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Undisturbed planetary alignment after galactic collision
How do planets still orbit around the empty space where a black hole was after a black hole absorbs the milky way's black hole and keeps moving through space without disturbing surrounding planets that were orbiting that black hole?

This would be on topic on Astronomy, but is currently unclear. Please clarify what you mean by "black hole takes it".

I recently edited the question.

Is that a hypothetical scenario or is this an actual situation you read about? Something's still not clear about it. If you have a source that talks about this, please link to or quote from it.

Here is what I currently understand from this question: If one black hole passed through a black-hole-centered system and absorbed the black hole without disturbing the planets (not possible) and kept moving on, how would the planets still orbit the central space? The answer is that they wouldn't, but I suspect something is still lost in communication here.

You mention the Milky Way's black hole, by which I think you mean to refer to the supermassive black hole at the center of our galaxy. If that were removed, the galaxy would continue to rotate, but that's because it is not gravitationally bound by the supermassive black hole in the first place. The supermassive black hole is not massive enough to hold the whole galaxy together--the galaxy is held together under its own mass. Outer systems orbit the mass of the center systems, not of the black hole alone.

The short answer is that they don't. If an even larger SMBH passed through the centre of the Milky Way and "swept up" the existing SMBH, all the stars and planets that are close to it (within a few tens of light-years, maybe more) would experience huge disruption and might end up orbiting the new black hole and being carried away, or flying free, or whatever.
The stars and planets much further away, which make up the bulk of the Milky Way wouldn't be much disturbed, because they are held in their orbits by their gravitational attraction for each other, and for the dark matter halo which suffuses the galaxy and only a small contribution comes from the gravity of the SMBH.


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Could an Applegate-like mechanism operate in AM CVn systems?
One of the controversies surrounding the detection of exoplanets around post-common envelope binaries is the possibility that the orbital period variations observed might be the result of the Applegate mechanism as opposed to reflex motion from an orbiting planet.
Would these concerns also apply to AM CVn systems or other close white dwarf+white dwarf binaries? The Applegate mechanism itself is explained in terms of the activity cycle of the main sequence component of a cataclysmic variable, could a similar mechanism operate or would the changes in the quadrupole moment of the star be too small to produce observable effects?

related but different: V471 Tauri's circumbinary brown dwarf non-observation; Applegate, or over-restrictive assumptions?

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How to convert magnitude to spectral luminosity?
In the Gaia catalogue I have the magnitude of several stars in the G band. I also have the distances to these stars (parallax is known). Additionally, I oftentimes have the effective temperature in Kelvins.
Is there any way to convert this magnitude to an approximate spectral luminosity (units of W/Hz) at some frequency band? You may assume that we are in the Rayleigh Jean's regime (Intensity proportional to frequency2).

There are dozens of astrophysics web pages which discuss this in detail.

@CarlWitthoft can you point me to one? I am not looking to convert to bolometric luminosity (W), rather, spectral luminosity (W/Hz)

I think you simply need to get the temperature of the Star and use Planck's Law to estimate the spectral distribution

@CarlWitthoft Ah I see. So from there I can get spectral radiance at frequency v. How can I then obtain the spectral luminosity? Is it just this number time 4pidist^2?

Hi @Jake and Welcome to Stack Exchange! This site works a little different than other forums or Q&A sites. It uses more formal Question posts and Answer posts. Questions should usually contain a bit of research and effort if possible, not just "how do I..." It would be better if you do some reading first, and make an attempt to do the calculation first, even if it's wrong. This way it is easier for other people to see what data you are starting with and how you plan to manipulate it. That makes writing answers much easier. Thanks!

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Cosmology : Formula for the bias of galaxies
first I have posted originally on the physics stackexchange but I have not had answers about my question, so I try my luck on this specific forum (if forums leaders want to delete the original post on physics exchange, they can do it without problems, I don't want to make duplicates).
From this article : https://arxiv.org/pdf/1611.09787.pdf , I try to deduce the equation that my teacher told me which links 2 quantities :

the global number density of galaxies

the local number density of galaxies

the contrast of Dark matter density

The relation that I would like to find (the relation given by my teacher) is very simple :
$$N_{1} = n_{1} b_{1}\,\delta_{\text{DM}}\quad\quad(1)$$
where $$N_{1}$$ is the local number density of galaxies in Universe, $$n_{1}$$ is the global number density, $$b_{1}$$ is the bias (cosmological bias of galaxies) and $$\delta_{\text{DM}}$$ the contrast in dark matter density. When I say "local", I mean in the volume of scale that I consider (in a cluster of galaxies for example, doesn't it ?)
for the moment, I can't find this equation.
Into the article above, they define the bias by doing the relation $(1.1)$ (equation reference on the article) :
$$\delta_{g}(\vec{x}) = \dfrac{n_{g(\vec{x})}}{\overline{n_{g}}}-1 = b_{1}\,\delta_{\text{DM}}(\vec{x}) = b_{1}\big(\dfrac{\rho_{m}(\vec{x})}{\overline{\rho_{m}}}-1\big)\quad\quad(2)$$
with $b_{1}$ the bias.
As you can see, in this article, authors are reasoning with the contrast of density number of galaxies ($\delta_{g}(\vec{x}))$ and the contrast of matter density of Dark matter ($\delta_{\text{DM}}(\vec{x})$).
I tried to modify this equation $(2)$ to get $(1)$ but I am stuck by the following difference : on one side, one takes number densities and on the other one, they take contrasts of density (with contrast density number and Dark matter contrast).
Multiplying the both by the volume $V$ is not enough since there is the value "-1" in the definition of contrast : I don't know if I have to write :
$$\text{Global Number of galaxies} = \overline{n_{g}}\quad V$$
or
$$\text{Local Number of galaxies} = \overline{n_{g}}\quad V$$
???
I think that I have to use the following relations : $N_{g}\equiv N_{1}$ and $\overline{n_{g}}=n_{1}$ in the relation of my teacher but I am not sure.
Anyone could help me to find the equation (1) from the equation (2) of article cited ?
EDIT 1:
If I take the relation eq$(2)$, I can write :
$$n_{g(\vec{x})} = \overline{n_{g}}\,b_{1}\,\delta_{\text{DM}}+\overline{n_{g}}\quad\quad(3)$$
As you can see, $(3)$ is not equal to the equation $(1)$ that I would like to get (since a second term $\overline{n_{g}}$)
With the notations of the equation$(1)$, in order to be coherent, I think that I have to assimilate $N_{1}$ to $n_{g}(\vec{x})$ (local density) and $n_{1}$ to $\overline{n_{g}}$ (global or mean density).
How can I circumvent this issue about the presence of this second term into eq$(3)$ compared to eq$(1)$ ?
I am near from the equality between both, this is frustrating . Maybe it is a problem of convention about the factor $b_{1}$ called "bias" ?
Edit: This is also cross-posted at https://physics.stackexchange.com/questions/489437/demonstration-about-large-scale-bias-of-galaxies

Cross posted in Physics SE


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Could the solar system survive in a void?
If the solar system were suddenly placed in the middle of a large void... would everything still function as normal?
Would we even notice any difference here on Earth apart from the night sky?
...and speaking of the night sky, would that even look much different to how it is already? I imagine just less colourful and and more sparse, but seeing whole galaxies instead of individual stars?
Or would moving the solar system into a void just mean the end of the world as we know it?

Hungry void, or just an ordinary 1 particle per couple per couple cubic meter void?
I don't think anyone's even considered the possibility of matter destroying voids yet. We would not do well there, if they exist.

The solar system could continue to exist outside a galaxy, and apart from a lack of stars in the night sky, I see no reason why anything much would change. It would be very unlikely to form there, even if some chance provided a sufficiently dense gas cloud. The cloud would be made up of almost nothing except hydrogen and helium (since there would not have been previous generations of stars to enrich it) so (a) it would be hard for a star as small as the Sun to form and (b) there would certainly be no solid planets.
A void is just a region with a relatively low density of galaxies. Relatively low does not mean zero, so the solar system could continue to exist, or even form, in a galaxy in a void with no problems, but in most parts of most voids it could not form.

If you took away the Milky Way except for the Solar system (but kept the MW's satellite galaxies), the  "naked eye" night sky would feature the Andromeda Galaxy, 2.5  million light-years away, the Large Magellanic Cloud (LMC), 160,000 light-years away and the
Small Magellanic Cloud (SMC), 200,000 light years away.

As far as we can tell, the solar system is in such a void already. The region of space outside of the solar system is thought to be pretty sparse but that is hard to tell. We know the nearest star is Alpha Centauri which is over 4 light-years away. If the solar system was moved to an even less dense area if would not be noticeable. There may be planetary bodies nearby to our solar system but we currently have no way of finding these.

You're undoubtedly right, but the answer would be the same: No noticeable impact other than a boring night sky. Maybe fewer new comets, since they are hypothesized as being partly caused by disturbances in the f/o/r/c/e/ Oort Cloud due to other stars.

What if the void means: in a non-expanding universe, no galaxy to turn with. Would really everything be the same?

i had no reason to think Void meant no expansion...

Could you provide a reference for "As far as we can tell, the solar system is in such a void already"? Our solar system is part of the Local Group, within the Virgo Cluster, within the Lanakea supercluster. You can't be in a supercluster and also be in a void: they are mutually exclusive.

https://hypertextbook.com/facts/2000/DaWeiCai.shtml.  You are correct but the matter density of interstellar space is about one particle per square cm.  It is thought to be as small as .1 particle per square cm in other parts of the universe.   My understanding of the void was interstellar space.

@jmh sorry for late response. I didn't mean to offend or say your answer is wrong or not complete. I just wanted your opinion on that "what if" case. I'm not an astronomer, just curious about what universe expansion does on the solar system scale. And a bit afraid to ask it as a question :)

@Koray Universe expansion is very weak, so it only affects things on the very large scale. There's no expansion inside a solar system, a galaxy, or even inside a cluster of galaxies, the gravitational attraction inside those systems overwhelms the expansion. So you only see expansion between galaxy clusters, and at larger scales.

Actually the further away the Solar System is from anything else the better for life. There would be less radiation and fewer mass extinction events from supernovas. We would not see galaxies instead of stars; their brightness would not increase. In fact if we were located in one of the great intergalactic voids the night sky would be completely dark to the naked eye.

to be honest it was actually intergalactic voids I was referring to, but going by the answers received it appears I wasn't very clear with wording my question! — nonetheless would the sky really appear completely dark in such a case? I heard that you would still see lots of tiny points of light, but instead of them being stars they would be entire galaxies

Yes, completely dark. Only 8 galaxies are visible to the naked eye and half of them only under ideal circumstances. In the middle of an intergalactic void the closest galaxy would be orders of magnitude too far away to be visible. In fact merely moving the solar system outside of the local group would have the same effect.

