Why does the sun rise north of east between the vernal and autumnal equinox? Can someone please explain this phenomenon? A solid explanation seems to be missing from the entire internet.

You know, the first time someone told me this, I was absolutely certain they were confused, ignorant, or otherwise mistaken, and I told him he had his facts wrong. To be precise, we were talking about an observer on the equator. I maintained that the Sun would rise due east and set due west, all year round. He said no.

What gave me pause, though, is that this person was far from ignorant, so maybe the other two appellations didn't fit either. I went home and tried to construct a convincing proof of my side, and in the middle of the night a very simple and convincing proof did come to mind. Except that it was a proof that I was wrong.

Imagine that you're on equator at sunrise on the summer solstice. Due east is along the equator. But the plane of the equator is tilted relative to the Earth's orbit, and the Sun is well above (north) of it. You cannot see it due east at sunrise. It has to be $23.5^\circ$ north of east. (Remember, this is for an observer on the equator.)

By the same token, the Sun must rise $23.5^\circ$ south of east on the winter solstice, and gradually transition between these extremes during the year.


Is it possible to completely see the edge of a half moon at night? Some days ago, about 1 ~ 1.5 hours after the sunset I could see a very bright half moon, and the interesting part is, I think I could see the edge of the dark half of the moon, i.e. a whole circle, but with very little contrast to the surrounding black space. It was very weak.
I'll use my impressive graphic design skills to try to explain it better:

The line I'm talking about is that grey one on the left, closing the moon.
My hypotheses are:

My brain made it up, as if it was trying to "close the circle"
The earth was lighting the moon up, as in: light from the sun was hitting the earth and being re-emitted to the moon (as a very bright moon can actually provide you enough light to see in the dark).

Thanks!

As uhoh has written in the comments, this phenomenon is known as "planetshine", and in this case more specifically "earthshine". The sunlight hits the earth and is reflected to the moon, illuminating what cannot be directly reached by the sunlight.

https://en.wikipedia.org/wiki/Planetshine
https://en.wikipedia.org/wiki/Earthlight_(astronomy)


Non-Newtonian motion of the moon I watched an interesting excerpt from a BBC show on the work of the McDonald Observatory in its laser measurements of the distance to the moon which are apparently accurate to within a few centimeters. The observatory used to make a measurement every day and by this means a plot of the Moon's orbit around the earth was obtained over a period of approximately 40 years from about 1970 to about 2010.

At the end of the excerpt Dr. Peter Shelus says that the measurements show that the shape of the moon's orbit is not quite as expected by "Newton's" theories, which I interpreted to mean that the orbit is non-Keplerian, but possibly he meant something else. It is hard to know because the excerpt just cuts off mid-sentence, so it is unclear what he was saying exactly.

In what way is the orbit of the moon irregular according to the MacDonald measurement series?

Also, as an aside, I notice that the observatory's funding was cut in 2009 so that these measurements are no longer being made, but two other observatories are making new series. Are these other observatories making daily measurements the way the MacDonald Observatory used to do?

The motion of the moon, as measured by the Lunar Laser Ranging data is not as predicted by Newton. As the scientist being interviewed says, Newton's model was correct to the accuracy of the data available to him, and was good enough to land the Eagle on the moon in '69. However the Lunar Laser Ranging data can position the moon to an accuracy of 3cm.

Relativistic effects cause the moon to deviate from the orbit predicted by Newtonian mechanics by about 1m, mostly due to special relativistic effects such as Lorentz contraction. General Relativity accounts for a variation of 10cm. This is much less than the perturbation due to Jupiter (about 1km). However, general relativity does fully account for the motion of the moon, to the accuracy of the experiment.

  All tests are in agreement with Einstein's General Relativity which is used for the numerical integration of ephemerides. No variation of the gravitational constant is discernible. Source

Lunar ranging is ongoing at the Apache Point Observatory Lunar Laser-ranging Operation.


Is broadcasting the location of Earth to potential extraterrestial civilization regulated? How many times have we broadcast the location of Earth to the outer space? Are there any regulation in doing so?

I know so far we have send the Arecibo Message, and any receiver can trace the message sender's direction by the direction with which the message is received. But have we ever pin point our exact location of the galaxy and send it as a message?

No, except in the sense that national broadcasting and so on is regulated. Our location can be fixed, as you suggest, from simply looking at where the signals come from.

Perhaps, though, you mean is it sent out "regularly" (i.e., at specified intervals). In truth we have been sending very powerful signals continuously into space since the dawn of the television age. Those signals may now start to wane in strength as we move to digital terrestrial television but that won't bother any aliens for some time yet.


Can you assume atmosphere height for the purpose of surface pressure calculation? Given that by definition of scale height an atmosphere thins by a factor of 1/e^x where x is elevation in terms of scale height multiples (See the table here: Definition of Scale Height), can we assume that the atmosphere is effectively non-existent at the elevation of 6H?

Density at elevation 6H
1/e^6 = ~0.00248 would mean about 0.2% of density at surface level

I know there is no real physical boundary but what I'm looking for is what is the assumed standard for simplifying calculations. Or is it just a bad idea to try and do it this way?

*This is a followup to another question:
How can you determine the initial volume of a planet's atmosphere?

It all depends on what you are trying to do. For some purposes 6H might correspond to an adequate vacuum  for others not. The density is not low enough to ignore its effects on a space craft (we are talking of $\approx 48$ km in the case of the Earth). Indeed $48$km is less than the altitude record for an unmanned balloon, though greater than the altitude records for air breathing aircraft.


Missing supernovae in the Milky Way Stars with more than 8-16 solar masses usually go supernova. These are OB stars, of which 0.2% currently exist in our galaxy. The average lifespan of said star ranges from 10 to 40 million years. So 200 million OB stars ÷ 40 million years = 5 per year. That's how many supernovae we should be seeing annually in our Galaxy alone. So why do we see only an average of one supernova every 50 years?
Clearly, I am not considering something else important here. What am I missing?

There are far fewer than 0.2% of stars that are as massive as $8M_\odot$. It is possible you've confused it with the percentage of stars born with $M>8M_\odot$.
The thing is, those massive stars die very quickly, but most stars, those with masses $<1M_\odot$, are still alive. So the fraction of massive stars in existence now is much lower. i.e. Almost all the massive stars ever born have already died, but almost all the low mass stars ever born are still alive.
What matters for the present supernova rate is the current birth-rate of massive stars.
The current star formation rate in the Milky Way is something like 1 solar mass per year. The way the birth distribution of stellar masses works out, most stars born are of order $0.3 M_\odot$, but about 0.2% are $>8 M_\odot$, which means, if you work through the numbers, that you need round about $150 M_\odot$ of star birth to produce one massive star - which means you get a type II (core collapse) supernova every 150 years. These numbers are uncertain by factors of two.
Another factor to bear in mind is that we don't "see" a supernova every 50 years in the Milky Way. In recorded history, it is more like one every 300 years. However, the majority of supernovae would have been totally hidden from an observer on Earth by dust in the Galactic plane (where star formation occurs). Thus the true rate in the Milky Way might be closer to 1 in 50 years.
Finally, note that some of these supernovae are caused by exploding white dwarfs, not the core collapse of massive stars. Thus the rough number of one every 150 years is probably consistent with the historical observational record.


Are stars beyond the asteroid belt less constant? If we graph the variability of a small bright star which is lined up with the asteroid belt, is it less constant than stars that are far away from the asteroid belt? To what degree does variability change relative to distance away from the most dense areas of the solar system?

I added the arrows to illustrate the question.

No, stars that are seen through the asteroid belt are not any more variable than stars in any other part of the sky.

All stars vary in brightness on measurable timescales, although some have been measured to have such small variations that they are recorded as "standard" stars and for most purposes can be considered to have a constant brightness and thus be used as reference for other brightness measurements.

What you are referring to, as noted, are occultations. There are many hundreds of thousands of known asteroids in our Solar System, and billions of stars in our galaxy, and so there could be hundreds of predicted occultation events every day, and there are many people interested in occultation events. For one thing, we can calculate the distance and speed of an asteroid quite accurately but we can't directly photograph it to determine its size. But when an asteroid passes in front of a star, we can measure the duration of the occultation and set precise limits on its size.

But for all the asteroids in the Main Belt, the space between them is huge and for any one star the chances of an occultation are quite (vanishingly) small, so the stars seen through the asteroid belt are not any more variable than anywhere else in the sky.


How to scale down solar system data to simulatable values Okay, I am seriously ashamed for asking this especialy when I geniualy study on physics but there is something that bugs me with the simulation I am working on.
I am re-creating the solar system in an n-body simulation that i programmed before. And I've a problem with scaling down the solar data. So even if I use kg and km for the metric units, the values are far bigger than the variables can hold in programming. Also as some of you know, bigger the value is, bigger the floating point error it makes. (error noise in data) Also it makes it slower to process.
I decided to scale down the data with a reference point, and for that, I took the earth's radius as 1 unit. And scaled down every other distance and radius according to it. (So a unit is 6371 km just to be clear)
But I am not sure whether if I should scale down the mass or not. My common sense says that I should scale down the mass so density of each body should remain same. So I took the density, and calculated a new mass value for each body, with the new scaled down radius.
But I am somehow couldn't conviced myself about If it's true or not. So here I am, asking to you :) Should I also scale down the mass?
PS.1: I used using $F = GMm/r^2$ equation for the calculactions as usual. (Iterating it through each body pairs)
If there are other programmers like me interested in making a simulation like this, how did you accomplish this data size problem? Are there any better solutions than scaling down the values?
PS. I have created an excel file that does the scale conversion. So I am sharing  the sheet in OneDrive. (http://1drv.ms/1NIekGo) If you can check my calculations and values, that I'd also be really helpful to me. Thanks for any help.

As barrycarter mentioned in his comment, you should be more concerned with units and less with scale.

Generally, it's best to stick to conventional units that people recognize. (It will keep your head on straight and make it easier for others to review your work.) In astronomy, these are a little different than the standard SI units, since things are—shall we say—big, and numbers get out of hand quickly (as you have noted). Here are some units suggested in the Astronomical system of units Wikipedia page:

Time: Day. This is probably not too important, so feel free to use seconds here. (After all, it's only a 5 order of magnitude difference.)
Mass: There are several conventions, but the one I've personally seen most is solar masses. Since you're modeling our solar system, it makes a convenient reference.
Length: I would definitely use astronomical units. (What an appropriate name!)

