Did I see a mentionable event in Germany early morning on 6th of February 2019? Around 6.15am (GMT + 1) on the 6th of February 2019 I
was looking out of my window somewhat north, when I saw something one would probably call a shooting star. The features were:

A magnitude I haven't seen before; it even had a soft blinding effect, that's how bright it was.
From my view frame it moved from my right to my left (so most likely from east to west).
The size was a quarter to half of what the moon usually has from my view frame.
It traveled almost through the sky half I was able to see within half a second before its glow was immediately gone again.

I am assuming the object must have been either very large and fast or quite close.

Was there any meteor worthy mentioning passing by Earth at that given time?
Or are the features I just described way more common than I am assuming, and I just spotted an event that happens quite regularly?

Fireballs are real but rare events. Meteors are formed when small rocks enter the atomsphere from space. Larger rocks or small asteroids will produce brighter meteors. The brightest meteors are called fireballs or bolides. It is not uncommon for parts of the rock that made the fireball to survive and can later be found as meteorites.
A couple of things can be confused with fireballs.  When satellites re-enter, they can burn up. Satellites move much more slowly than true fireballs, most fireballs last less than 10 seconds.
A comet is an astronomical, not atmospheric, object. It looks like a fuzzy star with a tail. It remains almost motionless in the sky, sometimes for months. Really bright comets are rare, and even the brightest comets are much much dimmer than a fireball.
There are several organisations that collect reports of fireballs:

https://fireball.amsmeteors.org/members/imo/report_intro
https://ukmeteornetwork.co.uk/fireball-report/

The advice given states:

Please, don't report sighting that lasted more than 30 seconds: the vast majority of fireballs are only visible for few seconds.
Please, don't report recurring events: seeing a fireball is extremely rare and often a once in a lifetime event.
Please, don't report slow blinking objects or lights crossing the sky going by 2 or 3: a fireball looks like a big shooting star.


Can protoplanetary disks form main-sequence stars? As has been pointed out by @Envite in the context of a more general discussion (see Generalised planets?), there seems to be a moderate possibility for protoplanetary discs forming main-sequence stars (MS stars).

Main argument here is that Jupiter has mass of about $10^{-3}M_\odot$, while in principle for an object to be a MS star, it would need the mass of at least about $0.07M_\odot$. Also, there is a handful of systems, such as HD 29587, which contain a star and a brown dwarf, possibly forming from a protoplanetary disk.

Hence, a few questions. Can low-mass MS stars form in protoplanetary discs? If no, why? If yes, how often does it happen? If yes, which implications would it have for the subsequent dynamics of the planetary system, if one of the objects became a MS star, as opposed to it being a brown dwarf?

Again, thanks to @Envite for the idea.

Yes, unless you want to get really particular with the "protoplanetary" part. For example, there are stars forming in the circumstellar disks around Wolf-Rayet stars [reference]. If we were picky, we might not call the circumstellar disk around this Wolf-Rayet star a protoplanetary disk (and instead refer to the planet-forming circumstellar disks of the newly formed stars as such), but there exists no clear line between the disks of gas within and the larger disk of gas around the Wolf-Rayet star of which they are a part.

Source:

Triggered star formation surrounding Wolf-Rayet star HD 211853 - Liu, Wu, Zhang, and Qin


What would you find within a void? I'm aware that voids are relatively empty regions of the universe, but just how empty can they be? Wikipedia states that voids have 'few or no' galaxies, but I can't find much else.

To make my question specific, suppose I were in the centre of a large void, way outside of the few galaxies that may still exist in this void. Would I occasionally encounter asteroids (if I may call them such), rogue planets, or even some occasional intergalactic star?

Remark: I'm aware that this question is a bit of a one-up of What is there in the intergalactic space?. If this is too much of a duplicate, let me know and I remove the question.

The Wikipedia article on voids is pretty good (though IMO unusually awkwardly written.) The key thing is that voids are not empty, they are just large volumes which have a lower density (typically around 10% of average) compared with the rest of the universe.

These low density areas still contain stars and galaxies, just fewer of them and the galaxies they contain tend to be smaller. The Lambda-Cold-Dark-Matter (LCDM) model, which is currently favored as our best description of the universe, predicts that the voids should not have many large galaxies in them, but should have a fair scattering of small, diffuse galaxies (SDGs).  (A small diffuse galaxy is, well, small compared with the Milky Way --
sized more like the Magellanic Clouds or one of the small galaxies which are being devoured by the Milky Way today such as the Sagittarius dwarf. They are also more spread out than the Magellanic Clouds.)

Here's an interesting senior thesis reporting on a search for small diffuse galaxies in voids. (They think they found some, but need confirmation.)

You ask what the night sky would look like from within a void. First, we'll assume you're in one of the SDGs, though this wouldn't make a huge difference.  What you'd see mostly is whatever planets share your solar system -- they'd be comparable in brightness to our own planets from Earth. You'd also see a few dim stars.  SDGs tend not to have much star formation, so there would be very few young stars and thus very few intrinsically bright stars. The visible stars would all be ones which happened to be nearby. And because SDGs are diffuse, there'd be many fewer nearby stars. At a guess there would be 20-200 naked eye stars, with only a handful of magnitude 3 or brighter. (Compare that with 6000 naked eye stars visible from Earth.)

There would be no naked eye galaxies. The SDG itself would not appear as a milky way, but more like the Zodiacal Light -- a vague glow dimly visible under exceptional conditions.

This would be a world whose astronomy would be heavily focused on the planets.

(Planetary formation might be quite different in SDGs, but I don't know even to try to speculate on how it would be different. So I assume that the planetary system would be similar to ours.  YMMV.)


Lack of planets in the Oort Cloud With the current news about the possibility of a planet in the Oort cloud, I was thinking, why is the matter in the Oort cloud distributed so? It makes sense that inside the planet region there is not a huge concentration of comets and meteors, but why is the matter inside the cloud itself not agglomerating? It seems like there is enough matter to form not one or two, but many more planets.

Simple answer: the material is just too spread out.

Forming a planetary-size object takes LOTS of collisions to build up enough matter to start gravitationally attracting the surrounding stuff.  The Oort cloud is very far from the sun (starting at ~1000 AU, compared to a maximum of ~50 AU for Pluto), meaning that the icy fragments are moving very slow in their orbits.  Furthermore, the material is spread out over a huge volume, especially since the Oort cloud is a spherical shell (the matter is spread in 3 dimensions), as opposed to the 2-d disk that makes up the vast majority of matter in the Solar system (Sun, Planets, Asteroid & Kuiper Belts).  The timescale for these objects to accrete to any graviationally significant size is far beyond the lifetime of the Solar system (or even the universe).


What other galaxies can be seen with the naked eye in the night sky? What other galaxies can be seen with the naked eye in the night sky? Can we see any other galaxy besides Andromeda and Magellanic Clouds?

We can see the Magellenic clouds easily, and we can see Andromeda on very dark, clear nights, but only when we are far from the city.

Some people claim to have seen the Triangulum galaxy (the third largest member of the local group, after Andromeda and the Milky way). Exceptional conditions, and very well trained eyes would be required to see it.

No other galaxies are visible, even under the best conditions.


How to create synthetic test images of stars? Is there any python/IDL code available to create synthetic test images of stars with and without background noise?

This doesn't need programming but the SkyMaker package will generate very realistic astronomical images from catalogs produced by Stuff. It includes realistic models of stars and galaxies and the point spread function including e.g. atmospheric blurring and noise effects in the CCD (blooming, saturation etc)


Is there a better explanation of Hawking radiation? I'm writing a piece on Hawking radiation, and find I have something of a problem. The "given" explanation which I find on Wikipedia and elsewhere is unsatisfactory:

  "Physical insight into the process may be gained by imagining that particle–antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself, but rather is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles[10]. As the particle–antiparticle pair was produced by the black hole's gravitational energy, the escape of one of the particles lowers the mass of the black hole[11]. An alternative view of the process is that vacuum fluctuations cause a particle–antiparticle pair to appear close to the event horizon of a black hole. One of the pair falls into the black hole while the other escapes. In order to preserve total energy, the particle that fell into the black hole must have had a negative energy..."

It relies on virtual particles and an negative-energy particles. However vacuum fluctuations are not the same thing as virtual particles, which only exist in the mathematics of the model, and we know of know negative-energy particles. So I'm looking for a better explanation. The Wikpedia article also says this:

  "In another model, the process is a quantum tunnelling effect, whereby particle–antiparticle pairs will form from the vacuum, and one will tunnel outside the event horizon[10]."

However that suggests pair production is occurring inside the event horizon, which seems to disregard the infinite gravitational time dilation, and that one of them a) appears outside of the event horizon and b) escapes as Hawking radiation when pair production typically involves the creation of an electron and a positron. Again it's unsatisfactory. So:

Is there a better explanation of Hawking radiation?

Andy Gould proposed a classical derivation of Hawking radiation in a somewhat obscure paper from 1987.  The essential argument is that a black hole must have a finite, non-zero entropy (otherwise you could violate the second law of thermodynamics with a black hole).  Moreover, the entropy of the black hole must depend only on its area (otherwise you could change the area of a black hole via the Penrose process and lower its entropy and make a perpetual motion machine).  If a black hole has an entropy and a mass, then it has a temperature.  If it has a temperature, then it must radiate thermally (otherwise you could again violate the second law of thermodynamics).

Of course, if you look at the Hawking radiation temperature, there's a Planck's constant in there, so it has to know something about quantum mechanics, right?  But it turns out that it's actually thermodynamics in general that knows about quantum mechanics, not general relativity --- Planck's constant is only needed to keep entropies finite (and therefore temperatures non-zero).  This is true of black holes and blackbodies alike.


