Sun as our Supergiant Let's talk about the time when our sun becomes a supergiant. We know that our solar system may get destroyed or it may get affected due the large mass of the supergiant and also due to the large temperature. But somehow let us consider nothing happened to our solar system by luck. It's obvious that life won't be possible on earth anymore(due to the extreme heat), but is there a possibility that, the events that took place during the formation of Earth, may take place in the far off planets? Can life formation take place in some other planet just like it happened in Earth??

When the Sun becomes a red supergiant (more accurately, an asymptotic giant branch star), it will have less mass than it does now. That is because mass is lost by an increasingly massive stellar wind as the Sun approaches the end of its life. As a result, the orbital radii of the planets would increase. The surface temperature of this future AGB Sun would be lower than the temperature of the Sun, but its luminosity would be much larger.

However, even if the Earth manged to avoid being engulfed in the outer envelope of the evolved Sun (which is still uncertain), the amount of energy received per square metre at the Earth's surface would be vastly increased over the 1.4 kW per square metre it receives from the Sun now. So your conclusion that life (as we know it) would be fried is quite correct.

The Goldilocks zone for the possible occurrence of life, uses the criterion of liquid water as the basis for allowing life to evolve. As a star becomes more luminous (and an AGB Sun would probably be about 100 times as luminous as the present Sun), the Goldilocks zone moves outwards roughly proportional to $\sqrt{L}$, so about a factor of 10 further away.

The problem then with allowing life to evolve on another planet (e.g. Mars after its orbit has expanded a bit, or Jupiter) is that (i) you need the ingredients to be there already, or some way for them to be delivered, but more importantly (ii) there is little time for life to evolve, because the Goldilocks zone position evolves really rapidly towards the end of the Sun's life as the AGB Sun's luminosity increases - the timescales would be measured in millions rather than billions of years. Furthermore, the luminosity of AGB stars can be highly variable as they undergo big pulsations, which may also be a problem. (iii) After the (comparatively brief) AGB phase has ended, you would end up with a very hot ($T \simeq 10^{5}$ K) white dwarf remnant that might effectively sterilise the whole system with ultraviolet light.

I don't think we know enough about how life gets started to say that these problems rule out your scenario; but they perhaps make it less likely.


When the sun "enters" a constellation, doesn't it make it impossible to see that constellation? I was reading on NASA's website about how the Babylonians came up with dividing the sky into different slices and picked a pattern of stars in each slice to represent it. This ties in to astrological signs, and that being born "under" a sign means that that particular constellation is aligned with the continuation of an imaginary line from earth to sun (and onwards out into space).
But here's what I don't get: when "the sun is in Scorpio", isn't the actual constellation "Scorpius" impossible to see? It would be "behind" the sun, right?
This seems backwards to me, but then again I'm not a Babylonian from 3000 years ago. Am I misunderstanding something, or is this really how it works? You're born "under Aries", yet Aries isn't even visible on the sky at that time?

It is true that, for an observer standing on Earth, the daytime sky blocks out our ability to see the stars (with the exception that stars are visible during a total eclipse).
It is possible to get pretty close.  In a different post, it was asked if the the center of the galaxy is only visible for 1/2 of the days of the year. That post is here:  Is it true that we see the center of the milky-way for only half of the year?  The center of the galaxy happens to be close to the ecliptic and located roughly between Scorpius and Sagittarius.  In that post I worked out that it is possible to see that spot in the sky (which happens to be near the ecliptic) roughly 10 months out of the year.
The point of the above post is that if we pick a specific location (for example... that spot in the sky between Scorpius and Sagittarius) you cannot see directly at that spot during the daytime.  But choose to observe merely a month earlier or a month later and you can see that spot either just prior to sunrise or just after sunset.  Since ancient observers would certainly have been able to observe these constellation side-by-side during months when they are visible at night, it would have been easy to derive (based on the first visible constellation after sunset or the last visible constellation prior to sunrise) where the Sun is located on the current day -- and to do this without needing to see the stars that beyond the Sun's current location.
Also consider that there are roughly 365 days per year and 360° in a circle.  Each day the Sun advances about 1° along the ecliptic.  The ancients would have understood this as well... and would have been able to pretty accurately plot the current location of the Sun without actually being able to directly observe those stars located in the sky beyond the Sun's current location.


Are all solutions of the 3-body problem restricted to 2 dimensions? I am not a physicist.
Suppose there are two massive bodies orbiting each other. Their orbit will presumably define a plane.  As far as I know, the existing solutions to the 3-body problem are limited to a plane.
Has any work been done on 3-dimensional orbits? For example two bodies are orbiting one another and a third body enters the system at right angle to their plane?
Specifically (1) Are there any solutions even for special cases of this? (2) Would such 3-D systems inevitability decay into 2 dimensions?

Are all solutions of the 3-body problem restricted to 2 dimensions?

The five stationary solutions to the three body problem are restricted to a plane. That does not mean that all solutions are restricted to a plane.

Has any work been done on 3-dimensional orbits?

NASA is proposing to build its Lunar Gateway as a vehicle in a "Near-Rectilinear Halo Orbit" (NRHO) about the Moon. These halo orbits only exist in the presence of a third body, in this case, the Earth. The orbit is nearly orthogonal to the Moon's orbit about the Earth.


How do astronomers estimate the total mass of dust in clouds and galaxies? As I understand it astronomers estimate the mass of stars by studying the orbits of their companions (other stars or planets).  This helps them to derive relationships between the luminosities and masses of stars.  This allows them to infer the mass of stars of known luminosity.

But how do astronomers estimate the mass of cosmic dust in particular clouds or in entire galaxies?

Dust absorbs stellar light (primarily in the ultraviolet), and is heated up. Subsequently it cools by emitting infrared, "thermal" radiation. Assuming a dust composition and grain size distribution, the amount of emitted IR light per unit dust mass can be calculated as a function of temperature. Observing the object at several different IR wavelengths, a Planck curve can be fitted to the data points, yielding the dust temperature. The more UV light incident on the dust, the higher the temperature.

The result is somewhat sensitive to the assumptions, and thus the uncertainties are sometimes quite large. The more IR data points obtained, the better. If only one IR point is available, the temperature cannot be calculated. Then there's a degeneracy between incident UV light and the amount of dust, and the mass can only be estimated to within some orders of magnitude (I think).

If lines from various atomic or molecular transitions are seen as well, the composition can be better constrained. The size distribution can be determined from fitting the theoretical spectrum of a given distribution to observed dust spectra. This information is often not available in a given high-redshift galaxy, so here we can be forced to assume that the dust is similar in nature to "local" dust, i.e. in the Milky Way and our nearest neighbors.

If you're interested in the relevant equations, they can be found many places, e.g. here.

Another way to estimate the dust mass is to measure the metallicity of the gas with which the dust is is mixed, either from emission lines or absorption lines if a background source is available. The dust mass is then found from an assumed dust-to-metals ratio, which is pretty well-established in the local Universe, and to some extend also at higher redshifts.


How is Cosmic Microwave Background's temperature measured? How do Cosmic Microwave Background missions/telescopes measure CMB temperatures? I understand that CMB has Planck's spectrum. So I was thinking one strategy could be to measure the spectral radiance corresponding to different frequencies coming from a certain direction and then compared these data points with theoretical Planck's law. Is this technique practically feasible?

That is what is done.  This is shown in an old xkcd comic https://xkcd.com/54/

The curve shows the distribution of frequencies in the CMB, and by using the marked value of the maximum you can determine the value of T, the apparent (red-shifted) temperature of the CMB


What is the planet relative to earth that shows the greatest change in apparent brightness? What is the planet relative to earth that shows the greatest change in apparent brightness?

I think the answer is Mars, and a quick google search reveals it is Mars.

I have my reasoning here, but I am not sure if I am correct.

One way to do this problem is to get all the planets and collect their perihelion and aphelion distances from the sun, and see the different.

However I want to be able to explain this without the extensive use of numbers.

Here is my explanation:

The closest planet to us is Mars, both in terms of its aphelion and its perihelion distances

The Earth-Mars distance varies between 2.66 AU and .38 AU

Mercury is almost always blinded by the sun, so we rarely see it in the sky

Venus is frequently viewed as a crescent, and thus its apparent brightness also dimishes

Jupiter and rest of the superior planets are simply too far away for us to notice a big differene

Are all my points correct here?

$
\begin{array}{cccc}
                   \text{Body} & \text{Max} & \text{Min} & \text{Delta} \\
                   \text{Sun} & -26.78 & -26.71 & 0.07 \\
                   \text{Mercury} & -2.45 & 5.58 & 8.03 \\
                   \text{Venus} & -4.89 & -3.82 & 1.07 \\
                   \text{Moon} & -12.87 & -3.76 & 9.11 \\
                   \text{Mars} & -2.88 & 1.84 & 4.72 \\
                   \text{Jupiter} & -2.94 & -1.66 & 1.28 \\
                   \text{Saturn} & 0.42 & 1.47 & 1.05 \\
                   \text{Uranus} & 5.31 & 5.95 & 0.64 \\
                   \text{Neptune} & 7.8 & 8. & 0.2 \\
                   \text{Pluto} & 13.75 & 15.96 & 2.21 \\
                   \text{Comet Halley} & 2 & 25.66 & 23.66 \\
                   \text{Tesla Roadster} & 6.66 & 29.29 & 22.63 \\
                  \end{array}
$
The answer is Mercury (as above), subject to the proecedure/caveats below:
Procedure

I used HORIZONS to generate daily brightness data for a century for:

all the planets as viewed from Earth (except Earth itself)

the Sun, the Moon, Pluto, Comet Halley, and the Tesla Roadster

I then noted the minimum and maximum brigtness, along with the magnitude difference of these brightnesses, in the table above.

You can use HORIZONS to compute the results yourself, or view the results in the *-brightness.txt.bz2 files in https://github.com/barrycarter/bcapps/blob/master/ASTRO/

Caveats

Although Mercury's brightness changes more than Mars, the Sun's glare makes it impossible to see Mercury though Earth's atmosphere when Mercury's angular distance from the Sun is small. Therefore, Mars may be a better practical answer.