One of the biggest is the Bootes void, which was discovered in 1981; it is 250 million light-years across, and until recently only two dozen galaxies had been found in it. That's a very void void. "If the Milky Way had been in the center of the Bootes void," says astronomer Greg Aldering of the University of Minnesota, "we wouldn't have known there were other galaxies until the 1960s."
-- https://web.archive.org/web/20071119011039/http://findarticles.com/p/articles/mi_m1511/is_n8_v16/ai_17253874


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What is the effect of this expanding universe on the energy density of it?
We know that our universe is expanding i.e.  volume of the universe is getting larger with time with total energy being conserved. My question is whether the energy density is getting smaller with time.
If this is true then what is the effect of this phenomenon on all the celestial bodies?

Total energy is not conserved. The conservation of energy does not apply on cosmological scales.

If I may: We don't know the volume of the universe is getting larger. The volume of the universe is much, much more likely than not, infinite. And I mean literally infinite (I have to say that due to the bewildering existence of 'multiple types of infinities' in theoretical astrophysics.

@WhitePrime That may well be true, but if you take any sufficiently large region and "anchor" its boundaries to the average trajectory of matter near them, it will be found to be expanding.

I am no expert and I am sure there will be a more rigorous answer later, but here my 2 cents:
It appears to me that the term "energy" and energy conservation becomes a bit more complicated in general relativity and cosmology, taking into account more abstract forms as the energy of the gravitational field or space curvature. As you seem to be more interested in the implications I just want to focus on the common forms of energy, i.e., radiation, matter and dark energy.
For a given expanding volume of space only the energy from matter (dark and normal) is dilutes as expected. You could say that with respect to expansion its energy is conserved. This means its energy density decreases with $\sim a(t)^{-3}$, where a is the scale factor (or just the diameter) of the universe.
Radiation additionally looses energy via cosmological red-shift, so its energy density decreases even faster with  $\sim a(t)^{-4}$.
Finally you have the dark energy density. What makes it special is, that its density does not depend on $a(t)$. One can think of it as the energy associated with space itself. So if the universe expands new space and dark energy "appears" and its density stays constant.
So over time the average (matter and radiation) energy density decrease and approaches the energy of empty space itself (or vacuum energy).
Note, however, that this is only true for the average densities. In regions of space, that are held together by gravity, as our local group, the matter density will not decrease. On the contrary, these region are still contraction due to gravity.
So to answer your question on the effect on all the celestial bodies:
Those which are gravitationally bound, won't notice much but others will drift apart never to see each other again.

Three points on conservation of energy: (1) It's an observed fact, not a primitive logical consequence of reality. It may or may not be true in all times and places and situations.  (2) As far as we know, energy conservation is a consequence of time symmetry by Noether's theorem. In an expanding spacetime, time symmetry probably does not hold, so energy conservation probably doesn't either. (3) Besides all of the energies mentiond, there is the global energy of space-time itself. Calculating that is non-trivial, but some have argued persuasively that changes in it balance gains in dark energy.

The experts say that as fast as the universe expands the increased space fills instantly with countless trillions of virtual particles which arrive from nowhere. Not only does this contravene the 1st Law of Thermodynamics (mass/energy can neither be created nor destroyed),but there is a discrepancy of 40 orders of magnitude (yes,40 orders of magnitude!) between the vacuum energy density required by Quantum Mechanics and cosmological observations. It has no effect on celestial bodies, though it should have, which is where the conflict with cosmological observations comes in. Supposedly, the 1st Law of Thermodynamics doesn't apply on a cosmological scale.

The first sentence doesn't make any sense, and could use some backing up.

Why does the 40 orders of magnitude confllct between  the alleged vacuum energy density and the results of cosmological observations not matter, and why is nobody willing to explain it? If there were a 40-orders-of-magnitude discrepancy between your official wages and your take-home pay, would you then hold the view that it doesn't matter?


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When was it realised that most major moons orbit in the equatorial plane of their parent planets?
Inspired by the discussion of the moons of Uranus providing a clue to the planet's axis of rotation in this question, I'm wondering when it was realised that the major satellites are typically located in the equatorial plane of their planets. Our own Moon is a counterexample to this trend, so presumably it would have taken a bit of detective work.

for me it was 2018-09-24 15:54:58 UTC

Since the question has sit unanswered for months, I'm trying to give an acknowledgedly incomplete answer.
In summary: By 1700 it would have been a reasonable guess. By 1800 it was becoming an observed trend.
TL;DR:
It's hard to decide when the consensus was created on the typical orbits of satellites, because there are only a few data points (planets with satellites) but we can know when those few data points where known.
Obviously, the only data point known in antiquity was the Moon, but by that time the whole idea of satellites was far from mainstream.
In 1610 Galileo discovered the four biggest moons of Jupiter and that they orbit Jupiter in roughly the same plane. However, it took until the 1660s for Cassini to discover features in Jupiter and therefore be able to realise that its moons orbited in the equatorial plane.
By 1655 Huygens had discovered Saturn's rings and Saturn's satellite Titan, both of them in the same plane. Between 1671 and 1684, Cassini discovered Tethys, Dione, Rhea and Iapetus all of them in the same plane of the rings except for Iapetus.
The rotation of Saturn is a bit harder to detect, and it wasn't measured until 1794 by Herschel.
There is an additional data point even before being able to see Saturn rotating: both Jupiter and Saturn appear elliptical in the 17th century telescopes. Although a causal relation with rotation wasn't established until Newton, as soon as Jupiter's rotation had been discovered it was known that the planet was rotating around the smallest axis of the ellipsoid.
No more data points were added for a long time, because the moons of Mars and the rotation of Uranus were discovered much later, although the first two moons of Uranus had been discovered by Herschel in 1787.
Additionally, it was known from antiquity that all planets were in similar planes, and when sunspots were observed (by Galileo) and the rotation of the Sun could be measured, the Sun's equatorial plane was close to the planes of all planets orbit. Therefore, when the idea of planets and satellites as "little solar systems" arrived, the idea of the central body rotating in the same plane of the system wouldn't have seemed strange.
Then we can summarise what an informed observer at the end of the 17th century would have known:

Jupiter's rotation, Jupiter's oblation, and Jupiter's moons were in the same plane.
Saturn's rings, Saturn's oblation and most of Saturn's moons were in the same plane.
There were only two known exceptions to the rule of "everything around the same planet in the same plane": Earth's Moon and Iapetus, which aren't in the equatorial plane - although they aren't very far away from it.

With all that at hand, I would say that guessing that most satellites orbit in the equatorial plane of its planet would have been a reasonable guess by 1700.
By 1800 the evidence would have increased with:

The rotation of Saturn had been actually observed, confirming that most satellites and the rings orbit on the equatorial plane.
Uranus had two satellites in the same plane.

With that, it seems even more reasonable to suppose that Uranus'satellites were orbiting on its (unknown) equatorial plane.

I agree that this would be expected.  However note Jupiter has very low axial tilt,so its equatorial plane and the ecliptic plane almost co-incide.  Saturn and Uranus have substantial tilt, but have much fewer features.  Jupiter's moons are very close to the ecliptic plane (as close as our moon) And while It would be clear  that Saturn's moons were not in the ecliptic plane it wouldn't be clear that they were in the equatorial plane until the 1800s (though I don't think this would have surprised anyone). There's also phobos and deimos to mention.

Yes, that satellites orbited near the ecliptic could have been another reasonable guess before Uranus. Phobos and Deimos weren't discovered until 1877.


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Why are there no photos of exoplanets?
There are a lot of photos of diffrent astronomial objects such as galaxies, nebulae and stars of which some are very far away.
However there are no real photos of exoplanets (by “real photos” I mean pictures that actually show something more than a blob of light). Why is that?
Distances of galaxies are by many orders of magnitude larger than distances to exoplanets and yet we have a lot of very detailed (and beautiful) pictures of them. Meanwhile, everything we got on exoplanets is limited to a few pictures with blurred blobs of light.

This may answer your question: https://space.stackexchange.com/questions/31768/are-images-of-exoplanets-surfaces-technically-possible

Also relevant: https://space.stackexchange.com/questions/35396/can-we-make-super-massive-telescopes-to-image-exoplanets/35397#35397
The largest exoplanets are ~10000 times smaller than the red giant stars that we are only now starting to be able to image.

Galaxies are far away but very big. Planets are nearer but very small.
Even for planets that are not completely swamped by the much brighter light of their host star, their angular size is much, much smaller than even quite distant galaxies and certainly smaller than the best achievable angular resolution by any telescopes on Earth or in space.
A quick example will suffice.
The angular resolution of the Hubble Space Telescope (HST) is about 0.1 arcseconds. That is, unless something subtends an angle of much more than 0.1 arcseconds then you are never going to get a good resolved image of it.
A big spiral galaxy is about 50 kpc in diameter. Such a galaxy will still have an angular diameter of 0.1 arcseconds when it is at a distance of $10^{11}$ pc. i.e. We can get images of such things (if they exist and if they are bright enough) right across the observable universe.
A planet the size of Jupiter subtends an angle of 0.1 arcseconds at a distance of 0.01 pc. i.e. We could not resolve the surface of Jupiter even at only a hundredth of the distance to the nearest star.

The reason is that an exoplanet only shines by the light of its parent star, and this reflected light is not only very, very faint, but often obscured by the dazzling light of the nearby star. Exoplanets are seldom if ever detected directly, for the reasons I've just explained, so we can never hope to have the sort of photographs we have of planets in our own solar system. However, there are some amazing new, state-of-the-art telescopes coming on line in the next few years, and they will probably be able to get photographs of planets which aren't too close to their stars. By analysing their spectra, astronomers will be able to find out quite a lot about them. The other things you mention are easily photographed because they are trillions of times as large and trillions of times as bright as an exoplanet.

As an example, we didn't have good photos of Pluto before 2015: https://i.imgur.com/ilpdzsC.jpg

NASA is working on it:

    Artist’s depiction of a possible image from a Solar Gravitational Lens (SGL) telescope.
  Credits: Slava Turyshev

Direct Multipixel Imaging and Spectroscopy of an Exoplanet with a Solar Gravitational Lens Mission

For an exo-Earth at 30 pc, ... one can reconstruct the exoplanet image with ~25 km-scale surface resolution, enough to see surface features and signs of habitability.

There are a few exoplanet photos:
http://www.slate.com/articles/health_and_science/bad_astronomy/2012/11/exoplanet_pictures_astronomers_have_photos_of_alien_planets.html
Given the usual size and brightness differences between a planet and its star, conditions have to be unusual for direct imaging.
-Sorry for link only answer, I'm mobile right now.


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What is the significance of minus sign in the expression of apparent magnitude?
The apparent magnitude of a distant object in the sky is measured as
m=(-)2.5 log (I/I0)
Where I is the intensity of the object whose apparent magnitude is to be measured and I0 is the reference magnitude. My question is why we put the minus sign in the expression.