Astronomical Term *Mezzocielo* What is the name of the point of intersection of the great circle passing through the celestial north pole P and the Zenith Z, with the celestial equator? (Point T in the figure)
In Italian this point is called "Mezzocielo" (English machine translation), which translates into mid-sky, since it is located halfway between the Zenith and the horizon for an observer in Italy ($\phi \approx 40^{\circ}$).

I've checked several sources. Where this term appears in dictionaries, it is translated with a definition, and not a gloss (usually if a direct translation exists in English only a gloss is given) For example https://en.wiktionary.org/wiki/mezzocielo intersection of the meridian of a given place with the celestial equator.

I conclude that English speaking astronomers have not had the benefit of single word translation of mezzocielo.


How do different formulas for calculating the mean anomaly of an elliptic orbit correlate? I need to calculate the mean anomaly of an elliptic orbit at a specific point in time. I found two different formulas for $M(t)$ and I'd like to know how they correlate and if they result in different ouputs.
The first formular is:
$M(t) = \frac{2 \: \pi}{T}\:(t-t_o)$
and its taken from this book.
The second formular is:
$M(t) = M_0 + \Delta t \sqrt{\frac{\mu}{a^3}}$
taken from this document.
Inputs:
$T$... Orbital Period
$t_o$... Starting point in time (epoch)
$t$... Point in time
$\Delta t$... Elasped time: $t-t_0$
$\mu$... Standard gravitational parameter $\mu=GM$
$a$... Semi-major axis
$M_0$... Mean anomaly at epoch

Kepler's Third Law including the constants of proportionality is:
$$GM T^2 = 4\pi^2 a^3$$
Substituting $\mu = GM$, this can be rearranged to give:
$$\frac{2\pi}{T} = \sqrt{\frac{\mu}{a^3}}$$
Which lets you rewrite your first formula as your second one and vice-versa.
Note that in the first formula the reference epoch $t_0$ is assumed to be the time of periapsis, which corresponds to the case $M_0 = 0$. Your definition of $\Delta t$ should also read $\Delta t = t - t_0$, i.e. time since the reference epoch.


How did LIGO verify that the gravitational waves originated 1.3 billion years ago from two specific black hole collision? So they finally found gravitational waves.  That's cool and I trust is the case, but how did they confirm it originated 1.3 billion years ago and that two blackholes merging caused it and what distance it came from?

I am not sure if the extent of variation in the laser wavelength can tell us all this above information..

Any clues ?

The original paper states the methodology used to determine the location of the gravity waves. The bottom line is that they were detected in two different locations. Using the difference in timing, one can estimate the location in the sky to an area of about 600 degrees squared where the origin of the wave could have been. The two signals had an offset of 7 ms. Pure speed of light differences would have been 10 ms. This can be used to effectively identify a circle in the sky where the merger might have happened. Of course, this doesn't reveal the distance.

The distance was figured by a number of things. First of all, the frequency pattern was used to estimate the mass of each black hole. Using these masses, one can find the estimated total power of the event. Using this estimate, and seeing the estimate of the amplitude from the signal, one can use the inverse squared law to estimate the distance to said event.


How would we detect a planet behind the Sun? Let's assume hypothetically that Earth has a twin planet on the opposite side of Earth's orbit. Its orbital period would be exactly the same as Earth's and it would always be behind the Sun so directly observing it from Earth would not be possible.

How would we detect the planet's existence short of sending out a spacecraft that could look "behind" the sun and observe it visually? Have we confirmed that there is no such planet?

According to the Wikipedia article Counter-Earth, this idea has been floated around for millennia.  However, modern day earth-bound techniques would easily detect a 'Counter-Earth' via the following (but not limited to) these methods:

Gravitational perturbations would have been observed to have occurred on the other planets (terrestrial)
As Earth's orbit is not circular, it is elliptical, it would not be consistently 180 degrees around the orbit, as below

(Image source)

Most of all, it would have been detected/observed by the satellites we have around the Earth, Sun and other planets (many examples here).


Textbook to explain measurements and calculations for heliocentric solar system I'm trying to identify a textbook I had briefly and then lost. I had a PDF file of an astronomy textbook, but the hard drive crashed irreparably before I could thoroughly read the textbook, or even record the author and title.

The textbook covered the history of geocentric theories, explained how to make basic astronomical instruments to measure angles, and how to make arguments for heliocentrism. As I recall, the book mentioned that given high-resolution photography, one can prove heliocentrism by measuring how some celestial bodies appear larger at different times of the year.

I am trying to at least work through the calculations for heliocentrism on paper, even if I don't end up crafting my own instruments. If no one knows of any one textbook that covers all these topics, I would still be grateful for relevant textbooks that cover some of these topics.

Update: It is not a complete answer, but a good start can be found with the ephemeris question asked at:

https://astronomy.stackexchange.com/a/7973/29149

It's possible the book you seek is by Christopher Graney.  I cannot find them now, but I believe he had put together instructions on how to build a telescope similar to what Galileo used and how it could be used to argue for and against a heliocentric universe.  But his overall conclusion is that it's poorly suited for the task, and scientists of the time had good reason to not accept it.  (E.g. the scope had a very small aperature, and thus a horrible diffraction limit (which they didn't understand at the time).  It could measure the planets just fine, but through it, stars also appeared quite large too.  Since stars exhibited no detectable parallax, they'd have to be comically huge for the measurements to make sense.)

That said, if you want to actually do it, you don't need any fancy equipment.  Almost any telescope off the shelf today will do.  You'll need a camera with removable lenses (e.g. a DSLR), but not anything "high res" by today's standards, the oldest used one you can find would be plenty.  You'll also need an eyepiece adapter for it (just search for your camera type, e.g. "Canon eyepiece adapter").  You won't need a motorized mount just for the planets.

With that you can take photos of the planets at different times over months, but most likely a least a year or two to get convincing results.  Mars will generally show the most dramatic results, and if you search on Google, you'll find plenty of people who've done it.

You can also prove Venus orbits the Sun just by watching its phases.  In order to go from full, to half full, to crescent, it has to be on different sides of the Sun than we are during some times.  As well as the size change.  You'll also find a lot of people have done that if you search for that.

While not showing evidence of a heliocentric universe, one solid piece of evidence Galileo provided in favor of at least a non-geocentric universe was the moons of Jupiter.  Even with the limits of his telescope, he was able to show that they orbited Jupiter, so the Earth was not the center of everything.  Any telescope of today will be able to see them.  Even most binoculars, or even the scope off a rifle.


Is the turn-off point of the main sequence the same as the terminal-age main sequence? The turn-off (TO) point is one tipping point on the Hertzsprung-Russell diagram (HRD) where low-mass stars start to evolve from the main sequence (MS) toward lower temperatures. If I understand it correctly, this is the beginning of the sub-giant phase when the stars no longer have hydrogen burning at the core.

The definition above suggests that the TO point is the same as the terminal age main sequence (TAMS), however, I have not found a source that explicitly states so.

Is the TO point the same as the TAMS?

TL;DR: Turn Off point (TO) means the location of the "knee" in the HRD of clusters of stars. TAMS is an imaginary line in the HRD that is defined by the location of stars of different mass at the end of hydrogen burning. It constitutes the upper boundary of the main sequence.

More details:

There maybe some confusion because the terms are not completely specific.

A Zero Age Main Sequence star (ZAMS) star is a star that, after formation, has reached hydrostatic equilibrium and just started hydrogen burning. But ZAMS can also mean an imaginary line in the HRD that is formed by the location of all ZAMS stars of different mass. This line constitutes the lower boundary of the main sequence.

A Terminal Age Main Sequence (TAMS) star is a star that has reached the end of hydrogen burning. TAMS can also mean the line in the HRD diagram that constitutes the upper boundary of the main sequence.

The main sequence can thus be considered a band bounded by the ZAMS and the TAMS in which Main Sequence stars may be located. In the following figure (Maeder A., Meynet G., 1989, Astronomy & Astrophysics, 210, p. 155) this band (and other areas of slow development phases) is shown as shaded area:

With time (and hydrogen consumption) a star moves from its ZAMS location (determined by its mass) up and right to its TAMS location. The most massive stars evolve more quickly, reach the TAMS earlier and then move away from the main sequence.

The Turn Off point is only relevant for clusters or groups of stars that formed at the same time. With increasing age of the cluster, its stars move away from the main sequence, starting with the most massive ones in the upper left area, leaving more and more of the upper area of the "band" unoccupied. The TO point can be thought of „sliding down" the TAMS with time. From the position of the TO the age of the cluster can be determined.

https://commons.wikimedia.org/wiki/File:Open_cluster_HR_diagram_ages.gif#/media/File:Open_cluster_HR_diagram_ages.gif


How to calculate the S/N of IFU or long-slit galaxy observations I would like to calculate the Signal-to-Noise ratio (SNR) of long-slit or integral field unit (IFU) observations of H$\alpha$ emission. I can calculate the SNR or each individual spectrum, just fine. However, how can I assign a single SNR value to an observation? Do I just quote the higher SNR or do I calculate the median/mean SNR of all spectra? Is there a convention for this? My data is low-redshift galaxy IFU observations, and thus the SNR is higher in the center of the galaxy and lower at larger radii.