Because I used daily brightnesses, it's theoretically possible I missed absolute (intraday) minimums or maximums, especially for the Moon, whose brightness changes rapidly. However:

I took 100 years worth of data, and the moon's brightness doesn't have a period that's a multiple of one day. Since the moon's synodic period is approximately 29.5 days, its brightness is almost periodic in 59 days (2 synodic periods), but it's far enough from 29.5 days that this shouldn't be too much of a problem.

The Moon isn't a planet: I just added it for reference

Unless the Moon's actual brightness difference was higher than 22.64 magnitudes (which is probably unlikely[?]), it would remain in 3rd place in terms of brightness change, so the exact value isn't as important.

Because I used only a 100 year period, I did not include a complete orbit for either Neptune or Pluto. This shouldn't be an issue because:

The synodic period of both planets is just over a year, and much of the brightness change comes from Earth's own orbit, not Neptune's or Pluto's.

Even if the maximum magnitude change were slightly higher than in the table, it wouldn't make much of a difference.

Note that "Max" and "Min" refer to brightness, which is ordered the opposite of magnitude: lower magnitude means greater brightness.

Because I used daily data, I missed rare events such as transits and eclipses. The table above is for an "average" orbit, excluding special cases.

In some cases, HORIZONS gives "n.a." for magnitude data. I ignore these "n.a." values.

Data for the Tesla Roadster is only available from 2018-Feb-07 03:00 UTC to 2090-Jan-01 23:00 UTC, not the entire century.

Complexity
The problem is nontrivial. As you correctly note, the planet's geocentric and heliocentric distance play into the formula, but there's more to it.
Quoting Oliver Montenbruck and Thomas Pfleger's "Astronomy on the Personal Computer" (https://books.google.com/books?id=nHUqBAAAQBAJ):

Paul Schlyter's http://www.stjarnhimlen.se/comp/ppcomp.html#15 provides similar nontrivial formulas.
EDIT: https://en.wikipedia.org/wiki/Phase_curve_(astronomy) provides more information on how planetary brightness varies nontrivially with planetary phase.


2019OK energy at impact The asteroid 2019 OK recently passed at 77.000km of the Earth, with an estimated size of 50-130m.

But if this asteroid had crashed onto the Earth, what would have been its kinetic energy on impact? And how would its mass have been affected by the atmosphere?

You can use the Purdue 'Impact Earth!' website to estimate the effects of impacts of objects of various size, speed and composition. As detailed in the accompanying documentation, which was published in Meteoritics and Planetary Science, the energy calculation and atmospheric drag during entry is relatively straightforward physics; things like crater sizes, overpressure and thermal effects are scaled from nuclear tests.

Plugging the numbers in for 2019 OK involves some assumptions since there is considerable uncertainty (~50%) on the diameter since we don't have an albedo and this leads to a large uncertainty in the volume. In addition, since nobody got a spectrum during the close approach, we don't have a good idea on the NEO's taxonomy and therefore what it's made of. This makes the mass, and therefore the energy of the impact, somewhat uncertain (since $Energy\sim\frac{1}{2}mass \times velocity^2$).

If we assume 80m diameter and density of 2000 kg/m$^{-3}$, which is mid-way between that of C- and S-type asteroids (most common types among NEOs, making up ~65% of NEOs; Binzel et al. 2015, in Asteroids IV), and an encounter velocity of ~25 km/s (rounded up from JPL Small Body DB close approach data) gives ~40 Megatons of TNT equivalent. ImpactEarth! also reports:

  The projectile begins to breakup at an altitude of 71800 meters = 235000 ft
  The projectile bursts into a cloud of fragments at an altitude of 3860 meters = 12600 ft.
  The residual velocity of the projectile fragments after the burst is 4.94 km/s = 3.06 miles/s.
  The energy of the airburst is 1.61 x 10^17 Joules = 3.85 x 10^1 MegaTons.
  No crater is formed, although large fragments may strike the surface.

If you put in a distance from the impact point, it will calculate the effects of the airblast for you, which will be the main effect (as it was for the ~0.5 Megaton 2013 Chelyabinsk impact)


Eclipses with SkyField I'm trying to compute solar eclipses for a long period of time, see Back-predicting solar eclipses

Initially, I wanted at several points on Earth whether an eclipse was visible at some point in time.
I have since changed the approach and I will now compute whether an eclipse could be seen from the center of the Earth at a given moment. Later, after knowing roughly the time of the eclipse, I will iterate over a grid on Earth to check where it could be actually seen.
So far, my first question is: is it right to check for eclipses at the center of the Earth?

Secondly, I wrote a small code in Python to test how good I am at predicting eclipses, checking against https://eclipse.gsfc.nasa.gov/SEsearch/SEsearchmap.php?Ecl=01501207
for these coordinates:

Lat.: 27.4782° N
Long.: 109.9299° W

#!/usr/bin/python

from skyfield.api import load, Topos
from skyfield.positionlib import ICRF

planets = load('Ephemerides/de422.bsp')
ts = load.timescale()

earth = planets['earth']
sun = planets['sun']
moon = planets['moon']

time = ts.utc(150,12,06,23,49,21.9)
place = earth + Topos('27.4782 N','109.9299 W')

observe_moon = place.at(time).observe(moon).apparent().position.au
observe_sun = place.at(time).observe(sun).apparent().position.au
distance = ICRF(observe_moon).separation_from(ICRF(observe_sun))
distance = distance.degrees

print distance

The distance I obtain is above 10 degrees, while at the time provided, if I did things correctly, the eclipse was at its maximum at those coordinates. What is wrong in my code or my understanding?

I believe the differences here are caused by the different timescales that are being used by SkyField vs NASA.

It seems that SkyField uses the proleptic Gregorian calendar for dates in the past. However, NASA used the Julian Calendar for dates before 1582, so for example, the eclipse on 0150-12-06 (Julian) falls on 0150-12-05 in Gregorian. Also, I would use ts.ut1 instead of ts.utc, as ut1 is the timescale used by NASA.

Using the geocenter can be a efficient way to check for eclipses worldwide. When observing from the geocenter, the penumbral shadow cone touches the Earth when the following condition is true:
$$\mu ≤ arcsin(\frac{r_m + r_e}{d_m}) + arcsin(\frac{R_s - r_e}{D_s})$$
Where $\mu$ is the angular separation between the Sun and the Moon, $D_s$ is the distance of the Sun, $d_m$ the distance of the Moon, and the three constant radii are:

Sun: $R_s = 695700 km$

Earth: $r_e = 6378 km$

Moon: $r_m = 1737.4 km$

This indicates that an eclipse, at least partial, is occuring somewhere on the Earth. The formula uses a spherical Earth as an approximation of its shape, so in some rare cases it can detect very near-misses by 20-25 km, but overall it is a good way to narrow down the search. Also, it is unnecessary to test this when the angular separation is outside the range 1.3° - 1.7°, since values below 1.3° always produce an eclipse, and values above 1.7° never produce one.

I would also like to point out that uncertainties, mostly in Earth's rotation speed, make it very difficult to know where eclipses occured on the surface of the Earth for dates in the far past. The uncertainty becomes very large outside the range BC2000 - AD3000. This is the reason why NASA hasn't computed eclipse paths outside that range. According to research, the uncertainty could be as large as over 4 hours (several thousands of kilometers for eclipse paths) for dates before BC4000. See this explanation on NASA's Eclipse Web Site.


Are telescope eyepieces ever made with a diopter adjustment (similar to DSLRs)? I am learning about eyepieces, and "eye relief" and what sorts of eyepieces an amateur may want for their purposes, and was wondering if there are eyepieces with diopter adjustments, or if, due to the optical systems in reflecting and refracting telescopes, diopter adjustment like those in a DSLR viewfinder are never found in the designs of eyepieces.

Are there eyepieces with diopter adjustments?  The answer is both yes and no.
Diopters are not needed on telescopes for simple spherical myopia (near-sighted or far-sighted vision issues) but may be needed if a person suffers from cylindrical myopia (astigmatism) and wants to be able to use the telescope without wearing prescription eyeglasses.
For near-sighted & far-sighted vision issues there is generally no need of a diopter adjustment on a telescope as their is on an SLR or DSLR camera for spherical correction.  Also, cameras generally don't a way to compensate for astigmatism ... but it turns out you can get eyepiece adapters that compensate when using a telescope (more on that later).
Why?  To answer the question, it's important to first recognize what the diopter adjustment does on a DSLR camera.
On a DSLR, the "SLR" stands for "Single Lens Reflex" because the camera is designed to allow the photographer to look through the same lens that will be used to capture the photo (as opposed to viewfinder or rangefinder cameras... or even twin-lens reflex cameras where the photographer looks through a separate lens used to "approximate" the image that will be captured.
If you remove the lens and inspect the top of the mirror chamber, you'll notice that the mirror is shining light up onto a piece of frosted glass.  The glass will often have etched marks to serve as aids when taking the photograph.
The distance that light must travel through the lens, into the camera body, onto the reflex mirror and up onto the frosted glass (focusing screen) is actually the same as the distance through the camera and onto the imaging sensor (when the mirror swings clear to capture a photograph).
This means that when you look through the viewfinder, you're technically not looking "through the lens" ... you're actually looking at a back-lit projection screen (the frosted glass).
This means you could have an image which is precisely focused on the focusing screen (and also precisely focused for the imaging sensor) but appears out of focus due to eyesight issues.
The Diopter Adjustment allows you to adjust for near-sighted/far-sighted issues (spherical myopia).  To adjust, manually put the camera out of focus and point at something with no contrast (such as a blue sky outdoors or a plain white wall if indoors) and adjust the diopter until the etched markings on the focus screen are as sharp as possible.  Once you've done this, you can be confident that if the image projected onto the focus screen is well-focused, your eyes will also perceive it as being well-focused.  It also means you may not need to wear your prescription eyeglasses when using the camera.
Telescopes
With a telescope, you aren't looking at an image on a focusing screen (such as frosted glass)... you really are looking through the telescope.  When doing visual observing, there is no camera imaging sensor involved.
This means if you are near-sighted or far-sighted (both of which are spherical corrections) you can just adjust the telescope's focus to compensate.  There is no need for a diopter adjustment.
Spherical myopia means the curvature of your eye is spherical... but is either slightly too flat ... or slightly too round to focus accurately at certain distances (resulting in a person being near-sighted or far-sighted).
But there is also cylindrical myopia (aka astigmatism).  This means that instead of having uniform curvature (spherical curvature) on your eye, your eye's curvature is a bit more barrel-shaped where the curvature in one direction is stronger than the orthogonal direction.
TeleVue (and so far as I know ... only TeleVue offers such a thing) makes a product called Dioptrx.  This is an adapter that fits onto many TeleVue eyepieces (not all TeleVue eyepieces are compatible with Dioptrx).  It is designed for those who suffer from astigmatism.
You would need to know your corrective lens prescription to get the correct Dioptrx adapter (and since your right vs. left eyes will each have their own prescription, you need to decide which Dioptrx to order based on the eye you prefer to use when viewing through the telescope.
If you follow the link to the TeleVue Dioptrx page, you'll notice that the instructions offer helpful information on how to read your eyeglass prescription and how to identify the cylindrical correction needed for your eye.  But you'll also notice it doesn't matter if the correction needed is positive or negative and also the axis can be ignored.  This is because you can't just rotate the lenses in your eyeglasses ... they have to mounted in your eyeglasses frames in the correct orientation to work properly.  But you can rotate the telescope eyepiece.  This is why the positve/negative and axis values can all be ignored (TeleVue marks the mounting ring of the Dioptrx with letters ... so once you decide which axis offers the sharpest view, you note the letter at the top and just make sure to always rotate the eyepiece to the same orientation relative to your eyes when using it.
The Dioptrx astigmatism corrector allows you to use your telescope without needing to wear prescription eyeglasses.  If your astigmatism if very weak (the cylindrical correction is a low value ... such as 0.50 ... then Dioptrx may not be needed.  But if your correction is very high (e.g. 2.00 or more) then Dioptrx will probably be a big help.