So if I read your question correctly, you're asking why as stars get dimmer, they are given higher magnitudes?
The reason is purely historical. The ancient greeks assigned stars with 6 brightness levels or magnitudes. The brightest stars were of first magnitude and the least bright stars (to the naked eye) were of 6th magnitude.
In the 1800's this system was formalised by an astronomer named Norman Robert Pogson. By then they were (presumably) able to more accurately measure the brightness of a star, and he decided to use a system that fitted well with the ancient measurements, but could be used to interpolate and extrapolate to brighter and dimmer objects. In order to do this, the formula has to have a negative in front of the log.
More information here


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How to interpret error of absolute magnitude of a star given data from wikipedia?
This Wikipedia page contains information about the star Spica in the rightmost table. A small snapshot of this table is below.

In this table, the absolute magnitude is given as −3.55 (−3.5/−1.5). Instead of seeing the value and error given as value ± error, I am thinking the error is given as it is because of absolute magnitude is given on a logarithmic scale (as opposed to linear scale). But I am not quite sure how to interpret this as an error reading. How does one interpret this reading?

Spica (α Vir) is a binary star.
The Wikipedia article cites Herbison-Evans et al. 1971, whose Table III says:

Absolute magnitude of primary (Mv1)   -3.5±0.1
  Absolute magnitude of secondary (Mv2)   -1.5±0.2

So the confusing "-3.55 (-3.5 / -1.5)" can be interpreted as "combined (primary / secondary)."


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Are red dwarfs (M dwarfs) known to be strong infrared sources?
If so, why? Are they known to be surrounded by large amounts of dust? Do they typically flare?

What do you mean by "strong"? Compared to what?

Yes, red dwarfs are strong in the infra-red. This is because they are much cooler than our sun, and just like a red hot poker which has cooled from being yellow hot, their emissions are redder. It takes less energy to emit red light than to emit longer wavelengths, and still less energy to emit in the infra-red. Being much smaller than the sun and with less fusion taking place, red dwarfs have less energy.They are known to emit flares, sometimes quite large ones.

Thanks for your answer. From the blackbody spectrum, we wouldn't expect red dwarfs to be strong emitters in the submillimeter correct? Is this true to observation?


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Could anything consume a small black hole?
Whenever I read about black holes, it is normally involving how they devour anything that comes too close... regardless of how big anything is.
But what if it were a really small black hole vs something really large? How about something like a large star or neutron star... could a black hole be small enough to get devoured / consumed by something like that or would it just not be possible?

Black holes don't "devour".  They can't "eat" things.  But things can "fall into" a black hole.  It really is just gravity, but really really intense gravity.
A really small black hole would have a mass of about 3 times the mass of the sun. All neutron stars are smaller than that (or they would turn into black holes).  A really large star could be more massive and would be much much larger in volume.  A black hole could orbit with such a star, but if it got close enough it would pull gas off the star which would then fall into the black hole, causing it to get larger.
There is no way that a large star can in any way "pull apart" or otherwise "consume" a black hole.
There is a theoretical notion of the quasi-star. These are similar to a Thorne–Żytkow objects.  In normal stars, the outer layers are prevented from collaping by energy from nuclear fusion in the core.  In a quasi-star, the core collapses into a black hole and the release of energy from matter falling into the black hole prevents the collapse of the star. A steady state can be achieved, as if more matter starts to fall in, more energy is released causing the star to grow, and reducesing the amount of matter falling in. Nevertheless, ultimately all the matter from the star would fall in.
Smaller black holes can't form from stars. We don't know if they exist at all, but if they do, very very small black holes could be too small to interact much with regular matter. A very small black hole could be smaller than a proton, and even if one fell into the Earth, it could pass through the gaps between and inside atoms. Such a black hole would be very hot, due to Hawking radiation.

What if it isn’t orbiting?  Wouldn’t a collision between a black hole (above some threshold size) and a large star result in a larger black hole?  Which one consumed the other would be moot.

FWIW, a 3 solar mass BH has a Schwarzschild radius of ~8.862 km.

That would be a bigger black hole. Black holes are indestructible but a big one can consume a smaller one.
The collision between anything and a black hole results in a black hole plus some optional debris from whatever the other thing is, so the result of a collision between two black holes is simply a bigger black hole.

It's "in principle" possible. For example if we had a black hole the size of a small mountain on Earth (this kind of black hole are possible in theory but cannot form from stellar collapse) then it'd be impossible to confine in a laboratory, and it would fall through the Earth to eventually settle at the Earth's core. We could then argue that the Earth has "consumed" the black hole.
It would be temporary, however, because eventually the particles at the Earth's center would fall into the black hole, it would gain mass, eat more particles, etc, until everything falls in and the Earth is consumed by the black hole.

A 1 kg black hole has a Schwarzschild radius of 1.48517E-27 m, far smaller than a proton. And assuming that Hawking radiation is real (and that some quantum gravity effect doesn't stabilise micro black holes) its lifetime is only 4.65118E-17 seconds. Please see https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator


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The energy density of phantom dark energy increases with time. Does it violate the concept of accelerating universe?
Dark energy with equation of state parameter $\omega<-1$ is called phantom dark energy. Its energy density increases with time.
My question is whether this model violates the concept of accelerating universe, because the energy density turns out to be negative. If we use these values in the acceleration equation, we see the acceleration of the universe (second derivative of the scale factor) is negative.
My derivation is as follows. Let's set $\omega=-2$, so the equation of state is $p=-2\rho c^2$. From the fluid equation,
$$\dot{\rho}+\frac{3\dot{a}}{a} \left(\rho + \frac{p}{c^2} \right)=0\Rightarrow\rho\propto a^3$$
From the Friedman equations, $a=-k_1 t^{-2/3}$, so the energy density is $\varepsilon=\rho c^2=-kt^{-2}<0$. Hence the acceleration is
$$\ddot{a}=-\frac{4 \pi G}{3} \left(\rho + \frac{3p}{c^2} \right)=\frac{4 \pi G}{3} \times 5\rho = \frac{4 \pi G}{3c^2} 5\varepsilon$$ Since $\varepsilon <0$, this implies $\ddot{a}<0$.

This question needs more work to be understood.

In the image it is seen clearly that the universe is decelerating. But in reality it is accelerating @ Rob Jeffries

I don't follow where your negative solution to the Friedman equation comes from (and it obvously doesn't make sense in an expanding universe) or why the energy density is negative, or where the $5\rho$ comes from in the last line.

It's very easy. Just integrating the Friedmann equation you can derive the relation that the scale factor is proportional to t^-2/3 as it is shown above. Now If you simply put the value of p in the acceleration equation you will get (5×rho) in the last line.

@SoumantiChakraborty Why would phantom energy density be negative ?

Why is DM (dark matter) relevant to this question?

For phantom energy model we have $$H^2 = H_0^2\Omega_pa^{-3-3w_p}$$ such that $w_p < -1$
$$\dot{a}^2 = H_0^2\Omega_pa^{-1-3w_p}$$
$$\dot{a} = H_0 \sqrt{\Omega_p} a^{-1/2(1+3w_p)}$$
$$\int a^{1/2(1+3w_p)}da = \int H_0 \sqrt{\Omega_p}dt$$
Now at this part we should define the integrals upper and lower bounds. When we set time from $t → t_0$
$$\int_a^{1} a^{1/2(1+3w_p)}da = \int_{t}^{t_0} H_0 \sqrt{\Omega_p}dt$$
We get something like,
$$\frac{2} {3(1+w_p)}[ 1 - a^{3/2(1+w_p)}] = H_0 \sqrt{\Omega_p}(t_0 - t)$$
$$a^{3/2(1+w_p)} =1 -  \frac{3(1+w_p)}{2} H_0 \sqrt{\Omega_p}(t_0 - t) $$
Let us take $w_p = -2$
$$a^{-3/2} =1 +  \frac{3}{2} H_0 \sqrt{\Omega_p}(t_0 - t) $$
So
$$a =(1 +  \frac{3}{2} H_0 \sqrt{\Omega_p}(t_0 - t))^{-2/3} $$

here $y$ axis is the scale factor $(a)$ and $x$ axis is the time $(t)$
For $C = H_0 \sqrt{\Omega_p}$ and $a=t_0$
As you can see scale factor goes to infinity
For the acceleration equation
$$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3c^2}\epsilon_p(1+3w_p)$$
For $w_p=-2$
$$\frac{\ddot{a}}{a} = \frac{4\pi G}{3c^2}5\epsilon_p$$
From the graph and also from here its clear that $\ddot{a} > 0$.

To explain more plainly what is going on here, the (now deleted) asker noticed that solving the first Friedmann equation for a phantom dark energy model leads to a universe whose size scales as a negative power of the time. For the equation of state parameter $w=2$, the dark energy density grows as $\rho\propto a^3$, where $a$ is the scale factor. The Hubble rate then grows as
$$H\equiv\frac{\mathrm{d}a/\mathrm{d}t}{a}\propto a^{3/2}$$
or
$$\frac{\mathrm{d}a}{a^{5/2}}\propto \mathrm{d}t.$$
Integrating this without paying attention to integration constants leads to $t\propto a^{-3/2}$ and hence $a\propto t^{-2/3}$. That is odd -- apparently the universe must shrink? Alternatively, if the proportionality constant is negative, the universe must have negative size and energy density?
The resolution is to pay attention to the integration constant. Integrating the equation above really leads to $a\propto (\tilde t-t)^{-2/3}$, where $\tilde t$ is the integration constant. This universe grows and accelerates for $t<\tilde t$, with $a\to\infty$ as $t\to\tilde t$. $\tilde t$ is the time of the Big Rip!


text:
Can someone calculate the age of the KOI-4878 star?
The star’s low metallicity and fairly high space velocity suggest that KOI-4878 is older than the Sun.
But I don't know how to calculate an estimation for the age.
KOI-4878 data on Simbad

You probably want to look if it has been picked up by an asteroseismic survey. Given that its a KOI, this could be true. Ages from asteroseismology are the most accurate ones, particularly that this star is a G4.

@AtmosphericPrisonEscape Where would one look for that?

@B--rian: I would think the relevant archives should be those with accurately sampled photometric time series data, i.e. Corot, Kepler, Tess. But I see you already posted the link to the exoplanet archive, which does link all those info together, if available. If standard methods (isochrone fitting, seismology) would allow the age determination, I suspect it would already have been done.

@AtmosphericPrisonEscape Thanks for the hints, made me realize that I do not understand the exoplanet archive well enough - all I see is that there is no obvious entry age.

According to Wikipedia on KOI-4878.01 nobody did yet calculate the age of that object, but says exactly what you quoted:

its low metallicity and fairly high space velocity suggest that KOI-4878 is older than the Sun.