Unless you in fact find a "standard" definition for the mean or collective SNR, there is no way to answer this.  As you noted, the signal strength varies with view angle (as well as wavelength), so there's really no single value.  If you have a defined algorithm which processes all your spectral data to produce a final image or analysis value, then you can use standard statistical techniques to calculate the SNR of the output as a function of the SNR of all the contributing factors. (Here "factor" means each independent input variable, i.e. spectral line).  I recommend getting a copy of Bevington to learn how to do this.   Note- that appears to be a not-necessarily-legal-copy; you can purchase the real thing anywhere, e.g. amazon


Could science be lost if a phenomena is observed before predicted? An observation could validate the predictions made by a previous theory. If something unpredicted is observed, then a new theory which is compatible with the observation should make predictions which later are confirmed by new observations.

But what if no such new predictions are possible to make? What is the scientific status of a theory which after the fact is only compatible with all data, but which cannot make any new predictions? If it would be considered weak, does this mean that discoveries through observation can actually destroy science?

Is this a valid argument for delaying the construction of better observatories until theories have matured given all existing data? Kind of comparable with the forward planetary protection argument which says that human spaceflight to Mars should wait until it is certain that there's no life there which could go extinct when it meets Earthly life.

The question leaves me confused on how to interpret it.  As some of the comments have noted, observations have routinely predated a theory that (accurately) predicts them.  In astronomy and cosmology this has tended to be the main driving force behind seeking new theories.  Yet I sense a more subtle question concerning how scientists would view a theory that predicts everything so far, but nothing new.

Ultimately I think there is a confusion concerning the typical mantra for adopting a new theory as the main theory for physics/science/whatever: if a new theory doesn't make any new testable predictions, and so can only reproduce what existing theories already predict, then preference is given to simpler and existing theories.  This has nothing to say about a single theory.  It's strictly about multiple theories being compared to each other.  It does not imply that a theory, viewed entirely on its own, that makes no new predictions is undesirable.  It simply means if we're going to replace theory X with theory Y, then theory Y better predict more stuff than X does.  It doesn't care if X does or doesn't predict testable things that have yet to be proven/disproven.


Minimum telescope needed to see the rings of the Saturn I am a total novice. My daughter, likes Saturn.  Since Saturn is so close I thought it was a good time to view it with her.
What are the minimum needed telescope specs?
Update: Can answer answers be put in terms of magnification? Or the specs?

Large binoculars are already enough. If you have some at home, try them. At our recent public outreach event, we had a 20x magnification on them.

You need a relatively heavy mount though, or at least have an attachable weight to a tripod. This is needed in order to damp shaking of your mount, which can severely impact the viewing experience. It would be not unusual for the mount to be more expensive than the binoculars.
Also the connector mount-to-binoculars should be some kind of smart design, because you might want to scan around the sky, i.e. change the binoculars orientation by a few degrees every so often during gobservations.

However if you don't want to use binoculars, then telescopes with ~ x100 magnification are a safe bet for bright objects. If you want to observe faint objects, you have to think also about the telescope diameter, field of view,...
For telescopes there is also the issue that small children often cannot properly view through the one-eyed telescope pieces, or are unable to process the instructions like "keep your eyes two fingers from the viewing thingy", and then report seeing nothing. This is considerably easier for kids with regular binoculars.


Is there a paper on galaxy mergers in clusters of galaxies? I'm looking for something to cite about galaxy mergers in clusters of galaxies.

I need numbers on the frequency of mergers in the outer parts of galaxy clusters vs. the cluster center. In the center number densities are higher but so is the velocity dispersion.

Are mergers of galaxies more likely to occur in the outskirts or the central region of galaxy cluster?

Galaxy mergers in clusters explain the large central galaxies in clusters.  In fact, the centers of clusters often host a special type of galaxy called a cD which is usually extremely massive because of galactic cannibalism.  Dynamical friction makes galaxies spiral down toward the center where they get eaten by the central galaxy and that in turn makes the central galaxy more massive and more efficient at this process.

In the rest of the cluster, the velocity dispersion is significantly higher than the escape velocities from individual galaxies.  So a) galaxies cannot gravitationally attract other galaxies because they go by too fast and b) even if there is a head on collision by sheer coincidence, they would pass through and not capture one another.  There may well be huge consequences to both galaxies, such as stellar rings and shells formed and tidal tails extracted, but the two remnants would separate and perhaps never meet again.

This is an oversimplification though because clusters often have subgroups of galaxies which have lower velocity dispersion wrt each other and these could merge much as galaxies in groups merge.  But the tidal field of the cluster actually works against this and so the rate would be lower.

There is a lot of literature on cD or central galaxies since it is one of the standard types of galaxies. Here is a recent one:

"Growth of brightest cluster galaxies via mergers since z=1", Claire Burke and Chris A. Collins, 2014, MNRAS, v. 434, p. 2856.
http://mnras.oxfordjournals.org/content/434/4/2856

A study of merger rates in the outer parts of clusters may take a major effort of going through the results of N-body simulations on your own and your answer may depend on which simulation you use.


Radial Velocity Clarification Sought According to Google, radial velocity is "the velocity of a star or other body along the line of sight of an observer".

The radial velocity of Sirius according to Google is    −5.50 km/s.

If the number is positive, does that mean it's moving away or towards us?

If you are referring to the Earth...

If the number is positive, the star is moving away from us. When the number is negative, the star is moving towards us.

Hope that helps!

Source with more info: https://cseligman.com/text/stars/radialvelocity.htm


Has the Earth-facing side of the Moon that we see today always faced us ever since the Moon got tidally locked? Or does it precess? Title. Does the Earth-facing side of the Moon slowly precess due to perturbations and torques exerted by other bodies? Or is the side of the Moon we see today the same as when it first got tidally locked?
P.S. I know we don't see the exact same side of the Moon   everytime due to libration. That's not what I'm asking.

The Earth-facing side of the Moon doesn’t change currently due to any of the forces on it by other bodies (excepting librations).
From Gladman et. al.: “tidal dissipation in the satellite drives it to a state where […] the spin axis of a satellite in a generalized Cassini state will appear stationary.”  The $S_2$ Cassini state is one of the two possible eventual stable states for a general moon system.  But, according to Gladman, our Moon has "$S_2$ as the only possible Cassini state".
Before achieving a Cassini state, the tidal forces exert an average torque of $-SC\cos\theta\sin\theta$, where $\theta$ is the obliquity of the moon's orbit measured with respect to the orbit normal, $C$ is the satellite’s moment of inertia about the spin axis, and $$S=\frac{3Gm_p}{2r^3}\frac{C-(A+B)/2}{C}$$ where $m_p$ is the mass of the planet, $G$ is the gravitational constant, $r$ is the orbital radius of the satellite, and $A$ and $B$ are the other two principal moments of inertia.
Note that $S$ has $r^3$ in the denominator (also as pointed out by Stuart Robins in a comment above), which means the torque exerted by tidal forces was much higher when the moon was closer to the Earth.  In other words, since the Moon’s face towards the earth is not precessing now, it certainly wouldn’t have in the past when the potential tidal torque was higher.
I think Gladman et. al. give pretty credible evidence for their claim of tidal lock, in that they account for wobbles, highly eccentric orbits, and oblate spheroids in both the planet and the moon.  They don’t specifically account for the gravitation of other bodies like the Sun.  The above equation also gives us a comparison of relative tidal effects on the Moon by noting $m/r^3$ is 9.2e7 for the Earth/Moon system, but only 5.9e5 for the Sun/Moon system. Gravitational perturbations from the Sun and other solar system objects affect the Moon's orbital elements, but don't have a significant effect on it's rotation rate.
There is plenty of lunar geological evidence supporting Gladman et. al.’s claim that the Cassini state is stable, in that we see a fundamental asymmetry between the near and far faces of the Moon.  The near face has a thinner crust and more volcanic maria.  The far side has a thicker crust and more visible asteroid craters:

Roy et. al. claim the asymmetry between the near and far faces is due to heat from the Earth causing the early tidal locked Moon to cool more slowly on the lunar face pointing to the molten cooling Earth. If their theory is correct, then the Moon's face towards Earth hasn't significantly changed since the Earth/Moon system infancy.
The near face of the Moon doesn’t appear to be precessing currently, but could it have changed orientation from its original tidal lock configuration? There is a strong suggestion from Kadono, that an asteroid impact could have an effect on spin rate.  Unfortunately, the paper is behind a paywall, so lets do a "back of the napkin" calculation to see how much angular momentum a big asteroid impact could have on the moon.
The largest known basin on the Moon is the South Pole-Aitken basin. The size and speed of the impactor was estimated by Potter et. al. One of their models assumes a chondrite impactor with density $\rho$ of 3580kg/m3 at 100km radius $r$ at an impact velocity $v$ of 10km/s. We can calculate the mass $m=\rho 4/3\pi r^3$ or 1.5e19kg. The equation for linear momentum is $l=mv$ or 1.5e20km*kg/s.
The moment of inertia of a sphere is $I=2/5MR^2$ and we can use $M=7.342e22kg$ and $R=1737.4km$ as the mass and radius of the Moon to get $I=8.86e28$.  If we set the angular momentum of the Moon $I\omega=1.5e20$ and solve for $\omega$ giving us a resulting rotation rate change of 1.69e-9 radians per second or about 3 degrees per year. These calculations assume an impact angle parallel to the lunar surface, which is impossible, but forms an upper bound for a momentum change.  Higher angle impacts still conserve momentum, but more of the momentum goes into changing the orbital parameters rather than rotation rate.
From When did the Moon become tidally locked to Earth?, we know the tidal locking time is $$t_{\mbox{lock,Moon}}=7.12753\cdot 10^{-25}wa^6 \frac{\mbox{kg}}{\mbox{Nm}^2 \mbox{km}^3}.$$ where $w$ the spin rate in radians per second, and $a$ the semi-major axis of the Moon orbit.
I calculated the above equation, supposing the Moon's orbital semi-major axis was less than half it’s current value so long ago. If the moon got struck in the way presented above, it would tidal lock again in less than 2 years, rotating less than 6 degrees in that time.  So, no, after the Moon was first tidally locked, the same face almost certainly always pointed to the Earth.
Notes:

It's possible that the asymmetries in the Moon's composition would decrease the time to tidal lock, but it isn't needed for any of the above calculations. Even if the Moon's composition was homogeneous, tidal locking would still occur.

