indicate significant dark matter was present very early in the universe, before the star formation that was necessary to create binary black holes in the first place.)

The GW energy from a binary-black-hole merger is enormous, yes; but those are very rare events. The current estimate from LIGO observations is about 50 or 100 such mergers per year per cubic gigaparsec, which translates into a rate of less than one per million years per galaxy. This paper by the LIGO and Virgo Collaborations has estimates of the current GW energy density in units of the current energy density of the universe (their Figure 1):

The green curve shows the estimated energy density from binary-black-hole mergers as a function of the frequency of the GWs; the red curve shows the contribution from binary-neutron-star mergers. The combined peak is always less than $10^{-8}$ of the critical energy density, so even summing up over frequencies, we have an energy density much smaller than that due to ordinary matter ($\Omega \sim 5$%), so it's probably safe to ignore its effects.


Is there any difference between rotation axis 90 degrees and 270 degrees? Is there meaningful reason to distinguish 90 and 270 degree rotational axis?

No, I don't believe there's any meaningful distinction.

A planet's axial tilt can be thought of as a 2-dimensional quantity: its angle relative to the ecliptic (or to whatever baseline you're using), and the direction in which it's tilted. When we talk about a planet's axial tilt as a number of degrees, we're only talking about the first component.

A planet whose rotation axis is perpendicular to the ecliptic would have a tilt of 0°. If its rotation axis is parallel to the ecliptic (rotating "on its side"), its tilt is 90°. If its axis is "vertical" but it's spinning backwards, its tilt is 180°. The range from 0° to 180° is enough to express all possible tilts, both prograde and retrograde.

Consider taking a planet with a 0° tilt and tilting its axis by 270°. The result is exactly the same as tilting it 90° in the opposite direction. Since the axial tilt is usually expressed as just the magnitude of the tilt, and not its direction, the distinction between 90° and 270° can be ignored.

And if we want to express the direction as well, we'll just say that it's tilted 90° in some specified direction.

We could have had a convention where the tilt ranges from 0° to just under 360°, with the direction specified more narrowly, but that would be less useful; the particular direction of a planet's axial tilt is less interesting than its magnitude. Also, the direction varies over time due to precession.

We could also have had a convention where the tilt ranges from 0° to 90°, and we also specify whether it's retrograde or not; then Venus would have a tilt of 3° rather than 177°.


Can we know the orbital planes of extraterrestrial planetary bodies? Astronomers can tell if there are any exoplanets orbiting its parent star using but not limited to the following methods:

Transit

As an massive object usually a planet passing in front of a star causing the amount of light emitted from the star to dim, by measuring the dimness the data can reveal how big the planet is.

Radial velocity

The star will seems to wobble as if there is a massive object usually a planet in orbit playing tug of war in a game of whose gravity is mightier. However since this wobbling effect is difficult to observe nerds prefer to measure the blueshift of the light (as star accelerates in our direction) and redshift (as star accelerates away from our direction) which seems to also tells how massive the planet is.

Note
Astronomers can derive the planet's density from its mass and radius(see transit).

Question

So far astronomers are able to determine the size, mass as well as density thus even its composition, is there any way to find the angle of the orbital plane of the exoplanet orbiting its parent star?

I presume what you mean is how does the plane of the orbit compare to the equatorial rotation plane of the star?

The answer is, you can sort of estimate this, by using something called the Rossiter-McLaughlin effect (see also Rossiter 1924; McLaughlin 1924).

You can find plenty of information on the web - I'll add a couple of links when I have a moment - but to summarise:

The rotation of a star broadens its spectral absorption lines. The hemisphere coming towards us emits light that is is blue shifted, the hemisphere receding is redshifted. If we now take a transiting planet, during the eclipse it crosses the disc of the star and obscures regions that are blue or red shifted by various amounts.

Now what you do is measure the line-of-sight velocity of the star. During the transit you would not expect this to vary due to the "Doppler wobble" caused by the exoplanet, except that if the planet obscures a blue shifted portion of the stellar disc, the net spectral absorption line that remains shifts to the red, and vice-versa. The pattern of red, then blue shift (or vice versa) as the transit progresses is known as the Rossiter-Mclaughlin effect.

A schematic showing how the Rossiter-McLaughlin effect works and how a different transit geometry with respect to the rotation axis of the star leads to a different line-of-sight-velocity signature in the spectral lines of the star. (Image credit: Subaru Telescope, National Observatories of Japan.)

If the planet orbits in the same plane and in the same direction as the stellar rotation (as the planets in our solar system nearly do), then the blue shifted limb of the parent star is obscured first, followed by an equal amount of redshift as the planets moves to obscure the receding stellar limb (see image above, left). Thus the stellar absorption lines show a redshift followed by a symmetric blueshift. If the planet was retrograde it would be symmetric but occur in the opposite order. A Polar orbit would show no RM effect. Inclined orbits would have an asymmetric RM effect - ie perhaps more blueshift than redshift (see image above, right).

The RM effect cannot give an exact orientation, it gives the projected angle between orbital and rotation axes on the plane of the sky. Nevertheless, that is sufficient for us to know that a lot of the transiting exoplanets have highly misaligned orbital and stellar rotation axes.


Does Big Bang red shift affect the frequency of both light waves and individual photons? With regard to the wave/particle duality of light, does Big Bang red shift affect both the frequency of light waves, and, the frequency of individual photons? I assume that the e=hf formula relates to individual photons, and not to light waves too?

Yes it affects both. If you treat light as waves then the wavelength from a source at a redshift $z$  is "stretched" by a factor of $1+z$.

Photons emitted from a source at redshift $z$ have their energies reduced by a factor of $1+z$.


Can New Horizons be used to measure the distance to Betelgeuse (despite its fickle photocenter)? The New Horizons spacecraft is almost 50 AU from Earth and near-simultaneous images taken by its telescope and one on Earth have demonstrated parallax without relying on the movement of Earth in it's orbit.
The simultaneity is important because measurements of Betelgeuse are confounded by its fickle photocenter†. It's apparent disk is lumpy and nonuniform; the "center of gravity" of intensity of its apparent disk moves around relative to its true center of mass. Simultaneous imaging from two widely separated positions using the same spectral bandpass could be used to try to address that problem.
Could this be done given what's known about its Long Range Reconnaissance Imager or LORRI and considering that the spacecraft has plenty of time on its hands?
Recent headlines saying that Betelgeuse is much closer than we thought point to Standing on the Shoulders of Giants: New Mass and Distance Estimates for Betelgeuse through Combined Evolutionary, Asteroseismic, and Hydrodynamic Simulations with MESA which proposes in the abstract

...a new, independent distance estimate of ${168}_{-15}^{+27}$ pc and a parallax of $\pi ={5.95}_{-0.85}^{+0.58}$ mas, in good agreement with Hipparcos but less so with recent radio measurements.

based on extensive modeling and photometric data.
What's truly remarkable is that I just copy/pasted that blurb and it self-formatted in MathJax, yay!
New Horizons is currently at X, Y, Z = 14.11, -46.42, 1.67 (AU) relative to the solar system barycenter in J2000.0 with ecliptic as reference plane using JPL's Horizons. I'm hoping that's orthogonal to Betelgeuse.