What will the last day of humanity look like? It's certain that human life on Earth will not go forever. Therefore there will be one last generation of humans on Earth.

How will life look like for last few generations approaching the end?
How will life end for the very last people, will they all die at once as if struck?

Assume the technology won't be much more advanced than today, no colonized planets, and we'd stay biologically very similar to now. Also assume that no meteor will struck us and end everything but that extinction will happen due to aging of the Earth and Sun.

The last humans in this scenario likely die of starvation and/or drought during a natural short-term climate fluctuation.

The basic issue is that as the sun heats up there will be a gradual reduction of carbon dioxide in the atmosphere (due to plant life but also silicate weathering, which speeds up in warm environments). There may also be a reduction in volcanic outgassing, although I think the timeframe is too slow. This is bad for plants, since they will get less CO$_2$ to photosynthesise from, and it will reduce the climate stabilising feedback so temperatures will start going up as the sun increases in luminosity. Eventually we get to a runaway greenhouse stage where water vapour alone triggers overheating, but this is possibly long after the biosphere is mostly dead.

The most complete model of this I have seen is this paper by Franck, Bounama, & von Bloh. They estimate complex multicellular life to go extinct in 0.8–1.2 Gyr.

Note that this will happen long before the sun becomes even a subgiant.

In any case, if this model is correct we should expect that slowly but surely the future humans would be forced to migrate to cool regions and would be especially limited by the dwindling biomass available. But beside this they will be subject to normal random climate fluctuations that sometimes create decade-long droughts or other misfortunes, and it is likely one of those that actually gets the last individuals.

Of course, the scenario assumed makes some very strong assumptions. In order to remain recognisably humans over millions of years these humans would have to engage in deliberate genetic control, otherwise genetic drift will make them mutate into new species. It assumes that nobody does anything about the long-term trend. And so on.


How to get from the earth to another planet/ the moon? What happens on the way? I have a problem explaining for my students how to get from the earth to the moon? Is there someone who can give a good explanation?
I Think these are words that might be included. Escape velocity, gravity, force, free fall, weightless, earth orbit, lunar orbit.(How do the spaceship get into the lunars orbit how do the spaceship get from lunars orbit to the surface of the moon?)

Short answer is easy.   Long answer, there's lots of maths involved, and I'm out of practice with my maths, but there's a few basic parts that your students should be able to follow.   It's a little confusing, not too bad.

We can ignore the Earth's motion around the sun for the most part but not the Earth's escape velocity.

Start with Newton's cannonball thought experiment.  That will get you into space, and that's an example of orbital velocity.  With a fast enough launch, the cannonball enters into and stays in Earth's orbit.

2nd, explain the difference between escape velocity and orbital velocity.   Escape velocity is greater than orbital velocity and the cannonball that orbits the earth hasn't really escaped the earth, it's still in the Earth's gravitational sphere of influence.   You need escape velocity (or very close to it) to get to the moon, about 11.2 KM/s.  You can get away with a bit less cause the Moon hasn't fully escaped the Earth either.

Astronauts can't get shot out of a cannon at high speeds cause it's too many G forces, so what rocket launches do is accelerate over a few minutes.  (you could probably look up the precise amount of G forces), but within G forces that humans can tolerate.   If they experience 4Gs, they're accelerating the equivalent of 3Gs (or 29.4 meters per second), at least at first, as they move away from the Earth, the gravitational acceleration towards the earth drops and after only a few minutes they can shut down the launch and drift towards the moon.   Most of the trip to the moon is done in this way, just drifting through space.   You don't want to go too fast cause you'd just need to slow down as you approach the moon again and the moon has no atmosphere to assist the deceleration like the Earth does.

The ships momentum traves through the combined gravitational field of, primarily, the Earth, Sun and Moon, and the Maths get pretty complicated, so the the launch needs to be quite precise.

The Moon also orbits the Earth at about 2,300 miles per hour and that orbital speed has to be matched as well as, since the rocket will often enter a partial orbit around the moon, that orbital velocity around the Earth needs to be matched too.

Once this is done, landing on the moon requires a simple enough deceleration to have a gentle enough touchdown to not harm any equipment.  (this probably isn't as easy as it sounds).

An orbit around the moon isn't strictly necessary but a partial orbit was used to assist landing on the desired spot.   A partial orbit around the Earth was also used in the launch to the moon.

They accelerate away from the Earth in 2 different times, first to leave the earth and get into Earth orbit.  That's the big one, then from a different spot (and not even an hour later) to accelerate out of Earth's orbit towards the Moon.

See diagram.

Source

At least, that's a pretty basic summary.   I welcome correction if I got anything wrong or missed anything important.


Best telescope for astronomy I am interested in buying a telescope and I would like to use it to look at like galaxies and nebulas.

Right now I'm looking at a 130 mm fast (f/5) Newtonian that comes with two eyepieces of 10 and 20 mm focal length. It comes with a tripod and German equatorial mount that includes fine motion controls for both axes.

This seems about right for what I'd like to do. It's going to be my first telescope and first time to try to look at galaxies and nebulas.

What are other considerations I should weigh before choosing the final one to purchase? Which features might be most helpful for a beginning observer looking for faint objects?

"Galaxies and nebula" are faint and spread out, and for that very little matters as much as aperture!  Back before light pollution was a a serious problem, 6 inches (150 mm) was generally acceptable, these days I tend to recommend 8" or 10" (200 or 250 mm) to just to get through the sky glow.

The 5" scope you are thinking about will be very nice on the Moon, and Saturn/Jupiter and some larger smudges should be detectable but... the view you'll see of "Galaxies and nebula"  is not likely to knock your socks off.


Mars and 115 mm - which eyepieces? Today I could roughly see Saturn with ring structure for the first time with my Seben 114-1000 telescope.

Then later I was dealing with Mars and found out that I must do something wrong, as with no eyepiece I could really see this planet closer.

So, whether I take 20, 12.5, or 6 mm eyepieces, the size as I set sharp focus remains a small light orange circle of solid uniform colour.  The app sky view shows for today 64% clear view, maybe an explanation why I can't see more details.

But is this possible to see Mars filling in the view field? And why there is no recognizable difference between 20 and 6 mm - do I need them in combination with something else?
Here it says, even the polar caps of Mars should be recognizable with this telescope model scale.

UPD

So far, it looks like that with up to 6 mm I can zoom in fairly well to such details as:

large moon craters
Jupiter (seen as small white illuminated blob) with the Galilean moons (seen as  stars)
Solid white-coloured Saturn with its ring and some space between planet and ring visible

Mars keeps being small to medium yellow-orange disk of solid color.

Using 4 mm gives little bit more zoom but less detail.

However, I can see with 6 mm that Moon contours are slightly jittering, maybe through light refraction in the night air (all observation done from the ground on a field). The temperature in the night has been 21°C.

Mars is currently experiencing a global dust storm. There are virtually no features visible.

Otherwise, some dark markings and the polar cap would be visible when conditions are favorable (meaning when it is close to us as it is now.)


How to determine the age of a star using asteroseismology? I keep reading that a standard way to determine the age of a star is asteroseismology, and I tried to learn more about it. I am wondering if somebody could help me to describe the method in simple terms for a primary school kid. That should be easier for experts of that field.
I found a review article Asteroseismology of High-Mass Stars: New Insights of Stellar Interiors With Space Telescopes a good starting point for myself, but that is not yet at the layperson level I am aiming for:

The study of stellar structure and evolution using stellar oscillations—asteroseismology—has undergone a revolution in the last two decades thanks to high-precision time series photometry from space telescopes. In particular, the long-term light curves provided by the MOST, CoRoT, BRITE, Kepler/K2, and TESS missions provided invaluable data sets in terms of photometric precision, duration and frequency resolution to successfully apply asteroseismology to massive stars and probe their interior physics. The observation and subsequent modeling of stellar pulsations in massive stars has revealed key missing ingredients in stellar structure and evolution models of these stars.

My question in other words: What observation variables does one need from a star in order to calculate the age using asteroseismology? Following what recipe will bring me from the observations to a number for the age? What accuracies are possible?
Related

Wikipedia on Stellar Age Estimation
Can someone calculate the age of the KOI-4878 star?

Asteroseismology effectively measures the sound speed inside a star by finding the characteristic oscillation frequencies of a star.
The sound speed depends on the composition because the pressure at a given temperature depends on the average mass of a particle in a gas.
As the star gets older, it turns its hydrogen into helium, changing the composition, the average mass per particle in the gas, and hence the average sound speed.
Asteroseismology tells you the age of the star.
The answer is somewhat model-dependent, so ideally you would have an effective temperature and surface gravity in order to constrain matters. Precisions of 10% of the main sequence lifetime are possible for stars of around a solar mass. Accuracy is hard to assess since we have an accurate age for only one star - the Sun.