This supposidely relies on the following sources:

https://exoplanetarchive.ipac.caltech.edu/cgi-bin/DisplayOverview/nph-DisplayOverview?objname=KOI-4878&type=KEPLER_TARGET
https://www.drewexmachina.com/2019/09/05/koi-4878-01-keplers-most-earth-like-planet-candidate/

A search with the keyword KOI-4878.01 on different publication databases like arXiv or scholar.google did not help either.
Related

How do scientists determine the age of stars? focuses on the general question how to determine the age of a star.


text:
Why is Mars cold?
The surface temperatures of Mars are about -87C to 5C, which is much colder than that of Earth's.
If Mars has 95% carbon dioxide, which is a Greenhouse gas, why is the surface of Mars so cold? Shouldn't it trap heat and render it hot?

You should be careful drawing conclusions from percentages. Mars's atmosphere is a lot sparser than Earth's.

Venus only has 1.1% more CO2, why's it so hot? (spoiler: it ain't because its closer) "Average surface pressure: 93 bar" ... that's 93 times that of Earth. - Density, not percentage.

@Mazura: Well, it's partly because it's closer; if Venus and Earth had exactly the same atmosphere, Venus would still be warmer (although not by nearly as much) due to its closer proximity to the sun (and it's been postulated that this might actually be the reason Venus got tipped into a runaway greenhouse effect, while Earth hasn't [yet]).

Sounds like a setup for a Chuck Norris joke.  "Chuck Norris has already been to Mars, that’s why there are no signs of life."  "Chuck Norris roundhouse kicked Mars and knocked the atmospere off."

@CrossRoads: Pretty much. The Sun is playing Chuck Norris, doing repeated roundhouse kicks, with Mars undefended due to low gravity and no magnetic field.

So, we're calling orbital revolutions a roundhouse kick now? Fine, why not.

I can't help thinking about this story bots clip

Firstly, Mars has a mean distance from the Sun of 1.524 AU, so by the inverse square law the energy it gets from the Sun is about 40% of what the Earth gets.
But the main reason that Mars is so cold is that its atmosphere is very thin compared to Earth's (as well as very dry, see below). From Wikipedia Atmosphere of Mars:

The atmosphere of Mars is much thinner than Earth's. The surface pressure is only about 610 pascals (0.088 psi) which is less than 1% of the Earth's value.

In comparison, the mean surface pressure on Earth is 101,300 pascals. So the atmosphere of Mars is barely more than a vacuum compared to Earth's.
So even though the Martian atmosphere is over 95% carbon dioxide, there simply isn't enough of it to trap much heat.
Although carbon dioxide is a greenhouse gas, the predominant greenhouse gas on Earth is actually water vapour. However, water is usually cycled in and out of the atmosphere very quickly in response to temperature and pressure changes. Carbon dioxide is a problem because it stays in the atmosphere for a long time, and its presence shifts the equilibrium temperature upwards from that of the plain water cycle.

But Earth's atmosphere only has 0.04% CO$_2$, so that's an order of magnitude less than Mars, in absolute values.

@pela True, and that's why I added that info about water. But I suppose I shouldn't make that a "BTW" section.

Ah okay, I only saw that now.

@pela  :)

Good thing I'm not an observer :D

I made a small edit, I missed the water part the first time through as well, since you led with absolute pressure, then went on to CO2.

@uhoh Good idea.

@pela: So what would the temperature of Mars be if it had no atmosphere, or if the atmosphere was composed entirely of non-greenhouse gasses like nitrogen or argon?

@jamesqf Probably almost exactly the same. The radiative forcing from carbon dioxide is tiny, and quickly drops off (e.g. the difference between 100 ppm and 200 ppm is much larger than between 1000 ppm and 1100 ppm). Water vapour and methane are much more important, but on a planet like Mars, water vapour would be "quickly" lost. On Earth, more carbon dioxide also lets the air to hold more water vapour, which also lets more carbon dioxide into the atmosphere, which is a bit of a positive feedback loop. But geologically speaking, carbon dioxide on its own essentially doesn't matter.

@jamesqf It would be roughly 5 K colder, which is the difference between Mars' effective temperature and its average surface temperature. For Earth, that difference is 33 K (see my answer below for references).

Thank you very much for the answer!

While we are talking percentages, Venus has as much or maybe more nitrogen than Earth. It is just a low percentage compared to the CO2 abundance.

I'm just going to expand and deepen on what the other answers already said.
In the following I contrast the atmospheric transmission ($T$) and absorption  ($A$, which is $A=1-T$) of Mars and Earth. The Mars plot (top) is from Prof. J. Irwin via this review by P. Read et al. 2015 and the terrestrial data (bottom) is from wikipedia.
The plots of $A$ and $1-T$ area easily matched by eye when comparing $T=0$ and $A=1$ for Mars and Earth. To make this superclear, I've added arrows to mark the $CO_2$ absorption bands.
Additionally, the plot for Mars has the energy spectra of the incoming solar (T=5800 K) and outgoing infrared (T=216 K) radiation inserted. Earth's outgoing infrared radiation from the surface would be at T=300K, and in this graph at nearly the same position. For a strong greenhouse effect, an atmosphere must absorb the majority of this outgoing peak.

From this we learn a few things:
1.) Looking at Earth, we see that there are several important greenhouse gases in the atmosphere, but the most significant one is water.
2.) A significant portion of Earth's greenhouse peak is absorbed by $H_2O$+$CO_2$, while only a small fraction is absorbed by the martian atmosphere, visually explaining what already @pelas answer noted, that Mars has a weak greenhouse effect.
3.) If comparing the width of the $CO_2$ absorption bands very carefully, then one will note, that the $CO_2$ bands for Earth are broader than on Mars, although Earth's atmosphere has much less total $CO_2$ in it! This is an effect of the so-called 'pressure broadening' of atmospheric lines, enhancing their greenhouse capability significantly. Pressure broadening becomes important in atmospheres at pressures of about $0.1-1 \; bar$ (rough rule-of-thumb). This is another reason why the low mass/pressure of the martian atmosphere cannot create a strong greenhouse effect, as the pressure is around $0.01 \; bar$ on the surface.

While it has little to do with the question, I'm curious why Methane is such a strong greenhouse gas with a relatively small absorption spectrum in the chart above.    Nice details by the way.

@userLTK There is not enough methane in the atmosphere to saturate those absorption bands (yet). The optical depth $\tau_{\lambda}$ at a given wavelength $\lambda$ and height $z$ is $\tau_{\lambda}=\int_{\infty}^{z} dz' n(z') \kappa(z')$, with $n(z')$ being the local number density and $\kappa(z')$ the opacity per mole, which is again a function of the temperature. Only at $\tau_{\lambda}>1$ is the photon mean-free path smaller than the atmospheric scale (which means the photon can't escape), and we see that both the absorption strength (given by the opacity) and the amount of gas contributes.

Am I correctly understanding that first graph, that even as sparse as the Martian atmosphere is, almost no light makes it through the CO₂ absorption spectrum in the middle of the outgoing infrared at ~15μm?

@MichaelS: Right, this band at $15\mu m$ is opaque, so the outgoing radiation gets trapped there. But this is only a small fraction of the total black body energy, so it doesn't contribute to a strong greenhouse effect.

It might improve this answer is you explicitly mention that the bottom graphs are Earth, the top is Mars, and the Mars absorption graph is "upside down" compared to Earth graphs (Earth graphs "absorption and scattering", while Mars measures "transmission").  Great answer!

@Yakk-AdamNevraumont: Thanks, but that's literally my first two sentences.

Mars does have a greenhouse effect, only somewhat weaker than Earth's.

Mars' atmosphere is very dilute, with a with a surface pressure only 0.6% of Earth's. So even if 95% of it is CO2, that's not a lot. However, it is actually a higher absolute abundance of CO2 molecules than on Earth, which only has a CO2 abundance of 0.04% (by volume; e.g. NOAA, corresponding to roughly 0.06% by mass).
The exact calculation depends on how the atmosphere decreases in density with height (the "scale height"), but for an order-of-magnitude calculation, we can use the total masses of the atmospheres $M_\mathrm{atm,Earth} = 5.15\times10^{18}\,\mathrm{kg}$ (Trenberth & Smith 2004), and $M_\mathrm{atm,Mars} = 2.5\times10^{16}\,\mathrm{kg}$ (NASA). The total amount of CO2 in Mars' atmosphere, compared to Earth's, is then
$$
\frac{M_\mathrm{CO_2,Mars}}{M_\mathrm{CO_2,Earth}} = \frac{95\%\times M_\mathrm{atm,Mars}}{0.06\%\times M_\mathrm{atm,Earth}} \simeq 7.9,
$$
i.e. almost an order of magnitude higher. The fact that Mars is smaller than Earth means that this number is probably somewhat larger.
Roughly 10–20% of the radiation emitted by Mars' surface is absorbed in the atmosphere (Haberle 2015). One way to quantify the greenhouse effect is the difference between the effective temperature and the planet's average surface emission temperature.
For Earth, the difference is 33 K, while for Mars it's a much lower 5 K (but must have been much higher in the past in order for liquid water to exist).
So why is the effect so much larger on Earth? Well, CO2 is not the only agent of the greenhouse effect. Other gases — e.g. methane, nitrous oxide, and ozone — add as well. But actually the largest contribution comes from water vapor. Exactly how much is a matter of heated debate, I think, but it's probably not controversial to say that at least 1/3 of the effect is due to water (RealClimate says 36–66%).
Water molecules are less reflecting than CO2, but constitutes 0.3% of Earth atmosphere. In contrast, only 0.03% of Mars' atmosphere is water vapor (e.g. Trokhimovskiy et al. 2015).
In summary, the smaller greenhouse effect, and the fact that Mars is roughly 1.5 times as far from the Sun as Earth, and hence receives less than half of the incident radiation, is the reason that Mars is so cold.

"However, it is actually a higher absolute abundance of CO2 molecules than on Earth, which only has a CO2 abundance of 0.06% (by mass)." citation needed?  AFAIK there is still less CO2 per unit area and in the air column on mars than on earth, though it may still be in the same order of magnitude, and actual measurements for the mass for both earth and mars for CO2 is scarce from google searches.

@opa References for the abundance of CO$2$ in Earth's atmosphere are plentiful, see e.g. NOAA (it's 0.04% by volume, so roughly 0.06% by mass). As for the comparison, I just did the calculation $\rho{\mathrm{CO}2,\mathrm{Earth}} / \rho{\mathrm{CO}_2,\mathrm{Mars}} \simeq 0.06 / (0.95\times0.006) \sim 10$. I used the Mars-to-Earth pressure ratio, which is slightly wrong — I should instead use the density ratio, which actually increases the ratio a bit.