†fickle photocenter You can try to say that three times fast, but whatever you do, don't say the star's name three times!

Related:

What will it finally take to accurately measure the distance to Betelgeuse?
What equipment and techniques were used to study Betelgeuse's diameter in 1920?
How often are radio telescopes used to measure parallax? When was this first done? Currently unanswered.
Was a “spare LORRI” camera available to take matching image on Earth the same time New Horizons took the parallax image from the Kuiper belt? No conclusive answer yet, I really thought that there was one, and it might come in handy for this.

In principle yes, in practice no. The telescope is good enough, but the CCD camera will saturate, preventing a good positional measurement.
Salient facts. The parallax to Betelgeuse as seen between Earth and New Horizons will be about 250 mas. The current distance uncertainty is ~20%, so to better that, the position of Betelegeuse measured by New Horizons would need to be more precise than ~25 mas.
Looking at the LORRI camera, it has a plate scale of 4.95 microradians/pixel and a point spread function with FWHM of 1.8 pixels, which equals 1.84 arcsec (Cheng et al. 2007).
A useful rule of thumb is that your positional accuracy in a resolved image is given approximately by
$$ \Delta \theta \sim \frac{ {\rm FWHM}}{{\rm SNR}}\ ,$$
where SNR is the signal-to-noise ratio of the image. Using this, it suggests that if one could get an image of Betelgeuse with a SNR of around 100, then you should be able to centroid the position to better than 25 mas using the New Horizons data. Combining this with a simultaneous Earth-based image (which one presumes would be much more precise), then the parallax could be determined to better than 10%.
The problem appears to be the brightness of Betelgeuse. Cheng et al. also give a photometric calibration formula for the number of analogue to digital units per second, $S$, recorded by the camera for a point source with a given $V$ magnitude.
$$V = -2.5 \log S + 18.94 + 0.4\ ,$$
where the extra $0.4$ is probably appropriate for a star of Betelgeuse's red colour.
Using this formulae we see that $S = 2.16\times 10^8$ ADU/s, assuming $V\sim 1.5$ for Betelgeuse.
The CCD camera on LORRI only returns 12 bits and has a frame transfer time of 13 ms, which means that 13 ms is about the smallest exposure time you could use and hope to get accurate positional information (Cheng et al. 2007), though in principle a 1 ms exposure time could be done. Even in 1 ms you would get $2.16\times 10^5$ ADU from Betelgeuse spread over perhaps 4 pixels. Given that a 12 bit ADC saturates at 4096 ADU, then Betelgeuse is more than 10 times too bright and will produce a completely saturated image whose centroid cannot be determined with any precision. If it were possible to remotely reset the gain on the CCD (currently 22 electrons/ADU) to something much larger, then maybe it would be possible.


Are the Leonid meteor storms a thing of the past? According to the Wikipedia article on the Leonids, "a close encounter with Jupiter is expected to perturb the comet's path, and many streams, making storms of historic magnitude unlikely for many decades". Does this mean there will not be another incidence of the famous storms which spawned thousands or tens of thousands of meteors per hour within our lifetimes? Or will the 33 year periodicity reestablish itself?

Wikipedia does not cite a source.

NASA used to make predictions, eg.:

  2009 LEONID OUTBURST FORECAST A significant shower is expected this year when Earth crosses the 1466-dust and 1533-dust ejecta of comet 55P/Tempel-Tuttle. According to J. Vaubaillon, the narrow (about 1-hr) shower is expected to peak on November 17, 2009, at 21:43 (1466) and 21:50 (1533) UT, perhaps 0.5 to 1.0 hour later

These are extensively studied studied, eg. List of all meteor showers  .

The great storms occur at ~33 year intervals. I can't find any prediction that this pattern will be unusually perturbed by Jupiter at this time.


Can a black hole undergo a Penrose process inside another black hole's ergosphere? I recently watched this video on black holes:
https://www.youtube.com/watch?v=ulCdoCfw-bY

And that statement should give you an idea of my level of understanding :)

I was wondering if a spinning black hole A could complete a Penrose process through another spinning black hole B's ergosphere. i.e. if it could travel through the ergosphere of another spinning black hole and emerge out of it the other side.  What would happen?  Perhaps this is related to the question of whether black holes can orbit each other (in two or three-body system)?  What happens if black holes orbit each other, but their ergospheres overlap?

Black holes can orbit each other (we know this from the gravitational waves from their eventual mergers) and there is no particular reason against a small black hole passing through the ergosphere of a large black hole.

The main problem is avoiding a merger. The the two-black hole spacetime has no analytic solution and numeric simulations are messy, but a merger will definitely happen if the event horizons overlap. In any case, a rotating black hole has an ergosphere with an equatorial radius of $2GM/c^2$ and event horizon somewhere between this and the limiting radius $GM/c^2$. If we use the extremal case, then we can presumably just barely fit in nearly an identical mass and rotation black hole in the ergosphere. More likely a much smaller black hole could fit in. The direction of rotation would matter for the dynamics, since counter-rotating black holes would tend to coalesce (see this page for a plot of the relevant radii and where the innermost stable circular orbits are).

A single black hole passing through the ergoregion would not be able to perform the Penrose process, but we can imagine a binary black hole pair that is ripped apart by tidal forces and then presumably the remaining hole would be ejected with a lot of energy. So, yes, we could get fast small black holes this way.

In practice this system would lose a lot of energy through gravitational waves, making it likely that the holes would coalesce anyway. The exact limits require numerical relativity simulations to estimate, and this one would be a tough one.

(Note that the gravitational waves also carry away energy, up to 1/8 of the total mass of two equal-sized holes merging, but this energy is very hard to capture. Penrose did note in his classic paper that Misner was first in proposing black hole mining using a variant of this.)


"True" motionlessness - red shifts and CMB This is a question I asked years ago in an astronomy course, but to which I never got a straight answer. Please feel free to correct me if any of my assumptions here stray from facts. It goes like this:

To any observer, at large enough scale the universe appears to be receding in accordance with Hubble's constant, which causes the light that reaches them from objects such as distant galaxies to experience a red shift. Technically, though, if the observer is standing on a planet, they are in orbit moving in a particular direction at any given time, and so the distant galaxies in the direction of that should have a(n extremely) slightly smaller red shift while those in the opposite direction should have one slightly more severe. This means that, relatively, the galaxies behind are receding at a faster rate from the observer than those in front.