If you stick to these units, you should be able to limit error due to rounding and reduce the computational load of large numbers.


Probability of planets in binary star systems? There is much fiction on binary stars with planets, for whatever reason, and this keeps supporting the popular impression that those systems are frequent. Indeed,  the methods for identifying exoplanets and binary stars are somewhat the same. Are there actual observations of a binary star system with at least another planet (habitable or not)?
I know that binary star systems are rather frequent, and I am wondering whether there is any hard-science based statistics on how probable it is that systems of two or more stars have planets? As non-expert, I have the impression that not all configurations of binary star systems seem to be in favor of planets, see e.g. Roche lobe.

There are only about a dozen circumbinary exoplanetary systems known - the first discovered was Kepler 16b.
The statistics of circumbinary planets is a relatively unstudied area. One of the major uncertainties is the inclination distribution of the exoplanets. If they are confined to the orbital plane of the binary system, then transiting circumbinary planets are easier to detect and so their observed frequency translates to a lower actual frequency than if their orbits are scattered around the binary orbital plane. Armstrong et al. (2014) use the Kepler survey results and find that if circumbinary exoplanets are coplanar with the binary, then the frequency of giant exoplanets within 300 day orbits is about 10%, and consistent with the frequency of similar exoplanets around single stars. On the other hand if their orbital inclinations were random with respect to the binary, then the exoplanet frequency could be a factor of 5 higher.
On the other hand we do know that circumbinary planets around very short period binaries (those with periods less than 7 days) are rare Fleming et al. 2018). This probably has something to do with orbital evolution of the binary system in these cases.
Exoplanets that orbit around one of the stars in a binary system are better studied. If the stars are widely separated then they have little influence on each other or on planetary systems that orbit within say 1/10 of the stellar separation. Exoplanets on orbits wider than about 1/5 of the separation are expected to be unstable.


Reference: First spectrographic observations by Fraunhofer In several books I find reported that the first scientist to perform spectrographic observations of celestial objects was Fraunhofer with the spectrographs he himself produced. I do not find though a reference, a paper, in which he presents this work.

Using the NASA astrophysics data systems I find only the papers concerning his technical work with glasses but not the observations of Venus, Mars, and other heavenly bodies, he performed.

I find books on Google Books reporting such observations
but unfortunately I can view limited parts of these books and not their bibliographies.

Anybody can find a reference for any of these observations?

With further research I found an Answer in "The Cosmic Century" by Longair (Cambridge).
When Longair says

  In 1823, Fraunhofer made further observations of the spectra of the planets and the brightest stars, anticipating by about 40 years the next serious attempts to measure the spectra of the stars (Fraunhofer, 1823)

the article referenced is Kurzer Bericht von den Resultaten neuerer Versuche über die Gesetze des Lichtes, und die Theorie derselben (A short account of the results of recent experiments upone the laws of light and its theory.) which was actually fetched by my research in the NASA ADS (https://ui.adsabs.harvard.edu/abs/1823AnP....74..337F/abstract). But the title did not mention any planet or star and I could not access the full text, so I thought it was not the right reference.


Universe becoming finite If sometime in the distant future, the Universe stopped expanding, would this mean that the Universe was no longer infinite?

If the density of the universe is high enough then the mass of the universe would be enough to eventually stop expansion and cause the universe to contract (the critical density is about $10^{-26}\, \mathrm{kg\, m^{-3}}$).

If this is the case (it's probably not, the universe is probably flat, with dark energy causing expansion to speed up) then the universe is finite in time and always has been. The universe doesn't stop being "infinite".


Does gravitational lensing cause a black-hole to be the main 'source' of light in a given area? Light gets bent around high-gravity sources, such as a black-hole. A source of light, say A, is directly observed in its 'true' position and some of its light is bent around the black-hole, X, which gives a second observation of A, called A'.

Does gravitational lensing cause the amount of emissions (detected as coming from the (point/position/direction) of X) to far exceed anything else in that vicinity?

Does this then make the 'black-hole' by far the brightest, noisiest thing to be observed? (kind of like waving & shouting in a quiet room)

[edit - based on astromax's answer]
I follow what is meant regarding super massive black-holes.
Consider the following scenario: A suitable massive sun on dying contracts into a (small) black-hole. This black-hole will now cause a lensing-effect, but since it is small the amount of emissions so 'lensed' around it is very small and maybe too small to detect? or, it is detected but erroneously ascribed to be a type of sun or other type of emission source?

So the question is, Can black holes be really be extraordinarily bright due to lensing of background objects?

Let's first also specify that these black holes must certainly be super massive. If not, we're talking about stellar mass black holes (or intermediate mass black holes), and the lensing signal would be much weaker. More mass = more lensing generally speaking (there are some exceptions, e.g. - microlensing, where the lensing is done by stars in our own Milky Way galaxy and the signal is enhanced because the source and lens are aligned nearly perfectly).

This is an important statement to make because virtually every galaxy has been identified with a super massive black hole at its center. The reverse is also true - every super massive black hole is associated with a galaxy. It's also important to note that, realistically speaking, galaxies have total masses of about $\sim10^{11} - 10^{12} M_{\odot}$, whereas super massive black holes typically have masses of millions to hundreds of millions (maybe even billions) of solar masses. This is a smallish fraction of the total mass. This means that most of the lensing will be done by the galactic halo and not the super massive black hole.

If we modeled the center of the lens galaxy as a point-lens, and the galactic halo as a singular isothermal sphere, the relevant question I would ask is what are their Einstein radii (which is a measure of how effective or efficient they are at lensing) individually, and what are they in combination? Essentially, how much does the existence of a central super massive black hole matter to the system as a whole. Strong lensing features (arcs, rings, or multiple images) generally occur at around the Einstein radius of the object.

My best guess:

Quite honestly I don't see the central super massive black hole mattering all that much when it comes to lensing. Many of these mass profile models for the lens galaxy halo are singular, or rise very rapidly to a central core. Furthermore, I've never really heard of a situation where a lone super massive black hole (not associated with a galaxy - call it a 'rogue' smbh if you will) has been found floating around in space to do this sort of lensing. They generally hide at the centers of galaxies, or show themselves only if they're actively accreting material. Correct me if one has been found (maybe it would come from a merging of two smbh's where one is kicked out of the system).


If you looked at Earth while travelling at relativistic speeds, will you see sped-up activities? Suppose you traveled off Earth and back at relativistic speeds, so that when you come back much longer time elapsed on Earth than you experienced. Suppose you had an excellent telescope through which you could observe the details of the surface of the Earth down to human scale, all the time.

What will you see?  Will you see people moving incredibly fast? Will you see people progressing from birth to old age in a matter of seconds?

It doesn't have to be a visible telescope, any electromagnetic radiation will act in the same way.

I'll give you the answer, but if you want the details you need to look at a description of the resolution to the so-called  twin paradox, for for example here.

Assume that the spaceship travels  outwards and back to Earth at the same speed, with a  brief period of acceleration in between. From the point of view of the observer on the spaceship, they will receive signals at at lower frequency on the outward half of the journey. The change in frequency to depends on how fast they go and is given by the relativistic Doppler effect formula.
$$f_{\rm out} = f \left(\frac{c - v}{c + v}\right)^{1/2} = \alpha f,$$
where $\alpha < 1$ and $f$ is the frequency of the light/signal on Earth. So the Earth would appear slowed down.

On the way back the observed frequency
$$f_{\rm in}= f \left(\frac{c + v}{c - v}\right)^{1/2} = \frac{f}{\alpha}$$ and the Earth would appear speed up.

According to the observer on the spaceship they spend half the travel time in each of these phases. The average frequency they receive is
$$\bar{f} = \frac{(f_{\rm out}+f_{\rm in})}{2} = \frac{f}{2} \left(\alpha + \frac{1}{\alpha}\right),$$ which is always $>f$.

This means that on average things on Earth appear speeded up, such that the Earth is older upon return, but what is seen from the spaceship is divided into distinct outbound and inbound phases.


Definition question: continuum-subtracted spectrum What is the exact definition of a "continuum-subtracted spectrum", such that one finds in SDSS spectra?

http://classic.sdss.org/dr7/dm/flatFiles/spSpec.html#specmask

Is there a definition somewhere? Thanks

A spectrum of, say, a star or a galaxy, consists of wavelengths emitted from various physical processes. In the case of the star, there is the blackbody spectrum that reflects the temperature of all of its gas, whether it is hydrogen, helium, or metals. On top of this, there are lines, i.e. emission from certain atomic transitions that are excited and then de-excite emitting photons in a very narrow wavelength interval. The blackbody spectrum is said to comprise the continuum of the total spectrum.

In the case of galaxies, the continuum consists of the combined spectrum of many different stars' different blackbody spectra, infrared light from dust, etc. On top of this, there are lines, e.g. from nebular processes such as my favorite line Lyman $\alpha$, which is emitted when hard UV radiation from hot O and B stars ionize their surrounding neutral hydrogen, which subsequently recombines emitting mostly Lyman $\alpha$.

If we want to measure the total flux in a given line, we are usually only interested in the flux that comes from a particular physical process, and not the part of the flux that comes from the "underlying" sources which contribute to the continuum. Hence, we need to subtract the continuum. This is usually done by fitting some functional form to the continuum (e.g. a straight line or a second order polynomial), substract this fit from the spectrum, and finally integrate from one side of the line to the other in the resulting "continuum-subtracted spectrum".

In the top figure below, the total spectrum's flux density$^\dagger$ is in black, and the fit is the red lines. Underneath is drawn the continuum-subtracted spectrum, and a few dashed green lines show some lines that rise above the continuum. The integrated line flux is indicated by light red.

$^\dagger$Note that many authors use the term "flux" where it should actually be "flux density". Integrating flux density in $\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Å}^{-1}$ over wavelength in Å$^{-1}$ gives flux in $\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}$.


Converting ecliptical to equatorial coordinates When looking for a formula to convert polar ecliptic geocentric coordinates of an object to equatorial coordinates I find various sources that give these formulae (like Wikipedia):

Declination $δ = \arcsin(\cos ε \times \sin β + \sin ε \times \cos β \times \sin λ)$
Right ascension $α = \arctan((\cos ε \times \sin λ - \sin ε \times \tan β) / \cos λ)$

Where
β = ecliptic geocentric latitude
λ = ecliptic geocentric longitude
ε = obliquity of the ecliptic

But when applying these formulae I get results like in the following list where β = 0° and ε = 23.4°:

 λ       δ        α
  0   0.0000   0.0000
 45  16.3095  42.5443
 90  23.4000  90.0000
135  16.3095 -42.5443
180   0.0000  -0.0000
225 -16.3095  42.5443
270 -23.4000  90.0000
315 -16.3095 -42.5443
360  -0.0000  -0.0000

The values for declination seem good, but right ascension values seem to lack some sort of adjustment to the quadrant of the full circle (just a guess). But nowhere did I find any mentioning of this. Can you help? Thanks.