There are many other theories for the lunar face asymmetry, but most of them rely on a steady tidal lock through the early part of the Earth/Moon system.  An exception might be Elardo et. al. who claim the asymmetry is due to a greater abundance of radioactive isotopes on the near side of the Moon, which add heat and lower material melting temperatures.  They call the mix KREEP.

We can't completely eliminate the possibility that the Moon was hit by an even bigger asteroid then referenced above, with the evidence later being covered by volcanism.  This answer is only based on currently available evidence.


Why can't we feel the Earth's revolution? I googled it and checked a few Q&A and there's only things about "Earth's rotation". But why can't we feel the revolution?

They say we can't feel the rotation because the Earth spins at a constant speed. Okay, I get what happens for the rotation, but isn't it different when it comes to the revolution?

At night we feel the sum of speed of green and blue, whereas at day we should feel the sum of speed of green and minus blue, shouldn't we? In other words, shouldn't we feel the changes in velocity by times?

You do, but it's too small to really notice

First, it's not correct to say that we don't feel Earth's rotation because it's rotating at a constant speed.

Think about driving a car, or riding in an airplane.  Whether you're cruising down the road at 90 kph, or soaring through the air at 900 kph, you don't really "feel the speed".

However, When you take a sharp turn, or take off from the runway, you definitely feel something.  That's the acceleration.  It doesn't matter if your speedometer stays steady - if you take a sudden 90 degree turn, you're going to feel it.

More relaxed turns, such as going through a roundabout, or when the airplane circles the airport before landing, are much less likely to spill your drink.

Even if Earth is spinning at a constant speed, the spin is a change in direction, which requires acceleration.

Acceleration is quite noticeable, depending on its magnitude.  Even just sitting down, you can feel the pull of Earth's 9.8 m/s² gravity - your body's "weight", as it were.

So how large is the acceleration keeping Earth in orbit?  About 0.0059 m/s².
What about the acceleration of Earth's rotation?  A ever so slightly larger 0.0339 m/s².

Small wonder that it seems like you can't feel these forces!


When the moon eclipses the sun what direction is darkened first: left or right? I saw the question and answer for the explanation of the direction of the moon's eclipse from left to right, but what about the direction of the moon as it passes over or between Earth and the sun?

Does the moon always pass from the right to the left or from the left to the right as it covers the sun; if you live above the equator?

Or does this change? Can it change? August 2017 the USA will have a total eclipse of the sun for the first time in decades.

Relative to the Sun, stars, and Earth's shadow, the Moon always appears to move from west to east.
How this looks relative to the horizon depends on your latitude and time of day.
If a solar eclipse occurs in the middle of the day, the Moon appears to move from right to left if you're in a northern latitude, or from left to right if you're in a southern latitude.
If you're in a tropical latitude, the Moon appears to move downward if the eclipse occurs in the morning or upward if in the afternoon.


What is the least dense exoplanet? An exoplanet with density 0.31 grams per cubic centimeter has been found. Is this the least dense exoplanet we know of?

The article you link to refers to Borsato et al. 2019, which attempted to rectify discrepancies in the measured properties of planets in the Kepler-9 system between transit timing variation measurements and radial velocity measurements. They arrived at $\rho\sim0.31^{+0.05}_{-0.06}\text{ g cm}^{-3}$ for Kepler-9c. However, Borsato et al.'s Figure 10 shows that there are other exoplanets in this mass regime with substantially lower densities, e.g. WASP-107b, which comes in at about $\rho\sim0.19\text{ g cm}^{-3}$:

Even WASP-107b, however, doesn't hold the record for the least dense exoplanet. The three planets in the Kepler-51 system, Kepler-51b, Kepler-51c, and Kepler-51d, may hold that record. Multiple groups (Masuda 2014, Roberts et al.) have found densities of around $\rho\sim0.03\text{ - }0.06\text{ g cm}^{-3}$ for both planets.


Is the equinox Sun always at the same zodiac sign in the ecliptic belt? During the year the Sun moves along the ecliptic passing all the signs of the zodiac belt, in my understanding. This means that it takes ~1 month to go from one zodiac sign to the next.

The equinoxes are approximately on 20th of March and 23rd of September. This means the vernal equinox Sun is in Pisces (or Aries) and the autumnal equinox Sun is in Libra (or Virgo).

But also I found that there is the observation of precession of the equinoxes, where the equinox Sun passes through all zodiac signs in a calendar round that lasts ~25,000 years.

How is this possible?

If my understanding about the Sun's yearly round across the zodiac is correct, the equinox Sun should always be at the same zodiac signs. How can this precession happen?

As you said, the Sun takes a year to cycle the Zodiac, and it takes a year between two spring or autumn equinoxes. Both statements are true, but "year" in them means two different things. It takes a sidereal year to cycle the Zodiac, but it takes a tropic year to cycle between equinoxes. Since the tropic year is about 20 minutes shorter, the equinoctial points move across the Zodiac cycling it in about 25,000 years.

The calendar year we use in the Gregorian calendar is designed to approximate the tropic year to have it synchronised with seasons.

The reason for all it is that the Earth's axis is not fixed but its orientation slowly rotates.


How does the solar wind destroy the atmosphere of a planet lacking a magnetic field? It is said that because Mars doesn't have a magnetic field its atmosphere disappeared, but how can solar wind cause that?

That the protons and electrons can changes molecules in an atmosphere I can understand, but how can solar wind cause molecules to escape from gravity?

Or is it that the atmosphere didn't escape in outer space but in contrary the molecules became too heavy, so they got solid and were pulled to the planet's surface?

What determines how long a planet will keep its atmosphere is the strength of the planet's gravity versus the random thermal motion of the molecules in its atmosphere.

The atmosphere is a gas, and all gasses have molecules that move around with velocities that are dependent on the temperature.  High-temperature gasses move faster, and low-temperature gasses move more slowly.  Each molecule of gas will have a velocity $v_{\text{gas}, \text{average}}=\sqrt{\frac{3k_BT}{m_{\text{gas}}}}$.

In addition, each individual molecule is pulled to the Earth by gravity.  Heavier gasses, such as water, carbon dioxide, and oxygen, are pulled more than lighter gasses, such as hydrogen and helium.  The escape velocity of the planet is given by $v_{\text{escape}}=\sqrt{\frac{2GM_{\text{planet}}}{R_{\text{planet}}}}$.

If a gas molecule has a high enough thermal energy, then if it is at the top of the atmosphere, it could fly off into space, the thermal motion overcoming gravity's pull.  The lighter it is, the less thermal energy it needs to have.  Of course, this all depends on probability and statistics, but, in general, if $v_{\text{gas}}\ge0.2 \times v_{\text{escape}}$, that species of gas will be reduced to 1/2 its initial amount after 1 billion years.

Solar wind consists of charged particles and high energy photons, which are deflected by a planet's magnetic field.  However, in the case of a planet like Mars, without a magnetic field, the particles interact with the atmosphere.  They tend to split molecules apart, resulting in lighter gasses (a process called "photodissociation").  These gasses, being bound less strongly by gravity, are more free to fly off into space under thermal motion.  As more and more gas leaves the planet, the atmosphere becomes thinner and thinner until, ultimately, only the heavy gasses are left.  In addition, when a molecule is split, the energy imparted to it by the impact may be enough to eject it from the atmosphere (a process called "sputtering").

Planets (and moons) that are closer to the sun are subject to more intense solar radiation, and also have higher atmospheric temperatures.  This is why distant moons, such as Titan, have thick atmospheres of hydrocarbons.  Planets close to the sun, such as Venus, can retain their atmospheres for a variety of reasons.  Firstly, Venus's atmosphere is primarily carbon dioxide, which is much heavier than Hydrogen, and more difficult to split via photodissociation than water.  Planets like Earth also have hydrogeological cycles which can recycle gas from rocks and water back into the atmosphere.

For more information on this, read up on Jeans Escape.  A good set of notes for beginners can be found here.


highspeed black holes or neutron stars on (almost) head-on collision course and kinetic energy First: I don't think this is a duplicate question. There are related question on mergers of black holes that orbit around each other etc., so please read on.

I'm having a hard time picturing what might happen in the following scenario:
Assume two black holes of the same size on an almost frontal collision course with a very high relative speed of x km/h (replace x with whatever would be necessary to enable the scenarios described below).

If their event horizons don't "touch" I'd assume their courses will change considerably but it would be a normal flyby - right ?

But what happens if their event horizons "touch", if their event horizons "overlap" just a few meters ?

On the one hand both objects have a massive kinetic energy with opposite directional vectors. And on the other hand nothing can escape below the gravitational radius.

Where does all the kinetic energy go ? Would the two black holes extremely deform, like creating a long string before they collapse into spherical form?

And what if in a similar scenario one of the two objects is a neutron star. Would the neutron star be ripped apart when it scrapes the black hole on high-speed flyby ?

I think it's a fun question, and I can answer some of it.

  If their event horizons don't "touch" I'd assume their courses will
  change considerably but it would be a normal flyby - right?

While the relativistic speeds and time dilation might change things a bit, a flyby is still a flyby. They would either fly past each other, the dense objects would change direction as a result, but the combined kinetic energy and momentum would remain constant.