@opa I added references and calculations :)


text:
Could we detect an extraterrestrial signal with the SKA radio telescope?
As I have not seen this question answered and, following the rules: https://stackoverflow.blog/2011/07/01/its-ok-to-ask-and-answer-your-own-questions/ , I would like to answer this important question.
As you know, by 2027 we expect to have the biggest radio telescope already working: the Square Kilometre Array.
According to the former Interim Director General of the SKA project, Michiel van Haarlam, it will be possible to detect the radar of an extraterrestrial airport 60 light years away, once the initial stage of the project is completed.
However, a recent study also points out that the SKA could detect extraterrestrial airports 200 light years away.
Source: https://www.youtube.com/watch?v=ayqyb8XCtE0

So what's your question? If the study points that out, then yes?

Yes, we're encouraged to answer our own questions, but you still have to follow the usual question & answer format. Please post the answer in the Answers section.


text:
What alternative facilities would be available in the event of JWST being destroyed?
Unfortunately, space launches can and do go wrong. Suppose that after all the delays and budget overruns, the launch of the James Webb Space Telescope fails and the telescope becomes a cloud of very expensive shrapnel raining into the sea.
Are there any upcoming/planned facilities that would be able to do similar science to what JWST would be capable of within the next decade, or are all our eggs in one basket on this one?

All eggs in one basket. If JWST fails, it would set back astronomy enormously. You can't do from ground what this telescope would otherwise do.

JWST is not an intentionally cheap mission to the moon (a type of mission which hasn’t had much success this year, thus there’s been lots of worrying headlines about failed missions, which shouldn’t worry you). We’re sending it out to a quiet place, floating in space, using a nice expensive rocket. Rockets are very reliable nowadays, we can land them vertically for crying out loud. JWST will be fine. It has to... It has to.

If SpaceX does indeed fly the Starship with its 12m diameter building a replacement could be much quicker without the need for complex folding of the mirror and sun-shield .

Heck , they even built two machines in the movie Contact. Couldn't they build extra parts and give them to the military if all goes well?

For all (like me) reading this question in 2021: The JSWT is scheduled for October 31st, 2021.

And XKCD just brought a cartoon on the topic: https://xkcd.com/2447/

@B--rian They have postponed it once again: https://arstechnica.com/science/2021/06/webb-telescope-launch-date-slips-again/.

It didn't explode after all.

Definitely It will be a huge setback for astronomy and science community.

All eggs in one basket. If JWST fails, it would set back astronomy enormously. You can't do from ground what this telescope would otherwise do. –
AtmosphericPrisonEscape

JWST is one of the most technological advanced telescope of this decade and it is now kept safely in the L2 point and will soon start to do its work. And it is very unlikely to fail or get destroyed. Suppose if there is any software error, attempts will be made to resolve the problem from the Earth itself, BUT.....
any Hardware damage will definitely cost 10 billion dollars. Because sending tech crew for repairing the James Webb telescope to a place 1.5 million kilometer would be impractical, if done, then it will the longest distance the humanity would achieved to travel at a stretch.
And if JWST fails anytime in the future (which is unlikely) there will not be any telescopes equivalent to JWST in service for at least the next 5 to 6 years until the Nancy Grace Roman Telescope maybe....
https://en.wikipedia.org/wiki/Nancy_Grace_Roman_Space_Telescope


text:
Force exerted by space debris on satellites
Hypothetically, if space debris were to collide with the currently orbiting satellites, how much force would they exert, and how much damage would they make?
Will the satellites be able to withstand that much force, especially if they don't have Whipple shields?
Since space debris travel at tremendous velocities, I am guessing the force is pretty large too. How do you calculate the force? Is it a simple $\rm{F=m \cdot a}$ calculation?

This question isn't particularly related to astronomy. This kind of question may be a better fit for the Space Exploration Stack Exchange.

Sorry I am a new user, didn't know Space Exploration stack exchange existed. Thank you.

I’m voting to close this question because the effects of the space environment on spacecraft is an engineering issue and not about Astronomy. This is however nicely on-topic in Space Exploration SE, so recommending migration when closed.

The force depends on the details of the collision. Depending on how hard the materials involved are, and the angles and everything, it might be a larger force for a shorter time, or a shorter force for a larger time.
What are relatively predictable is the impulse -- force times time, essentially (or the integral of force over time, if you want to be exact) and the energy released by the impact. For a small piece of debris these are determined by the mass and velocity of the debris with respect to the satellite.
Relative velocities are typically a few km/s, so, 1 mg piece of debris (a paint flake or something like that) will deliver about $10^{-3} kg m/s$ of impulse, and about $10--100J$ of energy.  The impulse will change the orbit of the satellite (typical mass perhaps 1 ton) by very little. Perturbations from tides and other effects will be bigger. The energy is more of a concern -- depending on what it hits something may get hot, or have a hole made in it.
A much larger piece of debris (say 1g -- a small screw) carries 1000 times the impulse and energy and may will wreck a satellite (although if it hits a solar panel or something it may just make a hole and go on through). If the satellite survived, then it the change to its orbit would be noticeable, but not a huge deal.

heliocentric orbital velocity circa 1 AU is 30,000 m/s, 1 ton vs 1 mg is 1E-09 so the delta-v imparted by an absorbed 1 mg impact is 3E-05 m/s which is almost a million times larger than the Pioneer Anomaly, so it would certainly be detectable by tracking over time! https://space.stackexchange.com/a/3594/12102


text:
Client-side JavaScript astronomy libraries which are not based on node.js?
Are there any client-side JavaScript astronomy libraries? Almost every library I see appears to be server side for node.js. I realize some node.js libraries can be adapted for client-side use, so I'd also be happy with a server-side library that happens to work client-side as well.
EDIT: I did find https://github.com/mourner/suncalc but it appears to be abandoned, has 35+ open issues, many related to accuracy, and, despite its description, appears to give only solar and lunar positions, not planetary positions.

What are your expectations/requirements to such a library? Calculating "things", animating objects? For solar systems, the Milkyway, the observable universe? Please be more specific.

For right now, it's just a general question. Obviously, I'd want "the most complete" library eventually, but, for now, I'll settle for anything.

I think it wouldn't be so hard to extract the computing part of the server-side node.js libs.

@peterhsaysreinstateMonica I actually tried this with orb.js and livestreamed my attempt at https://www.youtube.com/watch?v=v1ItMdi3_GY but it's harder than it looks, especially when file loading is involved. Of course, you're free to take a shot.

I found "Astronomy" from "cosinekitty" on Github:
https://github.com/cosinekitty/astronomy
It supports both node.js and browser.
Browser demos:
https://github.com/cosinekitty/astronomy/tree/master/demo/browser
Testing the library without downloading whole repository:
Save this file locally:
https://github.com/cosinekitty/astronomy/blob/master/demo/browser/positions.html
Into the file, replace line
<script src="astronomy.browser.js"></script>

by
<script src="https://cdn.jsdelivr.net/gh/cosinekitty/astronomy@1.0.1/source/js/astronomy.js"></script>

Reference: https://cdn.jsdelivr.net/gh/cosinekitty/astronomy/
Change this line:
let observer = new Astronomy.Observer(latitude, longitude, elevation);

to
let observer = new Astronomy.MakeObserver(latitude, longitude, elevation);

Remove line:
 <link rel="stylesheet" href="astro_demo.css" />

You should get the table in the file filled up with data calculated by the library.

Another one is astronomy-bundle-js:
https://cdn.jsdelivr.net/gh/andrmoel/astronomy-bundle-js@6.0.4/src

You should definitely checkout "astronomy-bundle"
https://www.npmjs.com/package/astronomy-bundle
This library covers most of the important astronomical calculations and has a great documentations.


text:
How does one find the electric conductivy of a planet's atmosphere?
I am working on an undergraduate project in which we study shaped structures in the atmosphere of the gas giants. We are investigating whether the magnetic field is the source of the instability generating the observed shaped structures. We therefore use spectral methods to make a stability study on the magnetohydrodynamics equations which in our case are
$$\frac{\partial{}\boldsymbol{B}}{\partial{}t}+\left(\boldsymbol{v}\cdot\boldsymbol\nabla\right)\boldsymbol{B}=\frac{1}{\mathrm{Re}_\eta}\boldsymbol\nabla^2\boldsymbol{B}+\left(\boldsymbol{B}\cdot\boldsymbol\nabla\right)\boldsymbol{v}$$
$$\frac{\partial{}\boldsymbol{v}}{\partial{}t}+\left(\boldsymbol{v}\cdot\boldsymbol\nabla\right)\boldsymbol{v}=\frac{1}{\mathrm{Re}}\boldsymbol\nabla^2\boldsymbol{v}+\boldsymbol{f-\nabla}p+\mathrm{N}\left(\boldsymbol{j\times{}B}\right)$$
where $\rm{}Re_\eta$ is the magnetic Reynolds number and $\rm{}N$ is the Stuart number expressed
$$ \mathrm{N} = \frac{\sigma B^2L_c}{\rho{}U}$$
where

$B$ is the magnetic field intensity;
$L_c$ is the characteristic length of the studied structure;
$\sigma$ is the flowing fluid electric conductivity;
$\rho$ is the fluid density;
and $U$ is the characteristic velocity scale of the flow.

Our model is now functional and the next step is to run the calculation with real values for physical parameters. We have all the data except the electric conductivity. Hence, I am  looking for this data (or a method to calculate an approximation). I thought that a first approximation could be made from the volume fraction of hydrogen and helium and their respective conductivity.
All ideas, avenues of research and scientific publication references are welcome. Thank you for help.
PS: I don't know much (yet) about astrophysics and the planet's atmosphere.

The conductivity of neutral gas is negligible, you need to estimate the ion fractions in your gas and a current density model. For the solar system gas giants and the dynamics of their visible atmospheres this is going to be completely irrelevant, as those are very cold. In Hot Jupiters that would be a different issue. But if you are determined on working on this topic, I can write up an answer.

Why would it irrelevant for the very cold atmospheres?
I am interested and I would appreciate your answer, thank you.

Because you're very cold, so your ion conductivity will be very small, and the ion conductivity is the one that counts. Hence your Stuart number will be very small.

Indeed it makes sense, thanks.

The conductivity of neutral gas is negligible, you need to estimate the free-electron conductivity in your gas.
For this, one could first consult a phase diagram, like referenced in this answer. There you'd see that for the atmospheric pressures and temperature for all gas giants, you will stay in the $H_2$-gas regime, so no exotic equation of state is necessary to describe the state of the visible atmosphere of the gas giants.
The updated models with and without Juno-data show that adjustments to structure models in the interior of the planet need to be made but the atmosphere, which you are interested in, is still an ideal gas.
To compute the amount of free electrons in the atmosphere, you need to compute the possible ions, and their ionization fraction. This gets really nasty for cold atmospheres, because molecules form. Mole fractions of those have to be computed by a chemical model, then ionized (possibly by a Saha-type ionization model) and translated into a conductivity (possibly by a Lorentz-Drude-type of model).
For your purposes that could get really involved, but I guess what you could do that is much simpler, is look at some published literature, e.g. here, just take and cite their conductivity profile.
However even in this source, the two models they use for the conductivity differ by 9 orders of magnitude in the neutral atmosphere, so you want to read carefully about their assumptions.