Convert commments to community wiki

This is a standard problem with using an arctan function (inverse tan function returns a value between -90 and +90). A lot of programming languages have two functions the simple one like you appear to have used and a more complex one that takes two parameters for when you are dividing (before the arctan function) as in this case. (This is described in further detail at (http://youtu.be/_JbkTK17MrY)

In Java you can use α=atan2((cos ε∗sin λ−sin ε∗tan β), cos λ) you may need to add 360 if the angle returned is negative.

An alternative is α=atan2(cosβsinλcosε−sinβsinε,cosβcosλ). This is equivalent in most cases but avoids an infinite term tanβtanβ at the ecliptic poles, costing one more trig function call.


How is the cost of JWST distributed on different parts of development and operations? The James Webb Space Telescope costs nearly 9 billion dollar. What parts of this project cost how much?

I suppose that almost every component has to be developed and is not available to buy off the shelf. So most of the costs should end up as salaries to engineers. I see that the top 10% US specialist aerospace engineers earn 163,000 dollar. The JWST budget covers 54,000 such yearly salaries. Are really thousands of the best engineers working on this project? Or are maybe some materials, like beryllium, very expensive? Like JWST's average cost of about $1½ million/kg. Launch costs are anyway negligible in this context.

So most of the costs should end up as salaries to engineers.

Even in a project that is solely people-based, most of the costs do not end up as salaries to engineers. Salary is typically a bit less than a half of the cost. Those engineers get benefits. Health care in the US isn't cheap. They get holidays, vacation, and sick leave. They receive matching funds on their 401K contributions. In the US, employers and employees both contribute to the cesspool lockbox sometimes known as "Social Security". The employer share is part of the cost of having an employee (and companies charge for it). The employer also pays for unemployment insurance and other items on behalf of their employees. Those engineers sit at desks in an office or a cubicle, and have a nice computer on which they do their work. They use the employer's servers, printers, copy machines, conference rooms, break rooms, and coffee machines. All of that costs money. These overhead costs are the first of several indirect costs.

The accountants who write those engineers their paychecks aren't charging to any project. They're still paid by the company, as are the executives of the company. The executives also typically don't charge their time to a project, and those executives can be paid quite handsomely. This leads to another category that adds to quantity that companies charge the government, "general and accounting". Finally, for-profit companies need to keep their shareholders / stakeholders happy. They need to earn a profit. This leads to a third category that companies charge the government, "profit".

When you add up all of those indirect costs, they come close to and oftentimes exceed direct costs (i.e., chargeable salary). Note that this is for a project where the only costs are the people working on it.

Building a spacecraft notches things up a lot. Cleanrooms are expensive. Control rooms are expensive. The flight computers are expensive. The bolts and fasteners are expensive. The custom, one-off sensors are expensive. The spacecraft's thermal control system is expensive. The spacecraft's structure is expensive. Launch is expensive. Every piece of hardware, every fabrication facility is expensive.

Finally, cost overruns, requirements changes, and delays can represent a significant portion of the cost of a poorly managed project. JWST has experienced numerous cost overruns, requirements changes, and delays. Big contractors have learned that the only way to win big government contracts is to underbid. They would lose their shirts if the government kept to that initial contract. But it never does. The government instead changes the requirements, changes the payments, and changes the schedule. Those changes are beyond expensive.


How quickly does a supernova heat up/expand? Suppose there's a star out there that's a lot more massive than the Sun.

Suppose further that orbiting this star is a planet not unlike Earth. Water, oxygen, civilization, and all.

Now the star decides to go supernova. How quick, or slow, is the process?

How long will it take it to heat up to make life on that planet impossible? A month? A year? A hundred years? A thousand years?

And how long will it take for the supernova to engulf the planet? A day? A month? A year? A million years?

Ricky, it is very rapid. The core collapse and initial neutrino burst takes seconds to tens of seconds. We don't normally think about neutrino interactions, but so many are released that even this might be a problem for a nearby habitable planet. It then takes a few hours for the shockwave from the core collapse and bounce to make it out to the surface, accompanied by an intense flash of UV light that would likely sterilise anything in its planetary system. The outer layers of the exploding star are hurled out at around 1000 km/s, so could travel an astronomical unit in a day or so.

The supernova continues to become more luminous for about a week thereafter, increasing in luminosity, from what must have already been enormous (thousands of solar luminosities) by another factor of $10^5$. The equilibrium temperature of any planet scales roughly as $L^{1/4}$, so temperatures would rise by more than an order of magnitude in a week.


Why do "they" portray colliding black holes like that? When I see an animation of two colliding black holes anywhere, it looks like two merging beads of water. But why would the event horizon form a bridge during collision? There is no material flowing in-between. Wouldn't the event horizons "push" each-other away as they merge?(the effect not the black hole itself) For what I understand the gravity would poorly, but surely, be canceled out. Wouldn't this "push" only disappear from view as they got significantly closer, where it still exists but is trapped in a much larger one?

Just to make sure, as I see the similar questions: I don't mean what happens as they collide, but rather why isn't this portrayed if it does this effect at all?

The event horizon is not a physical structure. Rather it is the boundary of a region of spacetime from which no information can escape. If you were (unfortunate enough to be) between two black holes, you may be in a position in which there is no net force, and yet you could not travel from your location to the point distant from the two black holes, no matter how fast you were able to move.

If you were in such a position, you would be inside the event horizon of the two black holes.

Finding the event horizon means solving the Field equations of General relativity. Exact solutions are known in several situations: A single mass concentrated at a point, a mass with a charge and a rotating mass. If there are two masses, then the Field equations have no exact solution, but can be solved numerically (with a fast enough supercomputer) It is the results of these calculations that you are seeing.

So, the shape you see is the event horizon, The event horizon of two nearby black holes can be calculated by a computer. Since you may be trapped between two black holes the event horizon will be deformed towards another black hole's event horizon, giving the "teardrop-> bridge->merger" appearance of the final moments of two back holes.


Is TESS replaceable by gound-based-small-telescope arrays? TESS is designed to perform an all-sky transiting exoplanet survey.

  TESS's survey will focus on nearby G, K and M type stars with apparent
  magnitudes brighter than magnitude 12. Approximately 500,000 stars will
  be studied, including the 1,000 closest red dwarfs.

Since its targets are very bright and its sensitivity is relatively worse than Kepler, I wonder whether it can be replaced by gound-based-small-telescope arrays. There can be no gaps in the time series too.

Depends on the wavelength ranges to be studied as well.  Quoting from the link you provided:

  The sole instrument on TESS is a package of four wide-field-of-view
  CCD cameras. Each camera features a low-noise, low-power 16.8
  megapixel CCD detector created by the MIT Lincoln Laboratory. Each has
  a 24° × 24° field of view, a 100 mm (4 in) effective pupil diameter, a
  lens assembly with seven optical elements, and a bandpass range of 600
  to 1000 nm

Atmospheric absorption has some serious peaks near 750 and 920 nm, which is one reason to place these cameras in orbit above the atmosphere.


Is this object an asteroid or comet, and how can it produce so many tails? The NASA Hubble Space Telescope Science Release HEIC 1320 - When is a Comet not a Comet? - Hubble astronomers observe bizarre six-tailed asteroid is self-explanatory, but it doesn't explain what actually happening.

Sometimes comets will show two tails, but one is usually identified with the path of the comet's orbit, and the other with the direction away from the sun. But this is an asteroid (or so it seems) and there are many tails with complex behavior.

edit: I've found this ArXiv preprint: The Extraordinary Multi-Tailed Main-Belt Comet P/2013 P5 where "main belt" refers to the main belt of asteroids, and yet the object is now called a comet.  Wikipedia lists both names and suggests both designations - asteroid or comet:

  311P/PANSTARRS also known as P/2013 P5 (PANSTARRS) is an asteroid (or main-belt comet) discovered by the Pan-STARRS telescope on 27 August 2013.(4) Observations made by the Hubble Space Telescope revealed that it had six comet-like tails.(5) The tails are suspected to be streams of material ejected by the asteroid as a result of a rubble pile asteroid spinning fast enough to remove material from it.(2)

It's been three point five years now - has this been figured out yet? Is it an asteroid or comet (or both), and how can it produce so many tails?

above: HST image of comet P/2013 P5 from here.

above: HST image (annotated) of comet P/2013 P5 from here.

A paper titled "Monte Carlo Models of the dust environment of a sample of comets from the Oort Cloud to the outer Main Asteroid Belt" (PDF also here) (2014) studied the behaviour of P/2013 P5 using models and concluded that

  The object has been subjected to an intermittent dust mass loss, most likely associated with a rotational disruption. This is confirmed
  from the analysis of both HST and GTC images. The total dust mass
  released was of the order of $10^7\ kg$, for particle density of
  $1000\ kg\ m^{−3}$ and geometric albedo $p_v = 0.04$.
  The model of rotational disruption, based on simulations of an object that loses mass from its equatorial region, and whose
  rotational axis is perpendicular to its orbit plane, reproduces to the
  last detail the observed complex brightness pattern at four different
  epochs of HST and GTC observations. For obliquities different from 0°
  or 180° , the fits get much worse. On the other hand, an isotropic
  ejection model does not fit the HST data because it produces much more
  diffuse tails than observed.
  The ejection velocities are very low, of the order of $0.02–0.05\ m s^{−1}$. This places the limit on the size of the object to be in the
  range $30–134\ m$ for assumed densities of $3000–1000\ kg m^{−3}$.

This seems to confirm what the original question cited as a possible explanation, i. e.

  The tails are suspected to be streams of material ejected by the
  asteroid as a result of a rubble pile asteroid spinning fast enough to
  remove material from it.

The asteroid is loosely bound (a "rubble pile") and spinning fast, so everytime some of the boulders which form it shift a bit, some material may escape into space and form a small tail.