If the observer were to possess instruments of sufficient precision to determine the discrepancy in the red shifts and the means to adjust their velocity in any necessary way, would it be possible for them to equalize the red shift in every direction and thus be motionless in respect to the expansion of the universe itself? The cosmic background radiation might be a good reference for measuring red shifts in every direction, since it should be equally distant everywhere and very, very red shifted.

Yes. This is is actually something that is done in the analysis of redshift data. If no correction to the redshifts are made, then the redshifts are known as "geocentric." When you correct for the motion of the Earth around the Sun, the redshifts are known as "heliocentric" (see, for example, the description of data in the 6dFGS database).

Less common is to correct redshifts for the motion of the Milky Way all the way to the cosmic microwave background's (CMB) frame. This is done by applying a shift to a frame where the CMB dipole is zero. There are two reasons why this is less common. First, the CMB dipole was only well measured relatively recently by the WMAP and Planck missions. Second, correcting for the solar system's motion relative to the CMB fixes only half of the problem. The target galaxy will also be moving with respect to the CMB at speeds that are probably comparable to the Milky Way. This fact leads to distortions in when we try to use redshift as a proxy of distance, some of which are called "Finger-of-God" for their tendency to make plots of galaxies appear to contain filaments that point at the observer.


How to convert factors of time dilation into something relatable? On this post,

Time according to the gravity of Sagittarius A?

They found the factor of time on the varying distances from the SMBH, I am curious on how to turn those factors into relatable terms such as:
At 1 km from the event horizon ____ time passes on earth in an hour.

I'm just not sure how to figure this out.

I am curious on how to turn those factors into relatable terms such as: At 1 km from the event horizon ____ time passes on earth in an hour.

The key here is to understand what the formula relates to.  The (all important) observer is in this case a notional observer so distant that curvature (and hence gravitational effects) are negligible.

That observer is the "base clock" we reference.  Everything is dilated relative to that clock.

That means that the dilation on Earth (not strictly speaking an ideal distant observer) has to be taken into account to get an answer in the form you mean.  So starting with the formula you had :

$$t = t_\infty \sqrt{1 - \frac{r_s}{r}}$$

We want this relative to Earth so we need :

$$t_e = t_\infty \sqrt{1 - \frac{r_s}{r_e}}$$

And combining youy get :

$$t = t_e \sqrt{ \frac {1 - \frac {r_s} r} {1 - \frac {r_s} {r_e}} }
= t_e \sqrt { \frac { r_e(r-r_s) } { r(r_e-r_s) } }$$

Now in the specific case of Earth and Sagitarius A*, $r_e$ is huge so the dilation factor for Earth is as close to one as makes no difference in practice, so you don't really need to go to all that trouble and you can use $t_\infty \approx t_e$.


Seeing a star near zenith around sunset I'm recalling while taking a university astronomy class the professor brought a few of us up to the observatory, around sunset.  He showed us a star near zenith with the telescope.  After we left the observatory and were walking back to our dorms we thought we saw the same star, with the naked eye.  The sky still seemed somewhat bright.

Is that possible?

Is that possible?

Yes!

You can see bright stars and planets through a telescope in the evening and in some cases during the day. As magnification increases the brightness of the sky or any extended object decreases with the square of the magnification. For example, a patch of sky will have a certain brightness per square degree, but at 100x magnification it will be 10,000 times dimmer.
A star will (roughly speaking) be simultaneously brighter by the ratio of the square of the telescope's diameter to your pupil (which is 6 millimeters or less), so an 6-inch (15 cm) diameter telescope will make the star look 625 times brighter.
When it got dark, the sky naturally became far dimmer, and your pupils dilated to larger diameters, making the star brighter.

So you can think of the two methods (viewing through telescope versus waiting for night) as similar, in that they both make the sky dimmer and the star brighter.

It's possible to start guessing what that star might be, but I don't know. Here are two predictions of the sky on November 15 at 5:30 PM and 8:30 PM local time at roughly 42 degrees North latitude.

That far North there are no planets that could be very close to the zenith, and it would be necessary to know the year as well to see how close one might have gotten.

Images are from in-the-sky.org's planetarium feature.


What are these objects near Mirach Night sky objects always interested me, nowadays Im trying to observe and photograph some of these objects. My phone is iPhone 6S, i used SkyGuide app to find positions, NightCap to photograph. Im sure brightest star is Mirach according to SkyGuide app.

It's photographed at 00:30 at my local time (Istanbul, London + 3), SkyGuide says Mirach's azimuth was 77° 11’ 28” .

If this is really Mirach, then I would say the point-like object above and a little to the left of it is the star mu And (another designation: 37 And) in the same constellation of Andromeda. Then the next point-like object slightly further above and to the left from mu And would be the star nu And (35 And). And a barely visible fuzzy object above nu And would be the Andromeda galaxy (M31). Then in the left lower corner we see another star Nembus (51 And), and above it the star phi And (42 And). And at the same height with nu And farther to the right from it at the edge of the building there is the star pi And (29 And).

I would strongly suggest you to use Stellarium (www.stellarium.org) for identification of nearly any night sky object visible at your or any other location on Earth at any time.


Evidence behind Giant Impact hypthesis The giant impact hypothesis is a theory to explain how the moon was formed. What evidence is there that supports the theory and led us to find the current theory?

The smoking guns:

The ratio of oxygen isotopes of lunar rocks are almost identical to those of Earth.
Lunar recession due to the tides which causes the Earth's rotation to slow down, means that just after the Earth formed, the Moon was very close to the Earth and the Earth was rotating very fast. This situation can be reached due to an oblique impact that transfers a lot of angular momentum to the Earth.
The formation of the planets is known to have proceeded via small objects growing larger by colliding with each other, this means that the last stages of planet formation would have involved collisions between large objects. It is to be expected that around the time the Earth was taking shape, there could also be a proto-planet at the Lagrange point 60 degrees away from Earth in the same orbit. On the long term this is an unstable situation which ends with a collision that according to simulations is of the sort that typically leads to the formation of the Moon.
The asymmetry in the geology between the far side and the near side of the Moon. (there are many more maria on the near side compared to the far side). If the Moon had indeed been very close to the Earth after it formed, then the near side would have stayed very hot due to the Earth's surface still being molten and the radiant heat affecting the near side of the Moon. An impact on the near side of the Moon would far more easily be able to penetrate the solid crust and cause magma to flow to the surface compared to the far side of the Moon.


How much mass will the Sun have when it becomes a white dwarf? In 4 billion years, when our Sun sheds all of its outer gas layers and turns into a white dwarf, how much mass will the white dwarf have compared to what the sun has today?

Will the planets still orbit in the same way, or will the reduced mass cause the planets' trajectories to change, so that they eventually leave the solar system?

