How can a rotation period of a planet's satellite/moon be "chaotic" (like Hyperion around Saturn)? According to wikipedia (see info below image) Hyperion's rotation period around Saturn is "chaotic".

What does that mean? The period (in days/hours) is different every rotation without any law helping to define it?

How can it be?

According to wikipedia (see info below image) Hyperion's rotation period around Saturn is "chaotic". What does that mean?

The wikipedia article refers to Hyperion's rotation being chaotic rather than its orbit about Saturn. I'll assume you are asking how Hyperion's axial rotation can be chaotic. (Its orbit is not.)

An object that is not subject to any external torques will have a constant angular momentum. This does not mean that such an object will have a constant angular velocity. Place a rubber band about a book and toss it in the air, giving it a nice spin rate before letting it go.

See the question Stability of rotation of a rectangular prism for details.

Hyperion is small enough and rigid enough that it's not quite capable of pulling itself into a roundish shape by its self gravitation. It instead has a markedly triaxial shape. Designating the principal moments of inertia as $A < B < C$ (corresponding respectively to the rotations in subfigures b, c, and a in the above image), the longitudinal libration parameter $(B-A)/C$ is about 0.21 for Hyperion. That's huge, and is a key factor that makes Hyperion's rotation chaotic. If Hyperion was a perfectly rigid body that wasn't subject to external torques and was instantaneous rotating about an axis close to that depicted as c in the above figure, it's rotation would essentially be chaotic.

None of the above conditions is true. Hyperion is not perfectly rigid and it is subject to external torques. Almost every regular moon in the solar system exhibits synchronous rotation and revolution. This is a result of tidal torques on the moon by the parent planet and the moon exhibiting a bit of plasticity (non rigid behavior) in response to those torques. The torques coupled with the plasticity are the mechanism by which those moons eventually became tidally locked.

There are other possible spin:orbit couplings than the 1:1 coupling exhibited by those tidally-locked moons. Mercury, for example, is gravitationally locked with the Sun in a 3:2 spin-orbit resonance. Mercury has a rather low longitudinal libration parameter of about $10^{-8}$. The high eccentricity of Mercury's orbit together with its nearly spheroidal nature makes it more likely to have been captured in the observed 3:2 spin-orbit resonance than in a 1:1 resonance.

With most moons and planets/exoplanets that orbit sufficiently close to the parent planet or star, the torque, plasticity, and low longitudinal libration parameter will eventually result in the orbiting body slipping into one of those small integer resonances. This is very unlikely to happen with Hyperion thanks to that very large longitudinal libration parameter. When Hyperion's spin and orbit happen to get close to a small integer ratio, the torque from Saturn is more likely to kick the moon into a rotation that is further away from the two stable axes than toward either one. It tumbles away from the resonant spin.


How to find Kepler/TESS data I have some basic understanding of the Kepler/TESS mission. The missions both look at the stars and try to find the dip in the light from the stars to find possible existence of exoplanets (exomoons as well). I was trying to find the dataset. I found one URL to retrieve the datasets. The URL is given below.

http://exoplanetarchive.ipac.caltech.edu:80/data/ETSS//Kepler/005/141/63/kplr010666592-2009131105131_llc.fits

But this URL contains the filename. But there should be several other files as well. To find the other files, I wanted to check the above URL without the filename (http://exoplanetarchive.ipac.caltech.edu:80/data/ETSS//Kepler/005/141/63/). But this URL shows error : "Forbidden
You don't have permission to access /data/ETSS//Kepler/005/141/63/ on this server."

I even created an user account in https://exoplanetarchive.ipac.caltech.edu/ but still I am receiving the same error. Please assist me, how I can get the keplar data.

Also I found that the file mentioned above is very small only a few KB and after converting it to csv I found that it contains only less 2000 rows. But I know that it needs several hours of observation to find out that an exoplanet, that means the dataset cannot be such a small dataset. Please explain why the dataset is so small. Is it only a part of the dataset, in that case please let me know how to get the full dataset.

It's usually a good idea to use the main page and follow links there instead of guessing URL; a search for "kepler telescope data" and "tess telescope data" quickly revealed these links:

Kepler: https://keplerscience.arc.nasa.gov/data-products.html which links for some data to https://exoplanetarchive.ipac.caltech.edu/index.html

TESS: https://heasarc.gsfc.nasa.gov/docs/tess/data-access.html

There no general access to derived data needed for a particular publication (though it's good practise and meanwhile becoming more common to offer that in supplementary online data).


What kinds of technologies are required to directly image exoplanets I have read that immensley powerful telescopes such as the European Extremley Large Telescope will be able to directly image the atmospheres of exoplanets, and even determine their rotation rates.