Second option is they enter into orbit around each other.   Pretty much the same with any two massive objects passing near each other.

  But what happens if their event horizons "touch", if their event
  horizons "overlap" just a few meters?

With classic (non Kerr) black holes, if the event horizons touch the black holes have no option but to merge.  The event horizon is directly proportional to the radius, so when the two event horizons touch it, as that happens, no longer two black holes but one black hole.   The mass of the two objects combined has an event horizon that circles both of them.

For Kerr black holes this question is more complicated and I'm not sure but I think it's the same. Once the event horizons touch it's no longer two separate black holes but one bigger one.

  On the one hand both objects have a massive kinetic energy with
  opposite directional vectors. And on the other hand nothing can escape
  below the gravitational radius.

When you're talking about event horizons, gravity always wins.  Kinetic energy cannot exceed the speed of light, so it loses.

  Where does all the kinetic energy go ? Would the two black holes
  extremely deform, like creating a long string before they collapse
  into spherical form?

A long string doesn't make any sense.  A temporarily non-round event horizon as the holes merge seems possible, but think of a merger of two stretched balls, not a string.   Each black hole is centered around it's "singularity", those shapes might warp some as the holes collide or spiral into each other.  As an FYI, direct collisions would be rare.  Spiraling into each other is much more common.

The kinetic energy gets absorbed inside the black hole.  It can't, by definition of the black hole, escape.   Total momentum is likely preserved.

  And what if in a similar scenario one of the two objects is a neutron
  star. Would the neutron star be ripped apart when it scrapes the black
  hole on high-speed flyby ?

A few points to consider regarding a black hole. Anything that falls within a black hole's photon-sphere is unlikely to ever escape, so one of the magical distances to consider is 1.5 times the radius of the black hole where all orbits are destined to fall into the black hole.   If the black hole and neutron star are of similar size and the neutron star passes close to the event horizon and relativistic speed, then there's an interesting scenario where some of the neutron star (at close to the speed of light) would be outside the photon-sphere and have the momentum to escape and some of it would not.   Neutron stars are bound together very tightly, but the tidal forces would be very strong and it's possible that some of the neutron star material could get torn off in this hypothetical scenario.  Would the entire neutron star be torn apart?  Maybe.  It's hard to say.  The math in that hypothetical is way above my pay-grade.

If the neutron star is much smaller than the black hole, then it's likely inside the photon-sphere and it will spiral into the black hole, perhaps breaking apart in the process into an accretion disk, perhaps not, but it will get absorbed into the black hole if it's entirely within the photon-sphere, even at nearly the speed of light.

The problem with this hypothetical is that it's pretty much impossible.  Objects the size of stars don't accelerate at nearly the speed of light relative to other near by stellar mass objects.  What actually happens is described in the first answer to your question, the two dense objects would either have sufficient relative velocity to fly past each other or they would get caught into a mutual orbit that would likely lead to a merger.  In the actual universe, a close to the event horizon fly-by between a neutron star and a black hole would result in a merger of the two.  In theory, with enough kinetic energy the neutron star could fly close to and still fly past the black hole, but that much relative velocity probably never happens.

People smarter than me should feel free to correct anything I got wrong, but I thought this one was a fun one to answer.


Terminology of Orbits A hopefully simple terminology question:  When you have an object that orbits another object in space, what do you call the object being orbited in relation to the object orbiting?

For instance, you might call the object orbiting the "satellite" of the orbited object, but what would you call the object being orbited?

Orbitee is all i've been able to come up with but it doesn't seem quite right ;)

When one body has a much larger mass, so it's motion is negligable, it can be called the central body (even for eccentric orbits, where the central body is a the focus of the ellipse, and far from the centre)

This is the terminology used by Wikipedia editors

  The central body in an orbital system can be defined as the one whose mass (M) is much larger than the mass of the orbiting body (m)...

Also by Glasgow university

  The most straightforward orbit calculations occur when the central body is much more massive than the orbiting body.

and elsewhere

  The angular momentum of a satellite depends on both the mass of orbiting satellite and central body as well as the radius of the orbit.

Normally, in a specific situation "planet", or "star" can be used.


Why there is no axis measurement when telling distance between objects in space ? What's in upper part in our galaxy ?
If space is 3 dimensional, what lies beneath solar system, or in upper part ? Why it seems from every graphical view of solar system / galaxy, everything is in one single axis ?

You might benefit from watching this Minute Physics video on Why the Solar System is Flat. A summary is due to the way objects in our solar system and particular galaxy are drawn around a central massive item by gravity. Through the conservation of this angular momentum, and the loss of vertical motion though collisions, the solar system and galaxy "look" flat.

Talking of "upper" and "lower" doesn't really make sense in space. Gravity feels like it's pulling us down on Earth, and that is why we have this concept, but in space we lack this reference. For our own purposes, humans have created an artificial means of determining points of reference for spacial coordinate systems. Due to the rotation of the Galaxy we have established a base plane called the Galactic plane, from which a coordinate system and sense of "up" and "down" might be inferred (I think our solar system tilts at a 63 deg to the galactic plane of the Milky Way). In the context of Earth and our solar system, we use the plane that the Earth "makes" as it circles the Sun as the point of context for the Earth's axial tilt; we call it the ecliptic plane.

As for the rest of the Universe you may wish to check out this blog post from Cornell called Why is the Universe flat and not spherical?. Basically the Universe has a slight curvature to it but since we are on the surface, and have a limited view, it is difficult for us to see anything but flatness. This is related to why graphical views appear flat. Graphical views are generally limited due to our point of view and the sheer astronomical size of distances in space. Even in "dense" super clusters there are vast distances between objects. Overall it's easier to linearly show things rather than as they truly might be.


Is it practical to build a giant telescope in Moon? (Considering the fact that the atmosphere of moon is very rare) China's FAST (Five-hundred-meter Aperture Spherical radio Telescope) is the largest radio telescope in the world.(This is huge!!!)

As we all know Space Telescopes are better than Earth-Based telescopes always.(Hubble space telescope was such a tremendous success)Even though ground-based observatories are usually located in highly elevated areas with minimal light pollution, but still they contend with atmospheric turbulence, which limits the sharpness of images taken from this vantage point.

What if all the space agencies collaborate together to build a telescope half the size of China's FAST in moon(Even smaller than half),even then it will be more powerful than the "FAST" right(because the atmosphere of moon is very less dense)?Will it be impractical?It may take decades to build one but still the outcome will be huge(We may even look and find aliens!!)

(Taking in account that Elon Musk is already planning Colonization of Mars)This project looks somewhat tamable than that,right?  Is there a possibility of this happening? Is it feasible enough?

It's very unlikely that large optical telescopes will ever be built on the Moon, because the Moon is almost the worst possible place to build them. (The surfaces any of the  planets other than Earth are worse.)  It has no particular advantages over orbit and costs a lot more to build there.

The Moon looked like a good location when observatory technology was film-based, because observatories required people and people needed a base to live in and work best under gravity.  But we now have the technology to shade an orbiting telescope from the Sun, point it exquisitely accurately, and take endless photos without changing the film -- all in orbit.  Orbit is easier to get to than the Lunar surface -- even the Webb (should it ever launch) which will be parked a million miles from Earth, is easier to get to than the surface of the moon.

If we had the space capabilities to build a giant reflector on the Moon, we could build an even bigger and better one in orbit for the same cost.

You spoke of FAST. FAST is a radio telescope. It's likewise much cheaper to build radio telescopes in orbit than on the Lunar surface. The one possible advantage of the Moon is that the far side of the Moon might have advantages of being especially radio quiet.

But even there, it seems likely that any space radio telescope would be a large array of radio telescopes joined to form a giant synthetic aperture, and that would almost certainly want to be much larger than earth-based instruments -- for which the Moon is already too small.


Rate of Mass Loss from the Solar Wind This is problem 1-4 from Principles of Stellar Evolution and Nucleosynthesis by Clayton:

Assuming at the Earth a characteristic velocity of 400km/s and density of 10amu/cm$^{3}$ for the solar wind, calculate the rate of mass loss for the sun.

There weren't any formulas about this in the section so I made a stab at it with dimensional analysis.

$$\frac{dM}{dt} = \frac{\rho V}{\Delta t} = \rho v A$$
$$\frac{dM}{dt} = \left( \frac{10 amu}{cm^{3}} \right) \left( \frac{400km}{s} \right) \left( \frac{4 \pi (6.96e10 cm)^{2}}{1} \right) \left( \frac{10^{5}cm}{km} \right) \left(\frac{10^{-24} g}{1 amu} \right) \left( \frac{M_{\odot}}{2 \times 10^{33} g} \right) \left( \frac{3600s}{hr} \right) \left( \frac{24 hr}{day}\right)\left(\frac{365day}{yr}\right)$$
$$\frac{dM}{dt} = 3.84 \times 10^{-19} M_{\odot} / yr $$

However, the answer given in the book is $0.4 \times 10^{-13} M_{\odot} / yr$. So, I'm off by about five magnitudes. Can anyone point out where I went wrong and/or point me in the correct direction?

$$\left( \frac{4 \pi (6.96e10\,\text{cm})^{2}}{1} \right)$$

This is the primary source of your error. Your value of 6.96×1010 cm is the radius of the Sun. The problem specifically said "Assuming at the Earth ...". You need to calculate the flux through the surface of a sphere whose radius is about one astronomical unit rather than one solar radius. The astronomical unit is 149597870700 meters (exactly), or about 1.5×1013 cm. This error alone makes your value low by a factor of about 50000. The remaining factor of two results mostly from using 10-24 grams per amu.