Short answer:

The Sun will lose about half of its mass on the way to becoming a white dwarf. Most of this mass loss will occur in the last few million years of its life, during the Asymptotic Giant Branch (AGB) phase. At the same time the orbital radius of the Earth around the Sun will grow by a factor of two (as will the outer planets). Unfortunately for the Earth, the radius of the Sun will also reach to about 2 au, so it will be toasted.

There is the possibility that the decreased binding energy and increased eccentricity of the Earth and the outer planets will lead to dynamical instabilities that could lead to planetary ejection. This is highly dependent on the exact time dependence of the late, heavy mass loss and the alignment or otherwise of the planets at the time.

Long answer:

Stars with mass less than about 8 solar masses will end their lives as white dwarfs on a timescale which increases as their main sequence initial mass decreases. The white dwarfs that are formed are of lower mass than their progenitor main sequence stars, because much of the initial mass of a star is lost through stellar winds (particularly during the thermally pulsating asymptotic giant branch phase) and final ejection of a planetary nebula.  Thus, the current distribution of white dwarf masses, that peaks between $0.6$ and $0.7 M_{\odot}$ and with a dispersion of $\sim 0.2 M_{\odot}$, reflects the final states of all main sequence stars with $0.9 <M/M_{\odot}<8 M_{\odot}$, that have had time to evolve and die during our Galaxy's lifetime.

The most reliable information we have about the relationship between the initial main sequence mass and final white dwarf mass (the initial-final mass relation or IFMR) comes from measuring the properties of white dwarfs in star clusters of known age. Spectroscopy leads to a mass estimate for the white dwarf. The initial mass is estimated by calculating a main sequence plus giant branch lifetime from the difference between the age of the star cluster and the cooling age of the white dwarf. Stellar models then tell us the relationship between main sequence plus giant lifetime and the initial main sequence mass, hence leading to an IFMR.

A recent compilation from Kalirai (2013) is shown below. This shows that a star like the Sun, born with an initial mass of $1M_{\odot}$ (or maybe a per cent or two more, since the Sun has already lost some mass), ends its life as a white dwarf with $M = 0.53 \pm 0.03\
M_{\odot}$. i.e. The Sun should lose approximately 50% of its initial mass in stellar winds and (possibly) planetary nebula ejection.

A comprehensive treatment of what happens to solar systems when the central star loses mass in a time-dependent way is given in Adams et al. (2013). The simplest cases are initially circular orbits where the mass loss takes place on much longer timescales than the orbital period. As mass loss proceeds, the gravitational potential energy increases (becomes less negative) and thus the total orbital energy increases and the orbit gets wider. Roughly speaking, $aM$ is a constant, where $a$ is the orbital radius, which is a simple consequence of conservation of angular momentum: so the Earth would end up in a 2 au orbit.

However, in the presence of a non-zero eccentricity in the initial orbit, or in the case of rapid mass loss, such as that which occurs towards the end of the AGB phase, then things become altogether more unpredictable, with the eccentricity also growing as mass loss proceeds. This has a knock-on effect when considering the dynamical stability of the whole (evolved) solar system and may result in planetary ejection. The faster the mass loss, the more unpredictable things get.

The radius of an AGB star can be calculated using $ L = 4\pi R^2 \sigma T_{eff}^{4}$. Stars at the tip of the AGB branch have luminosities of $\sim 10^{4} L_{\odot}$ and $T_{eff} \simeq 2500\ K$, leading to likely radii of $\sim 2$ au. So it is quite likely that unless the Earth is ejected or has its orbit significantly modified by some dynamical instability that, like the inner planets, it will end up engulfed in the outer enevelope of the AGB star and spiral inwards...

Even should it narrowly escape this immediate fate, it is then quite likely that tidal dissipation will rapidly extract energy out of the orbit and the Earth will spiral in towards the envelope of the giant Sun... with the same result.


How was Io not torn apart by tidal forces during its formation? Jupiter's moon Io is arguably one of the most volcanically active bodies in the Solar System.  The reason, according to NASA's page Scientists to Io: Your Volcanoes Are in the Wrong Place is believed to be caused by Io being

  caught in a tug-of-war between Jupiter's massive gravity and the smaller but precisely timed pulls from two neighboring moons that orbit further from Jupiter – Europa and Ganymede. Io orbits faster than these other moons, completing two orbits every time Europa finishes one, and four orbits for each one Ganymede makes. This regular timing means that Io feels the strongest gravitational pull from its neighboring moons in the same orbital location, which distorts Io's orbit into an oval shape. This in turn causes Io to flex as it moves around Jupiter.

So, how did Io form in the first place, given the tidal stresses acting upon it? Does this suggest (and what evidence is there) that Io 'migrated' into its present orbit?

No, it is not just a matter of migration. You need to take into account two facts.

One is that (as experience shows) Io's own gravity is enough to avoid it breaking by tidal forces. It has been like that through all its history: Io could not have been formed if it started aggregating today, but it was formed at the same time Europa and Ganymede did: they three were growing in parallel.

Other is that of the orbital resonances, which makes precisely that orbit with so simple integer number relations with those of Europa and Ganymede a stable one. Io could not have been formed in another place.


Is energy required for gravity to have effect, or is it an energy itself? If magnetism requires some energy to exert change of motion to attract ferrous metals, then, does gravity require energy to exert attraction, or is gravity energy itself, creating attraction through centrifugal force from the earth?

Gravity is an effect of the stress-energy tensor. Put differently, energy and momentum (they're related concepts) are the cause of gravity in general relativity. A fully charged electric battery has a very slightly stronger gravitational field than a depleted one. You may have heard that $E = mc^2$ or in as the terms are more descriptively defined in modern usage, $E^2 = m^2c^4 + p^2c^2$ . Just thinking about these equations from special relativity, you can see that more energy means more mass, which means more gravity.

Beyond that, gravitational waves carry energy. You may remember the recent discovery of gravitational waves from a black hole collision, in which about three solar masses worth of energy were calculated to have been given off as gravitational waves.

Getting back to plain old Newtonian mechanics, when masses move around in a gravitational field energy is converted from one form to another as Dean noted in their comment on the question. For example when a comet is near the sun on its highly elliptic orbit, it has high kinetic energy and low potential energy. It loses kinetic energy as it moves outward, being slowed by gravitational energy, but gains gravitational potential energy. This potential is, quite simply, a representation of the fact that it can get its "lost" kinetic energy back when it falls in again on the inbound leg of its orbit, being helped along by gravity instead of hindered when inbound.


Why the comets can cross the orbits of planets? Why are comets able to cross planetary orbits without colliding with the planets - Why is the comets' orbits elliptical, intersecting all the other planets' orbit.

Why don't they get attracted by the planets when they are in their orbit.

Or, why doesn't the force of gravitation from the planets not attract or compel the comet to rotate around them.