I am wondering, what are the obstacles that prevent current telescopes from achieving this; how will these obstacles be overcome exactly by the new telescopes such as the EELT; and tentativley when can we expect these kinds of telescopes to come online?

Also, is it theoretically possible to build a telescope (earth or space based) that can discern fine structures such as clouds, or surface features of exoplanets (or is this in fact the level of detail that will be achieved by telescopes such as the EELT)?

When the light from two telescopes is combined to make an interferometer, the resolution of the combined instrument is equal to the resolution of a single telescope with a diameter equal the the distance between the two (or more) smaller telescopes. This is the technology you need to view exo-planets to any resolution you want.

This has been incredibly successful in radio astronomy. Long baseline radio astronomy gives the resolution of a radio telescope the size of the diameter of the Earth. Optical telescopes in space can be even bigger. Arrays of relatively small telescopes with specially calculated spacings can do even better. Big mirrors are still better in order to get enough light on a sensor to be useful.

The twin Keck telescopes in Hawaii can form an interferometer but funding has dried up at the moment. It has an effective size of 280 feet. I don't know why there is no funding, but at this point it may be likely that a space-based interferometer would work so well that finishing the ground based instrument is not cost effective. Or the time required to set up and get data is a poor trade-off compared to other demands for telescope time. Also, longer IR is better for the planet imaging and space is a lot better place for that.

For some reason, this is a hot topic among PhD candidates, but a delay in getting pictures of these objects is not going to mean much for the rest of us.


How do elements with only one shell emit light? Electrons move to higher energy levels and drop back (electron migration). When they drop levels, photons are emitted and the wavelength/frequency emitted is based on how many levels are dropped. My question is, how does this work with elements with only one shell, such as hydrogen? Also, what determines how many energy levels are dropped?

Actually, hydrogen can have as many electron levels as you want. However, they only exist at specific energy levels, with the potential energy given as:

$$E=-\frac{R_H}{n^2}$$

Where $R_h$ is the Rydberg constant, and $n$ is any integer. When electrons are exited to a higher energy lever, they will fall back after some time and emit photons (not protons, by the way). So how many levels do they drop? Well, in the case of hydrogen, it will eventually go back to the basic energy level, but this is not guaranteed to happen in one jump. This may or may not take multiple jumps, that depends on probability. For a simple atom like hydrogen, those probabilities may be possible to calculate analytically. For more complex atoms however, this becomes an extremely difficult task, usually approached by simulations.


How to plot orbit of binary star and calculate its orbital elements? I have a set of dates, position angles ($\theta$) and angular separations ($\rho$) for visual binary star. For example:

1994.3132  57.0    0.097
1996.1783  73.4    0.051
2006.1938 182.0    0.126
2009.2597 192.9    0.152
etc.

What are algorithms to calculate orbital elements and to plot the orbit of the binary star?

Are there available codes (or programs) for this calculations?

If it's a binary, it's fairly simple (in comparison with a ternary system), because the stars of the binary orbit about their common barycenter in Kepler ellipses.

An orbit simulator, see here; replace the central star by the barycenter of the binaries.

Calculation of an Orbit from Three Observations. Kepler problem on Wikipedia.

If you don't find a ready-to-use software, try to solve it numerically by simulating the orbit of the binaries according to Kepler's law, and vary mass, distance, excentricity assumptions until they match to the observations. Use optimization methods, e.g. hill climbing algorithms, or gradient methods.


Does Earth changes angular velocity in 24 hours evaluation? I am trying to detect tidal forces influence on a pendulum. I use a laser interferometer that is able to detect very tiny space changes. The pendulum never stops even after long times quiet and protected from wind by a kind of plastic box. The pendulum is 2,60 m,  long, 10Kg and one of the mirrors of the interferometer is placed in the Iron cable near to the pendulum hanging point. I can see that the top energy of the pendulum changes but I am not sure of the causes.

There are a series of resources on the leapsecond.com site which specifically talk about detecting tides with pendulum clocks. It has been done by e.g. the Shortt pendulum clocks which had a master pendulum in a vacuum to remove disturbances synchronized to a slave pendulum. From the first link in the resources at 1, the order of magnitude of the change in pendulum rate caused by the change in gravity is ~1e-7 so your pendulum clock needs to be stable to a few milliseconds per day (few x10^-8) if you hope to see any tidal effects. You will need a more stable source of time such as a GPS Disciplined Oscillator (GPSDO) and normally a time interval counter to measure the stability with.