Does bending of light around the Sun depend on the wavelength? If the energy of light is high, does its curvature differ from that of low-energy light around the Sun? In other words, if the wavelength of the light is shorter than another wavelength of light, then does the bending of the two lights differ around the Sun?

The amount of "gravitational light bending" is independent of the photon energy (light wavelength).
The reason is that the light follows a path through spacetime that is appropriate for a massless particle and this is unique for a given set of initial conditions.
That this is so is amply demonstrated by the consistent angular displacement of "stars" near the limb of the sun whether observed at optical or radio wavelengths.
As pointed out in comments - there are small effects that must be taken into account, associated with the well-understood phenomenon of refraction in the corona of the Sun. However, these do not affect observations of lensing taken well away from the solar limb - which is easily possible at radio wavelengths and now becoming possible for the same sources using Gaia data.
Further evidence comes from the wavelength-independent nature of gravitational lensing and microlensing seen outside the solar system.


How does Titan maintain its atmosphere? Titan, which is smaller than Mars, has an atmosphere but Mars is not able to maintain its atmosphere. Even Luna (the Moon) doesn't have an atmosphere. Why is Titan able to hold an atmosphere, despite being smaller than Mars?

You are right that it's surprising that Titan, being just a moon, has a thick atmosphere. Usually, the answer includes magnetism: Earth has an atmosphere because the liquid magma inside the planet produces a magnetic field. This magnetic field changes the paths of the particles in the solar wind, thus preserving the volatile gases intact. Mars did use to have an atmosphere, just like Earth, but it being farther away from the Sun, the magma froze and lost its magnetic properties.
Titan itself doesn't have a magnetic field, but Saturn does. Saturn's magnetosphere is produced by the motion of the super-compressed hydrogen gas inside the planet (metallic fluid). The field is so strong it encompasses the satellites, including Titan.
Saturn doesn't have the same magma as Earth. Earth is a rocky planet. This means that it was formed so close to the sun that light gases (such as hydrogen) could not coalesce due to the high temperature and the solar wind. Therefore, the inner planets (Mercury to Mars) are composed mostly of rock and metal. It's the liquid iron in the outer core that makes Earth magnetic.
Gaseous planets instead have a small solid metal/rocky core (therefore, no magnetic field from there), and a huge layer of light gases (hydrogen and helium). The gases are not usually magnetic, but under such an enormous pressure they take a "metallic" structure, which means they can conduct electricity like a metal. This same property allows them to generate a magnetic field.


Is there any orbit at which the Roche limit can be "felt"? Do any of the planets have a Roche limit that is strong enough to be felt by an astronaut whilst in orbit?

Roche limit happens where the gravity of the object, trying to pull the object together, becomes smaller than the tidal force (trying to pull the object apart).
But the astronaut is bound by not gravity, rather by the electromagnetic interaction between his/her atoms. The own gravity of the astronaut is negligible, compared to the electromagnetic interaction.
However, the tidal force affecting an astronaut, should require a little calculation. We can derive the formula of the gravitational acceleration around a point-like body ($F=\frac{GM}{r^2}$), we get
$$\frac{dF}{dr}=\frac{2GM}{r^3}$$
(We can ignore the sign on obvious reasons.)
Here $G$ is the gravitational constant, $M$ is the mass of the body, and $r$ is the distance.
Substituting the values of the Sun, we get $\frac{2\cdot 6.67 \cdot 10^{-11} \cdot 2 \cdot 10^{30}}{(7\cdot 10^8)^3} \approx 7.78\cdot 10^{-7} \frac{\mathrm{m/s^2}}{\mathrm{m}} \approx \underline{\underline{8 \cdot 10^{-8} \frac{g}{\mathrm{m}}}}$.
More clearly, if we are orbiting the Sun just above its surface, a roughly 2m long astronaut feel that his head and foot are pulled apart by around $1.6\cdot 10^{-7}g$ weight. In the case of a $70\:\mathrm{kg}$ astronaut, it is around the weight of $0.0112$ gram on the Earth.
The astronaut wouldn't feel it, but not very sensitive sensors could already measure it.

This calculation sometimes used $\mathrm{g}$ for "gram", as unit of mass, and $g$ as the (non-standard) unit of acceleration.


Which constellations are straight up and straight down? Pretend you're in our galaxy near Earth and you're standing upright wrt the flat galactic disk.

If you look directly towards the center of the galaxy, that's Sagittarius. If you swivel your head and look directly away from the center, I believe that's around Auriga.

In fact, if you look straight up, about which constellation is that?  And if you look straight down?

(If I'm not mistaken, could be wrong, this is equivalent to asking about...

Wikipedia explanation of galactic coordinates

... galactic coordinates of longitude zero, latitude 90° (up overhead) and -90° (down through your feet).

Crap: I've just realized the answer is given on that Wikipedia page!

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

Looking up from here: Coma Berenices

Looking down from here: Sculptor

Hope this helps someone googling.


If Mercury and Venus where to collide would it make a habitable planet? I have been wondering for a while now about a scenario of where if you combined Venus & Mercury would it be habitable if located in the goldy locks zone? I have found in my little research that Venus doesn't have an internally generated magnetic field. But Mercury does indeed have an internally generated magnetic field. Though weaker than Earths of course.

I don't really know what other factors for making it to whether or not this new plant would be able to sustain life.

So my question is, would the new planet be habitable, or are there more factors needed to determine this?

The answer is "Probably, but not very quickly."

The energy of a collision comes from two sources, the kinetic energy of the collision and the gravitational energy. If Mercury were to impact Venus, even if it began by creeping up on it very, very slowly so as to have as little kinetic energy as possible, it would fall into Venus's gravitational well and release gravitational energy enough to melt both bodies.  The resulting mess would take ca. 10-100 million years for a crust to resolidify.

The resulting body could evolve towards habitability.  One theory of planetary development says that active plate tectonics are critical to habitability.  We know that new rock absorbs large amounts of CO2 -- new volcanic rock in Indonesia has been estimated to absorb a quarter of the CO2 released each year.  So the continuous exposure of new rock by plate tectonics may be a major force in preventing the runaway greenhouse effect that has made Venus uninhabitable.

We don't know for sure how plate tectonics starts or ends, but if the resulting planet had a particularly vigorous case of it, this might compensate for its closeness to the Sun.

On the down side, Venus is already short of hydrogen and thus water and a collision with Mercury would eject Venusian volatiles into space and Mercury would not add any new, so the resulting planet might be rather barren.

In any event, whatever happens, it's going to take a very long time for the resulting body to cool enough to be livable.


How can gravity lead to the Big Crunch scenario? According to modern cosmology, space is expanding, causing proper distances (but not comoving distances) to increase between galaxies. In the Big Crunch hypothesis, gravity halts and reverses the expansion of the Universe, causing all matter to collide and eventually form a single black hole. This gives way to other oscillating universe hypotheses, which generally propose that the conditions in a compressed Universe would be the same as those during the Big Bang, leading a cycle of expanding and contracting universes.

Ignoring the problems with entropy in reverting the Universe to Big Bang conditions, how can gravity be the cause of a Big Crunch in the first place? Specifically, gravity (to my knowledge) only curves space; the idea that it can revert the Universe to the conditions of the Big Bang seems to imply that gravity actually can contract space. Is this actually the case?

If not, the gravitating objects should be moving through a comoving coordinate system, so space itself would not actually be contracting. As far as I can tell, we would have all the Universe's matter compressed within a single point in space, instead of space itself contracting. This should be completely unlike the Big Bang, when space was far less dilated than it is now. If this is in fact what the Big Crunch hypothesis describes, than I'm utterly confused as to how an oscillating universe could work in such a situation.

Am I mistaken, or do the Big Crunch and oscillating universe hypotheses imply that gravity actually contracts space (as in, the comoving distances of gravitationally attracting objects would not change)? If not, how could gravity possibly lead to these scenarios?

The amount of matter in the universe is directly related to the curvature of space itself.  We can look at the Friedmann Equations to see how this works:
$$
H(t)^2 = \frac{R'(t)^2}{R(t)^2} = \frac{8\pi G}{3}(\rho_{m} + \rho_{r}) + \frac{1}{3}\Lambda - \frac{c^2}{R^2\mathscr{R}^2}
$$
This is the equation for determining the scale factor for finding distances in the universe at any given time.  Here, $H$ is the Hubble Constant, $R$ is the scale factor, $G$ is the gravitational constant, $\rho_m$ is the density of matter (both dark matter and baryonic matter), $\rho_r$ is the density of radiation (photons), $\Lambda$ is the cosmological constant due to dark energy, and $\mathscr{R}$ is the radius of curvature of space.

Next, we define a quantity we call the "critical density" ($\rho_c$).  This is the density of matter needed to make the universe go from an "open", hyperbolic geometry to a "closed", spherical geometry.  $\rho_c=\frac{3H^2}{8\pi G}$. We create a value $\Omega=\frac{\rho}{\rho_c}=\frac{8\pi G \rho}{3H^2}$ to take into account this critical density, making our equation:

$$
H^2 = \Omega_m + \Omega_r + \Omega_\Lambda - \frac{c^2}{R^2\mathscr{R}^2}
$$

From here, we input our present-day values for everything.  We define $R$ to be 1 and $H$ to be $H_0$ at the present day.  We also choose to neglect the density of radiation and dark energy.  Simplifying the equation a little gives us:

$$
\frac{c^2}{\mathscr{R}^2} = H_0^2(\Omega_{m,0} - 1)
$$

From here, we can see that the curvature of the universe is dependent on $\Omega_{m,0}$, which is directly related to the density of matter in the universe.  Specifically, for $\Omega_{m,0}\gt1$, $\mathscr{R}$ will be positive, meaning this will be a spherical, closed universe.  If $\Omega_{m,0}\lt1$, $\mathscr{R}$ will be negative, meaning the universe will expand forever, having hyperbolic curvature.  If $\Omega_{m,0}=1$, then $\mathscr{R}=\infty$, which is a flat universe.