Thank you! I will have at look at all of this.

@Loic: Sure, if you run into trouble, just ask.


text:
What is the cycle for when the maximum amount of the Moon's apparent area is sunlit?
The Moon is defined as full when its ecliptic longitude is 180$^{\circ}$ distant from the Sun's.
But what is the cycle for when the maximum proportion of the moon's apparent area is sunlit, and for observers at what latitudes of the Earth's surface is this maximum realised?
Notes
1) After first guessing (on this expert Q&A site) that the answer is "one lunar month" and asking me whether there was "a reason" why the said cycle might be different from it, user UhOh has requested clarification of the term "moon's apparent area", pointing out that the moon has a rough surface. The difference of the apparent perimeter of the Moon's disc from a circle does not have a bearing on the answer to this question that is of anything like the magnitude of the contribution made by the 5$^{\circ}$ tilt of the plane of the Moon's orbit around the Earth relative to the plane of the Earth's orbit around the Sun. Please ignore mountains and craters. Please assume either that the Moon is a sphere, or if you wish then take it to be a spheroid with an eccentricity of about 1/900. The question is about the relative movements of the three celestial bodies concerned. (If further clarification is required of any other term used, please let me know.)
2) Wouldn't it be great if there were a website where you could input your latitude and longitude and desired start date and end date and get a graph of the proportion of the moon's apparent disc that is sunlit from your location? But I don't think this exists (yet!)

Naively I would guess it's also one lunar month, at least on average. Is there some reason that it might be different?

If there wasn't a reason I wouldn't have asked. The plane of the Moon's orbit around the Earth is inclined at an angle of 5$^{\circ}$ to the plane of the Earth's around the Sun. To get a handle on why that affects the Moon's apparent fullness, imagine if it were 90$^{\circ}$. An article on the possible origins of the lunar inclination appeared in Nature in 2015. Given the inclination I don't think the proportion ever reaches 50%. The 5$^{\circ}$ angle is not completely constant. This is not a simple problem.

Local maxima occur once per synodic month, yes, but the cycle is more complicated.

okay, then how exactly should "the maximum proportion of the moon's apparent area is sunlit" be defined? The Moon isn't a sphere, it has a rough, contoured surface. And what if when all the orbital-mechanical details and approximations for the geography are taken into account one near-maximum has a 0.001% less illuminated proportion than the next? Would it no longer be considered a maximum? Or if 0.1% less? If you want a detailed answer you'll have to define your terms better and constrain the scope of the answer more clearly.

I have amended the question to help you. You may be unfamiliar with superposition of cycles. If there is a 10-month cycle across which the monthly local maximum varies from 49.300% to 49.400%, then yes, the overall maximum would be 49.400%, reached every 10 months, and when the local maximum was 49.300% or 49.372% the quantity would not be at a global maximum. Of course there are probably a lot more than two cycles involved. (To extend the example, the 10-monthly maximum might vary from 49.398% to 49.402%, and its period might be 10.141 months.) A good answer will be enlightening.

I should add that I do not know what cycles are involved. I hope a responder can go into this in some detail, explaining contributions to the overall cycle, but even if they only reference one cycle other than the synodic one and are able to give a maximum figure for the sunlit proportion of the Moon's disc (which I suspect may not even reach 49.9%), their answer will be very welcome. Is there anything else you do not understand about the question?

Thank you for the edit! I've reversed my down vote to up again and removed my vote to close. Asking for clarification is routine in Stack Exchange and the goal of clarification is to make it easier for people to figure out what kind of answer that the question's author is hoping for.  Sometimes people are afraid to put a lot of work into writing in-depth answers to two-sentence questions to then receive a comment "that's not what I meant". Now that you've added much more detail, your question is much clearer and readers have a better idea what to take into account in an answer.

I asked the same question on the Usenet group sci.astro and received the following helpful reply from Barry Schwarz that goes some way towards an answer, reposted here with permission. But the question remains hard. I have not seen this cycle addressed in any astronomical textbook, although it may be in one that I haven't looked at.

Since the Sun's diameter exceeds the Moon's, the Sun always
  illuminates slightly more than 50% of the Moon's surface.  (To
  visualize, draw the external common tangents from a larger circle to a
  smaller one.)  Using 98M miles as the distance, 432K miles as the
  Sun's radius, and 1K miles for the Moon's, the Sun illuminates almost
  7 miles into the opposite hemisphere.
Any observer a finite distance from the Moon can only ever see less
  than 50% of the Moon's surface. The further away the observer is, the
  closer the observable percentage is to 50.  (To visualize, draw the
  two tangents from an external point to a circle.)  However, when the
  Moon is in the plane of the ecliptic, an observer at the north pole
  can see approximately 12 miles over the Moon's north pole into the
  opposite hemisphere.  So he can see some 5 miles beyond the
  terminator.  Similar geometry causes his view of the Moon's south pole
  to fall short of the terminator.
However, for a person standing in line with the ecliptic, his view in
  any direction falls 4 miles short of the terminator and that is the
  maximum you seek.
When the Moon is within 4K miles of the ecliptic, an observer on Earth
  at the same offset from the ecliptic will have the view described
  above.
When the Moon is more than 4K miles away from the ecliptic, all
  Earth-bound observers see some non-illuminated portion and therefore
  do not see the maximum you want.
Not only is the Moon's orbit tilted with respect to the ecliptic, but
  the direction of the tilt wobbles like a child's top.  I have no idea
  how to factor that into an equation that would compute the  time
  between one visible maximum and another.

The illuminated fraction k is related to the Sun-Moon-Earth phase angle i by
$$ k = \frac{1 + \cos i}{2} $$
At full moon, $i \approx |\beta|$ where β is the Moon's ecliptic latitude, so k is between 0.998 and 1.000.
The Moon crosses the ecliptic twice each draconic month.
Synodic (29.53 day) and draconic (27.21 day) months together have a beat cycle length of 346.6 days.
Maxima of k occur twice as often, 173.3 days apart on average.
The same interval separates eclipse seasons.
Using the DE430 ephemeris,
Skyfield's almanac functions,
and Matplotlib,
here are illuminated fractions vs. dates of several full moons,
compared with a sine wave of amplitude 0.001 and period 173.3 days.

This assumes a point of view at the center of the Earth.
Far northern or southern observers would see the Moon displaced up to 1° south or north,
biasing the mean topocentric β away from 0.
In the plot of k vs. time, alternate maxima would shift left and right, and alternate minima would shift up and down.


text:
How are Y-dwarf spectral subtypes defined?
It is often stated that the spectral subtype indicates the relative position between spectral types, e.g. G5 being halfway from the start of spectral type G to the beginning of spectral type F.
Given that we haven't observed a spectral type cooler than Y, how can subtypes Y0, Y1, Y2 be defined?

Spectral class Y is also used on sub-brown dwarfs. Sub-brown dwarfs don't have enough mass to ignite Deuterium. An example for such Y-type body is https://en.wikipedia.org/wiki/WD_0806%E2%88%92661

@Greenhorn - this is a question about spectral types, i.e. the classification of spectra, not about how to define "sub-brown dwarfs".


text:
"Juxtapositions" of moon,stars and planets in the sky,
What are "Juxtapositions" of the Moon, stars and planets called? When the Moon, stars and planets take exactly the same position after few or many years, how to describe the positions and what is called to describe the position?

There are several words to describe two celestial bodies being in the same place on the celestial sphere.
A conjunction is when two bodies seem to pass close to one another, as seen from the Earth.
A transit is when a celestial  body passes between the observer (us) and a bigger body. For instance, Venus can transit in front of the Sun, when it passes between the Earth and the Sun.
An occultation is when a celestial body in the foreground hides a celestial body in the background. For instance, the Moon occults the stars it passes in front of.
An eclipse designates what happens when one body passes in front of the other for an observer, as a result of a transit or an occultation.
A syzygy is simply when three celestial bodies are roughly aligned. One of those three bodies can be the Earth, in which case, the two other bodies will either be in the same place on the celestial sphere (a conjunction) or at the antipodes.

Wikipedia says the conjunction symbol is not not used in astronomy (and even that's not properly sourced, and I believe it's false), but the word conjunction certainly is used in astronomy: https://blogs.nasa.gov/Watch_the_Skies/tag/conjunction/


text:
Can Kepler's 3rd Law be applied in higher dimensions to describe orbits (quadric surfaces)?
Motivation for asking:
The 2-D orbital path of a celestial body revolving around another celestial body due to the force of gravity can be described by a conic section; this is the curve formed by intersecting a plane with the surface of a cone. This generalized conic section is given by:
$A x^2 + B xy + C y^2 + D x + E y + F = 0$
The ellipse is one example of a conic section (others being parabola and hyperbola); this happens when $B^2 - 4 A C < 0$.
Kepler's 3rd Law can be applied to elliptical orbits; this relates the orbital period $T$ and the semi-major axis $a$ of the orbit such that:
$\frac{a^3}{T^2}$ $=$ $\frac{G (M + m)}{4 \pi^2}$
As I understand it, Kepler's Law holds when the orbital path is 2-D. What happens when the orbital path is 3-D? I am aware that the conic section in 3-D is referred to as a quadric surface, which is given by:
$A x^2 + B y^2 + C z^2 + D xy + E xz + F yz + Gx + Hy + Jz + K = 0$
Say one were to rotate the 2-D orbit by its polar angle in spherical coordinates - that is, the altitude (as opposed to azimuth) - by a non-zero amount such that the orbit was in 3-D (changing z-coordinates as a function of time). Doing this, we can generate an ellipsoid, which is given by: $(\frac{x}{a})^2$ + $(\frac{y}{b})^2$ + $(\frac{z}{c})^2$ = 1.
Question:
Could the logic behind Kepler's 3rd Law be used to relate the orbital period $T$ of a body on an ellipsoidal path to any of the ellipsoids parameters $a, b, c$? What if the path weren't ellipsoidal, but some other quadric? Can the quadric surface always be reduced to a rotate 2-D conic section? Or are ALL orbits considered Keplerian?

You may enjoy this article by John Baez: Planets in the Fourth Dimension.

Kepler's is a special case of two-body problem using Newtonian. Trajectories of two bodies orbiting each other are on a 2D plane. 3D trajectory happens when there are perturbations, or more-than-two-body problem. So, you can still use Newtonian to derive any complex many-body problem.