How bright is Andromeda's center, in magnitudes per square arcminute or something similar? I'd like to know how bright the Andromeda galaxy's core is as an extended object, not the total integrated brightness.

Ideally a contour map or scan across the center might show brightness in magnitudes per square arcminute or square arcsecond. Or a low power digital photo with some kind of brightness scale per pixel. Older images such as the one below are from emulsion and so if they are just scanned and uploaded they will not be linear.

I don't even know where to begin to look for such a thing. Thanks!

Below: "This picture shows M31 (NGC 224) and its small companions M32 (NGC 221), lower center, and NGC 205 (sometimes designated M110), to the upper right. The image was made by combining three separate frames derived from photographic plates taken in 1979 at the Burrell Schmidt telescope of the Warner and Swasey Observatory of Case Western Reserve University (CWRU)." From here.

The surface brightness number depends on a few things: how large an area of the image is averaged, how close to the galactic nucleus it is, and in what wavelength band.
Some papers give a brightness profile along a line bisecting the galaxy image.

Sandage, Becklin, and Neugebauer 1969 report, for a 7.6 arcsec diameter circle centered on the nucleus and measured in the $V$ band, a total magnitude of 11.3 and an average surface brightness of 15.5 mag/arcsec$^2$.
The same aperture 30 arcsec east or west of the nucleus was 1.5 magnitude fainter.

Kent 1983 images the central bulge in the $r$ band and finds a surface brightness of 14 mag/arcsec$^2$ at the nucleus and 16 mag/arcsec$^2$ 10 arcsec away along the major axis of the galaxy image.

Walterbos and Kennicutt 1987
measure a 60 arcsecond circle and find a surface brightness,
1 arcminute away from the nucleus along the major axis,
of 18.5 mag/arcsec$^2$ in the $B$ band.


Why do black holes choke? I read that a black hole can sometime "choke" on a star:

  "...the disrupted stellar matter was generating so much radiation
  that it pushed back on the infall. The black hole was choking on the
  rapidly infalling
  matter."

Why does a black hole choke? How do astronomers manage to observe this rare event?

There is a limit to how quickly a black hole can accrete matter. As a star gets close to a black hole it may be tidally disrupted and pulled apart. The material from the star will then fall toward the black hole and as it does so, the gravitational potential energy will go into heating it up. Now the stellar material would not normally be able to fall directly into the black hole - angular momentum must be conserved, so first it will form an "accretion disk". Viscosity in the disk can transport angular momentum outwards, allowing material to then fall through the event horizon. The ultimate rate at which the black hole can swallow mass will be controlled by the rate at which material can reach the inner parts of the accretion disk.

If the accretion disk is very hot and luminous, there could be a limit to how fast material can be added to it. This is known as the Eddington limit. Basically what happens is that the radiation from the inner parts of the accretion disk is so intense that it can "blow away" any new material falling towards the black hole. This is then a self-regulating phenomenon because the luminosity of the disk is derived from the material falling onto it - and so an equilibrium will be set up that limits the accretion rate to that which produces only enough luminosity that allows the accretion to proceed. This limit on the accretion rate is what the article you refer to means by "choking".

An estimate of the maximum accretion rate is set by equating the accretion luminosity and the Eddington luminosity (the maximum luminosity that an object may have before radiation pressure will halt accretion). This in turn depends on the efficiency $\epsilon \sim 0.1$ with which gravitational potential energy is turned into radiation as material falls into the black hole.

The journal paper that reports this discovery - Vinko et al. (2015) - analyse the light curve of the event and suggest this is black hole of $10^{5}$ to $10^{6}$ solar masses, tidally disrupting and swallowing a star of about a solar mass.

The Eddington luminosity (for spherical accretion) would be $\leq 3\times 10^{10}$ solar luminosities , but this thing had a peak luminosity (even in X-rays) of $10^{11}$ solar luminosities indicating that it must be in the regime where radiation pressure would reduce the accretion rate. To supply this luminosity would require $\epsilon \dot{M} c^2 \sim 10^{11} L_{\odot}$.
For $\epsilon \sim 0.1$ this gives a peak mass accretion rate $\dot{M} \sim 2\times 10^{-4} M_{\odot}$ per day. Thus it would take more than 10 years to "swallow" the star even at this peak rate.


Are there any reliable optical measurements of the radius of a neutron star? I know that there is plenty of theory to predict the size of a neutron star:  my question is whether or not there are any reliable size estimates based on observation. Please note that, to be reliable, such an observation would need to have also a good estimate for the neutron star's distance from us, which must necessarily be reckoned from some independent source, i.e., it must not be reckoned from the apparent size of the object, or from theory, otherwise one is guilty of circular reasoning!