Gamma ray detection for SETI. How far away could we detect an alien nuclear powered rocket? I suspect our best bet for finding advanced civilisations is to look for gamma rays characteristic of nuclear/antimatter propulsion.

If an alien civilisation was using something similar to the Orion drive, how close would they have to be for us to be able to detect the gamma rays?

In my mind, this question has two parts:
What are the magnitude/frequency limits on the gamma ray telescopes in orbit? (Specifically Fermi, assuming that has the best detection capabilities of an orbital gamma ray observatory)

What's the magnitude and frequency profile on, say, a 1 megaton nuclear bomb?

I've had a bit of trouble finding these out myself from a casual search, and am wondering if maybe there are military secret issues surrounding the detection of gamma rays.

It's unclear if nuclear energy would be a choice propulsion for a space faring civilization.   It's possible, but it's not the only option.   That said, if it was the method of choice, it's not at all clear to me that it would be easily detected.

Let me preface this with stating that I'm not an expert.   There's two kinds of basic nuclear engine that could be used for space travel, closed or open.  An open nuclear propulsion is much simpler.   Basically explosions are made outside the ship that propel the ship by force from the explosion.   Tiny nuclear pelots are detonated and that's what pushes the ship.

The other method would be an internal nuclear engine, well insulated that generates electricity and electricity generates magnetic fields and presumably ion thrusters that push the ship.   We wouldn't detect nuclear energy if it's internal and insulated.

The Orion ship that you mention in your question wouldn't make one megaton explosions, but many smaller explosions.  It would carry more fuel than 1 megaton, but it would use it for tiny explosions over it's journey.    Space is also full of Gamma radiation, so I don't think gamma radiation would stand out.

You can't look for gamma-rays from Earth's surface because the Atmosphere blocks most of them, so you need a space telescope and our best space telescope for that is the Fermi Gamma-ray Telescope.  https://fermi.gsfc.nasa.gov/  (having trouble with linking, not sure why).

The Fermi scans the entire sky, so it's really more of a survey than a telescope, and gamma ray sources are still being studied, with 31% (per chart) listed as unknown sources.   https://www.nasa.gov/mission_pages/GLAST/news/gamma-ray-census.html

I don't think our current equipment is close to detecting a 1 megaton nuclear bomb a few star systems away, much less an orion type spaceship.   I think it's too few gamma rays at several light years distance.  Perhaps with better equipment we'd have a shot, but I don't think we're anywhere close to that now.    That's largely a guess though, I'm not an expert on the subject.

Edit

There's something wrong with my account, I can't comment either, so I'll post it here.

Your assumption is that 1 megaton is large enough to be visible from lightyears away - by gamma ray telescope.   I'm not sure that's true.

1 Megaton = 4.18 x 10^15 Joules.
1 Joule = 6.24 x 10^18 electronvolts.

So, 1 Megaton in explosive energy (and not all of it would go to gamma rays, but a decent fraction).   But ignoring that fraction, 1 Megaton = 2.61 x 10^34 Electron volts.

The surface area of a sphere 4.1 light years in radius (the distance to the closest star), 4 Pi r^2, where the radius is 3.88 x 10^16 meters, is 1.89 x 10^34 square meters.

So a single megaton blast 4.1 light years away would average out to a bit over 1 electron volt on a 1 meter gamma-ray telescope.    A typical gamma ray is thousands of electron volts, so unless you have a very large telescope, there's a good chance you wouldn't pick up a single gamma ray from a one megaton blast 4.1 light year away, or maybe you'd pick up one if you were lucky.

4.1 light years is very far.  Far enough that a telescope, unless it was enormous, would have a hard time seeing a nuclear blast that far away, and that's the closest star system.

Corrections are welcome but that's what the math looks like to me.


Can one calculate the size (i.e. the radius) of our universe? If one can estimate the mass of our universe, might one also calculate its ultimate size, (after it fully expands), based on the Schwarzschild formula for the radius of a black hole? Please note that this simple formula
$$R=\frac{2GM}{c^2}$$
informs us that larger black holes are less dense. So, if it turns out that our universe is in fact a gigantic black hole, then one would expect it to be extremely low-density!

If I understand the question the answer is, no, you can't estimate the final size of the Universe by calculating the schwarzschild radius of the universe.

The mass of the Universe is not known, and may be infinite.

It is not know if the observable universe will ever stop expanding, however the evidence suggests that it won't, and so won't have an "ultimate" size.

The mass of the observable universe is of the order $10^{53}kg$, mostly comprised of dark energy and dark matter.

A black hole with that mass would have a radius of $2MG/c^2 = 10^{26}m$, which as has been observed is on the same order of magnitude as the (comoving) radius of the observable universe. Whether this has any cosmological implications is not clear. It doesn't imply that the Universe is a black hole. It doesn't say anything about the Universe at all, as this calculation was for the observable universe. It doesn't say anything about the "ultimate size" of the universe or the observable universe (which is expected to continue expanding without limit)


Are there any mirrors in space? We have encountered galaxies that act as lenses, magnifying the light coming from behind them. This is super fascinating and it makes me wonder what other optic tools we might have at our disposal.

The question of what we know about our own galaxy is obviously fascinating because it's kind of the the one we have the worst view at. There's so much noise around us it's hard to get a macroscopic view of the galaxy. It occurs to me that if we could stick a mirror out and look at our galaxy from the outside, maybe we have a better view.

Of course, we don't have the technology to make a very useful mirror. And it wouldn't be very efficient. If anything, we would simply launch a probe out there and send back pictures and science data.

But then I got to thinking maybe there's a naturally occurring mirror we could leverage. We know there are entities that can bend light, are there any that bounce it? It would be incredibly advantageous if there was one that allowed us to see our own galaxy, but also it seems unlikely. But are there -any- objects out there that reflect light like a mirror allowing us to see around them or behind them or their "backside"?

Very likely not.

If it's a gravitational process like galaxy lensing, then a "mirror" would require a very large deviation. That's extremely unlikely with this process.

With other processes, e.g. classic reflection, it seems even more unlikely. We don't know any natural processes that could craft mirrors at cosmic scale.


Why would tidal forces on planets become more intense when a star becomes a white dwarf? I was puzzled by this press release by the Royal Astronomical Society: "Small, hardy planets most likely to survive death of their stars" - phys.org, May 14, 2019

  Astrophysicists from the Warwick Astronomy and Astrophysics Group modeled the chances of different planets being destroyed by tidal forces when their host stars become white dwarfs and have determined the most significant factors that decide whether they avoid destruction.

According to the shell theorem, the gravitational forces of a spherically symmetrical body's regions, in terms of how they act on external objects, sum up to the functional equivalent of a point mass. A star therefore acts like a point gravity source both before and after becoming a white dwarf (albeit a lower-mass one after, due to casting off matter that becomes a planetary nebula — a drop in attraction that I'd expect to cause any remaining planets, those not close enough to have been slowed and swallowed by the red giant phase, to migrate outward).

So why would a star, after turning into a white dwarf, exert stronger tidal forces on its planets?

From the paper, p.2:

  The star eventually transitions into a white dwarf, which is comparable in size to the Earth but has a Roche radius which extends outward to about one Solar radius ...How objects can be emplaced at one Solar radius from several au – and then circularised – is still subject to debate.

They are considering planets which have been observed in close orbits (orbital period as short as 2 hours) around white dwarfs.


Where did Mars' water go? Many sources indicate that there is ice water on Mars today, but it is mainly traces, here and there, but not that much.

However, solid hypothesis state that there were oceans on Mars, see Wikipedia's Mars ocean hypothesis or at least large aquifers on early Mars. They all concern early Mars so around 3.5 to 4 billions years ago. These theories suggest that Mars oceans could have formed Mesa landscapes (just likeGrand Canyon or Monument Valley).

So my question is simple, if all these hypotheses are true, where did that water go?

Is it possible that so much water could have been evacuated from the planet? I guess it would have needed A LOT of energy.

So it may have been transformed on the planet.

Is it possible that the water has oxidized iron, creating Iron oxide (ie rust) and that it is the main reason for Mars red color?
The Mars ocean hypothesis page suggests some leads about the fate of that ocean, but none suggests the oxidation of iron consuming the entire oceans. Moreover, oxidation would imply the presence of water and dioxygen...

http://en.yibada.com/articles/18182/20150309/red-planet-once-held-large-shallow-ocean-mean-positive-signs.htm

Some of the water evaporated, some of it froze and another good part of it went into the surface and is now subterranean. For more info read Mars' Vanished Water & Atmosphere --Where Did It Go? and MAVEN’s Quest – Where Did Mars Water Go?.

There's much still to learn and experiments in the future will help us to better understand the evolution of Mars' atmosphere.


Calculate the time when Arcturus passed through the meridian If the observer observes Arcturus passed through the meridian about 23:00:00 of 28 April. Find the time that Arcturus passed the meridian in 29 April. And what day does the Arcturus will pass the meridian about 01:00:00?

I think that when Arcturus passed through the meridian, the hour angle will be 0 degree. Sidereal time equal hour angle plus right ascension. right ascention of any stars always be the same. So Sidereal time will be as same as every times while the star located at the meridian. And it doesn't matter because solar time equal sidereal time minus 12hr. Do I understand something wrong? And how do I do this question?

Adding question;

Your answer was very clear. But if I consider about the moon instead ex; the observer is at the equator and saw the moon which has a declination of 0 degree set at 11:00 PM yesterday, what time the observer will see the moon set today? I try to use the thinking process like this again. I know that the sidereal month of the moon is 27.32166 days. I calculate the solar month of the moon from 365.24219 of solar year divided by 12, and I get 30.43684 days per solar month. From subtracting both of these numbers, I get 3 days plus about 2 hours 45 minutes and 52 seconds. So to answer this question I will get 2 hours 45 minutes 52 seconds behind. My answer is that I will see the moon set at 13:45:52 PM. Do I right? I don't understand why the time I get is behind instead of earlier as the question about the Arcturus I asked before. I'm also not sure that does the declination of the moon and the position of the observer relate or not too. Thank you very much:)

Find the time that Arcturus passed the meridian in 29 April.

With respect to the Sun, the Earth rotates once every 24 hours.  But because the Sun's apparent position changes in the sky due to Earth's revolution around the Sun, the Earth rotates once every 23 hours, 56 minutes, 4 seconds with respect to the stars.  Because of this 3 minutes, 56 seconds difference, all stars rise and set 3 minutes, 56 seconds earlier each succeeding day.  This applies to the time of a star passing through the meridian as well.