Dimensional analysis can only take you so far. While your result is dimensionally correct, you didn't think enough about the nature of the problem.


Would a black hole passing next to a star create a deadly focal point due to gravitational lensing? Black holes bend all kinds of radiation thanks to their giant gravity. Now imagine that a black hole passes close to some star (we know of cases when black hole devours a star). Star emits massive amounts of radiation in all directions, so naturally the radiation just outside the event horizon will be bent by the black hole. Many rays will be focused into a single point.
Wouldn't that create a "certain death zone" on the opposite side? Could this be a real threat for celectial bodies from far-away black holes (for example for earth)? If this effect in fact happens, would different radiation types be focused farther away - for example UV, visible, X-Ray etc. in it's own zones (EDIT: meaning basically - do black holes work similar to prisms by lensing different types of wavelengths differently)?

In the draft above the yellow circle is a star, black dot is black hole, gray circle is event horizon and red circle is the "death area where rays are focused by the black hole

tl;dr: seen at a distance, the lensed object will appear as an annulus or "ring" around the lensing object, and while that will be brighter than if there were only empty space, sadly it won't be death-ray bright!

Let's first think about what makes a familliar lens a lens. Near the center the thickness doesn't vary much, but as you move farther from the center the thickness changes faster and faster.
If we measured the slope or angle of the surface, we'd see that the angle increased roughly linearly with distance from the center.
In the thin lens approximation the angle that the lens bends the light $\Delta \theta$ is proportional to the distance from the center of the lens $r$ where the light hits it. The bending power increases linearly with distance.
$$\Delta \theta \approx \frac{r}{f}$$
where $f$ is the focal length of the lens.
How do concentrated points of mass bend light?  Wikipedia's Gravitational lens give us
$$\Delta \theta \approx \frac{4 G M}{r c^2}$$
and that's a problem because now $r$ is on the bottom!
Single concentrated objects like black holes do not act like the familliar lenses we use to focus. They do have some ability to concentrate somewhat compared to empty space, but nowhere near as much as real lenses do. From a source that's a certain distance away, there will be only one angle that gets bent parallel into a "beam", angles slightly larger or smaller passing slightly farther or closer from it will be bent much less or much more, either diverging or converging from the axis afterward.
Thus seen at a distance, the lensed object will appear as an annulus or "ring" around the lensing object, and while that will be brighter than if there were only empty space, sadly it won't be death-ray bright!
In the case of your star/BH pair, exactly how bright it will be along that axis depends on the details, but one can get a good estimate by ray tracing, either with a few lines of Python or pencil and paper for estimate purposes.
If the star were close then it is an extended object, a wide disk, and those can't be focused to points even by lenses (we can't concentrate blue sky with a magnifying glass) so there wouldn't be much of a death effect. If it were far, you could concentrate it better (because it's more point-like) but it would be a lot dimmer to begin with because it would be farther away.

A Horseshoe Einstein Ring from Hubble  The bright thing in the middle is the lensing mass, the ring is the lensed object behind it. It's not a "beam" but just a distorted view, but it is brighter than if the lens weren't there.


Collision with Andromeda Galaxy According to Hubble's law, the universe is expanding and this expansion is proportional to distance between the two objects and thus objects are moving away at a very fast rate,  then why is it predicted that after 4 billion years the Milky Way shall collide with Andromeda galaxy?

As you say, expansion is proportional to distance. And in fact, compared to the size of the (observable) Universe, Andromeda and our Milky Way are close together. Indeed, farthest objects are now more than 20 billion Ly from us, whereas Andromeda is only 2.5 million Ly away. That is a 1/10000 factor.

The space expansion between them is not enough to set them apart, because of their general motion. On small scale (until galaxy cluster scale), the motion is primarily ruled by gravity. Galaxy cluster are in fact the largest gravitationally bound systems we know.

  They typically contain no more than 50 galaxies in a diameter of 1 to 2 megaparsecs

Since Andromeda is the closest galaxy, it is inside "our" galaxy cluster, and stays bound to ours by gravity.


How many moons in the solar system have water? I've heard of a few moons in the solar system that have water, like europa and enceladus. However, I can't find information about whether there are other moons as well. I'm guessing there are more, but I'm not totally sure.

How many moons in the solar system have water (either liquid or frozen)?

It's probably easier to consider the reverse question: which ones don't have water.

Talking about the major satellites of the planets, there are only two which are not ice moons: our own Moon, which has ice deposits at the poles, and Io. As far as I am aware, the evidence for water on Io is tentative at best (e.g. Douté et al. 2004), so of the major planetary satellites, the answer is all of them have water (usually in the form of ice, but some of them also have liquid subsurface oceans) with the possible exception of Io.

The small satellites of the giant planets are probably icy but I'm not sure the possibility that one or two of them are dry rocky asteroids that have been captured has been fully excluded. Phobos and Deimos have rocky surfaces but the possibility of ice in the interior of the Martian moons has not been ruled out.

There are probably lots of ice-free asteroid moons but finding out which ones they are would require going there and checking whether there's any ice in the subsurface, so confirming this is not currently possible. Small solar system bodies in the outer system are probably going to be icy, so dwarf planet satellites (Charon, Vanth, etc.) all likely contain ice.


Line of lights moving in a straight line, with a few following My husband and I have just seen a line of about 15 lights moving in a straight line, quite high above the cloud level with a gap, and then 5 lights, a gap, another few, a gap, 1 light, a gap, and a few more.

They were moving at the same sort of speed as I would expect a plane to move at - almost like a train in the sky with a banner flying behind. They were in the sky for at least a minute.

I had seen a shooting star a few minutes before and it wasn’t one of them... what could it be?

Could it be Elon Musk’s satellites?

Given the date and timing, this could be most like the Starlink satellites in their "stacked" configuration. They are currently in a line, but they will later move to separate orbits. Dr Marco Langbroek has a video of a line of 60 satellites.

They won't stay in a line; they have ion engines that they will use to spread out into separate orbits.


From color index to luminosity Given two colorindices LX and LY, such that the color X-Y is given by

X-Y = -2.5 log(LX/LY)

is there a way to infer the luminosity in X (or Y)? I'm think not because any information on units has fallen out through this definition, but I could be wrong. I'm just not sure how to compute the luminosity straightforwardly otherwise, as this color index is the only output (along with the distance modulus) from an interpolation scheme of stellar templates.

Help would be much appreciated.

You can never tell the luminosity of a star from knowledge of its color X-Y alone, that only tells you something about the surface temperature.  The problem is not solved by knowing the distance modulus, that only tells you how far the star is, but not how big it is.  If you want luminosity from those things, you need to know how bright the star looks, not just its color and distance.  Or you could look at the width of the spectral lines to infer luminosity class, but that brings in line resolution.  But if you know the distance modulus and Lx and Ly, not just Lx/Ly, then you can get the luminosity because you know its color, how bright it looks, and its distance.


What exactly is "the rotating lighthouse model" in the context of a double pulsar? The introduction to Testing the rotating lighthouse model with the double pulsar system PSR J0737-3039A/B (open access) says:

The double pulsar system PSR J0737-3039A/B was discovered by Burgay et al. (2003) and Lyne et al. (2004). This system consists of a 22-ms pulsar (hereafter A) and a 2.8-s pulsar (hereafter B) with an orbital period of 2.4 h. This discovery has provided a laboratory for the study of relativistic gravity and gravitational radiation (Kramer & Wex 2009). The system has several strange features that challenge the current understanding of pulsars and provide an uncommon opportunity to improve pulsar theories. One of the most interesting properties is the observed modulation of each pulsar's signal by the energy flux from the other, as evidenced by each pulsar's modulation period being approximately equal to the other pulsar's pulse period (McLaughlin et al. 2004a,b). Freire et al. (2009b) proposed a technique for analysing the arrival times of the pulsars’ pulses and their mutual modulations which could yield the sense of rotation of each pulsar with respect to its orbital motion, among other quantities.** In this paper, a complementary technique is presented with the same objective, but using measured periods rather than arrival times. The principal benefit of our approach is that it is simpler and more intuitive. If the validity of either of these techniques is confirmed, not only will new insights be gained but also the correctness of the lighthouse model will be empirically assessed beyond dispute.

I think there is a lot going on here, more than I can take in all at once. So I'd like to simply ask:
Question: What exactly is "the rotating lighthouse model" in the context of a double pulsar? Modulation is mentioned in this and the other papers, and an interaction with each others' magnetospheres is suggested to explain it, but is that part of "the rotating lighthouse model"?

It seems that the authors are just referring to the accepted model of a pulsar, i.e. a neutron star spinning and emitting beams of radiation at its poles. In that sense, the term is used here no differently than it would be in the context of an isolated pulsar. It's just a very simple way of visualizing why an observer far away appears to see periodic pulses rather than constant emission.
If the lighthouse model is correct, then the beams from one pulsar may interact with the other pulsar, which in turn would cause changes in the second pulsar's emission each time the beam sweeps across it. The modulation period would then be approximately equal to the rotational period of the second pulsar - not precisely, though, because of orbital motion; hence, the comparison between Earth's sidereal day and solar dat.
Modulation of pulse strength and features therefore isn't a feature of the lighthouse model but rather a consequence of it, present in some binary pulsar systems but of course not in isolated pulsar emission.


Where is the materials of comet after it crashed into the sun? If a comet crashes into the sun, where will the substance (eg: water,silicon) be? Will they be inside the sun? Or they will be ejected from sun again?

It depends on the size of the comet. If it's only $0.1$–$1\,\mathrm{km}$, it will evaporate before even reaching the Sun (Li. et al. 2013; Knight & Walsh 2013), and the released gasses will be ejected by the Sun's radiation pressure. Larger comets, so-called Sungrazing comets, can survive close passages, although they probably will be torn apart by tidal forces.