Yes, the neutron star RX J1856.5-3754 has an observed* radius of 17 km. After accounting for general relativity, its actual radius is calculated as 14 km. It is not the only neutron star whose radius is known.

* It's worth mentioning how astronomers arrived at this radius. Technically, fitting the X-ray data to a blackbody arrives at a calculation of a 5 km radius, which seems unusually small, so for a time it was believed that this star may be something more exotic, like a quark star. However, after atmospheric modeling based on temperature, a more accurate estimate of the emission radius was derived to be 17 km. Once gravitational redshift was accounted for, the value of 14 km was yielded. See Magnetic Hydrogen Atmosphere Models and the Neutron Star RX J1856.5−3754 - Wynn C. G. Ho, et al.


Does the universe have an edge? I've heard the universe is just an empty cavity. Does the universe have any edge or stop or is it just an infinite empty cavity?

As far as theories I've heard of, we can't see the edge of the universe (if truly exists) because the farther you look out into space through a telescope, the farther back in time you are actually looking. For example, if you are looking at a star 20 light-years away... you're actually currently observing what was happening on that star 20 years ago... the theory is that there is only so far you can look out into the universe because there is only so much history post Big Bang.

Image above: What is the furthest we can see? In 2003, NASA's WMAP satellite took images of the most distant part of the universe observable from Earth. The image shows the furthest we can see using any form of light. The patterns show clumps of matter that eventually formed into galaxies of stars. Credit: NASA/WMAP Science Team


How thick is the cosmic microwave background, including the part we cannot see within the observable universe? What I want to know is how thick the observable universe is from the point of the cosmic microwave background and beyond.

It appears the thickness of the cosmic microwave background itself (the part we can see) is above 100,000 light years, per the following article:
http://scienceblogs.com/startswithabang/2013/06/19/5-facts-you-probably-dont-know-about-the-cosmic-microwave-background/

However, I want to know the thickness of that, plus what lies beyond that we cannot see, another way of looking at it would be the distance between the surface of last scattering (cosmic microwave background end) to the beginning (e.g. Big Bang).

According to the following article, it appears that this time from the beginning to the surface of last scattering is about 300,000 years:
https://ned.ipac.caltech.edu/level5/Glossary/Essay_lss.html

That would imply that the thickness should be about 300,000 light years, but that doesn't take inflation into account.

What is the thickness (in the observable universe) between the beginning (e.g. Big Bang) to the surface of last scattering (Cosmic Microwave Background), including inflation?

If I understand you correctly, you want to know the distance from the point from which we observe the CMB, to the edge of the observable Universe.

During inflation, the observable Universe expanded from ridiculously small to some ten meters in radius, so that part can be safely ignored compared to the distances now$^1$.

The distance$^2$ to the "CMB shell"$^3$ is 45.4 billion lightyears ("Glyr"), and the distance to the edge of the observable Universe is 46.3 Glyr. Hence, the shell of the observable Universe that lies beyond the CMB shell has a thickness of only 0.9 Glyr.

Here's a sketch of how I interpret your question (not to scale):

If you want, I can add details on how to calculate these numbers.

$^1$The relative expansion during inflation was huge, however: The Universe expanded roughly by the same factor that it has expanded afterwards, namely a factor of $\sim10^{26}$.

$^2$Here, "distance" corresponds to the comoving distance, which is what you would measure if you froze the Universe right now, and started laying out meter sticks.

$^3$This shell is not infinitely thin, but has a thickness of some 60 million lightyears, so let's ignore that.


How to calculate the position to be to see ISS passing in front of the moon or the sun? I would like to know how to calculate the time/date of the ISS passing in front of the moon (between the observer and the moon) or the sun, or to calculate the position in which the observer have to be to see the ISS passing in front of the moon/sun.

Th goal is to make a photograph of the ISS in the moon/sun.

Is it possible with tools such as Stellarium?

The ISS Transit Finder offers such predictions up to 30 days ahead.
If a query yields no transits at the specified location,
the "Show on Map" button for a close pass shows where a transit would be visible.
As the documentation notes,
predictions beyond 10 days are rather uncertain.

Recent versions of Stellarium can show ISS's apparent position at a given place and time,
offering an alternative prediction for any given event.


Why do objects all orbit in the same direction? In a solar system or a planet that has many moons, why do their orbiting objects all orbit in the same direction?

Is it possible for two orbiting masses to orbit in the opposite direction of each other?

This video provides a visual answer: https://youtu.be/MTY1Kje0yLg?t=3m24s
objects orbiting in different directions are more likely to collide with each other. By the process of elimination, eventually one direction of orbit dominates.