So, we can see that having a dense enough universe, gravity will curve space and make it spherical.  You can also use the Friedmann Equation to calculate the deceleration parameter of space - how quickly the expansion is slowing or speeding up:

$$
q_0=\frac{\Omega_{m,0}}{2}
$$

Here, we can see that in a positively curved universe, $q_0>0$, meaning that the universe's expansion will be decelerating.  Eventually, the expansion rate will be negative, and then it will begin collapsing in on itself.

One thing to keep in mind when thinking about this is that gravity can, in a way, be considered a 'fictitious force'. Gravity is the force objects feel as they attempt to travel in straight lines through curved spacetime.  Matter is what is responsible for the curving of spacetime, so by increasing the density of matter in the universe, you increase the curvature, and thus make objects want to move closer together, which will increase the density, which increases the curvature.  Thus you have a feedback cycle, where you will ultimately collect all matter at a single point, and $\mathscr{R}=0$, so spacetime will have a 0 curvature radius, meaning spacetime has collapsed as well.


Why aren't globular clusters disk shaped I thought that solar systems and galaxies are disk-shaped because that is the most stable shape under gravitation. Globular clusters are very old, often times older than their host galaxies, so why haven't they flattened out?

In order to result in a disk-like stellar system there are two conditions that need to be satisfied. (a) The initial gas from which the stars form must have a significant ratio of rotational to gravitational energy. (b) The star formation would have to occur slowly enough, that the gas collapses to a disk before star formation is complete.

The formation of a disk from a spherically symmetric state requires a reduction (more negative) gravitational potential energy, whilst conserving angular momentum. This can only happen if there is some means to lose energy from the system in the form of dissipative interactions.

A pure stellar system is almost collisionless and there is no means to dissipate energy. That means once a spherical system of stars has formed, there is no way for it to become more disk-like. Thus if star formation takes place in a spherical way, then those stars will end up in a spherical system. Only in cases where the gas collapses to a disk before star formation occurs do we end up with disk-like systems. This is true of the disk of our Galaxy (and of the solar system).

In the case of globular clusters (and other types of star cluster in the Galaxy) it appears the star formation is rapid enough that the gas cannot have collapsed to a disk before most star formation is complete. NB. Many globular clusters do have measurable rotation.


Can a person at a photon sphere of a black hole decide where the black hole is? A person at the photon sphere of a black hole will observe the following:

The black hole surface will cover exactly half of the visible sky (say, the left half), and cosmic horizon will cover the other half.
Looking in any direction between these hemispheres the person will see oneself.
The person will observe no centrifugal force however fast he moves. Any object the person thowing to the left will experience the centrifugal force pulling it further left, while any object thrown to the right also will experience the centrifugal force pulling it to the right. Approaching the black hole in such circumstances may look quite like actually escaping it to the crew.
The space to the right as well the space to the left seemingly includes objects of any length and dimensions, the further objects are, the more they are time-dilated and length-contracted.

Is there any way for a spacecraft commander to decide (by an experiment conducted in a small$^1$ amount of time) which hemisphere is the sky and which hemisphere is black hole in such circumstances?

--

$^1$ Here small means small compared to characteristic time scales of the problem.

For an observer hovering [I say hover because by definition there are no orbits for an observer at the photon sphere] at the photon sphere for a non-maximal de Sitter-Schwarzschild black hole, a simple accelerometer will tell an observer which horizon is which. The reason for this is that in order to hover at the photon sphere the observer must take up a non-inertial local frame with a directional bias decided by the direction of the BH event horizon.

In the maximal de Sitter-Schwarzschild spacetime the BH singularity disappears and the observer will not be able to tell the difference between the two horizons as there is an automorphism of that spacetime which will map one horizon to the other.


Time Dilation and Particle Decay I am teaching myself astronomy and I'm working on a basic problem regarding time dilation around black holes.

The question is basically this: if a neutron were ejected from a nucleus at a distance of 3km from a 1 solar mass black hole, how long would it appear to take for the neutron to decay (the assumption here is that the neutron would decay in 15 minutes... right on the half-life) to an outside observer?

My first thought was that we'd never see it decay, because 3km is the Schwarzschild radius of a BH of that mass, so the frequency would basically be $0$ and we'd never observe the decay event.

The text notes the time dilation

$$
\frac{\Delta t'}{\Delta t_\text{obs}} = \frac{\nu_f}{\nu_i}$$
where $\Delta t'$ is the the time difference between two events at the source and $\Delta t_{obs}$ at the observer.

We can use the relativistic red shift equation to get the wavelength difference (9.2 in my case using 1 solar mass and a radius of 3km) so that wouldn't be infinite, just big. For a 15 min $\Delta t'$ then, the $\Delta t_{obs}$ would be 15 min * 9.2, yeah?

So my two answers obviously contradict each other. What am I missing in my approach here?

If you use the equation of the Schwarzschild metric.
$$ds^{2}=-(1-\frac{2M}{r})dt^{2}+(1-\frac{2M}{r})^{-1}dr^{2}+r^{2}d\Omega^{2}$$
(equation 11.1 from Schutz's book A first course in GR).

If we think the neutron stay at rest in $r=r_{S}=2M$ then between the born and dead of the neutron $dr=0$ and $d\Omega=0$, and $ds$ will be the proper time, which I will call $\tau$ (I'm folowing the convention $c=1=G$).

Putting this on the metric we have

$$  d\tau^{2}=-(1-\frac{2M}{r})dt_{obs}^{2}$$

$$  \frac{d\tau^{2}}{dt_{obs}^{2}}=-(1-\frac{2M}{r})$$

and if we put $r=r_{s}$ (where the neutros was), we have

$$  \frac{d\tau^{2}}{dt_{obs}^{2}}=0$$

Then $\Delta t_{obs}=\infty$ .

I couldn't write a comment that's  why I answered, but I think you had some mistake using the red shift equation to get the wavelength difference.

Can you show your calculus? May be I can help.


Star Map Vs Star Catalogue vs Planetarium Software What is the difference between a star map, star catalogue and the sky provided by planetarium software? I have read the wiki but was having a hard time understanding how the data, accuracy and detail for the star sky created out of each of these sources vary.
For example catalogues have properly studied specific stars, I assume, this means they will have the highest accuracy. But they are having only a small number of stars compared to a software like Stellarium. But I couldn't understand about the data accuracy or precision details of such software. Looking to understand more on the data accuracy and precision of skies. Thanks in advance.

A catalogue is just that what one would understand usually: it's a table of stars with their coordinates and possibly other properties like relative or absolute brightness, proper motion etc. The accuracy depends on how it was created, thus mostly is expression of the available instrumentation, processing pipeline and its limitations. They usually are output of a particular mission or survey and represent the observations in tabular format and help to make sure different people talk about the same thing. Often the stars in a catalogue are thus also assigned a number to make it easy to refer to.
The most recent star catalogues are those from the Gaia mission, the older yet still often used catalogues are like Hubble Guide Star Catalogue which hold MUCH fewer stars, but the brighter ones only. There's also specialized ones which focus on particular type of stars like the General Catalogue on Variable Stars - but all those are only a selection.
Star maps and planetarium software need catalogues as input - they simple read the table and then display it somewhat differently. Both need to know where to place stars. As such their accuracy depends for one on the accuracy of the catalogue(s) they use - and on the accuracy of their projection which is at least limited by the point or pixel resolution but also the need to draw stars with more than one dot or pixel, especially to also indicate different brightnesses.


Why can't light escape from a black hole? I've heard that light can't escape from a black hole. Can it? If not, why?

A black hole has an event horizon which 'marks the point of no return'. So yes, light cannot escape from a black hole.

Why? Well, think of a 'spacetime fabric'. It's the easiest way to understand the physics at work here, in my opinion.

Usually, the fabric would look like this:

(source: whyfiles.org)

However, a black hole has so much gravity that one could say it 'rips' the spacetime fabric:

(source: ddmcdn.com)

When the light hits this area of amazingly intense gravity, it simply cannot get out - the light travels 'along' the fabric, and since there is a rip in the fabric, one could say it simply goes away - it becomes part of the singularity.

This is a simplification, of course, but it's enough to understand at least part of the physics behind this phenonenom.


How Wide Would the Habitable Zone Be in a Singular Red Giant System? Astrobiologists looking for signs of life outside Earth usually take excitement over planets orbiting red giant stars because a larger, brighter star usually means a farther-away but still wider habitable zone, a slice of space where liquid water would be possible.

To put this in perspective, a single red giant can have a luminosity ranging anywhere between 1,000 and 10,000 times brighter than our sun.  Would that mean that a habitable zone in a solar system orbiting only one red giant would be 1,000 to 10,000 times as wide as our own?

Habitable zones have a width that scales as the square root of the luminosity. So a 10000 solar luminosity star will have a zone about 100 times wider.

However, the temporal width is another thing: the luminosity changes relatively fast, so planets will not be in the zone over long timescales.

(The reason for the square root dependence is that a piece of surface at semimajor axis $a$ gets $L/4\pi a^2$ Watt of incoming energy, and if a blackbody with emissivity $\epsilon$ radiates away $\sigma \epsilon T^4$. Energy balance gives $L/4\pi a^2 = \sigma \epsilon T^4$, or rearranged, $a = \sqrt{L/4\pi \sigma \epsilon T^4}$, proportional to $\sqrt{L}$. The life zone is given by the range where $T$ allows liquid water. In practice this is more complex and fuzzy since it all depends on complex atmospheric properties, but as a first approximation it feels OK.)


Hypothetical CMB space telescope design problem, received power from extended thermal source versus receiver front end NEP? Consider the example of a large radio dish antenna in space equipped with a heat shield protecting it from the Sun. Let's say the amplifiers front-end effective temperature (for purposes of noise calculations (i.e. NEP) is 2.7 Kelvin which is coincidentally the temperature of the Cosmic Microwave Background that it will look at.
We'll ignore thermal radiation from the dish itself, because although it is somewhat warmer it's highly reflective metal and so the emissivity is very low. (see this answer for more on that)
The receiver system is fixed in central wavelength near the peak of the CMB, let's say 2 mm, and has a fixed 1% bandpass.
Now it's time to choose the diameter and focal length of the dish.
I know that the characteristic "sky temperature" of the CMB for which I'm trying to image non-uniformities and the temperature of my receiver's front end are both 2.7 K but what is the ratio of their powers that I'll be measuring?
Question: If I have a large f/no. will a given diffraction-limited feed horn receive less power and so the signal will be weaker compared to thermal noise of the receiver, or will they end up somehow being roughly equal no matter the diameter or focal length of the dish?