So, on the next day, April 29, Arcturus will pass the meridian at 22:56:04.

  And what day does the Arcturus will pass the meridian about 01:00:00?

Over the course of an entire year, about 365 days and 6 hours, these 3 minutes and 56 seconds adds up to an entire day, so that exactly one year later, the same star will pass through the meridian at the same time of day.

For Arcturus to pass the meridian exactly 22 hours earlier, it should be 22/24 of a year before then, or about 335 days later, or about 30 days earlier, or March 29.

  solar time equal sidereal time minus 12hr

Generally not correct.  Solar time is based on the average time it takes for the sun to cross the meridian (24 hours), and sidereal time is based on the time it takes for a star to cross the meridian.  Sidereal time counts 24 "hours" in 23 hours, 56 minutes, and 4 seconds, meaning that it runs slightly faster than our normal solar time.

There is one day in which solar and sidereal time are 12 hours off, on the northward equinox in March.  The Sun is on the northward equinox, but solar time is measured since midnight where sidereal time is measured since passing the meridian (local noon).

There is one day in which solar and sidereal time coincide, on the southward equinox in September.  The Sun is on the southward equinox, 180 degrees in the sky from the northward equinox, so the 12 hours off from midnight and the Sun being 12 hours off from the northward equinox cancel.


Is the angular resolution of a telescope irrespective of used eye-piece? Suppose a telescope has aperture $D = 20$ cm. The angular resolution of such telescope, according to the Rayleigh criterion (if I have understood it correctly), is given by

$$\theta = 1.22\cdot\frac{\lambda}{D} \text{ rad}$$

where $\lambda$ is the wavelength of light and $D$ is the aperture. Suppose $\lambda = 550$ nm (mid-point of visible light). Then

$$\theta = 1.22\cdot\frac{\lambda}{D} \text{ rad}
= 1.22\cdot\frac{550 \text{ nm}}{20\text{ cm}} \text{ rad}
= 1.22\cdot\frac{5.5\cdot10^{-5}\text{ cm}}{20\text{ cm}}\cdot 206264.8''
\approx 0.69''
$$

In other words, the angular resolution of the telescope is $0.69$ seconds of arc for $\lambda=550$ nm.

Now, is it really so that the eye-piece of the telescope has no play in this at all? Wikipedia says that

  and D is the diameter of the lens' aperture.

which I would understand to be taken as the size of the primary mirror alone in a Newtonian telescope.

So if there is for example a double (binary) star, I will see them as one no matter what kind of magnification/eye-piece I am using if the angle between the stars (from my point of view) is less than $0.69''$? And vice versa, if the angle between the stars is greater than $0.69''$ I will see them as separate stars with any eye-piece or magnification?

(In theory, I know there's a lot of limitations from optics, seeing, etc.)

The resolution of a telescope is the resolution of the image created by the primary mirror at the focal plane.  It provides the minimum separation between two equal brightness stars that appear separate in that image.  Often, astronomers put a photographic plate or a CCD at the focal plane, create long exposure photos and these can then be examined at arbitrary resolution (with a magnifying glass, microscope, or zoom software on a screen).  But, no matter how much you magnify the image, stars separated by less than the resolution will appear merged.

It is true that if you use, say, a wide-angle lens when observing directly with a telescope the resolution that you get can be limited by the resolution of your eye.  What is happening is the telescope is forming an image on your retina that has the full resolution, but your eye cannot make use of the finest details.  On the other hand, a hawk looking through the same eyepiece may get full resolution benefits since hawks' eyes have better resolution than humans'.


Collision of Galaxies According to the big bang theory, the universe started from a small intial point and is essentially expanding. However, my question is that if the universe is expanding how is it possible for galactic collisons to occur? Are the galacies moving away from a relative position but not moving away from each other? If so, how do we know this?

The expansion of space is something that happens on the largest scales.  At small scales, such as distances between nearby galaxies, other forces, such as gravity, dominate.  Galaxy clusters are held together by the attractive force of gravity between these galaxies.  Space in these regions is still expanding, but the gravity pulls on these galaxies much more than space's expansion moves them apart.

Galaxies that are close enough together, such as the Milky Way and Andromeda, are pulled together by the force of gravity between them.


What Is The Maximum Distance Our Finest Instruments Could See When They're Perfected? Radius of the observable universe is 47.5 billion light years,which means we are seeing things which are about 48 billion light years away. At a rough estimate,at what distance would our finest instruments and telescopes be unable to see or detect anything at all,& is there anything in the idea that if we were able to see far enough,we might see our own galaxy as it appears from an immense distance,its light having circled the universe? And as space is not a perfect vacuum,wouldn't the light be significantly retarded by the end of such a long journey?

The furthest we can "see" is the cosmic microwave background at a redshift of about 1100.

The proper distance of the CMB-emitting gas that we see today is about 46 billion light years.

If you are talking about galaxies, then the first are thought to have formed at redshifts of about 20 (current distance 36 billion light years) and beyond that are the universe's "dark ages" where no stars or galaxies should exist; the most distant galaxies observed at the moment are at redshifts of about 10 (current distance 31 billion light years).

All current distances depend on the adopted cosmological model. The numbers are for a flat $\Lambda$CDM universe with the Planck cosmological parameters.

There is no chance of seeing our own Galaxy, since it only formed about 12 billion years ago and the universe is a lot bigger than 12 billion light years.


Have there been successful attempts at spotting Neptune with the naked eye? As pointed out here, the limit of naked eye visibility for people with extremely good night vision is about magnitude 8.5. Neptune at opposition is much brighter at magnitude 7.7, so it should be visible to people with good eyesight. Astronomer Brian Skiff, made an attempt about a decade ago, he failed to spot it. He said that it should be relatively easy to spot from the Southern Hemisphere as it is high in the sky viewed from there.

But surely many other people with good eyesight who are capable of seeing faint stars of magnitude 8 or even dimmer who work at the Paranal observatory will have given this a try?

Well, having seen Neptune and identifying Neptune are two totally different things.

Let's tackle this one step at a time. Your link says:

  From these observations it would appear that, if seen upon a perfectly black background, a star of magnitude approximately $8.5$ would be at the limit of unusually good vision.

The problem with this is that there are stars, diffused sunlight and airglow that prevent the night sky from being near-perfectly black. This has always been a problem, and it's only gotten worse with air pollution — meaning that we would be less likely to see Neptune unaided nowadays.

My guess is that in the past, there have been a few people who have seen Neptune, but none that would take note of it. Do you know how many objects of magnitude less than $7.8$ are there in the sky? The answer is, approximately 30 thousand! Good luck finding Neptune.

Even with telescopes, Neptune can be hard to point out. Galileo actually discovered Neptune 200 years before it was recognized as a planet, but he mistook it for a star. Point is, even aided eyes often cannot identify Neptune.

Anyway, the reason Curtis could see such dim objects is because he had exceptionally good vision and was aided by tools:

  Without using the diaphragms and apertures, Curtis was able to see stars of magnitude $6.5$ directly against the background of the sky. With their aid he easily observed stars in the magnitude range $7.2$ to $7.4$. Four stars between magnitudes $7.9$ and $8.1$ were usually seen with little difficulty...

Without these tools, he probably wouldn't have seen it at all. Not to mention, Neptune's brightness fluctuates — at its brightest, it is $7.8$; at its dimmest, it is $8$. That might not seem like much, but when finding Neptune, it could make a big difference.

Regardless, even in ideal conditions, it is generally agreed that Neptune is the only planet that cannot be seen with the naked eye. Most sources state that you need to at least use binoculars to see Neptune.


Is it possible for a moon to only be out at night? (For the sake of this question, let's assume that all other factors are malleable.)

Essentially creating a permanent lunar eclipse, could a moon orbit around a planet so slowly that it is always opposite the sun from its planet?

If the Earth were alone in space and the hypothetical moon had negligible mass, you can calculate the distance at which the period would be one year using:

$$ a = \left( \frac{GM_E T^2} {4 \pi^2} \right)^{1/3} $$

where $GM_E$ is the standard gravitational parameter of the Earth (about 3.9860E+14 m^3/s^2) and $T$ is the period (about 365.2564 x 24 x 3600 sec). You should get about 2.2 million kilometers.

It turns out that because the Earth is orbiting around the Sun, the added gravitational effect results in a location a bit closer, about 1.5 million kilometers from the Earth, where a low mass object would tend to remain at the far side from the Sun of the Earth for a while.

This location is one of the five Lagrangian points, and has the designator Sun-Earth L2. There is a similar point L1 between the Earth an the Sun as well.

However, this is not a stable orbit, even in a perfect circular restricted three-body problem (CR3BP or CRTBP) scenario, the least little displacement would cause it to eventually wander away into it's own, somewhat chaotic orbit around the Sun within a few years. The reality of the more elliptical orbits of Solar System bodies plus gravitational perturbations would only reinforce this instability.

So find a way to give the hypothetical moon a propulsion system to do station keeping maneuvers to keep it there, and you've got almost what you are looking for. The problem is that this point is in the same neighborhood as the point where the Umbra of the Earth's shadow disappears.

Since the Earth's orbit is elliptical, the point where the umbra terminates moves a bit, and since there is no exact definition of a Lagrange point's location for a messy, realistic Solar System, the question of whether a satellite in the vicinity of L2 would be in partial or total eclipse is debatable.

If you already have the propulsion to do station keeping, you could possibly move it in to the umbra and maintain it there as well.


Why are stars more metallic closer as you move closer to the galactic bulge? As I see it, most of the stars in the galactic bulge are Population I stars. However, as one moves farther from the galactic bulge, star metallicity decreases. In fact, halo stars are almost entirely Population II stars. Why is this?

It has to do with the formation of the Milky Way.

At the beginning, the Milky Way was much more spherical than it is now - perhaps closer to what an elliptical galaxy is like than a spiral galaxy. Population III stars would have formed first, then quickly died out. Next came Population II stars. They formed when the galaxy was still somewhat spherical, and so they tend to inhabit the galactic spheroid/halo.

Eventually, the rotation of the Milky Way flattened out much of the remaining gas and dust, and some of the stars. When younger stars formed, they formed in the flatter disk, nearer to the center. The disk itself became smaller that the spheroid/halo. Thus, the younger Population I stars are found in the galactic disk, and are closer in.

No more stars will form in globular clusters; they are relatively dust-free and contain old, Population II stars. The same is true for the galactic halo.