Cannot calibrate “upgraded starpointer finderscope” for Celestron AstroMaster 114EQ My wife just bought me an AstroMaster 114EQ which came with an "upgraded starpointer finderscope" -- it looks different than the one depicted in that link. I have tried following the instructions in their documentation to calibrate the finderscope with the primary telescope image and, unfortunately, the small red dot in the finderscope simply will not align with the main image of the telescope.

In attempting to calibrate it, I pointed the scope at a building about a mile and a half away and got it clearly focused in the main scope. I then looked through the starpointer finderscope and tried to adjust it but the range of adjustment would not allow me to line up the red dot with this building -- the dot was high and to the right and the adjustment knobs simply would not turn any more -- they were at the end of their adjustable range.

The supplementary documentation for the starpointer finderscope that came with the scope doesn't tell you which end points toward your target and which toward your face. I cannot really tell from the images in their website which orientation is correct. The mount doesn't seem to enforce any particular orientation. I assumed that the lense should be at the far end of the scope toward your objective and the power/brightness knob and the up/down adjustment knob, should be close to your face.

Can anyone suggest how I might remedy this? I did search here first and this other post did not seem to have the information I was after. I'm supposed to go camping the day after tomorrow and would love to be able to locate Jupiter. I'm concerned that the scope may have been incorrectly assembled...i.e., the mount is off or something.

I do not have one of these scopes, so my answer reflects general knowledge of astronomical telescopes.

Yes, the lens should be at the far end, away from your eye.

As for the alignment question, you can probably get a bit more movement in "left/right" direction by loosening the screws that hold the base of the finder to the scope, and twisting the mount and then re-tighten it.  For the up/down adjustment, you may need to use a shim (a thin bit of metal) to adjust the angle, but the adjustment built into the finder is normally adequate.

I doubt if there is a problem with the scope - if you can get a good focus, that's fine.


Why Helium could be produced in first minutes of big bang but not heavier elements At what extent could be sure that no heavier elements than helium could be produced, if we know that matter in universe today is about 25% composed on helium, why could be so sure that the kind of reaction He-He-He diden't took place?

The triple alpha process requires very high densities, very high (but not too high) temperatures, and a good amount of time. Large stars have all three. In particular, they have millions of years during which the very rare triple alpha process can proceed.

The temperatures in the very early universe were extremely high, too high to create anything more complex than protium. By the time the universe had cooled enough so that it could produce deuterium (and thence helium), its density and temperature had already dropped too low for the triple alpha process to proceed. All of this happened in a few hundred seconds. Compare that with the millions of years during which large stars burn helium.


Is normal matter always accompanied by dark matter and vice versa? Is normal matter always accompanied by dark matter and dark matter by normal matter?

Was ever observed effect of gravity lensing without normal matter?

We lack the precision to say that there aren't regions where there is matter without dark matter or vice-versa. But what is clear is that the ratio of dark matter to normal matter, which is (or needs to be) around 5 on average to explain the flatness of the universe, varies by orders of magnitude from place to place.

The reason for this is that matter interacts with matter in a completely different way to that in which dark matter interacts (weakly) with normal matter or itself by any other means than gravity. Gravitational interactions between normal/dark matter are not dissipative.

By that I mean that the sum of gravitational and kinetic energy is preserved. When normal matter interacts with itself there are normally dissipative consequences. For example, interacting matter may become hot and radiate away energy.

That is why the normal matter in our Milky Way galaxy is concentrated into a plane. Dissipative interactions during the gravitational collapse of our Galaxy, combined with the requirement for conservation of angular momentum result in disc formation. The same is not true for the dark matter associated with the Milky Way, which is expected to have a much more spherical distribution.

Likewise, when we look at the required density profiles of normal and dark matter in order to explain the kinematics of stars and gas in our Milky Way, we see that the normal matter is much more centrally concentrated than is the dark matter - hence we talk about a "dark halo". Again, dissipative interactions in normal matter are the reason.

Another famous example of the separation between dark and normal matter is the bullet cluster. The image below shows where the hot gas is (normal matter, observed by an X-ray telescope) compared to where it is inferred that the dark matter is (shown by contours) on the basis of gravitational lensing. This pair of galaxy clusters have recently interacted, passing through each other. The hot, intracluster gas has been stripped out of the pair and shock-heated, so that it appears concentrated between the two galaxy clusters. The dark matter content of each cluster has sailed on oblivious and remains concentrated around the visible cluster galaxies.

Yet another example would be dwarf spheroidal galaxies, these are relatively small galaxies that are nevertheless almost totally dominated by dark matter - they have mass-to-light ratios of $\sim$1000, compared with a "normal" galaxy, where this ratio might be $\sim$ 10.