There are several reasons, the main ones are:

The orbit of some comets may appear to 'cross' a planetary orbit when viewing the solar system in 2-dimensions, but their orbit does not lie in the ecliptic plane (that the planet orbits follow).  An example is Comet Hale-Bopp:

Analogous to why Pluto won't hit Neptune despite their orbits appear to 'cross' in 2D representations.

Image source: http://spiff.rit.edu/classes/phys235/comets/comets.html

Some comets do have orbits on the ecliptic, particularly some short-period comets, such as Comet Encke:

Same image source as above

However, it is important to note that the chances for the comet and planet to be in enough of a vicinity to each other to result in the comet being 'captured' by the planet is low - so if the planet is not nearby, the comet can just keep on going on its normal orbit.

Though we have witnessed if a comet gets too close to a planet, when Comet Shoemaker-Levy 9 got caught in Jupiter's gravitational field, tidally fragmented and eventually collided in 1994.

in terms of whether a comet could be captured in orbit around a planet, this has been addressed as answers to the question "Can a comet orbit a planet?".


Is mass+energy conserved when a new universe forms inside a black hole? My understanding is that there are credible theories out there in which the formation of a singularity in a black hole also represents the beginning of a new universe via a big bang. We can't see the new universe because we can't see anything from within a black hole. Further, the apparent size of the black hole to us does not have to equal the size of the new universe from the perspective of someone inside that universe. My question is, is mass+energy conserved? That is, is the total amount of material within the new universe limited to how much stuff has fallen into the black hole, or how much stuff has reached the singularity? If so then it would seem these black hole universes are only a tiny fraction the size of our own universe.

Under the assumption that there is some sort of "Big Bang" scenario, then we are assuming an expanding universe.  General relativity does not require the conservation of mass-energy on a universal scale in a changing spacetime (arguably, GR doesn't even have a concept of energy on a universal scale), so there is no a priori reason why this black-hole-universe cannot be comparable to our own (in whatever senses we pick).

A simple way to see why energy conservation must be violated, even in a vaccuum, is to first observe that the vacuum energy density is fixed.  So if spacetime is expanding (or contracting), then we are changing the total volume without changing this vacuum energy density.  Keeping the density the same but increasing the volume means more total energy; decreasing the volume with the same density means less total energy.

We can also easily observe this violation.  You may be familiar with the concept of redshift: light coming to you from a source that is moving away from you (or emitted from a gravitating source) is shifted towards the red side of the spectrum.  This stretches out the wavelength, so the light has less energy.  The energy doesn't go anywhere, it's just gone.${}^*$

We might also note that, with Noether's theorem in hand, we know that time translation invariance implies conservation of energy.  But a non-static spacetime is inherently not time translation invariant, so conservation of energy is not required.

${}^*$ Well, some may try to regain conservation of energy by asserting that there is an energy of the gravitational field to consider.  This isn't adopted by everyone, but it can be done if you're careful with your definitions.  It just doesn't work as generally and easily as you might naively hope for.


Eyepiece projection photography Nothing is showing up in the viewfinder of camera when I try to do eyepiece projection photography.could I have something set up wrong?

The problem may be focus and or a faint image in the viewfinder, try looking at a distant object in daytime.

Of course unless you are using a dslr type camera the viewfinder will not be looking through the main optics anyway so wont show anything.


Can white dwarfs have accretion discs too? Neutron stars and black holes can have discs of accreted matter for they have an extremely high gravitational pull. While white dwarfs don't have such high gravity theirs is of course also extremely strong. Can an accretion disc form around a white dwarf too?

Yes. They are called cataclysmic variables.


How is distance in light-years converted to years back in time? When I look at the Alnitak,the left most star of Orion's belt, it is 736 light years away. How could I convert this distance to an estimate of how long ago what I am seeing happened.

Would it be simply be 736 years ago?

Are there instances of when this isn't as straight forward?

Yes, for stars in our galaxy, if a star is 736 light years away, the light took 736 years to reach here.

For very very distant objects, you need to account for the expansion of space. The light might have been travelling for 10 billion years, but in that time space has expanded and so the distance that you would find if you froze time and measured would be greater (to get the proper distance at the current time) it would be larger.  This effect is not significant at distances less than about a billion light years.


Do neutrinos have as much information as photons do? If neutrino detectors keep improving so that a fair number of neutrinos can be observed, would they be as informative for astronomy as photons are?

They are of course a very valuable complement to photons, but I'm thinking about neutrinos in and of themselves. Photons have wavelength, spectral lines, redshift, diffraction, polarization which reveal their origin and interactions on the way to us. Do neutrinos say more than just what direction they come from?

This is a very broad question and though a comprehensive answer lies outside the scope of this simple Q&A format, I give you a couple of examples where "neutrino telescopes" would revolutionary.

There is a predicted cosmic neutrino background, analogous to the cosmic microwave background. Neutrinos decoupled at 1 second after the big bang, filling the universe with neutrinos that should have now cooled to a temperature of about 1.9 K. Confirming this would be a(nother) spectacular validation of the big bang model. However, there is a wrinkle; now that we know neutrinos have mass, it turns out that these neutrinos are non-relativistic at the present epoch. That means they are capable of being diverted and concentrated by gravitational structures - potentially making them fantastic probes of such structure and perhaps more sensitive than photons. So neutrino telescopes would make a fantastic contribution to our understanding of cosmology.
Neutrinos arise from energetic processes in the cores of stars. These processes are otherwise invisible, we certainly can't see them directly with light and can only probe them indirectly using techniques such as asteroseismology or looking at the mixing of chemical elements from the core to the surface. Neutrinos potentially tell us much more; for instance giving a direct estimate of the number of nuclear reactions taking place per second. Neutrino emission is also the predominant way that cooling takes place in supernovae, and in hot neutron stars and white dwarfs. A neutrino telescope could therefore be fundamental  for a deep (literally) understanding of stellar evolution and particularly the late stages of stellar evolution.

In terms of what can be measured; you can measure the flux of neutrinos and the neutrino energy spectrum. There is also of course the timing information that you might get from transient events like supernovae. There is also the question of flavour oscillations and the mix of neutrino types that are observed. I don't understand enough about this to make any predictions about how astronomically useful this property is - but I bet it will be.


How can I calculate 27 tropical/solar year since my birthdate? Let's say I was born on 27th November 1993 at 03:15:15 AM. Is there any way I can calculate exact 27 tropical cycle since my birthdate/birthtime.

What date and what time it will be after exact 27 tropical years since my birthdate/birthtime?

From the Wikipedia page on the Tropical year,

  The mean tropical year is approximately 365 days, 5 hours, 48 minutes, 45 seconds.

Starting from the Tropical year proceeding after your birth date, you can calculate 27 Tropical years forward by adding that approximate amount of time 27 times over.


How much infrared radiation is emitted by a bow shock? Has the amount of energy being emitted from R Hydrae's bow shock been studied?

If so, is there any information how much infrared radiation is emitted in any means of measurement?