The feed horn is an antenna which has an antenna pattern  Source
However, not all feeds are horns.  Here is an image of a parabolic dish with a Yagi feed:  Source
Ideally, you want the beam width of the horn antenna to cover the entire dish and only the dish.  So if the beam width of the horn antenna is, say, 30 degrees, you want the dish to subtend 30 degrees of the view from the center of the entrance of the feed.
Since the dish is parabolic, it can be described by the equation y=a(x-h)^2 Source.  Assuming the vertex of the parabola is at the origin, the feed antenna will be 1/(4a) units away from the vertex.  This is the focal length of the dish.  The aperture of the dish is its diameter.  Using the values from the above equation, the f-number of a dish antenna is (1/(4a))/x or 1/(4ax).
Now that we know where the focus is, we can compute the beam width of the feed required to cover the dish, then design a feed antenna meets that requirement.
Notice that the f-number doesn't contribute to the solution to the problem as we would expect if this were an optical system.  The most important thing is to match the beam width of the feed antenna to the radius of the dish, which is set by the value of "a".  So a feed antenna will receive the same amount of power regardless of f-number as long as the feed antenna is at the focus of the aperture.  (Whether one can design a feed antenna for a given "a" is another question.)


If an orbit is shifting due to orbital precession, is it still a Keplerian orbit? I was thinking about orbits a few days ago, and realized that orbits shift/precess naturally. Given that a two-body problem with a star and a planet, if the planet has an eccentric orbit that precesses, would we still call it a Keplerian orbit?

No orbits are strictly Keplerian, unless you really monkey about with your coordinate systems. However, a precessing orbit is pretty darned Keplerian.

The relationship between the various orbital elements can be a little complicated, and precession is traditionally thought of as the rotation of the periapsis of the orbit around the more massive body.  If we look at the orbital elements, we can think of it instead as a changing term to the longitude of the ascending node.  So we could capture the longitude of the ascending node, $\Omega_0$ at some initial time $t_0$.  Then, at any time t, the longitude of the ascending node would be $\Omega=\Omega_0+(t-t_0)\omega_p$, where $\omega_p$ is the rate of precession.  The rest of the orbital elements can be treated as constant.


Why do the rings around the Gas Giants get their 'ring' shape? In our Solar System, all the gas giants have ring systems. However, why do these rings form rings and not a cloud of debris around the planet?

There are different theories as to how planetary rings form, and so there are different answers to your question.

One theory of planetary ring formation is that a small (or large, depending on the ring size) moon wandered into the gas giant's Roche limit (the radius from the celestial body inside which no second celestial body can stay intact due to tidal disruptions by the larger body). The moon would have been orbiting the planet, and would thus have been moving in a flat . The resulting debris continued to move in a plane; if it had been angled, collisions would have flattened it.

Another theory is that the debris is left over from the early Solar System. A protoplanetary disk formed around the Sun; it contained all the material that makes up the planets. Over time (and through a lot of collisions!), bits of the material slammed together and formed planets. The trouble is, not all of that material became planets. Some became asteroids, some became comets, and some became moons (although the theory of the formation of the Earth's moon is a bit different). Some of this material could also have coalesced around planets.

Now, the protoplanetary disk would already have been in a disk shape. Why? Think of a rotating object - like a planet. You might have read that the Earth is not a sphere. Aside from the obvious deviations of mountains and valleys, some of the deformation comes from the Earth's rotation. Any rotating body will bulge a little around its equator; a spherical body will become an oblate spheroid. The same thing happened for the protoplanetary disk - it became a disk because it was rotating! Any material that coalesced around planets would already be in a plane - that's why the planets' orbits aren't at odd angles to each other. So this material would be in a flat shape if it was captured by the planets, and would become a circular ring because of rotation. If, at that point, there was material that wasn't in a flat plane, it would become flat by the same process that made the protoplanetary disk flat.

There are stellar systems where the protoplanetary disk isn't in the same plane as the star's rotation. This could affect planetary orbits; however, such systems have not been thoroughly studied because of their rarity.


How much more atmosphere is the sun passing through at sunset as opposed to at its peak in the sky? I was looking at the sun during a sunset earlier today and wondered how much atmosphere you need to be able to look directly at the sun without damaging your eyes. The thought crossed my mind because I was continuously driving into an extremely oppressive sunset, my eyes were battered by the sun continuously for about a half an hour. Then, suddenly, the sun became more and more dull as it lowered into the sky.

My question is:

What is the approximate ratio of atmosphere between the sun and me, when setting as compared to when its positioned at a peak point in the sky? If you have the math behind an approximation like this from a topological observer, I would love to see that worked out.

The amount of atmosphere that light from outside Earth has to pass through before it reaches the ground is given by the air mass, $X$. It is normalized to unity for a source directly overhead (at "zenith"), and increases as the source "sets".

If you don't look too close to the horizon, you can use the formula $X = \sec z$, where $\sec \equiv 1/\cos$, and $z$ is the angle from zenith. For instance, if the Sun is $30^\circ$ above the horizon, so that $z=60^\circ$, then $X = 2$, i.e. its light passes through twice the amount of atmosphere compared to when its at zenith.

The formula above makes (at least) three approximations, however. Firstly, it assumes a plane-parallel atmosphere (an infinite, flat Earth), and so $X\rightarrow\infty$ as the Sun sets. Secondly, it assumes that the atmosphere is homogeneous, whereas in fact its density decreases with distance from Earth. This not only causes a non-uniform absorption$^\dagger$ along the ray, but also causes refraction which increases the path length for high $z$. For astronomers, these approximations don't matter much, since they're quite accurate out to $z\sim70^\circ \text{–}\,80^\circ$ and we usually don't observe at greater angles anyway, precisely because of the large air mass. But several more exact formulas based on various models of the atmosphere have been created to fit observed air masses. These models converge roughly to $X \simeq 38$ for $z=90^\circ$. In other words, the amount of atmosphere that light goes through for a source at the horizon, compared to a source at zenith, is roughly 38. However, note that just $10^\circ$ above the horizon, $X$ is already $<6$, so it's the last few degrees that really matter.

The exact value depends on many factors, including temperature (and its change with height), position on Earth, and the altitude of the observer.

$^\dagger$Most photons are actually not absorbed, but rather scattered out of the line of sight. Effectively, the result is the same; photons are removed from the ray. The scattered photons then become part of the background (and since blue photons scatter more than red, the sky is blue, and the Sun becomes redder as it sets).


70-year star in Talmud confusing sailors? I don't know if this question is a good fit for here - it probably wants the non-existent "History of Astronomic Navigation SE", or maybe just "History SE" - but I'll start here:

As described in this article here by Dr Mario Livio:  The Talmud (an 1500-2000 year-old Jewish book of laws and traditions) describes how a Rabbi anticipated hs ship getting lost, because “There is a star that appears every 70 years and induces navigation errors.  I thought it might appear and cause us to go astray.”.

Various theories are put forward to the identity of this star - with the 70-year period suggesting Halley's comet, although another theory posits that it was a variable star (see article for details).

My question is one that the article doesn't seem to address: How could a comet or a variable star confuse sailors? Comets don't even look like stars, so as long as they don't cover up the Pole Star or any other important navigational bodies why should they confuse sailors?

Comets can look like stars. Sometimes the outside of the coma and tail of the comet are too faint to see with the naked eye, and only the central part of the coma is bright enough to be seen. In this case the comet will appear like a (slightly fuzzy) star.  Look at the Great Nebula in Orion. It appears star-like, perhaps a little fuzzy.  Other clusters and nebulae were initially classified as "stars".

Moreover in the language of the first century AD (and until relatively recently) the word "star" meant "any heavenly light" and included comets, planets, and meteors. A navigator at that time would consider a comet to be just a "hairy star", while the learnéd debated whether comets were astronomical or meteorological stars.

At the time, many people also believed in "astrology" (the belief that the stars could have magical influences on life on Earth). It would seem quite reasonable, then, to think a star that appears every 70 years could affect navigation in mysterious ways.

As with many such ancient texts, the correct interpretation is probably lost. I'd consider it to be a parable: A man may be knowledgable and wise (as evidenced by his knowledge of the stars) but still be poor. I see no reason to try to identify the "star", no more that I would try to identify the farm in Isaiah 28. Trying to prove the story is "scientifically correct" seems to miss the point.


How to calculate B-V colour index value percentage difference I need to calculate a percentage difference of a B-V colour index between its estimated and actual value. So I tried doing this by difference between values/actual value x 100. However as B-V values can be on either side of zero this did not give representative answers.

So if anyone knows a way of calculating percentages of values close to zero or on either side of zero their help would be appreciated.

Thanks

$B-V$ corresponds to the base 10 logarithm of a flux ratio.

$$B-V = -2.5 \log \left(\frac{f_B}{f_V}\right)$$

So trying to guess what you are trying to calculate, it is the percentage change in the blue to visible flux ratio?

In which case the percentage change is
$$ p = \frac{ 10^{-(B-V)_2/2.5} - 10^{-(B-V)_1/2.5}}{10^{-(B-V)_1/2.5}}\times 100$$

A percentage change can of course be negative.


Total apparent magnitude of eclipsing binary system This is a question from the 2016 USAAAO national exam, for which no solutions are given:

  An eclipsing binary system has a magnitude $m_p$ = 14.2 during the
  primary transit and $m_s$ = 13.7 during the secondary transit. Find
  the normal (non-eclipsed) apparent magnitude of the system.

I assumed that, as is usual for eclipsing binary systems, the higher magnitude/primary transit refers to the complete eclipse of a smaller, hotter star (say Star B), and the lower magnitude/secondary transit refers to Star B passing in front of a larger, cooler Star A.

To find the normal, non-eclipsed apparent magnitude of the system, usually we achieve a relation like $F_A = kF_B$, where k is a constant. Then, as we know the apparent magnitude corresponding to $F_A$ alone, we could easily calculate that for $F_A + F_B$. However, this is where I ran into trouble. I set up the following equation to attempt to find a relation between $F_A$ and $F_B$: $$m_p - m_s = -2.5log(\frac{F_A}{F_A(1-(R_B/R_A)^2)+F_B})$$

The denominator in the argument of the logarithm, $F_A(1-(R_B/R_A)^2)+F_B$, expresses the flux observed at the secondary transit. But I have no way that I can see to figure out the ratio between radii, and I can't think of any reasonable assumption that would simplify the equation further (such as $R_B << R_A$).