Source: Populations & Components of the Milky Way

The galactic bulge itself contains several populations of stars. Some may have come from the halo and thick disk (thus being metal-poor) while others may have formed together more recently from the thin disk itself (thus being metal-rich).


Is it possible for planetary rings to be perpendicular (or near perpendicular) to the planet's orbit around the host star? Is it possible for planetary rings to be perpendicular (or near perpendicular) to the planet's orbit around the host star?

Yes, the plane of the rings of Uranus are at about 98 degrees to the plane of its orbit around the Sun.
This means that the ring system looks as in your picture twice per orbit. As the planet orbits the Sun, the rings, although still inclined at 98 degrees to the orbital plane gradually become "face-on" when viewed from the Sun. This will happen about quarter of an orbital period after the configuration illustrated in the picture. Then another quarter of an orbital period later, Uranus will be on the other side of the Sun, but with its rings tilted as shown.


How imperceptible will the 4-July-2020 penumbral lunar eclipse be? Wikipedia's July 2020 lunar eclipse; Visibility says:

It will be visible from some parts of North America during moonrise, South and North America, the extreme part of Namaqua Land in South Africa and Western Africa (completely visible), and in southwestern Europe, New Zealand, some parts of The Pacific Ocean, Central, East Africa (except Somalia, Djibouti, Eritrea and a strip of North Eastern Sudan,Egypt and Ethiopia) and some parts of North Africa and some parts of The Indian Ocean during moonset.

But on the sidebar there is an image that shows the outer edge of the penumbra cutting across the center of the Moon at maximum, which means half of the Moon will be full brightness and the other have only partially shadowed by the Earth.
Question: How much darker with the eclipsed half of the Moon get at maximum? Just how "imperceptible" will the 4-July-2020 penumbral lunar eclipse really be?

Source and caption in the main Wikipedia article:

The moon will imperceptibly dim as it passes through the Earth's southern penumbral shadow

Completely imperceptible except to instruments.
Space.com reports

In general, most people don't notice the penumbral shadow projected on the moon until at least 70% of its diameter is covered. Some people who have very acute vision and better-than-average perception might notice an ever-so-slight shading when only 50% of the moon is inside the penumbra.

As your image shows, this is a lot less than 50%. Nothing will be observed by the naked eye.  "Visible" in your first source only means "the moon will be above the horizon at this time". It doesn't mean you will be able to see a change or dimming of the moon.


Can Astronauts Aboard ISS See the Shadow of The Moon on The Earth During an Eclipse? I was wondering if the astronauts on the ISS are able to see the shadow from the moon on the earth at all during the eclipse.

Yes. Here's an ISS photo from the 2002 eclipse over the Indian ocean.
This year, 2017, the ISS will see the moon's shadow on earth once, over southern Illinois, at 43.9% totality.


Calculate eccentricity from apside altitudes? I need to calculate the eccentricity of a satellite orbit with only the apside altitudes (height above the surface of the Earth), 400 and 1500 km. However, I don't know the semi-major axis ($a$) or semi-minor axis ($b$).

$$\text{eccentricity} = (1-(b/a)^2)^{1/2}$$

The major axis is just the sum of the distance at perigee plus the distance at apogee. This is 400 km + 1500 km + the diameter (not radius) of the Earth, and the semi-major axis is half of that.

However, the semi-minor axis is harder to determine. Therefore, why not use an alternative method to calculate the eccentricity?

$$e = \frac{r_a-r_p}{r_a+r_p}$$

Here, the $r$'s are the distance to the center of the Earth, not the apside altitudes. Can you take it from here?


How to calculate when Mars was in retrograde around 129 BCE? I am trying to make an accurate illustration of the sky from when Ptolemy was observing the planets in Greece.

I am trying to find out when Mars was in retrograde during those years. I have read both SE questions and answer about how to calculate it mathematically but both of those are a little over my head.

Some of the silly astrology sites have the retrograde dates listed back the 1930s, I want to know what were the dates Mars was in retrograde around 129 BC.

https://physics.stackexchange.com/questions/249493/mathematically-calculate-if-a-planet-is-in-retrograde

https://physics.stackexchange.com/questions/288023/mathematically-calculate-if-a-planet-is-in-apparent-retrograde-motion

Mars' retrograde motion happens in the summer of 129 BC.

The JPL Ephemerides are an excellent source of solar system predictions, and are widely used. DE422 is about 600 MB and covers your period of interest.

Since your profile shows that you can program a computer, I've used the Python package Skyfield to download it, do the interpolation and calculate the apparent positions of Mars and the Sun from Alexandria in 129 BC. If I understand correctly, this would be year -128 in UTC, since there is no year zero.

You can also get similar data online using the JPL Horizons web interface.

I'll leave the problem of checking when the Sun was below the horizon but Mars was above the horizon in Alexandria on each day to you, but I've included the altitude methods in the script.

Mars is red, Sun is yellow:

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Loader, Topos

load = Loader('~/Documents/SkyData')  # avoids multiple copies of large files
ts   = load.timescale()

data    = load('de422.bsp')
earth   = data['earth']
mars    = data['mars']
sun     = data['sun']

ts      = load.timescale()

Alexandria  = earth + Topos(latitude_degrees  = 31.2,
                            longitude_degrees = 29.9,
                            elevation_m       = 5.0)

# NOTE! The year 129 BC is year -128 in UTC I believe
# SEE   https://en.wikipedia.org/wiki/Year_zero

# Check when Mars is "up" and Sun is "down" on a given day

hours    = np.arange(0, 24, 0.1)
one_day  = ts.utc(-128, 7, 1, hours)  # July 1st

mars_obs = Alexandria.at(one_day).observe(mars)
sun_obs  = Alexandria.at(one_day).observe(sun)

mars_alt, mars_az, mars_dist = mars_obs.apparent().altaz()
sun_alt,  sun_az,  sun_dist  = sun_obs.apparent().altaz()

# Check RA and DEC in 129 BC

days     = np.arange(1, 366)
one_year = ts.utc(-128, 1, days)

mars_obs = Alexandria.at(one_year).observe(mars)
sun_obs  = Alexandria.at(one_year).observe(sun)

mars_RA, mars_Dec, mars_dist = mars_obs.apparent().radec()
sun_RA,  sun_Dec,  sun_dist  = sun_obs.apparent().radec()

if True:
    plt.figure()

    plt.subplot(3, 1, 1)

    marsalt, sunalt = mars_alt.degrees, sun_alt.degrees

    plt.plot(hours, marsalt, '-r')
    plt.plot(hours, sunalt,  '-y')
    plt.plot(hours, np.zeros_like(hours), '-k')
    plt.title('altitude of Mars and Sun from Alexandria, July 1, 129 BC')
    plt.xlabel('hours (UTC)')
    plt.ylabel('altitude (degs)')
    plt.ylim(-10, None)
    plt.xlim(0, 24)

    marsra, marsdec = mars_RA._degrees, mars_Dec._degrees
    sunra,  sundec  = sun_RA._degrees,  sun_Dec._degrees

    # get rid of ugly lines
    dsunra, dmarsra = sunra[1:] - sunra[:-1], marsra[1:] - marsra[:-1]
    sunra, marsra = sunra[:-1], marsra[:-1]
    sunra[dsunra<-100]   = np.nan
    marsra[dmarsra<-100] = np.nan

    plt.subplot(3, 1, 2)
    plt.plot(days, mars_RA._degrees, '-r')
    plt.plot(days, sun_RA._degrees,  '-y')
    plt.title('RA of Mars and Sun from Alexandria, year of 129 BC')
    plt.xlabel('days')
    plt.ylabel('RA (degs)')
    plt.xlim(0, 365)

    plt.subplot(3, 1, 3)
    plt.plot(mars_RA._degrees, mars_Dec.degrees, '-r')
    plt.plot(sun_RA._degrees,  sun_Dec.degrees,  '-y')
    plt.plot([0, 360],  [0, 0], '-k')
    plt.title('RA vs Dec of Mars and Sun from Alexandria, year of 129 BC')
    plt.xlabel('RA (degs)')
    plt.ylabel('Dec (degs)')
    plt.xlim(0, 360)

    plt.show()


Calculating length of summer solstice day using tilt and latitude I want to be able to predict how long the summer solstice lasts at any part of world given the maximum tilt angle $φ = 0.4101524 \; \mathrm{rad}$ and the latitude $θ \; \mathrm{rad}$. I specifically want to do this using a geometric approach. The problem is, I never had experience with spherical trig before, and the values I got are somewhat off from their real values. So I want to know if that's because I didn't do the math correctly, and/or because of physical reasons (e.g. the Earth isn't perfectly spherical, the tilt changes slightly etc.).
Here is how I did it:

Let the Earth be represented as a perfect sphere of radius $R$ moving at a constant angular velocity. Let the Earth's rotational axis be in the z direction, and the axis of the circle forming the dark/light boundary be tilted by angle $φ$ relative to the Earth's axis. During the summer solstice, our location of interest (say, in the northern hemisphere) will trace out a circular path of radius $r < R$ perpendicular to the z-axis. These three circles, $C_{\text{rot}}$, $C_{\text{bound}}$ and $C_{\text{path}}$ create a spherical triangle. Using the arc length $s$ of this triangle along $C_{\text{path}}$, we can calculate the length of the day in hours using: $$24 \times \frac{πr + 2s}{2πr}$$

The point where $C_{\text{rot}}$ and $C_{\text{path}}$ meet is located at $$P(\text{rot/path}) = (R\cosθ, 0, R\sinθ)$$

To find the point where $C_{\text{bound}}$ and $C_{\text{path}}$ meet, we first observe that $$y = \sqrt{R^2 - x^2}\sinφ$$ and $$z = \sqrt{R^2 - x^2}\cosφ$$
Which we can solve for $x$ by setting $z$ equal to the $z$ from the previous point
$$x = R\sqrt{1 - \frac{\sin^2θ}{\cos^2φ}}$$
Plugging this back into $y$ and $z$, we finally get
$$P(\text{bound/path}) = \left (R\sqrt{1 - \frac{\sin^2θ}{\cos^2φ}},R\sinθ\tanφ,R\sinθ \right )$$
Now these two points are on the same plane in the z-axis, which is the plane of $P_{\text{path}}$. The distance $d$ between these points is
$$d = \sqrt{\left (R\sqrt{1 - \frac{\sin^2θ}{\cos^2φ}} - R\cosθ \right)^2 + R^2\sinθ^2\tan^2φ}$$
From this we can get $s$ by using the arc length formula:
$$s = 2\arcsin \left (\frac{d}{2r} \right )r$$
I used Mathematica to compare the calculated and expected values in 2017:

New York: $θ = 0.7105724077$; length (calculated) = 14 hours 56 minutes; length (expected) = 15 hours, 5 minutes

London: $θ = 0.8989737191$; length (calculated) = 14 hours 25 minutes; length (expected) = 16 hours, 38 minutes

As you can see, there is a noticeable difference (10+ minutes).