If the infrared energy from the Sun, measured 1 AU from the Sun being approximately 680 watts per meter square, how would the infrared energy of the bow shock compare at the same distance measured in the same units?

Thanks to AtmosphericPrisonEscape for pointing me in the right direction.  I found this table from the DETECTION OF A FAR-INFRARED BOW SHOCK NEBULA AROUND R HYA: THE FIRST MIRIAD RESULTS.

with the CSEs being the circumstellar envelopes.  Does this mean the CSE (bow shock?) is about 57% (19.7 compared to 33.9) the flux of the star at the 70 microns (mosaic) band?

Has the amount of energy being emitted from R Hydrae's bow shock been studied?

Yes, here are a few papers:

"Detection of a Far-Infrared Bow-Shock Nebula Around R Hya:
the First MIRIAD Results" (July 13 2006), by T. Ueta, A. K. Speck, R. E. Stencel, F. Herwig, R. D. Gehrz, R. Szczerba, H. Izumiura, A. A. Zijlstra, W. B. Latter, M. Matsuura, M. Meixner, M. Steffen, & M. Elitzur.
"Detached shells as tracers of AGB-ISM bow shocks" (July 21 2006), by C. J. Wareing, Albert A. Zijlstra, Angela K. Speck, T. J. O’Brien, Toshiya Ueta, M. Elitzur, R. D. Gehrz, F. Herwig, H. Izumiura, M. Matsuura, M. Meixner, R. E. Stencel, and R. Szczerba
"No Shocks for This Bow: IBEX Says We’re Wrong" (May 14 2012), by Susanna Kohler

  Does this mean the CSE (bow shock?) is about 57% (19.7 compared to 33.9) the flux of the star at the 70 microns (mosaic) band?

The CSE is not a "bowshock". An envelope surrounds an object, while bowshock preceeds the direction of movement.

Reference: "The chemistry in circumstellar envelopes of evolved stars: Following the origin of the elements to the origin of life", by Lucy M. Ziurys, PNAS August 15, 2006 103 (33) 12274-12279:

  "Chemically rich circumstellar shells are found around asymptotic giant branch (AGB) stars and red supergiants (RSGs). ...
  Given sufficient contraction, the helium core will eventually ignite as the H shell continues to burn. The fusion of helium produces $^{12}$C, and alpha capture on $^{12}$C yields $^{16}$O in the core. Eventually, the central helium is exhausted, leaving an oxygen/carbon core surrounded by He-burning and H-burning shells. The appearance of this two-shell structure marks the beginning of the AGB phase. The primary source of luminosity in an AGB star is the H-burning shell, but periodically the He shell ignites for short periods of time. The energy released in this so-called “thermal pulse” creates a convective zone extending from shell to shell, which mixes nuclear products from the He-burning zone to the stellar surface. This process, called the “third dredge-up,” brings carbon and other elements into the detaching envelope. The thermal pulses also aid in driving the outer material from the star, creating a circumstellar shell. In fact, AGB stars lose up to 80% of their original mass in the form of an envelope, with typical rates of 10$^{−6}$ to 10$^{−4}$ M$_\odot\cdot$ yr$^{−1}$.".

A useful but rough analogy is: a plane's paint fumes are the CSE and the bowshock is caused by a difference in the speed of travel that is supersonic.

The image you included is used on Wikipedia's webpage "Bow shocks in astrophysics - Around other stars" and appears to have originally come from: Caltech's website, dated 2006-11-30. Another copy is on JPL's website: "Red Giant Plunging Through Space" (December 8, 2006), Spacecraft: Spitzer Space Telescope, Instrument: Multiband Imaging Photometer (MIPS). Simbad query for R Hya.

Within the webpage's text it says: "This bow shock is 16,295 astronomical units from the star to the apex and 6,188 astronomical units thick (an astronomical unit is the distance between the sun and Earth). ... The mass of the bow shock is about 400 times the mass of the Earth.".

The webpage: "About the color mapping table", which explains three of the wavelengths used in the image,
says: "$\bbox[#E8D8D0, 0.15pt]{\color{yellow}{\text{Infrared}}\, 3.6 \mu m}$, $\bbox[#F8E8D0, 0.15pt]{
\color{orange}{\text{Infrared}}\, 4.5 \mu m}$, $\bbox[#E5DBD2, 0.15pt]{
\color{red}{\text{Infrared}}\, 8.0 \mu m}$". They don't explain the purple band on that webpage. The raw data is far more useful than a color mapped image prepared for public consumption.

The paper that you linked to says on page 6:

  "At 165 pc (Zijlstra et al. 2002) with the -10 km s$^{-1}$ radial
  velocity (Knapp et al. 1998), the space velocity of R Hya is 50 $\pm$  1 km s$^{-1}$ into the direction 12° away (receding side) from the plane of the sky. The asymmetric CO profiles (Knapp et al. 1998; Teyssier et al. 2006) can then be explained by this inclined bow shock, in which the receding and approaching shock fronts intersect with the line of sight at distinct angles.
  The physical dimensions of the observed bow shock (0.03 pc thick located at 7.9 x 10$^{-2}$ ahead of the star) are consistent with the stellar-wind bow shock models for AGB stars (Wareing et al. 2006a). Adopting the mass-loss rate of 3 x 10$^{-7}$ M$_\odot$ yr$^{-1}$ (Zijlstra et al. 2002), wind velocity of 10 km s$^{-1}$ (e.g., Knapp et al. 1998), and formula for $l$ derived
  from the momentum conservation across the shock (Van Buren et al. 1990, eq. 2), the ambient $H$ density is 0.4 cm$^3$ and the
  shock surface density is 2.5 x 10$^{-7}$ g cm$^2$. Thus, the estimated amount of matter contained in the bow shock is about 1.3 x 10$^{-4}$, assuming a paraboloidal bow shock of $y = x^2 /3l$ with a 300" opening radius.".

In the paper: "Exact Analytic Solutions for Stellar Wind Bow Shocks" (March 1996), by Francis P. Wilkin, Astrophysical Journal Letters v.459, p.L31:

  "Stellar wind bow shocks are cometary structures due to the supersonic passage of wind-blowing stars. They sweep up interstellar matter into thin, dense shells, which may be revealed by their postshock emission or by scattered light, and provide a means of studying winds that might otherwise go undetected. ...
  The theory of momentum-supported bow shocks was first developed by Baranov, Krasnobaev, & Kulikovskii (1971, hereafter BKK), who were motivated by the problem of the interaction of the solar wind with the local interstellar medium. ...
  This method allows one to obtain simple, exact solutions for all quantities in the numerical model of BKK: the shell’s shape and the mass and velocity distributions within it. The shocked fluid in the shell has momentum that is precisely the vector sum of the momenta imparted to it by the wind and ambient medium, integrated over the surface of the shell, so there is no need to divide the problem into normal and tangential parts in curvilinear coordinates. Moreover, the tail of the bow obeys a simple, momentum-conserving snowplow model in two dimensions.".