Should this problem be attacked from a different angle? Any help would be much appreciated!

Without information about stellar radii, I think it's reasonable to assume
$R_A \approx R_B$.
Then your equation becomes

$$ m_p - m_s = -2.5 \log \frac{F_A}{F_B} $$

and you can compute $k$ and the non-eclipsed total magnitude.


What places on Pluto and Charon are facing each other? What places on Pluto and Charon are facing each other? Are Pluto and Charon perfectly tidally locked, or is there a slow rotation over time? And how long after formation should they have become tidally locked? Before or after the late heavy bombardment?

EDIT to add the image series of Pluto below, found at University Today

As New Horizons past through the Pluto system, it travelled so that the spacecraft was on the far side of Pluto, as seen from Charon. It did this so both Pluto and Charon were roughly in the same direction, so the space craft would not have to rotate 180 degrees between imaging Pluto to imaging Charon.

It means that in images of Charon, the face that you can see is roughly the face that faces Pluto, However the it is the far side of Pluto (the side with the "whale" and the "dots" that faces Charon.

Tidal locking is pretty stable. See our own moon: it is locked, and has been for a long time. There is likely to be some wobble, just like the libations of the moon, but no overall rotation of either Pluto or Charon.


Understanding expected TESS data structure I'm interested in trying to use the data that will be returned by TESS to apply some machine learning/computer sciences concepts. I'm not sure yet what I want to try to do with it (though I have some ideas - see end comments). This because I don't really understand the data structure it will be using.

So essentially I would like to first some help to get a high-level view of how the TESS data is structured. Then a bit of guidance as to how I could visualize it/get started with some basic example

Additional info/what I know at this point

I've checked out the MAST page. It details what type of FITS documents will be returned.

I have downloaded samples of some of these data products and I see they all come in FITS format. On Ubuntu, you can open that in Aladin. However Aladin in itself is a fairly involved piece of software. Do I need Aladin for this, or are there simple alternatives to just view the data quickly to help get a feel for what it means? In other words what kind of navigator do you use for FITS files?

I'm pretty new to astronomy - I'm coming at this rather from a computer science perspective. Some basic suggestions as to what kind of observation an amateur astronomer might want to look at would be welcomed if you have some.

The data linked above (MAST page) are simulated data (real data will only be available in a couple months). This is fine - mostly interested in being functional with the data when they arrive.

Data analysis ideas:

Using kmeans or similar to try to extract  what are some of the "typical" transit light curves

Generating those transit light curves from the raw data, say from a target pixel

An insightful option is to open in a python IDE using Astropy. I find it useful because you can manipulate the files as your learn their structure.

It is also possible to use matplotlib in conjunction with Astropy and then basically have a FITS file viewer that you can code yourself.

Something like:

import astropy
import matplotlib.pyplot as plt
from astropy.visualization import astropy_mpl_style

    def open_fits_file(fits_path):
        for file in Path(fits_path).iterdir():
            #Not really using the with, but it'll manage closing the file...
            with fits.open(file) as opened_fits:
                #printing header infos for that file
                opened_fits.info()
                plot_image(fits.getdata(file, ext=2))

Then to plot the image:

def plot_image(fit_image_data):
    #astropy has better plot settings for this use
    plt.style.use(astropy_mpl_style)
    plt.figure()
    plt.imshow(fit_image_data, cmap="gray")
    plt.colorbar()
    plt.show()

I used that to plot the pixels  from these simulated data linked in the question.

As mentioned by Mike G, HEASARC as a list of viewer, including FV (fits viewer). It works fine and it has some built-in support to link up with other astronomy database. It has an apollo-era feel in terms of appearance, which is nice... I guess?


How to determine the formula to calculate this position of reflected Sun? I created a geometrical simulation of how the Sun is reflected on a wall along the day and along the year, using Geogebra:

https://www.geogebra.org/classic/dckhbc8h
Playing with parameters, I found an oddity: in some specific cases, changing the "season angle", i.e. inclination of su circle w.r.t. horizon, I can see the "sun reflection line" always passing in same point on the wall along whole year:

Overlapped:

Is this a real behaviour or just an error in my simulation?
Does it happens always at same time of the ay? My simulation currently does not show the times along the "sunLine" segment.
How could I build a Geogebra formula to directly draw the "convergence point" of all the sunLine lines along the year?

When the "sun angle" is zero, the position of the "sun" doesn't depend on the season angle, so the lines will cross at the point that is the image of the sun at "sun-angle 0"
One important aspect in your model is that the season doens't change the "angle" but the position of the sun.  In summer, the sun rises and sets in the north East and north west and describes a small cirle around the north celestial pole. In winter it rises and sets in the southeast/west. And only at the equinox does it describe a great circle.


Relationship between photometric colour and redshift In this paper, the authors describe the theoretical relationship between photometric colours, particularly in Figures 1 and 6:

Since the colour is the difference between two magnitudes, and the redshift is defined as
$z=\frac{\lambda'-\lambda_0}{\lambda_0}$

I would have thought that the colour-z relationships would simply be a downward curve, since the light gets redshifted (goes down in frequency), and the ratio of two magnitudes is constant with redshift.

Why do the colour-z relationships in this paper (and in others I've seen) show such a complex relationship?

The main reason is that the intrinsic spectra of galaxies are complex and therefore a redshift of their spectrum, whilst leading to a redder spectrum overall, does not necessarily lead to reddening in all colours.

For instance if there is apeak in the intrinsic spectrum, then as that peak moves redward, then colours formed from bands on the same side or straddling the peak in wavelength would behave differently.

The dramatic redward turn up at high redshifts (note, a large colour is redder) is caused by the "Lyman break" moving through the colour bands. Basically, the intrinsic spectrum is self-absorbed at wavelengths shortward of 91.2 nm by neutral hydrogen. This absorption edge is redshifts into the visible region at redshifts bigger than 3, causing the bluer bands to essentially disappear and any colour formed with them to become very red.


Is this gas cloud or lens effect? I was going through the zoomable image of Milky way and after zooming to this particular location, we can see a reddish area on the left side of a probably large star. Here is a screenshot I took after zooming to that particular area:

So is this gas cloud or some weird lens effect ?

Edit: Location ( center of the blue rectangle )

This particular object I tracked down by comparing the visual with the infrared image of the sky. It was a little tricky because no very bright star appears at that location. But after making comparisons, the star in the original image is HD 155161, also known as V* AH Sco, indicating that it is a variable.

In fact, Simbad lists it as a red supergiant at coordinates [17 11 17.02114,  -32 19 30.7132], and visual magnitude of about 7.5 Red supergiants are often unstable and are semi-regular varible stars.

That explains why it is much brighter in the VISTA infrared image than in the optical. Red supergiants have cool outer layers that give off much of their radiation as infrared, not as visible light.

Looking at the star on Aladin Lite, even in the infrared, there is not a hint of nebulosity around the star. The conclusion is that this is an optical effect, lens flare or similar.


Why is sun's photosphere a million times less dense than air at the surface of the Earth? Sun's photosphere is about 19 times as hot as Earth's surface.
(Source: http://solar-center.stanford.edu/vitalstats.html)

But Solar surface gravity is about 28 times terrestrial surface gravity.

So, shouldn't the density of gases at the photosphere of the sun be comparable to that on the surface of the Earth?

Why is the density of Sun's photosphere a million times lesser?
(Source: http://solar-center.stanford.edu/vitalstats.html)

Essentially, what I am asking has been aptly worded in userLTK's answer, "how can the photosphere be so light and almost vacuum like under such high gravity (28 earth gravity)"?

Some of the answers have suggested that this may be because the Sun's photosphere is ionised. This doesn't seem to be accurate. The Hydrogen ionisation at optical depth = 1 is only 0.04%.
(Source: http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1968SoPh....3....5G&data_type=PDF_HIGH&whole_paper=YES&type=PRINTER&filetype=.pdf)

I think there is a misconception underlying your question. You write

But Solar surface gravity is about 28 times terrestrial surface gravity.
So, shouldn't the density of gases at the photosphere of the sun be comparable to that on the surface of the Earth?

It does sound natural that a stronger gravity would imply a thicker atmosphere, but no matter how strong the gravity is, there is a point where the atmosphere fades away almost to nothing. At the height of the ISS, earth's gravity is around 80% of the gravity at sea level. But yet the ISS is orbiting in the vacuum of space. Similarly, above Jupiter's atmosphere, the gravity is still more than twice that of earth, but there is still vacuum. With the sun, it just so happens that the part of the atmosphere that produces the light we see is close enough to the (imprecisely defined) edge of space that its density has gone down almost to nothing.
Certainly, if it were cooler, it would be denser. But some day the sun will be a white dwarf, about the size of the earth but with not that much less mass than it has now. Its gravity will be hugely strong, but there will still be a top of the atmosphere were it is less than a millionth of earth's atmosphere. As it cools, that atmosphere will cool so that it is no warmer than earth's atmosphere, but there will still be a point where the atmosphere is that thin. And even above that height, gravity will be much higher than even at the current surface of the sun.


dull rock/potato-shaped object seen with naked eye, moving in straight line during night. What could it be and is size and speed estimation possible? Between years 2009-2011, I was located in Greece, watching the perfectly dark night sky around 21:00-23:00 during summer, when I suddenly noticed a very dull dark-brown object that looked like a solid rock/potato having a perceived size around half the diameter of the moon in the sky, moving in a perfect straight line looking up
 and perceived as fast (it took 3-4 seconds to disappear in the horizon) - so total exposure if I had seen it from start to end would have been less than 10-12 seconds.

One could barely notice it, it had some texture like a rock but blurry. It had almost no light reflection and it wasn't rotating or anything. It wasn't spherical and it was somewhat elongated along the axis of movement. I was barely able to detect it above me with naked eyes, as it was passing from left to right. I only did when I saw it was hiding a small region of the night sky stars. This way I perceived its irregular shape.