If the comet comes so close that it crashes into the Sun (I'm not sure this has ever been observed, despite various blogs discussing this, but there's no reason it couldn't happen), it would reach the escape velocity of the Sun, i.e. $\sqrt{2 G M_\odot / R_\odot}\sim600\,\mathrm{km}\,\mathrm{s}^{-1}$. This New Scientist article claims that the impact would cause an explosion and possibly a coronal mass ejection, but the underlying scientific article by Brown et al. (2015) seems to indicate that the large ablation of the comet will result in less dramatic velocities.

Anyway, some of the released gas would stay at the surface of the Sun, just like some of the Sun's "own" gas stays. And since the convection cells, or granules, are quite shortlived (of order 10 minutes; Bahng & Schwarzschild 1961), the gas will soon be engulfed by the Sun, being carried deeply into the body of the Sun.

The convective zone of the Sun is in the outer ~30% of the radius, and there isn't much mixing of material between that and the inner, radiative zone. In the last phases of the Sun's life, however, when core hydrogen burning stops and the Sun becomes a red giant, convection reaches so deeply into the Sun that some mixing takes place. This is called dredge-up. Eventually, roughly half of the Sun will be ejected due to the large mass-loss during the red giant phase, while the other half becomes a white dwarf. If the dredge-up didn't happen, the comet material would all be a part of the ejected material, but due to the dredge-up, some of it will end up in the white dwarf.


Why are the moons of Neptune sorted by diameter? The five moons closest to Neptune are sorted by increasing size. Naiad < Thalassa < Despina < Galatea < Larissa. The trend continues with Proteus and Triton, only interrupted by Hippocamp, which is smaller than the closer moons.
Is this ordering due to pure chance? Is there a reason why moons closer to Neptune should be smaller? Is it possibly an observational bias?

Likely there's two different explanations, which can even work in conjunction:
a) An explanation along the same as goes for the size sorting of the planets in the solar system. For those regular moons which formed along with Neptune or in orbit of the protoplanet we see that the further outward the moon forms in the accretion disk around proto-Neptune, the more material is present due to the greater circumference of the orbit. Thus it can grow larger in size.
b) Neptune exerts tides onto the moons. Typically no moons can exist within the Roche limit. How close a moon can exist depends on the rigidity of the body, though. Big moons (mass $M_m$) cannot exist as far inward (distance $d$ to Neptune) as a small moon (Neptune's radius $R_N$ and mass $M_N$):
$$ M_m = \left(\frac{R_N}{d}\right)^3\cdot2M_N$$ or easier expressed as ratio of the moon's radius $r_{moon}$ and the moon's distance to Neptune $d$ as well as its density $ \varrho_{moon}$: $$ \left(\frac{r_{moon}}{d}\right)^3\cdot\varrho_{moon} = constant $$
Thus the further out, the larger the moon can be without falling apart due to tidal forces. Wikipedia has a page on Neptune's moons and references which indicates that the inner moons are somewhere at the tidal limit for bodies which are of somewhat granular nature as is expected for at least the smaller moons or maybe even all with the exception of Triton.


What will "the Earth being consumed by the Sun" look like? Sometime soon (astronomically speaking) the Sun will, as a main sequence star, expand into a red giant and, potentially, "consume the Earth".
If we take it as read that the surface of Sol will expand past the aphelion of Earth's orbit, what will Earth inside Sol entail, physically speaking, considering Earth as a rock, not an ecosystem, both as the surface of Sol expands past the orbit of Earth, and as the stellar envelope is ejected, leaving Sol as a white dwarf?

The exact fate of Earth as the sun becomes a red giant is somewhat uncertain because of two factors.
The first is that as the sun expands, its surface gravity $GM_\odot/R_\odot$ will decrease and gas can more easily escape as solar wind. This will reduce the mass of the sun, making Earth's orbit spiral outwards. But how much expansion depends on details of the far future solar wind we do not know yet, and this makes it slightly uncertain whether Earth can get into an orbit beyond the maximal solar radius. (The gas will also produce some drag, but it is likely negligible in this case)
The second factor is tidal interactions. As the sun expands it rotates more slowly, and in particular will rotate slower than Earth's orbital speed. That means that the tidal bulge induced by Earth will always be trailing the planet, creating a backward pull slowing it and dragging it inwards. This effect is small when the surface is far away, but grows rapidly as it approaches. Hence the sun can gobble up Earth even if the orbit is larger than the solar radius - but much depends on somewhat weakly understood tidal interactions.
Once the planet is inside the envelope it will rapidly get dragged in. To get a timescale, consider the time to push aside an Earth mass of gas: $\tau = M_\oplus /  \pi R_\oplus^2 \rho v$. For $\rho=$ 0.1kg/m$^3$ I get $\tau=0.4972
$ years. Maybe a bit longer if the orbit got wider and slower. Since the density of Earth is much higher than the gas, the planet will not be tidally disrupted but instead leave a hot plasma wake (that paper has more detailed calculations of infall timescales etc.).
The shocked gas will form a plasma sheath that erodes the planet at some rate, leaving a trail of metal-enriched material. Eventually the gravitational potential energy is released as heat, but this is mixed up with the rest of the solar luminosity over maybe a million years or so.
If Earth survives the giant phase it will orbit the white dwarf core (at a distance set by mass-loss expansion of the orbit minus gas drag inspiral; about twice the current orbit if drag is small). At this point it will become an airless (due to luminosity-induced atmosphere loss earlier), frozen rock over a few hundred million years or so as it cools from earlier heating.


If earth were the size of a marble would it be smoother than a marble? As the title states, If earth were the size of a marble would it be smoother than a marble?

No.

This What-If page describes the case for 'smoother than a bowling ball"

  These scans (along with various measurements of ball roughness1 tell us that a high-end bowling ball is quite smooth. If blown up to the scale of the Earth, the ridges and bumps[2] would be between 10 and 200 meters high, and the peaks would be between one and three kilometers apart:

  By Earth standards, this is quite smooth; our highest mountains are 40 times higher.

And it is simple to continue the scaling process to show that the same holds true for marbles. (Although I have not found a good source of data from people who scan marbles...)


What's the largest non-spherical astronomical object in the universe? Some asteroids and comets are non-spherical. But is the nature of big things and gravity so that large things in the universe are always spherical? What is the biggest astronomical object in terms of volume out there that's not spherical?

Note: Astronomical object is a defined term. See Astronomical object - Wikipedia.

The largest sub-galactic astronomical object (in volume) that we know of is the Carina Nebula, which is a non-spherical diffuse nebula. The Carina Nebula has a radius of about 100 parsecs.

Image credit: ESA

If you consider astronomical objects of the galactic scale, then the galaxy IC 1101 is the largest astronomical object (in volume) that we know of. From Wikipedia:

  The galaxy has a diameter of approximately 6 million light years, which makes it currently (as of 2013) the largest known galaxy in terms of breadth. It is the central galaxy of a massive cluster containing a mass (mostly dark matter) of roughly 100 trillion stars. Being more than 50 times the size of the Milky Way and 2000 times as massive

IC 1101 is on the left in the image below.

Here is an image from the Sloan Digital Sky Survey (SDSS) of IC 1101:

I'll leave it to you to decide if an elliptical galaxy counts as "spherical" or not.

If you don't count elliptical galaxies, the largest spiral galaxy is NGC 262, pictured below. It has a diameter of 1.3 million light years.


What happens to oxygen produced on the Sun (or other stars)? Through nuclear fusion, the Sun can (or at the very least, someday will) produce atoms of all elements up to and including oxygen.  And in terrestrial chemistry at least, when you combine oxygen, hydrogen, and a small amount of heat you get water.  Or lithium and oxygen and heat will produce lithium oxide.  Those individual components are all readily available in a star like the Sun.

So my questions is, does the oxygen produced by the Sun (or any other star, now, in the future, or in the past) react chemically (for instance, through oxidation or combustion processes) with the other elements present in the star's atmosphere?

Or I suppose more generally, do the elements formed through nuclear fusion in a star chemically interact to produce more complex molecules (and if not, then why not)?

The Sun is a small main sequence star. It does not produce oxygen via fusion. It can't. The temperature and pressure in the Sun's core are too low. Fusion in the Sun is currently limited to production of helium. This will remain the case for several billion years.
That said, there is oxygen in the Sun, about 1% by mass. This oxygen was produced long ago by other stars at the end stages of their lives. Our Sun is a third generation (or more) star. Most of the Sun is far too hot for those oxygen atoms to combine chemically. One exception is sunspots, relatively cool areas on the Sun's photosphere. (Relatively cool means less than 4500 kelvins, so still quite hot.) Molecules can form at these lowish temperatures, and scientists do see signatures of many different molecules in the light coming from the Sun.