+1 your question is a bit long but it's a good question and should certainly have a well-founded mathematical answer. I'm a little confused by the 3D rotation part though, I can't visualize around what axis exactly you are rotating. You would get an ellipsoid, paraboloid or hyperboloid if you rotated a conic section orbit around its line of apses; would that be a more suitable term than "its polar angle in spherical coordinates"?

@uhoh This is the first time I’ve heard the term “line of apses”. Basically, the the xy-plane can be rotated towards/away from 3rd dimension (z). The “polar angle” that links to the wikipedia page is theta for physicists and phi for mathematicians.

okay I'll give that some thought...

@uhoh The polar angle convention is described in the top-right figure of the wiki page in the link.

Oh, I missed seeing the "(z)" in your comment, I started to think that "away from the 3rd dimension" meant into the 4th dimension! Okay I'm almost up to speed now, thanks!

A picture is worth a thousand words. Unfortunately, I'm not the best at drawing (and am not quite sure how to implement this in a computer animation), but maybe this picture will help.

Not sure about the conic section part, but one can demonstrate that Kepler 3 still holds with multiple space-dimensions by applying scale invariance to the Lagrangian describing the problem. If you're interested in seeing this, I can give it a shot.

I forgot to mention Cris Moore (who I learned about from a John Baez post). His figure-eight orbit is relatively well-known, but he also found a bunch of interesting 3D orbits in rotating frames, like this. See his gallery for more examples.

@AtmosphericPrisonEscape I am very interested to see this!

On the risk of not fully answering your question:
Trajectories of a system through some solution space, are in physics always solutions of some set of differential equations. However using the right formalism, one can often derive properties of the solution, without actually knowing it.
For the Kepler-problem, there is such a trick.
Let us consider the Lagrangian $L=E_{\rm kin} - E_{\rm pot}$ of the Kepler-problem:
$$L = \frac{1}{2} m (v^2_x + v^2_y + v^2_z) - \frac{\alpha}{\sqrt{x^2+y^2+z^2}},$$ but to simplify the notation we shall abbreviate this to
$$L = \frac{1}{2}mv^2-\frac{\alpha}{r}$$
The fact that this simplifies to a 1-D problem is of importance later.
The equation of motion of the Kepler-problem can be derived from the corresponding Euler-Lagrange equation and results in Newton's law plus $E_{\rm pot}$.
Now we can multiply the Lagrangian by a constant $L'= c L$, and the resulting equation of motion, hence the solutions of the equation of motion, will remain identical. If we choose $c$ carefully, then we get the desired information about the solution of the equation of motion, by only touching $L$. This principle is called the principle of scale-invariance, and it doesn't apply to all Lagrangians or all systems in physics.
Now introduce a different idea, that of rescaling of space and time: Imagine a 'zoom' in space and a 'fast-forward' or 'slow-down' in time, such that $x$ and $t$ are rescaled according to
$x'\rightarrow c_x x$,
$t'\rightarrow c_t t$,
where $c_x$ and $c_t$ are some new constants, that are arbitrary initially. From those rescalings follow now, for the scaling of v that $v' \rightarrow c_x/c_t v $ and the potential, because it only depends on one length, scales as $\alpha /r' \rightarrow \alpha / (c_x r)$.
Let us plug-in all those results from the scalings into $L'$ and relate this to the unprimed, original quantities,
$$L' = \frac{1}{2}mv'^2 - \frac{\alpha}{r'} = \frac{1}{2}m v^2 \cdot \frac{c_x^2}{c_t^2} - \frac{\alpha}{c_x \cdot r}$$.
Now the latter looks a little like a mess. But we want to write this as $L'=cL$, which is only possible, if we can factor out those factors containing all the $c_x$ and $c_t$s. The condition for this is $\frac{c_x^2}{c_t^2} =  c \cdot \frac{1}{c_x}$ or  $$c_x^3/c_t^2 =c$$,
which is one statement of Kepler's third law, namely that the spatial-scale determining and the temporal-scale determining constants have the above relation w.r.t each other. Note that as long as the potential remains a central potential, it doesn't matter how many coordinates we have, i.e. if $r=\sqrt{x^2+y^2+z^2 + a^2+b^2+...}$ this statement retains validity.

I am still trying to fully understand the last part of your answer, but the question is technically answered. Regarding the scale factor $c_x$ in the lagrangian $L'$, why are $c_y$ and $c_z$ not included (the rest makes sense)? And given $c$ in terms of $c_x$ and $c_t$ is constant, how can this account for 3d- parabolic and hyperbolic orbits?

@allthemikeysaretaken $c_x$ and all the other spatial constants, are the same when all spatial dimensions are isotropic w.r.t each other, i.e. there is rotational symmetry between them. This can be broken in general relativity, where also Kepler 3 doesn't hold anymore, and no closed loop orbit solutions exist. As to the hyperbolic and parabolic solutions, they will always exist, because they're the result of effective centrifugal potential plus gravitational potential. The latter happens in the orbital plane, which is conserved no matter how many dimensions there are.

The Lagrangian is $L = T - U$, whereas the total energy $$ is given by the hamiltonian $H = T + U$. To my understanding, the orbit is elliptical if $E < 0$, hyperbolic if $E > 0$, and parabolic for $E = 0$. Does it follow that the constants $c_x, c_y, ...$ determine the sign of E?

@allthemikeysaretaken As long as you can factor them out, it doesn't matter if you're rescaling $T-U$ or  $T+U$

Very interesting answer! +1 Slightly related in that the orbit solution is 1d : Can the radial oscillations of an elliptical orbit be solved using a fictitious centrifugal potential?


text:
Are stars expected to become dimmer before a supernova?
With the recent news about the "fainting" of Betelgeuse and the speculation that this might be a precursor to a supernova, I'm wondering if there is any theoretical/observational basis for this interpretation, or whether this is a case of reading too much into a greater-than-average change in an intrinsically variable star. Why would a star become dimmer in the run-up to a supernova?

The connection between the dimming and a putative supernova relies on the interpretation that the decrease in luminosity may be due to circumstellar material, ejected in the years/decades/centuries immediately preceding a supernova. There are several mechanisms that could lead to this sort of mass loss (see slides 24-25), including

gravity-wave driven envelope loss (in red supergiants, during neon/oxygen core burning)
pulsations due to pair-instability in the days to decades prior to the supernova
turbulent eruptions à la luminous blue variables

These then lead to Type IIn supernovae, with narrow lines arising from interactions with the previously ejected circumstellar material. If the dimming is due to extinction by circumstellar dust from eruptions due to any of these mechanisms, it could mean that a supernova is imminent on a timescale of days to years.

Wouldn't the star become brighter before the supernova as more energetic fusion reactions happen forming bigger atoms and the star expands, then get dimmer as iron forms and fusion stops, and then brighter again from the gravitational collapse right before the supernova?

@Caters, the gravitational collapse at the end of the few days where the nickel-56 core grows to 1.4 solar masses lasts only a few milliseconds.  Any brightening from the gravitational collapse is due to neutrinos interacting with the outer envelope, and not due to that envelope contracting in any measurable way, as there are no pressure waves in the star that could travel that fast (and indeed, the first pressure wave that reaches the envelope is the visible part of the supernova event).


text:
Why does the star formation rate correlate with the total stellar mass in star forming galaxies?
The so-called "Main Sequence" of star-forming galaxies (e.g., Brinchmann et al. 2004; Noeske et al. 2007) clearly points to a linear relation between the stellar mass of galaxies and their star formation rate.
This relation holds at all redshifts, with a time-evolving normalisation.
The question is: what is the physical reason why, at a given time, the total stellar mass (which is an integrated quantity, i.e., the integral of the past star formation history if you want) should be linearly related to the rate at which stars are forming?


text:
Has J been used by itself as a spectral type?
While answering the question "Why have brown dwarf classes been dubbed L, T and Y?" I noticed that in the paper Kirkpatrick et al. (1999) "Dwarfs Cooler than 'M': The Definition of Spectral Type 'L' Using Discoveries from the 2 Micron All-Sky Survey (2MASS)", the letter "J" was ruled out as being in use as a standard carbon star type.
So far I have not seen a source that uses "J" as a carbon star type by itself, I've only seen it in combination with "C", e.g. "C3,3J" or "C-J4 C2 5" (both from XHIP). This contrasts with the other carbon star types "R" and "N" which have definitely been used as separate identifiers.
Has "J" ever been used as a stand-alone spectral type, e.g. "J3"?

As far as I can tell, the J-type was introduced by Bouigue (1954), and have subsequently often called "J-type stars" or "J-stars", e.g. Abia & Isern (2000). This likely led to the J-type designation being put as "in use", though the spectral types themselves appear to always have been used in combination with "C".


text:
Electrical potential difference between planets
In dealing with electricity, we usually refer to Earth as the electrical neutral or ground. Is there any evidence that other planets' ground is at the same electrical potential level as the Earth's?
If somehow, there was a lossless conductive connection between two planets, would current flow accros It? Would this be theoretically a source of energy?

If you connect electrically neutral and electrically neutral, then you get...

Different but related: What are the experimental limits to the residual charge of the Sun? This is an interesting question; the net charge on a planet is not necessarily zero; since the mobility of electrons and ions is so different, there are magnetic fields involved and a constant supply of ionized particles from the Sun. You might consider asking specifically about the Earth's net charge as a separate question in Earth Science SE.

I have also found a somewhat related question here: https://physics.stackexchange.com/questions/253475/do-stars-remain-electrically-neutral.