There are also examples of dark galaxies. These may not be galaxies made of non-baryonic dark matter, but dark matter in the sense that for whatever reason, these galaxies do not contain many bright stars. Thus although they may be massive, they do not emit much light and contain a lot of "inert", non-starforming gas.


Is it possible to use the vacuum of space to help with pollution Would it be possible to use a pipeline from the ground and uses the vacuum of space to dispose of things that are hazardous to the environment? (Liquid waste in particular comes to mind) Essentially acting as a syphon that shoots the waste off into the universe?

No.
It is hard to get stuff into space.  If you built a pipe that went to space, it wouldn't suck stuff up for the same reason that the atmosphere doesn't get sucked off the Earth: Gravity pulls things down.  Instead you would have to pump stuff up the pipe, you would need lots of power to do this.  And when it reaches the top it would just fall down, unless you make your pipe 36000km long, so that the rotation of the Earth becomes strong enough to push it away
What you have essentially built is a space elevator. There are huge engineering challenges in making a space elevator.  It is honestly the very hardest way to get rid of pollution.


Is the radius used in the formula for the escape velocity the average radius of the celestial object or the radius at the starting location? I learnt that the escape velocity is given by $$v_e = \sqrt{\frac{2GM}{r}}$$

Say I want to launch a rocket from the earth into space and want to calculate the escape velocity $v_e$ (I guess without air resistance). Which value of $r$ do I have to use:

The mean radius of the earth...
... or the distance between the center of mass of the earth and the launch location?

Thanks in advance.

(I have basic classical (and quantum) mechanics knowledge, but I am very new to orbital mechanics.)

Just to provide an official answer, the radius $r$ in that equation is formally the distance between the center of masses between the two objects. That does come with some assumptions of course, but for most standard physics problems that concept is perfectly fine. As you stated, the equation does not assume anything about air resistance, and to extend that, it does not assume anything about continued propulsion. If you're at some radius $r$ from the center of the Earth and you have some velocity $v>v_e$, then just coasting along in your trajectory will allow you to eventually escape the Earth's gravitational pull (ignoring the obvious case where you intersect the Earth with your trajectory).


lousy mirror corrected by software Would it be possible to get a terribly shaped mirror, like a large sheet of something reflective, give it a general shape like a telescope mirror, but not anywhere near as precisely shaped as in a proper mirror, then use software to correct the image collected by the CCD into likely images?

The software could "learn" by being pointed at various objects , sun, moon, planets, stars, houses, etc, to learn it's own mirror errors, which you could correct, with the correction data saved as a filter,,much like facial and object recognition software operates, then use X number of high end graphics cards to correct the image in real time.
 Seems as though the software correcting for the terribly made mirror would eventually create a filter that would give a good image. Even if the errors required a lot of storage space on drives to create the filter, it should be able to correct the images over time into something recognizable and maybe accurate.

Another thing would be editing software for the image , you could manually or automatically bend the observed objects lines, correction software would straighten or curve lines in initial images, and the correction you apply would be applied to your terrible mirror's error correction filter.   Your terrible mirror could be an old silvered Pyrex plate, or a large piece of Mylar balloon material mounted on a shaped frame, or an intentionally formed piece of sheet metal, or such common things with an inaccurate mirror shape that's not ground to a wavelength of light as current mirrors are. Information not received and reflected by the lousy mirror could be filled in by the   software after it learns for a while.  The cheaper but much larger mirror might fill in some of that lost information~~~

There are a couple different ways this could go, depending on where in the optical train (of other powered items such as mirrors and lenses) this particular mirror is.

In a simple case, such as when you view a "funhouse mirror," the mirror causes pure distortion, and not any defocus or aberrations.  This is easily corrected either with a mirror of opposite shape, or by remapping the image based on the distortion parameters.  Remember: distortion does NOT blur anything.

If the mirror is in an intermediat focus plane, then you essentially have created the heart of an adaptive optic system -- albeit your mirror is not "helping" .  In this case, the mirror's shape is causing the optical wavefront (phase) to get degraded. The phase delay, which is a function of optical path length, varies across the mirror.  In this case, you will need some sort of phase sensing device, such as a Shack-Hartmann sensor, to identify the errors and possibly drive correction systems.

I suspect your overall question is more closely related to the first of these cases, so a simple remapping will suffice.  The typical approach taken by us optickers is to image a large picture consisting of dots on a grid.  We expect the image dots to be on a rectangular grid, so when they aren't we know how much to "move" each pixel in the camera to get the correct image.