**Update, in response to edits to the question**
Molecules cannot form inside of a star. The temperatures are just too high. Molecules decompose (split apart) into their constituent parts at high temperatures. The Sun's photosphere is about 5800 kelvins, which is already too hot to sustain very many molecules. Temperature rises rapidly with increasing depth below the photosphere. The Sun's core temperature is about 15 million kelvins (27 million Fahrenheit), and the Sun is a small star. Larger stars have even higher core temperatures. At 15 million kelvins, there aren't even atoms, let alone molecules. There are instead atomic nuclei and electrons. Atoms are stripped of their electrons at those extreme temperatures.
In five to seven billion years, our Sun will have fused all of the hydrogen in the core into helium. That's when our Sun will become a red giant. Even then, it still will not produce oxygen. The first stage a one solar mass star experiences after leaving the main sequence is the red giant phase, where the core is an inert mass of helium surrounded by a shell of fusing hydrogen.
Eventually (after another billion years or so), the temperature of that helium core will rise to the point where the helium starts fusing into carbon, plus a little bit of oxygen via the first step on the alpha ladder. At this point, the Sun will leave the red giant phase and join the horizontal branch of the Hertzsprung–Russell diagram. This is a rather short-lived phase of a star's life. The carbon and oxygen produced by helium fusion quickly (in stellar timeframes) form an inert core. At that point, our sun will become an asymptotic red giant.
The red giant and asymptotic red giant phases are rather messy affairs, wracked by convulsions where the star expels lots of gas. Our Sun will lose about half its mass to such convulsions. Molecules do form when this expelled gas cools. This results in some of the prettiest pictures in astronomy, shown below.

(source: inspirehep.net)


What is the origin of the term "plane of the spirit level"? I have heard the term "plane of the spirit level" e.g. the sun's shadows is X degrees to the plane of the spirit level to signify the surface of the earth and I see it is a common term used in surveying and as an instrument to show if there is a curvature.
My question is: where is this term coming from?

A spirit level is an instrument used to find a horizontal or vertical plane. In surveying, the spirit level is used as a reference. Surveying instruments like the theodolite have a spirit level built in. This is used to set up the instrument, after which you can measure angles relative to that reference plane.
This makes sure you have a repeatable reference, and not one that depends on the incline of the local surface.
As to the origin of the name:
it's called a level because it is used to find a level surface (i.e. one that's horizontal). It's called a spirit level because the liquid inside is usually a spirit (i.e. a mix of alcohol and water).

Early tubular spirit levels had very slightly curved glass vials with constant inner diameter at each viewing point. These vials are incompletely filled with a liquid, usually a colored spirit or alcohol, leaving a bubble in the tube. They have a slight upward curve, so that the bubble naturally rests in the center, the highest point. At slight inclinations the bubble travels away from the marked center position. Where a spirit level must also be usable upside-down or on its side, the curved constant-diameter tube is replaced by an uncurved barrel-shaped tube with a slightly larger diameter in its middle.
Alcohols such as ethanol are often used rather than water. Alcohols have low viscosity and surface tension, which allows the bubble to travel the tube quickly and settle accurately with minimal interference from the glass surface. Alcohols also have a much wider liquid temperature range, and won't break the vial as water could due to ice expansion. A colorant such as fluorescein, typically yellow or green, may be added to increase the visibility of the bubble.

why "spirit":

The term "spirit" in reference to alcohol stems from Middle Eastern alchemy. These alchemists were more concerned with medical elixirs than with transmuting lead into gold. The vapor given off and collected during an alchemical process (as with distillation of alcohol) was called a spirit of the original material


How (un)stable are the Lagrangian points 1, 2 and 3? A couple questions, please: I know that the Lagrangian points 1, 2 and 3 are unstable and special Lissajous orbits plus some station-keeping are required to place a spacecraft around them. But I was wondering if they are so totally unstable that they can't temporarily "capture" a passing natural object (let's say an asteroid or a cloud of cosmic dust, in a stationary way or entering some kind of not-very-stable orbit around them)? Or could they?

And if they were able to, how much time would (approximately) be required for this/these object(s) to go "off-orbit" and abandon the L(1,2,3) area? In the case of the cloud of cosmic dust, would it just slowly (or quickly) drift away, or would they be "launched" towards another direction or orbit?

Thanks in advance!

All five Lagrange points are unstable

L1, 2, and 3 are "saddle points" in the effective potential formed from the combination of gravity and the centrifugal force of a rotating frame of reference. An object which is in front or behind in the orbit would tend to approach the Lagrange point, and then move away, either towards or away from the sun, following the lines in the picture.

However note that in the picture, the lines around the Lagrange points are not close together, this means that the motion towards, or away from the point is very slow (in the rotating frame) An object near the Lagrangian point would not be "ejected" at high speed. Any cosmic dust that was near the Lagrangian point would drift slowly away.

However although the L4 and L5 points are unstable, there are quite stable orbits that (in the rotating frame) orbit around the L4 and L5 point, allowing for Trojan satellites.


What's the highest frequency that's been imaged by a radio telescope? In this answer to Should we update definitions and remove the 100 GHz hard limits on radio astronomy related tags? I wrote the following partial answer:

Yes, there are plenty of dishes that focus into waveguides or horns that feed antennas connected to electronic amplifiers using transistor amplifiers connected to heterodyne down-conversion with balanced mixers and then analog-to-digital conversion, where intensity images are produced by interference implemented mathematically in a computer (rather than in wavefronts incident on material producing intensity signals in the form of electrons (CCDs) or phonons (bolometers)) which work up to circa 1,000 GHz, so the 100 GHz limit is obviously wrong!

With the idea in mind that a radio telescope might be defined as using conversion to electrical signals before conversion from amplitude to intensity (image formation), what's the highest frequency where this has been done?
One number I know is 230 GHz, which was used by the Event Horizon Telescope

The question How does ALMA produce stable, mutually coherent ~THz local oscillators for all of their dishes? suggests this is at least 950 GHz, but I don't know if that's a record or not. Narrow band optical emission could be mixed with a laser in a nonlinear crystal producing a microwave signal that could be detected via radio, so potentially the answer might be visible or near infrared light, but I don't know.
update: So I've gone ahead and asked this separately: Has optical interferometry been done at radio frequency using heterodyning with a laser in a nonlinear material?

Some extensive search e.g. with scholar.google.com led me to a manuscript from October 2020 entitled Design and Characterization of 275-500 GHz Corrugated Horns and Optics for a Wideband Radio Astronomy Receiver which already has all the information in the title: It looks like there is current development towards another 500 GHz receiver.
The question mentioned a 0.95 THz receiver which is even above that.
I also found an online course at nrao.edu which argues that 1 THz is the absolute upper boundary to far infrared in astronomy.


How to calculate parallactic angle from a fixed alt-az position? Say we have an observer at a specified latitude and longitude on Earth who would like to observer a specific fixed alt-az sky position (rather than RA and dec), for example to look at geosynchronous satellites.  Since the alt-az position of the observation is not changing, it seems to me that parallactic angle also would not be changing and so should be derivable from just alt-az and lat-long, but I haven't been able to figure out how to do it.

All the calculations I've seen online use RA or hour angles in calculating parallactic angle, which intermediary values I would like to avoid if mathematically possible.

I found an equation on this website,

$\require{cancel} \cancel{\sin{\theta_p}=\sin{\theta_{az}} \cos{\theta_{lat}}/ \cos{\delta}}$,

with $\theta_p$ being the parllactic angle, $\theta_{az}$ being the azimuth, $\theta_{lat}$ being the latitude of the observing stations, and $\delta$ being the declination.  The website cites this equation as being from "Spherical astronomy, Small pg. 49."  I am unable to find such a book (based on my searches, I'm wondering if the correct reference might actually be "Textbook on Spherical Astronomy, Smart", which I do not have ready access to), but in my experience so far this equation has been correct.

The above equation, my original answer, seems to provide something similar to the parallactic angle (something like the complement in each quadrant, with an extra negative sign, but I haven't investigated in detail).  However, when testing against values gotten from astropy, it did not match the astropy values.

Instead, I found the following, taken from Astronomical Algorithms by Jean Meeus, which does match the astropy values:

$\theta_p = \text{atan2}( \sin{\theta_H}, \tan{\theta_{lat}}\cos{\delta} - \sin{\delta}\cos{\theta_H}),$

with $\theta_H$ being the hour angle, calculated via

$\theta_H = \text{atan2}( \sin{\theta_{az,mod}}, \cos{\theta_{az,mod}} \sin{\theta_{lat}} + \tan{\theta_{el}}  \cos{ \theta_{lat} } ),$

Where $\theta_{el}$ is the elevation and $\theta_{az,mod}$ is the azimuth, but defined differently from $\theta_{az}$ as used originally and elsewhere in my answer.  The zero-point $\theta_{az}$ is due north, increasing clockwise, and the zero-point $\theta_{az,mod}$ is due south, increasing clockwise.

Declination can be gotten from latitude, elevation, and azimuth:

$ \sin{\delta} = \sin{\theta_{el}} \sin{ \theta_{lat} } + \cos{ \theta_{el} } \cos{ \theta_{lat} }  \cos{\theta_{az}} $,

Thus indeed, given an alt-az (or az-el depending on your preferred terminology) position and the latitude of the observing station, it is unnecessary to use RA or an independently determined hour angle in a calculation of parallactic angle. I say "independently determined hour angle" meaning that although hour angle is used in the calculation, it can be determined entirely from az/el and latitude, so one does not need to provide an hour angle for this calculation to work.


Is night sky getting brighter day by day? Night sky is dark because light from distant stars has yet not reached our earth. Light emitted by stars formed after the instant of creation i.e. 13.7 billion light years ago is on the way to reach us. As light from these distant stars reach us, night sky must keep getting brighter.
How true is this? And to what extent? Can one day night sky become as bright as a day?

The rate of star formation in the universe peaked about 3.5 billion years after the Big Bang, or 2.7 billion years after it, according to a different study—in any case, for most of the universe's lifetime fewer and fewer stars have been forming each year. Though I don't have precise data on the death rate, on the whole the number of stars is decreasing as more old stars die than new ones form.

Meanwhile, the light from old stars is being redshifted out of existence by the expansion of the universe, which is itself accelerating.

I'm sure there's a mathematical model that can quantify this properly, but I think those two facts, combined, must be the answer to why the sky isn't getting brighter.