If you are reducing everything to plain geometry, you can compute the fraction of the "sunny" arc of the circle of latitude for a given location. (Hint: draw a diagram showing the tropic of Cancer, the circle of latitude, the terminator and the axis, all as straight lines.) The height of this segment of a circle of latitude (at $\lambda$) is

$$h = \frac{\cos(\lambda) + \sin(\lambda)\times \tan(\epsilon)}{\cos(\lambda)}$$

The angle of this segment is

$$a = 2\times \arccos(1 - h)$$

I get

New York: 14.9297092807291
London: 16.4195095586004

which is fairly close to the "expected" numbers you have provided. The difference is very likely due to Mathematica using a geoid rather than a sphere and including atmospheric refraction.


Astrophysical black holes deviating from Kerr black holes? Rotating black holes are formed due to the gravitational collapse of massive spinning objects. And, it is generally believed that Kerr black hole solutions are valid for the empty space outside of the astrophysical black holes.
My question: Are there some pieces of evidence (observational data) for astrophysical black holes indicating deviation from Kerr black hole solution?

Evidence of "new physics" seems to be lacking (in my research into this question.) Black holes can be described fully by known consequences of general relativity as far as our ability to observe them.  There are known deviations from the Scharwchild metric, for example in black holes in which the accretion disc is inclined relative to the spin of the black hole, and in which the effects of frame-dragging can be observed as the accretion disc transtions to the spin plane of the black hole.
High energy cosmic rays are a poorly understood phenonomum. They seem to be charged particles (ie protons) that have been acclerated to very high velocities (much higher than 99% c) by black holes.  But the particular mechansim that does this is unclear.  A paper, https://arxiv.org/pdf/1904.04654.pdf, proposes that high energy protons may be an consequence of black holes with a Kerr-Newmann metric, that is a rotating charged black hole; as even a small charge would have a singificant effect on protons in the neighbourhood. This could be interpreted as evidence that some black holes have a Kerr-Newman spacetime metric.


What is meant by the bifurcation of hydrostatic equilibrium shapes in planetary formation? An intriguing comment below this question links to this Quora question and points to the answer by Robert Walker.

The sentence that particularly intrigued me is:

  It turns out that there are two possible solutions, as the spin rate increases. You can get an oblate spheroid, or a triaxial ellipsoid - the solution "bifurcates". But the triaxial ellipsoid is the most stable of the two as Jacobi found out in his paper published in 1834.  Figures of Equilibrium - Historical Account* - Chandrasekar

It seems to refer both to an 1834 paper by Jacobi and it's reference in the 1964 paper Ellipsoidal figures of equilibrium—an historical account by S. Chandrasekhar, Communications on Pure and Applied Mathematics, 20 (2), May 1967, 251–265. While the latter is paywalled, there seems to be a readable version here in google in the book A Quest for Perspectives: Selected Works of S. Chandrasekhar : with Commentary By Subrahmanyan Chandrasekhar, Kameshwar C. Wali, Volume 1. Imperial College Press, 2001.

Mathematical bifurcation is a fascinating topic and there is even an entire mathematics journal dedicated to the subject.

In this case, where exactly in the calculation of hydrostatic equillibrium of rotating bodies does this bifurcation (mentioned in the Quora answer) occur? I'm wondering if it simply means that for a given volume and density, below a certain rotation rate the hydrostatic equilibrium shape is an oblate sphere, but above it, the hydrostatic equilibrium shape will be a triaxial ellipsoid? Or is the bifurcation part of the evolution of the shape over time, involving viscocity or external effects?

If the triaxial ellipsoide is always(?) the most stable, then is a oblate spheroid actually an intermediate shape, and not really the shape of hydrostatic equillibrium?

I think all this means is that up to some critical rotation rate, the equilibrium shape of a rotating, self-gravitating fluid is an oblate spheroid. That is, this shape defines a unique global minimum of energy.

Above this threshold there are two possible equilibrium solutions. Apparently one is an oblate spheroid, whereas the other is a triaxial ellipsoid. These will define local energy minima, but the triaxial ellipsoid is the global minimum.

The gory mathematical details are no doubt contained in Chandrasekhar's textbook on equilibrium ellipsoidal figures. Pages 3-5 of Iurato (2014) appear to summarise the historical development of this bifurcation in the equilibrium solutions.


Smallest mass of star to be a black hole? It seems to me that I forgot the smallest mass of a star and its angular momentum in order to form a black hole.

So I know that electron degeneracy pressure is overcome if the core is 1.4 solar masses, and neutron degeneracy pressure is overcome if the core is 3 solar masses, but if the neutron star is spinning fast enough, it can have a maximum mass of about 4 solar masses. But then, this is just the core of the star. How large does this star have to be in order to form a black hole, and how fast would its spin need to be in order to avoid forming a blitzar?

I've looked on google and bing but to no avail.

Help is appreciated!

Thank you!

Max0815

Please stop trying to make this a duplicate.

Why: How large does this star have to be in order to form a black hole, and how fast would its spin need to be in order to avoid forming a blitzar?

Stars with 6 Jupiter Masses to 90 Jupiter Masses but are not able to heat up sufficiently to enable fusion become brown dwarfs, and die brown dwarfs.

White dwarfs form when stars have a mass range of about 85 ± 4 Jupiter Masses to 8~9 Solar Masses, or when a star sufficiently heats up to enable fusion.

Neutron stars generally form when the star is >9 Solar Masses to about 24~25 Solar Masses.

Also, as mentioned in the comments, neutron degeneracy pressure doesn't support the neutron star as a whole. Although it has some effects, a neutron star is supported by not only neutron degeneracy pressure but also by repulsive nuclear forces as a neutron star is comparable to that of an atomic nucleus.

As for your blitzar definition, I doubt 4 Solar Masses is the limit. Let's say that the rotation of a given neutron star is a little less than the speed of light. Supposing that if the neutron star doesn't fling itself apart, the mass limit would be much higher, because the centrifugal force forcing the neutron star is much, much more stronger. A blitzar's maximum mass is 18% more than the maximum mass for cold, nonrotating neutron stars.
For your question on blitzars, the star would need to spin rather slowly to avoid forming such, because a blitzar's spin is from the amplification of the original star's spin as is shrinks.

Black holes form when a star is greater than 25 Solar Masses to the point where stars are so massive that they form a black hole itself before even forming a stable star. Such is an example of a quasi-star and is theoretically real in the early universe.


Tuning radio telescopes In a paper called Further Observational Evidence for a Critical Ionising Luminosityin Active Galaxies, Section 2.1, the authors talk about using the ICRF2 to find a tuning frequency in search of the 21-cm absorption (of HI):

The Second Realization of the International Celestial Reference Frame by Very Long Baseline Interferometry (ICRF2, Ma et al.2009), constitutes a sample of strong flat spectrum radio sources,of which 1682 now have known redshifts (Titov & Malkin 2009; Titov et al. 2013 and references therein), yielding a tuning frequency in the search for 21-cm absorption. Being VLBI sources, all have significant compact flux, thus maximising the chance of a high covering factor and thus optical depth (Curran et al. 2013a).The original aim of the survey was to form part of a large observing campaign to search and quantify the incidence of associated ${\rm H\,{\small I}\,21}$-cm absorption over all redshifts, although observing time was only granted for the high redshift $(z \gtrsim 2.6)$  proposals.

Since we already know the frequency of HI (1420.40575177 MHz), why is this so important? What am I missing about the functioning of radio telescopes?
This question doesn't really answer my query.

The very first sentence of that paper makes it clear that it is concerned with observations of the 21cm line in the redshifted universe, i.e. on objects so far away from us that redshift becomes significant. And just as any other electromagnetic emission, radio waves are affected by this. So, in section 2.1, the authors describe selecting a sample of radio sources with known redshifts so that they would know how much they had to tune their receivers downwards in frequency.

This is actually quite common and relevant in today's 21 cm radio observations. Redshifted signals are observed even down to 140 MHz and below as stated on LOFAR's Redshifted 21cm hydrogen Line page.


What exactly is the cosmic definition of ‘now’ is? There is no cosmic definition of now because of special relativity, because of rate at which time flows for a particular object depends upon your point of view and your velocity. For a rocket traveling at speed of light, journey of alpha centauri might take a week. For us it would seem like 4 years. As a result of special relativity even the meaning of ‘simultaneous’ is related to your point of view

The cosmic "now" is well-defined: It is the time for an observer that has always been at rest in the Universe's comoving coordinates, i.e. the coordinates that expand along with the Universe. Although this reference frame is no more special than any other frame, it makes sense to use this. For instance, it is the only frame in which the cosmic microwave background is (statistically) isotropic, and it is the frame in which you'll see the same number density of galaxies in all directions.

You could (rightfully) define "now" in another frame, but it wouldn't be of much use. For instance, for a neutrino released 1 s after Big Bang, the age of the Universe is some ten thousand years, but it doesn't really tell you anything about the Universe.


Can Jupiter's bands be made out using a 15x70 pair of binoculars? I have been getting mixed opinions on this. I have a 15x70 Pair of Binoculars. I have tried staring at Jupiter for several minutes with it, I was able to see the Galilean moons, in fact the last time was so long that I noticed the Moon's position changing a little bit.

But the glare of Jupiter is so high that I could not see the bands. Is it really possible to see any of Jupiter's bands using a 15x70 pair when the angular size is about 45 arcseconds ?

I use a 20x80 binocular on a tripod and I cannot make out Jupiter's bands. I believe that a 15x70 binocular will not be adequate.


Do we know how old the matter that makes us is? Last weekend, I had the pleasure of visiting an observatory where there was a series of lectures to cover up the fact that there was too much cloud cover to actually observe much at all.

One of these lectures involved the life-cycles of stars, which mainly involves Hydrogen and Helium.

Out of curiosity, I raised the question

  How old is the matter that make us and where does it come from?