How much higher is the radiation around Jupiter than at Chernobyl, after the explosion? I suppose the radiation levels are higher at Jupiter, but how much? Would the Juno probe have been able to "operate" on the Chernobyl roof in $12\,000\, {\rm Roentgen/hr}$?
After watching the short series Chernobyl, I was curious how the radiation levels compare with that around Jupiter (generally), and what the Juno probe in particular have to withstand, but I couldn't directly compare the numbers I found.
In the Chernobyl accident, the reactor explosion causes pieces of graphite to fall on the roofs of the reactor building, and they describe that they can't push them off with remote-control machines because the radiation damages the circuits.
In the series, the value given for the most dangerous roof is $12\,000\, {\rm Roentgen/hour}$.
The Juno probe's radiation vault is a titanium box with ${\rm 1.75cm}$ thick walls, and the article says

But this orbit would still expose an unprotected probe to a radiation
dose of more than 20 million rads [...] over the course of its mission.

The mission was planned for 14 Earth months but the probe only spends short periods of time around Jupiter.
Perhaps someone has an idea how to fill in the gaps and help to make a comparison.

Let's first quickly estimate whether the numbers by HBO's TV series make sense:
According to gesundheit.ch/strahlung the liquidiators were exposed to $2 \ldots 20 {\rm Sv}$ which compares to a life dose of $\approx 100 {\rm mSv}$ for usual people. I remember from my education that each worker was supposed to spent only ${\rm 40s}$ on the roof of the reactor. Assuming a dose of ${2 \rm Sv}$ in ${\rm 40s}$ corresponds to ${\rm 180Sv/h}$, so the value of ${\rm 12\,000R/h=120Sv/h}$ makes sense.
On the other hand, you give an absorbed dose of $20 \cdot 10^6 {\rm rad}=2 \cdot 10^5{\rm Gy}$ which is a physical quantity. However, the dose equivalent in Sievert quantifies the biological effect (of the deposit of a joule of radiation energy in a kilogram of human tissue), see https://www.ncbi.nlm.nih.gov/books/NBK230653/

${\rm 1 Sv}$ is the amount of radiation necessary to produce the same effect on living tissue as ${\rm 1Gy}$ of high-penetration ${\rm 200 keV}$ x-ray

Only if we incorrectly assume that the composition of the radiation on the roof Chernobyl in April 1986 and on Juno's flight path are comparable, we could proceed with cross-multiplication by setting ${\rm 1 Sv= 1 Gy}$.
But this (wrong) assumption is not enough, we also need a time frame for your Gedankenexperiment: If we assume that the radiation dose would remain constant (aka reactor core being exposed) at the lower bound of ${\rm 120 Sv/h}$, the maths gives $2 \cdot 10^5 \frac{\rm J}{\rm kg}/ (120 \frac{\rm J}{\rm kg \cdot h}) = 1\,666.7 {\rm h}$ which is a bit more than 69 days of survival for the Juno probe on the roof of a burning nuclear reactor. Realistically, I assume the probe would stop working much earlier.
Further reading
You might also find the presentation Charged Particle Fluxes and Radiation Doses in Earth-Jupiter-Europa Spacecraft’s Trajectory by M. V. Podzolko et al. interesting which gives quite a few estimates on radiation dose around Jupiter.


How many satellites orbit their planet faster than the planet rotates? Phobos is so close to Mars that it orbits Mars much faster than Mars rotates. This means that it rises in the west and sets in the east, even though its orbit is prograde.
Are there any other known natural satellites that orbit their planet faster than the planet rotates, or is Phobos unique in this regard?

Uranus rotates once every 0.718 days. 11 of its satellites have a shorter orbital period. These are inner satellites of Uranus which are roughly in the equatorial plane of Uranus. I don't quite understand what direction they orbit in compared to the direction in which Uranus spins.
Neptune rotates once every 0.671 days. 5 of its satellites have a shorter orbital period and orbit prograde.
Jupiter rotates once every 10 hours. 2 of its satellites have a shorter orbital period and orbit prograde.
Saturn rotates once every 0.4 days and none of its satellites orbit faster than that.
So Phobos is not alone in its case. Around Mars, Jupiter and Neptune (and maybe Uranus), there are moons that orbit so fast that they rise in the west and set in the east.


Can we calculate the average temperature of the heliosphere? It is my understanding that one can measure the temperature of the universe by measuring cosmic microwave background radiation. Could we also use this method for calculating the average temperature of the heliosphere?

The average temperature of the universe has supposedly been measured in at 2.735 degrees above absolute zero. Presumably this measurement takes into account the massive amounts of 'empty' space which I imagine has a huge effect on the number.