In particular, on page 3, it says:

  "The tangential velocity and mass surface density are plotted in
  Figure 4 for several values of $a$. For most systems, $a$ is expected to be small, and the tangential velocity should prove a more sensitive way of obtaining a than the mass column.".

  FIG. 4.—Internal shell velocity in units of V$_*$ (top) and mass surface density in units of $R_{0\rho_a}$ (bottom) for several values of $a$. Curves for $a$ = 0 and 2 are labeled, and intermediate curves correspond to $a$ = $\frac{1}{4}$, $\frac{1}{2}$, and 1. The mass per
  unit area $\sigma$ is defined for a line of sight that is normal to the shell. The normal direction to the shell may be obtained from eq. (6), using the relation $\hat{e}_n = \hat{e}_t$ x $\hat{e}_\phi$. If viewed from angle $\vartheta$ to the normal, the mass column $\sigma$ must be modified by a projection factor sec $\vartheta$ so long as the path length of the line of sight through the thickness of the shell is short compared to the shell’s curvature.

The exact amount of infrared radiation emitted is a function of speed and resulting temperature, it's not a means to calculate the energy from the bowshock nor is it consistent. It's a great means to visualize that there is bowshock, much as smoke or clouds help to visualize a plane's bowshock; at best it would be an indirect means to measure speed and energy.

In the monthly notices of the Royal Astronomical Society, Volume 441, Issue 1, 11 June 2014, Pages 809–812 article titled: "Detecting floating black holes as they traverse the gas disc of the Milky Way", by Xiawei Wang and Abraham Loeb they explain that bowshock can be unaccompanied by CSE and that the radiation isn't confined to the infrared bands.

  "In this paper, we propose an additional observational signature of floating BHs, using the MW gas disc as a detector. As the BHs pass through the MW disc supersonically they generate a bow shock, which results in synchrotron radiation detectable at radio and infrared frequencies. ...

  Observations of the Galactic disc can be used to infer $n_0$ and $T_4$. The BH speed $V_•$ can then be estimated from the Mach cone angle. The maximum Lorentz factor $\gamma_{max}$ can be inferred from the peak of the synchrotron spectrum. This, in turn, yields $B_{−5}$ based on equation (3). From the slope of the synchrotron spectrum, the power-law index $\rho$ can be estimated. Finally, with the above parameters constrained, the synchrotron flux can be used to calibrate $M_•$. The above interpretation can be verified by observing the properties of the star cluster carried by the floating BHs (O'Leary & Loeb 2009, 2012). The diffuse X-ray emission from the BH and synchrotron emission from the bow shock are supplemented by stellar emission from the star cluster around it. Since the total mass of the star cluster is much smaller than $M_•$, gravity is dominated by the BH, and thus the stars do not affect the bow shock. One can measure $M_•$ spectroscopically from the velocity dispersion of the stars as a function of distance from the BH, and verify consistency with the synchrotron flux estimate.".


Would "layered" radio interferometry work? tl;dr - Is splitting up the process of interferometry as shown in the diagram possible, and if so, is it more efficient and/or easier than traditional methods?

I have been doing some research into radio interferometry, and I had a question about it - I know that we can combine the signals from multiple telescopes into one "image" using interferometry, but what if I was able to do that multiple times?

Let me describe this a bit better. Say I have nine dishes. I arrange three of them in an equilateral triangle, and then I do the same with the other 6, and then, I arrange those three groups of three into a larger equilateral triangle. Then, I combine the signals from each three of the telescopes, and then I combine the three resulting signals. Here's a diagram that might help - the blue dishes are the "telescopes", and the red and green boxes are the "processors" or where the signals would be interfered.

In principle, would this work? And, in the context of radio astronomy, would it be easier to combine or interfere only three signals rather than nine?

Absolutely this can work as can be seen in  The Murchison Widefield Array or MWA in Western Australia. This is essentially exactly how it works. You definitely do loose information (the first stage of combining antennas significantly reduces the field of view, and the number of interferometric baselines you get is reduced).
From this page:
Frequency Range            70-300 MHz
Number of Receptors        2048 dual-polarization dipoles
Number of antenna tiles    128
Number of baselines        8128

There are 32 receptors per tile (2 polarizations for $4x4=16$ elements), and $N = 128$ tiles. The number of baselines is $N(N-1)/2 = 8128$.
The advantages is a reduction in the data rates, so reduced digital processing. Note the first stage would be treated as beam forming - you cannot do interferometry between the first stage and somehow do interferometry between the results. Ie the first stage is summing the voltages (with appropriate phase corrections) and the second stage would be interferometry.

One of the tiles making up the 32T, a prototype instrument for the Murchison Widefield Array (source)

An MWA antenna consists of a four by four regular grid of dual-polarisation dipole elements arranged on a 4m x 4m steel mesh ground plane. Each antenna (with its 16 dipoles) is known as a "tile". Signals from each dipole pass through a low noise amplifier (LNA) and are combined in an analogue beamformer to produce tile beams on the sky. Beamformers sit next to the tiles in the field. The radio frequency (RF) signals from the tile-beams are transmitted to a receiver, each receiver being able to process the signals from a group of eight tiles. Receivers therefore sit in the field, close to groups of eight tiles; cables between receivers and beamformers carry data, power and control signals. Power for the receivers is provided from a central generator.


Which planet or moon has all resources that can be used to sustain life in a controlled biosphere? Now that we are making it to Mars, continuing onward, in an effort for humanity to civilize other planets, not just setting up residential colonies, but to mine minerals and carry on geological and scientific research, which planet or moon has all resources that can be used to sustain life in a controlled biosphere?

And how can the available resources on that body be converted to suit human requirement?

No known planet except Earth can be colonized by a human civilization. There are at least three serious issues: temperatures at around 300K, an atmosphere of appropirate pressure, and damaging cosmic radiation (low gravity is also a worry for long-term human presence). Minerals are less of a problem (and water can be synthesized). Mars and the Moon are close enough to the Sun for pressurized greenhouses to provide an environment where humans could stay without space suit. However, the problem of cosmic radiation remains and such "biospheres" must forever dependent on supportive technology. Venus's inhospitable atmosphere makes it completely unfeasible.

Using robots to mine planets and moons in the Solar system is certainly a realistic and interesting option. Sending humans there is very expensive and not necessarily sensible. Moreover, it is not clear whether humanity should send any live (including microorganisms) to such places, as this would inevitably forever change their nature (like sending rabbits/cats to Australia); all ongoing space missions are sterilized before being sent on their way.


What is the fastest spinning rotation of a Neutron star? What is the fastest spinning rotation of a Neutron star? I have heard that Neutron stars have a specially fast spinning rotation. What is the fastest?