Kavka, don't overlook the option of posting your own answer to this question if you find the links in the other comments lead you towards a solution. Self-answering is actually encouraged at Stack Exchange - in fact there's a specific badge you can earn! :-)

I dunno how or why a planet would have a significant net charge, but supposing it did, here's some initial calculations.
If the planets have a different net charge, you can treat them as capacitor plates.
Capacitance in its simplest form is
$C = \frac{\epsilon *A}{d}$
The permittivity $\epsilon$ of vacuum space is 8.85 x 10-12 farad per meter .
Let's choose a nearby (nonexistent) planet at 5 LY = 5 * 9.461e+15 m .
Looks like the capacitance is pretty close to zero.  So the energy available,
$E = \frac{CV^2}{2}$  is going to be nonexistent unless there's a seriously large voltage.


text:
Force to fix $\sigma_{8}$ or $A_{s}$ for Forecasts
In the context of Forecasts with Fisher's formalism, I make vary cosmological parameters to compute the elements of the Fisher matrix.
First, I generate with CAMB code a linear power spectrum. Then, from this, I am computing $\sigma_{8,\text{linear}}$.
Secondly, Before relaunching the code CAMB in non-linear regime, I apply a correction on $A_s$ (amplitude of primordial power spectrum), by a simpe relation of proportionality on the 2 $\sigma8$ (fiducial and linear), to get a new $A_{s,\text{modified}}$  that will procude the same $\sigma_{8,\text{fiducial}}$ after the non-linear regime execution of CAMB code.
For example, if $\sigma_{8,\text{linear}}$ is higher than $\sigma_{8,\text{fiducial}}$, I do the following correction before launching the non-linear CAMB regime :
$A_{s,\text{modified}} = A_{s,\text{fiducial}}\,\bigg(\dfrac{\sigma_{8,\text{fiducial}}}{\sigma_{8,\text{linear}}}\bigg)^2$
So, $A_{s,\text{modified}}$ will be smaller than $A_{s,\text{fiducial}}$.
My main issue of understanding :
1) Why have we got to do this correction ?, I mean to compute a new $A_{s,\text{modified}}$ that will give a new $\sigma_{8,\text{non_linear}}$ equal to $\sigma_{8,\text{fiducial}}$ (or maybe $\sigma8_{linear}$, I am not sure from the relation of proportionality above) ?
For the moment, I think we want to keep a fixed value for $\sigma_8$ to be consistent with observations data where $\sigma_8$ doesn't change (does $\sigma_8$ always correspond implicitly to amplitude of fluctuations at $z =0$ ?) : but I am not sure to grasp this subtility.
2) By the way, why we don't compute only once directly the non-linear regime instead of launching 2 times the code CAMB (first in linear regime and second in non-linear regime with this correction between both) ?
Thanks in advance for your help, even a succinct answer would be nice.
Regards

you shall considerate linearization https://en.wikipedia.org/wiki/Linearization, many computer programs like mathlab use matrix "linear" to solve problems


text:
How much rock has Jupiter "swallowed"?
During its formation approximately 4.5 bya, Jupiter passed through what is now the asteroid belt,gravitationally deflecting some spacewards and some sunwards.
Logically, Jupiter, due to its massive gravity, must have swallowed up a fair portion as well.
Is there any way to calculate how much rock Jupiter contains? Have any estimates been made?

I don't think the question makes much sense.  At the very least, might want to research the timeline of Jupiter formation vs. asteroid belt formation vs. settling into orbits.

In general, that's not how gravity works. Without gas drag, it is hard for objects to hit a massive object. The most frequent outcome in close encounters between two bodies are deflections. Also it is not clear that Jupiter went through the asteroid belt. The standard Grand-tack scenario you're referring to would leave too high eccentricities and inclinations in the asteroids, which we do not observe.

The total mass of the asteroid belt is roughly 4% that of our Moon. Jupiter was (probably) formed with far more rocky material than that, so any swallowed asteroids would only be a small fraction of its total rock content.

Groundless criticism of this question. In the Grand Tack model (which still is the consensus despite problems), Jupiter did indeed disrupt the rocky asteroid belt and many asteroids will have ended up inside Jupiter.

@PM2Ring the basis of an answer.

A few recent examples of Jupiter and Saturn eating stuff: 1, 2, 3

@RobJeffries Good point. :)

Jupiter is a gas giant, with a mass approximately 317.8 times that of Earth, and its atmosphere has a composition that's typical of gas in space: almost 75% hydrogen and almost 24% helium, (barely different to the primordial gas created by the Big Bang). However, its interior also contains rock and metal components.
Until quite recently it was thought that Jupiter's rocky & metallic core was fairly small and compact. Various models gave estimates of a core mass ranging from 7 to 18 Earth masses. But gravitational data from the Juno mission has changed the picture rather radically. It seems that Jupiter's core is diffuse rather than compact, extending fuzzily to almost half the planet's radius! At present, we aren't quite sure of how such a core was formed, but the leading hypothesis is that Jupiter's primordial core was shattered by an extreme impact event.
From Nature, DOI: 10.1038/s41586-019-1470-2 The formation of Jupiter’s diluted core by a giant impact by Shang-Fei Liu, Yasunori Hori, Andrea Isella:

The Juno mission has provided an accurate determination of Jupiter’s
  gravitational field, which has been used to obtain information about
  the planet’s composition and internal structure. Several models of
  Jupiter’s structure that fit the probe’s data suggest that the planet
  has a diluted core, with a total heavy-element mass ranging from ten
  to a few tens of Earth masses (about 5 to 15 per cent of the Jovian
  mass), and that  heavy elements (elements other than hydrogen and
  helium) are distributed within a region extending to nearly half of
  Jupiter’s radius.
Planet-formation models indicate that most heavy elements are accreted
  during the early stages of a planet's formation to create a
  relatively compact core and that almost no solids are accreted
  during subsequent runaway gas accretion. Jupiter’s diluted core,
  combined with its possible high heavy-element enrichment, thus
  challenges standard planet-formation theory.
A possible explanation is erosion of the initially compact
  heavy-element core, but the efficiency of such erosion is uncertain
  and depends on both the immiscibility of heavy materials in metallic
  hydrogen and on convective mixing as the planet evolves. Another
  mechanism that can explain this structure is planetesimal enrichment
  and vaporization during the formation process, although relevant
  models typically cannot produce an extended diluted core.
Here we show that a sufficiently energetic head-on collision (giant
  impact) between a large planetary embryo and the proto-Jupiter could
  have shattered its primordial compact core and mixed the heavy
  elements with the inner envelope. Models of such a scenario lead to an
  internal structure that is consistent with a diluted core, persisting
  over billions of years.
We suggest that collisions were common in the young Solar system and
  that a similar event may have also occurred for Saturn, contributing
  to the structural differences between Jupiter and Saturn.

In contrast, the total current mass of the asteroid belt is roughly 4% that of our Moon, so just under 0.05% of the mass of the Earth. Undoubtedly, Jupiter has swallowed numerous asteroids and comets over its lifetime, but even if it ate all of the asteroids in the asteroid belt that would only amount to less than 1% of its total rock and metal content, compared to the primordial rock and metal that it has had since its formation, including whatever rock & metal it inherited from the core impactor hypothesized by Liu et al.

The problem with comparing current Jupiter mass with current asteroid belt mass and then inferring conclusions about past Jupiter mass and past asteroid belt mass is that past Jupiter could be much smaller and past belt much heavier.  One of the things the Grand Tack hypothesis was designed to explain is why there is so very little mass in Mars and the asteroid belt today, and that explanation was: Jupiter migrated inwards and ate or dispersed most of it.  How much was ate vs. dispersed/ejected would be the relevant quantity.


text:
How long could an ocean last in space?
If all of Earth's oceans were suddenly moved into outer space, how long would it take for them to evaporate?
My understanding is that water evaporates straight away in space... but what about when it's a quantity as large as an ocean?
Would it be a matter of seconds? minutes?
Would it be possible to swim in an ocean, that was floating in space, for a short while before it all evaporated?

For swimming in water in space there's a variety of questions and answers in Space SE. There's even a reduced-gravity-sports tag!

@uhohs some very humorous questions / answers in there haha! and here's me thinking I was the one with the bizarre imagination :D

Consider you have a capsule with water at $T=300$ K. Water will start boiling right after you remove containing pressure. Evaporating water will take out the heat, required for phase transition into vapor, so remaining water will cool off. It will boil until it will reach $T_0=273$ K temperature. Then part of water will start to transit into ice. The remains are ice and vapors. Solving equation $c_p m_i + c_p m_i (T-T_0) - c_L m_v = 0$, where $m_i$ is mass of ice, $m_v$ is mass of vapor, $c_L,c_p$ --- energy required for phase transitions, you can obtain fraction of ice which will remain. Whole process takes up to few minutes (depending on what size is the hole of container). Quantity of water does not change the rate of evaporation, which mostly depends on the heat transfer. The heat transfer in this case is the exchange between vapor and ice, i.e. between water and water itself, which does not depend on quantity or the size of free surface. Water will boil in every point, over all volume at once.
If the hole of container is small, then process will be slower and remaining water will solidify into a solid brick of ice. If the hole of container is large, then all water will jump out immediately and break down into small drops. Similar to liqiod nitrogen breaking down when you pour it on table.
Atmosphere is the exact reason for oceans to exist on Earth, so without it the oceans have no meaning. Jupiter and Saturn have oceans of their own, but those are hidden beneath very thick atmospheres.

I don't think it makes sense to say Jupiter and Saturn have oceans in any normal sense. In Jupiter's case the hydrogen below the upper atmosphere is above the critical pressure and temperature, so there is no gas-fluid distinction (until maybe when one reaches the metallic hydrogen fluid - which again stretches the concept "ocean" rather badly).


text:
How scientists estimated the number of near-Earth supernovae over the last 11 million years?
According to this article on wikipedia,

An estimated 20 supernova explosions have happened within 300 pc of
  the Earth over the last 11 million years.

How did scientists determine that 20 near-Earth supernovae were within 300 pc over the last 11 million years? If this was empirically observed, then how do we know that there are not more supernovae that we haven't seen? If this estimate is obtained by some model, then what is the model and why do we trust this model - have there been enough observed supernova to trust that the accuracy of this model?
The source for the statement quoted above is cited here. I skimmed through the introduction and conclusion section of the article. It appears to me that the authors of this paper use the geologic record of radioactive elements to determine this "magic" number of 20. To my understanding, if one knows the initial/final concentration of a radioactive isotope, the decay rate of said isotope, and the time-interval to be considered, then one can deduce the final/initial concentration. Using this logic, how was the initial concentration tied exclusively to supernova (as opposed to a combination of things)? I am inclined to think that no other mechanism exists that can produce these isotopes over these time-frames, but even if this is the case - why?  I don't know if this is correct, and would like to know more about this topic. If I double the time-interval from 11 million years to 22 million years, would the estimated number of near-Earth supernova also increase by a proportional amount (maybe double 20 --> 40, or by a factor of 2 * r^2 assuming spherical distribution), or would the "within 300 pc" metric change this number due to the acceleration expansion of the universe?

The paper talks about 23 supernovae within 300pc, within the last 300,000 years. The more secure result appears to be that there were 4 within the last 50,000 years.


text:
Is there a place where one can see the beginning of the universe?
If the night sky Is dark because we’re seeing the world as it was a long time ago, when there wasn’t much to see, is there a place on the universe where we can see the universe at the  beginning of time? Where we can look at light from the moment of the Big Bang ?

"we’re seeing the world as it was a long time ago, when there wasn’t much to see," That's not quite right. Most of the stars we can see with the naked eye are relative close, less than a few hundred lightyears away. With dark skies (and good eyes) we can see more distant parts of the Milky Way, and our near neighbour galaxies: the Magellanic Clouds, Andromeda, Triangulum, but we can't see individual stars in those regions.

There is The Restaurant at the End of the Universe but I don't think that that's necessarily what you're looking for.

The moment you think of is seen as cosmic background radiation and within the current understanding it is seen wherever you go.