Unfortunately I cannot tell it's actual size or how fast it was going. Is it an object of a kind known to astronomers? I couldn't find any similar video footage or anyone seeing something similar talking in the internet. Could it be a space rock (asteroid?) that missed the earth atmosphere and at the same time be seen with naked eyes? Any estimates on space rock type, size and speed based on the time of sky exposure? Any astronomical id from a database or some guidance on how I could search it?
Could it be an asteroid occultation like this one: http://call4obs.iota-es.de/wp-content/uploads/2011/08/palma_south.gif
Thanks

Rock-shaped doesn't mean very much, rocks come in all shapes and sizes. It couldn't possibly have been an asteroid. It must have been quite large for you to see it at night, but it is surprising you could see the colour of it at night, when colour vision fades. Large, fast, elongated in the direction of motion - it sounds as though it could have been a dirigible, aided by a tail wind. Dirigibles are airships vaguely similar to a zeppelin. The ones with no rigid skeleton are called blimps. They have come back into fashion for certain tasks in the last 20 years or so; there are some in Europe and in USA and probably other countries too, Russia for example. Whether there are any in Greece, I couldn't say, but one thing I can say is that it is far more likely to have been an airship than an asteroid. One possible reason for its rapid disappearance is that it could have disappeared into or behind a cloud.


How can I contribute to the scientific community using my telescope? I've been using my Meade 90 ETX for a few years trying to become a better astronomer and am familiar enough with the basics to navigate my way around. I was wondering in what way an amateur astronomer such as myself can contribute to the scientific community. I would like to try and do more than just stargaze and take pictures.

Amateurs do useful scientific work by being many-handed and widely spread.

Many Supernovae are discovered by amateurs. You need a set-up that can image one galaxy after another and look for any "new stars" that appear in them. As you are looking for the appearance of a Mag-14 star you need a big mirror; 90mm might not be enough. However look to Backyard observatory supernova search for more information. With the right equipment, you might find 1 SN every 5000 galaxies imaged.

You can also contribute by observing asteroid occultations. These occur when an asteroid passes in front of a star, blocking out its light. By timing the occultation you can get an idea of the size of the asteroid in one cross-section. If multiple people observe the same occultation from different locations you can get a picture of the shape of the asteroid. See the Asteroid occultation site which links to FAQs and a page for submitting reports.


What is direction of Sun's axis of rotation relative to the Earth's orbital plane around the Sun? The present northern direction to the nearest 0.5 degrees with degree zero being the start of January and measured counterclockwise (same as Earth's orbit relative to celestial north pole).

If the assumptions below do not allow a precision of 0.5 degrees then I imagine this question is unanswerable without specifying a specific date and time, in which case the specific date and time would be 1 January 2017, 00:00:00 hours.

Assumptions: (1) That the Earth's orbit around the sun is a perfect circle (I know it's very slightly elliptical but intuitively imagine that won't significantly affect the precision I need). (2) That the Sun's center is at the center of Earth's circular orbit (I know the Sun's center isn't exactly but don't imagine that will affect the accuracy I need either).

Note 1: I've found three general answers from three different sites, but, two of them appear to me to conflict and the other is too imprecise.

Note 2: This is my first post to this site and I did not plow through the hundreds of questions related to these tags. So if this question has already
been answered I apologize and please be merciful and point me in right direction.

Thank you

Well, Wikipedia says that it's 7.25°to the ecliptic and gives a reference.
Looking there clarifies the value in 3 dimensions:

North Pole of Rotation
Right Ascension: 286.13
Declination    :  63.87

Since those coordinates are relative to the Earth’s axis, being a practical measurement system for Earth ground observers to use in pointing their telescopes, you need to combine that with Earth's orbit.
As a useful visualization, mark that point on a model of the celesial sphere.  Compare that with the ecliptic pole: the difference between thise points is what you want to know.
I expect you’ll find they are separated by 7.25°, but by marking a model sphere you’ll understand the actual direction.


What Causes the Large Radiation Fields Around Jupiter? I had heard that the large radiation belts around Jupiter may be formed by liquid metallic hydrogen in (or around) Jupiter's core (which Wikipedia says haven't been observed in labs yet due to the immense pressure required), but I don't understand how the liquid metallic hydrogen can form a planetary magnetic field. The article also says that the hydrogen can act as a conductor, is this somehow related to moving electric fields forming magnetic fields?

Source: Magnetosphere of Jupiter - Wikipedia

As you stated, we have not been able to simulate the pressure and temperatures required to generate those that are believed to exist in Jupiter's interior other than in short-lived shockwave experiments, according to the NASA webpage A Freaky Fluid inside Jupiter?, observing that

  "Liquid metallic hydrogen has low viscosity, like water, and it's a good electrical and thermal conductor," says Caltech's David Stevenson, an expert in planet formation, evolution, and structure. "Like a mirror, it reflects light, so if you were immersed in it [here's hoping you never are], you wouldn't be able to see anything."

Going further, according to the article Jumpin' Jupiter! Metallic Hydrogen (Lawrence Livermore National Laboratory), discuss the shock wave results, finding the level at which hydrogen metallises as being

  from 0.9 to 1.4 Mbar, resistivity in the shocked fluid decreases almost four orders of magnitude (i.e., conductivity increases); from 1.4 to 1.8 Mbar, resistivity is essentially constant at a value typical of that of liquid metals. Our data indicate a continuous transition from a semiconducting to metallic diatomic fluid at 1.4 Mbar, nine-fold compression of initial liquid density, and 3,000 K.

The findings from the researchers above are summarised in the diagram below

The source is the Jumping Jupiter link above.


Can it be inferred that our cosmological horizon has increased over time? If I am right, we can see only those stars that lie within our cosmological horizon, and there may or may not be any stars beyond that. Given last 150 years of using telescopes, and since then our cosmological horizon must have increased by about 150 light years, have we seen any new galaxies, stars? If given enough time, say 500 years, new stars are found, can we be sure they have become visible against not being formed, say, 150 years after big bang?

No. Space expansion doesn't slow down, unlike the expansion of cosmic horizon. That means the cosmic horizon actually swallows galaxies as they travel away from us due to space expansion, reducing their number instead of increasing it. Nevertheless, we still haven't gathered all cosmic radiation from Big Bang that occurred within our current event horizon, so it's still a good while before it begins restricting distance of observable universe for us - currently the speed of light (+space expansion rate) vs age of the universe limits it to a considerably smaller radius.
Yes, we discovered more stars and galaxies, but that is due to improvement of technology, which is still quite far (roughly a third of the way) from observing stars near the Hubble Sphere as the universe formed (ones whose light would be observable today; as opposed to these at the space horizon, whose light will reach us eventually, in infinitely distant future.)

Currently we're still quite a bit away from leaning against the cosmological limits of observability. Currently our limit of observability is the budget for building better devices.


Is there or will there be any search after a primordial black hole 500 au from the Sun? Since 2019 it is proposed that the hypothetical Planet IX may in fact be a primordial black hole. Black holes are being looked after differently than planets. You can look for a gravitational lens or for an accretion disc. However, the black hole theory doesn't seem to be taken serious by many.
Is there any ongoing search for such black hole rather than a planet, or are there any plans to look after a primordial black hole if the planet theory would get disproved?

Maybe, Harvard scientists have proposed a way to determine, once and for all, whether Planet Nine actually could be a black hole. Specifically, the new method would scour the outer solar system for evidence of telltale flares that are emitted when a black hole devours a comet or other distant object. Such flares, they say, should be detectable by the upcoming Vera C. Rubin Observatory in Chile, which is expected to begin a 10-year survey of the southern sky within the next few years.
I suppose if funding is available for this it will be done when the telescope comes on-line. This may solve the problem of if it's a black hole but a large planet could still be possible but unknown.
Ref: https://astronomy.com/news/2020/07/is-planet-nine-a-black-hole-or-a-planet-harvard-scientists-suggest-a-way-to-find-out


How did UV from the earliest stars 'alter the state of the 21 cm line' such that it shows up in CMB today? In this question I discuss the recent (open access) paper in Nature An absorption profile centred at 78 megahertz in the sky-averaged spectrum at length. The abstract begins:

  After stars formed in the early Universe, their ultraviolet light is expected, eventually, to have penetrated the primordial hydrogen gas and altered the excitation state of its 21-centimetre hyperfine line. This alteration would cause the gas to absorb photons from the cosmic microwave background, producing a spectral distortion that should be observable today at radio frequencies of less than 200 megahertz(1).

where reference 1 is Jonathan Pritchard and Abraham Loeb (2012) 21 cm cosmology in the 21st century. From there I've found Leonid Chuzhoy and Zheng Zheng (2007) Radiative Transfer Effect on Ultraviolet Pumping of the 21 cm Line in the High-Redshift Universe.

I understand some basics about the hyperfine transition in hydrogen, and that the "spin temperature" of a gas in space can differ from the temperature of other partitions if it's being pumped, but these papers are more than a bit hard to read.

Is there a simple way to explain the basics behind how the exposure of hydrogen to the UV light produced in early stars would cause the "blip" in the (now red-shifted) 21 cm part of the Cosmic Microwave Background radiation spectrum we see today?

The Intergalactic medium at the relevant redshift is made of neutral hydrogen. What we can measure is the brightness temperature (the temperature that the IGM would have if it emitted as a blackbody) relative to the CMB. This quantity depends crucially on the following expression:

$\frac{T_S - T_{CMB}}{T_S}$

where $T_S$ is called spin temperature.

The spin temperature is just a measure of the ratio between the number of hydrogen atoms in the first excited hyperfine state (spin parallel) and ground state (spin antiparallel).
It turns out that the spin temperature can be modified in three ways: one of them is Lyman resonant scattering (the other two are collisional coupling and scattering of CMB photons).
Using atomic physics it's not difficult to see that $T_S$ is a weighted mean of the temperature of the CMB and gas temperature.

UV photons can change the spin temperature because a hydrogen atom in the lowest state, n=1 (if you are familiar with chemistry, a s state) with antiparallel spins, can absorb an UV photon and jump in a p (n=2) state, and then fall again in a s state but with parallel spins.
This mechanism is known as Wouthuysen–Field coupling.

Before the emission takes place, the gas is at the same temperature of the CMB and both are equal to spin temperature, so the relative brightness temperature is zero.
When Lyman alpha emission begins the spin temperature decreases resulting in the observed absorption peak.
At some point the peak stops because the first stars emit X-rays and the gas becomes hotter than the background radiation.