As it turns out, the fastest spinning neutron star found yet is a pulsar 18000 light years away in the constelation of Sagittarius which scientist catalogued as PSR J1748-2446ad. Pulsars are neutron stars that rotate, are highly magnetic and emit a strong perpendicular beam of electromagnetic radiation.

This pulsar's speed is such that:

  At its equator it is spinning at approximately 24% of the speed of light, or over 70,000 km per second.

PSR J1748-2446ad rotates a little over 700 times a second, and scientists have this to say on the theoretical limits of the rotation speed of a pulsar (from here):

  Current theories of neutron star structure and evolution predict that pulsars would break apart if they spun at a rate of ~1500 rotations per second or more, and that at a rate of above about 1000 rotations per second they would lose energy by gravitational radiation faster than the accretion process would speed them up.


Which galaxy is receding from the Milky Way the fastest? What is known of the mechanism behind its recession? Galaxies are always in motion relative to the Milky Way. My question is, which galaxy is receding the fastest from our viewpoint?

What is the theorised mechanism that causes this?

This question may be a duplicate to: On what scale does the universe expand?

The mechanism for the expansion of the universe is gravity, manifested through what's known as Dark Energy. Dark Energy is a component to the universe which today, composes 68.3% of the energy density of the universe. You may be confused as to how gravity could be responsible for the expansion of the universe, but the equation of state of DE,

w = pressure/density

is negative (w_DE = -1), which means that it has anti-gravitational properties. Additionally, its density remains constant as the coordinates of the universe change, which is the reason why it dominates later on in the universe's history (while all other components, matter (dark or otherwise) and radiation/relativistic matter, become diluted with the expansion of the universe).

To answer your question, because the expansion of the universe is happening at all points in space, it looks like the "center" of expansion is at every point in space (see: Space expansion in layman terms), the galaxies furthest away (highest redshift), are the ones moving quickest away from the Milky Way Galaxy.


How would the barycenter of a double-planet system affect the two planets' surface gravity? Assuming two similarly sized, tidally locked planets orbit a common barycenter, how would the gravity of the barycenter affect the surface gravity? For example, on the inner sides of the planets, what would be the effect of the multiple gravity centers (the planets' and the barycenter), especially to a being on the surface?

There's basically no gravitational pull towards the barycenter.  The two objects are in free-fall around each other and objects in free-fall experience (almost) no gravity from the object they are in free-fall around.  Anyone standing on either planet would basically only feel the gravitation of that object alone, as if it were sitting still in space (though nothing actually "sits still" in space), everything in space is flying through it's local gravitational field.

There can be small variations to the local objects gravity, generated by any rotation.  Tidal locking might be common with a dual-planet system and Tidal locking is usually a relatively slow rotation unless the objects are very close, but while it's a slow rotation, it's still greater than zero, which creates a tiny tidal bulge and a very slight centrifugal force which would effectively lower the gravity.   This has nothing to do with the barycener, it is a product of the objects rotation.

As the objects get closer there is a tidal bulging which has some effect on gravity.   Parts of the planet that bulge outwards experience lower gravity and parts that are effectively stretched in closer to the center experience an increase in gravity.   This effect is pretty small and it's an indirect effect, caused by tidal forces from the other object, not caused by the barycenter.

At the Roche Limit, gravity is effectively cancelled out by the tidal force and objects can drift off the surface both on the near and far side of the smaller of the two tightly orbiting objects.

The amount of gravitational variation would be small, unless the orbit was very close.

I'd like to post a map of the change in gravity on Pluto's Moon Charon, or Jupiter's innermost Moon Io due to the tidal forces and tidal bulging, but I couldn't find any written results on that and I'm reluctant to try to do the math myself.  Even under very strong tidal forces like those two, the effect is likely still quite small, perhaps in the range of 1% or less.

Here's a question where the tidal force and it's effect on Phobos' gravity is discussed.  You need very strong tidal forces before the gravity gets significantly effected.


Can a material other than glass be used for making telescope mirrors? Although I do know of a few instances where professional or observatory mirrors have been made by alternate materials, it was primarily for targeting specific nuances, like weight reduction and low thermal expansion etc.

I was more keen on finding a material whose primary unique selling point is low cost ;P , keeping in mind a budding amateur astronomer.

Front-surface glass mirrors weren't developed until the 1850s.
Before that, the most common material was speculum metal, a high-tin bronze.
Notable examples include Isaac Newton's 1668 prototype and William Herschel's late 18th century telescopes.
Since speculum metal tarnished easily, it was common to have a spare mirror to continue observations while the other was being reconditioned.
Reflectivity and thermal characteristics were not great either.


Pulsation Modes of Cepheids There are various modes of Cepheids like

single modes:
   fundamental (F),
   first overtone (1O), and likewise.

And then we have

double modes: F/1O, 1O/2O etc.
triple modes: 1O/2O/3O etc.

What are these modes actually? What do they mean and which frequencies do they correspond to?

Stellar pulsations are caused by expansions and contractions in the outer layers as a star seeks to maintain equilibrium. What kind of equilibrium is being referred to here?

Some classical Cepheids pulsate simultaneously in two or even three modes.
Their lightcurves can be explained as a overposition of fundamental plus overtone modes.

The terminology you used is also found in this paper (where you have also other references):
"the discovery of many double-mode Cepheids (DMCs) pulsating in both the fundamental and first-overtone modes (F/10) and the first and second-overtone modes (10/20)"

Other reference: Classical cepheids


What earth organisms would survive if they arrived on Mars? Contaminating other planets is a concern in space exploration. My question is what organisms, or small collections of organisms, from earth would be able form a self-sustaining population on Mars? I'm not worried about how they might get there, so deep sea microbes and the like which would have to be purposely transported are in scope.

This is not the same as this earlier question, as that one does not consider sustainability.

On the surface of Mars probably none, since it's too dry or too cold, or both, to stay active.

Spores or other dormant forms probably could survive for centuries, until radiation will gradually destroy the organic molecules necessary to get back into an active state.

But there are "Mars Special Regions", where either Earth microbes or potential Martian microbes - if some exist - cannot be ruled out to be able to spread.

Things might look quite different in the underground, especially in warm and wet zones, which may exist by geothermal heat and ground water.
There, at least lithotrophic bacteria/"chemolithoautotrophs" could survive and spread.
Those could then form a basis for a food chain of other organisms.

Plants dependent on sunlight, or animals dependent on oxygen would hardly survive.

For a list of species, see table 2 on page 895 of the above reference paper. The following species and genera are mentioned in the table:
Psychromonas ingrahamii, Planococcus halocryophilus strain Or1, Paenisporoarcina sp. and Chryseobacterium, Rhodotorula glutinis (yeast), Colwellia psychroerythraea, Nitrosomonas cryotolerans, (lichen) Pleopsidium chlorophanum.