The "where" is covered by Nucleosynthesis, but the "how old" was shrugged away with "maybe 2-3 sun lifetimes".

Is it possible to more accurately estimate the age of the matter that made up our solar system's accretion disk?  Or is the likely source of the supernova/collisions involved unlikely to be known?

The material (heavier than helium) that makes up the solar system was made in millions, if not hundreds of millions of stars that lived and died  in the ~7 billion years between the formation of the Galaxy and the birth of the Sun. This material has been thoroughly mixed in the interstellar medium and so the heavier elements arise from countless individual stars of different ages. You can say something probabilistically, based on what we understand about the build up of chemical enrichment in our Galaxy.

Starting with hydrogen - the vast majority of hydrogen is primordial. That is, it hasn't been inside another star.

The carbon in the solar system is made in a variety of stars. Some will come from high mass stars and dispersed either when they exploded as supernovae or via winds from their surfaces during the Wolf-Rayet phase; some will have been manufactured in lower mass asymptotic giant branch stars and expelled in their winds. The distinction is important - massive stars enrich the Galaxy very promptly (see below, when discussing oxygen), near the beginning of the life of the Galaxy, long before the Sun was born. The enrichment rate by supernovae and massive stars tailed off considerably at later times. On the other hand, enrichment by lower mass stars takes longer because of their longer lifetimes. Thus the carbon in the solar system will have a spread of ages from about 13 billion (the oldest stars in the Galaxy), right up to the age of the Sun.

Elements like oxygen, phosphorus, magnesium, sulphur and silicon are primarily made and dispersed by massive stars and supernovae. These are likely to be at the older end of the range discussed above. On the other hand, nitrogen is also made in a mixture of stars, both high and low masses, with a consequent range of ages.

Iron and nickel are mostly dispersed into the interstellar medium by type Ia supernovae. These are exploding white dwarfs, which in turn are the remnants of comparatively long-lived stars of a couple of solar masses or so. Most likely, these elements were made not long before the Sun was born.

See also Parent stars of our Sun - Where are its remains? and How can there be 1,000 stellar ancestors before our Sun?

A more technical description of this differential enrichment history can be found in Pippino & Matteucci (2008).


Do solar systems have to evolve in a galaxy? Is it possible for a solar system like ours to exist and evolve apart from all other star systems and galaxies. Sort of like a stand alone solar system or rogue solar system for lack of a better name. Or is there some requirement that a solar system capable of managing and growing planets especially ones that will occur in the habitable zone originate in a galaxy?

Probably not. You'd need a could of gas dense enough to contract under its own gravity. In the space between galaxies, matter is spread too thin for a solar-system sized cloud to exist. A much larger cloud could be more diffuse and still collapse, but being much larger, it would collapse into a galaxy.

Even within a galaxy, stars seem to form when shock waves from nova or some other event initiates the gravitational collapse.


Identifying cluster of exoplanets between Deneb and Vega I noticed an interesting patern of exoplanets.
Is this a known phenomenon?

That's the field the Kepler telescope was observing during its primary mission (before the second reaction wheel failure). The shape is due to the detector array.
I.e. this is not a real pattern in the actual distribution of exoplanets, it's a consequence of how the observations were done.


How do I get from Star Atlas to real life? This question is about finding the elevation and bearing of a given star at a particular time at Latitude 52N.

I look up the RA and Declination of my star. How do I calculate its bearing and elevation at hourly intervals?

The Right Ascension of a star is fixed (ignoring precession of the equinoxes). What you need or want to know is the "sidereal time" - that is to say the "time" based on which RA is at your local meridian at a given time. The sidereal day is slightly shorter than the mean solar day (ie the 24 hour day) and so sidereal and solar times drift apart - only being equivalent at the vernal equinox.


The position of Polestar Despite the revolution of the earth around the sun, why does the Polestar appear to be at the same place all through the year?
My daughter, a class 8 student, asked me this question.

The Pole star, and all other stars, are a long long long long way away.  Space is big. Also, the pole star just happens to be very close to the axis of the Earth's rotation.
As the Earth goes around the sun, it would cause the position of the pole star (and other stars) to move with a regular annual motion.  The more distant a star is, the less it appears to move. The pole star in particular moves by about 2.1 millionth of a degree.  That is not enough to be noticed by eye.  It is mind boggling (to me) that we can measure that with telescopes!
There is another motion, and that is the apparent motion of stars across the night sky each night.  That is not due to the orbit of the Earth but the rotation of the Earth. The Pole star is close to a line drawn through the Earth from pole to pole, this means that, (unlike most other stars) it also moves relatively little from hour to hour each night; it always appears in the Northern sky.
So the motion due to the Earth's orbit is very small (for all stars)  The apparent motion due to the Earth's rotation is small (due to the Position of the pole star) and so the pole star does not appear to move, it is always in the same position.


When is the belt of Orion slanted from right to left, going up? (Egypt) I am trying to match up a detail of the Senemut ceiling to the time of the year it might involve. In the detail (shown below), the belt of Orion is slanted upwards from right to left:

Notice that the Orion Nebula is to the right. When I try to duplicate this in Stellarium, I can only get the belt to slant left to right, going up as shown below:

Is there some particular time of the year when the belt of Orion is visible from Egypt (Banî Suwayf) and slants upward going right to left, as in the Egyptian ceiling?

One answer below suggested that Orion's belt is like this when it rises, but I cannot duplicate it:

In Stellarium, I found by experimentation that Orion seems to rise on the morning of July 18th at the latitude in question and date range in question. However, as can be seen in the screen shot the belt still goes left to right.

EDIT: entirely new answer based on comments

You may actually be on to something here.

The most obvious (and thus least interesting) answer is that they were
depicting Orion shortly after sunset, not shortly before sunrise (all
times UTC):

Here is the helical rising of Orion today, and it does indeed go right
to left:

If we go back 1000 years, it's much closer to being a straight line:

And if we go back to 1473 BCE (which is -1472), the rising is clearly
left to right:

Interestingly, if we go far enough south, the rising is right to left
even in -1472:

Note that position in which Orion rises doesn't change much from day
to day. In other words, the position of Orion's stars at helical
rising are the same as the position of Orion's stars at rising any
other day of the year. Because of precession, the position does
change over long periods of time, but not day to day.

The only problem with the "most obvious" answer above is that we know
the Egyptians were very interested in helical risings, so it would be
odd to see a depiction of a "helical setting".


Why does black hole draw off some mass from orbiting star instead of sucking it? Researchers saw repeated outbursts — every 114 days, on average — in a galaxy some 570 million
light-years away from Earth. Here outburst because of black hole draws mass off from the galaxy?
Location: Galaxy ESO 253-3.
Why does black hole draw off some mass from orbiting star instead of sucking it? As we all know black hole sucks things instead of weird behavior like this.

Black holes don't "suck things in", they pull on things gravitationally with their mass, same as any other object. You might as well ask why the Sun doesn't suck the Earth in. The natural thing is for objects to orbit, not collide with, stars or black holes.


Difference between quasar and Active Galactic Nuclei? Are there any differences between quasars and Active Galactic Nuclei (AGN)?  As I understand it, they are both caused by extremely luminous electromagnetic emission from material accreting onto a super-massive black hole.  If they have the same source, is there still some fundamental difference between them (and if so what is it) or are they two names for the same thing?

All quasars are AGN, but not all AGN are quasars.

AGN is a terminology that comes later than quasars. Quasars is the term applied at the beginning, when the first objects of this type have been discovered. They were radio-loud and point-like (the so-called quasi-stellar radio sources). This characterization still holds nowadays. Another property is that quasars are cosmological, that is they are objects in the distant universe (redshift $z > 1$).

AGN is the most general term we can use to refer to active galaxies. AGN include all of them (Seyfert galaxies, Quasars, Blazars, etc.). In the end, it seems that the difference among various types of AGN is due to a mix of angle-of-sight/epoch-of-observation.


Word for blue ring around the Earth From certain positions in outer space the Earth will appear to have a blue ring around it when the viewer is in the shadow of the Earth and sunlight is passing through the atmosphere.

Here is a simulation from Google Earth:

Sample sentence: The astronaut looked down at the blue ring formed by the sunlight passing through the atmosphere around the Earth.

Is there a word that describes this ring?

I had assumed the ring would be blue, based on this photograph from the Tumblr site Col. Chris Hadfield:

There are actually a number of words / expressions related to this.

  Aureole

[Bright edge of object - sometimes with colour of light source vs object] This is when an object is silhouetted against a bright light source and it gives a ring of colour. This typically occurs when the light source is smaller than the sillhoutte. I think this is the closest meaning because its meaning implies the colouration caused by diffusion (or even refraction - in this case) of light around the sillhoutte. For example, a girls head in front of a candle: the fine granularity of the hair diffuses the light, bending it and causes a bright circumference of light around the extremities of her hair. This word is associated in optical physics with a number of other diffusion phenomina too, such as the aureole effect.

  Corona

[Colour of the light source, shape of light source] This is a ring of light when the sillhoutte is much smaller than the light source (or even no sillhoutte at all!), and there is practically no interaction between the light and the extremities of the sillhoutte, as a result most of the colour is from the star.

Noteable mentions:

  A diamond ring

[Angle of light source to object] is where the eclipse has passed either side of total and there's a flare on one side.

  Bailey's beads

[Shape of object to shape of light source] are when there is a total eclipse and there are flares all around the circumference, caused by imperfections in the planet's surface and / or the irregularity of the star's corona - entirely diffraction based.

  Glory

[Entirely colour of diffraction / refraction of light source around object] This is an optical phenomina which is less common, and is almost entirely based on refraction, giving colour bands in the vicinity of the object, typically in concentric colour bands. A bit like a rainbow. This only occurs in the context of where the observer and object are in an atmosphere so is least applicable to this context - although technically the whole solar system is within the atmosphere (or heliosphere) of the sun, right up to the helioshock, where the heliosphere interacts with either other stars or forms a boundary with outter space, but it's not inconceivable that this could occur in a nebula, or even give rise to a

  Brocken spectre

Incidently: the blue aura, in this case, is actually caused by a physics phenomina called Rayleigh Scattering which is where the atmosphere has the effect of appearing to change the wavelength of light as it passes through. Often this can be blue, however in some situations it might be red.

There is a whole family of words for these kinds of meanings, and a many of them are poetic. Some of the words relate to colour, some to shape, some to diffraction, some to refraction. This is one of those situations where knowing a little science could help you chose the right word.