My question: Can we measure the average temperature of the heliosphere? If so, then what is it?

This might be quite hard to define as how do you determine when our solar system ends? So for the purpose of this question the following assumptions should be made:

The distance from the Sun to the edge of the heliosphere is approximately 100AU (Roughly the size of the heliosphere as measured by Voyager 1).
Assume the heliosphere is a perfect sphere.
Assume we are measuring the black body equivalent temperature (as this is what was measured to calculate the temperature of the universe).

If we don't know how to measure this average temperature (which is likely if we can't differentiate between microwaves from in and outside the heliosphere) then how could we theoretically measure this value assuming a perfect environment?

The heliosphere is mainly defined by the region dominated by solar wind against the interstellar medium.

"The solar wind is divided into two components, respectively termed the slow solar wind and the fast solar wind. The slow solar wind has a velocity of about 400 km/s, a temperature of $1.4–1.6×10^6 K$ and a composition that is a close match to the corona. By contrast, the fast solar wind has a typical velocity of 750 km/s, a temperature of $8×10^5 K$ and it nearly matches the composition of the Sun's photosphere."

Hence with 1 million Kelvin we get a good estimate of the average temperature of the heliosphere.

The temperature is the freeze-in temperature derived from carbon ion charge states measured by SOHO. Hence it's not the black-body temperature, but a reasonable definition of temperature for solar wind.

"Coronal temperatures are inferred from density ratios of adjacent charge states. The freeze-in temperature derived from a given density ratio is the electron temperature that reproduces this ratio in a static situation. For the analyzed iron charge states we use the ionization and recombination rates of Arnaud and Raymond [1992]". More details about SOHO's temperature measurement here.

If the heliosphere would be in the thermodynamic equilibrium, according to Planck's law or Wien's law, you would get the peak at about 3nm, in the soft x-ray range. But the equilibrium assumption probably doesn't hold.

For the hot interstellar medium extreme ultraviolet radiation measurements have been tried (CHIPSat), some successful (EUVE).
Hence, in an ideal environment we would measure the radiation intensity across the electromagnetic spectrum, look for the peak intensity, apply Wien's law, and retrieve the  effective temperature, which would be identical to the black body temperature in an ideal world.

The plasma of the heliosphere is very rare, therefore almost transparent. That's why we don't feel that heat of about 1 milion degrees. Microwaves play a minor role in that temparature range.

A temperature profile of Sun's corona and heliosphere is shown e.g. on pages 42 and 47 of this survey.


How likely is it that the Sun will destroy our electric society? Given the studies of the Sun and similar stars and the most impressive understanding of its physics, is there any kind of probability distribution for its dangerous behavior? There are also some historic data about events in 1859 and 775. What is the estimated probability of solar activity destroying our electric grid within this century?

I should give credit here to @honeste_vivere, who pointed out to me today that there have been recent studies excited by some extremely large coronal mass ejections that were classed as "near misses" in terms of causing major disruption.

Of particular interest to you would be the event of July 2012 discussed by Baker et al. (2013). I quote from the paper

" Had the season and time of day for this CME passage been right on striking the Earth, the world would have witnessed a storm larger (possibly much larger) than the 1859 Carrington event. This most likely would have had devastating consequences for many technological systems "

and then in the discussion

"
It is the opinion of the authors that our advanced technological society was very fortunate, indeed, that the 23 July solar storm did not occur just a week or so earlier.
Had the storm occurred in mid-July 2012, the Earth would have been directly targeted by the CME and an unprecedentedly large space weather event would have resulted. In fact, there is very legitimate question of whether our society would still be “picking up the pieces” from such as severe event [see NRC, 2008].
"

It looks likely that such  events are perhaps things that occur every few decades. Riley et al. (2012) suggest something as powerful as the Carrington event has a 12% chance of occurring in the next decade, with a 1% chance of something several times bigger. Even more powerful events are seen on other stars, where both flare and CME energies follow a power law relationship of the form $dN/dE \propto E^{-\alpha}$, where $\alpha \sim 2.5$ (Drake et al. 2013). Thus much larger, but rarer flares are possible and also seen on (usually) much younger and faster rotating stars. There may however be a very significant sporadic tail to the solar flare energy distribution too. Recent observations of candidate "superflares" on otherwise unremarkable solar-type stars have been reported, though the reality of these and the mechanisms are still being explored. Shibayama et al. (2013) There appear to be a group, consisting of 1% of stars that show (repeated) superflares more than 100 times the size of the Carrington event. These appear a little less likely around stars rotating as slow as the sun, but the bottom line is that, if there is nothing "special" about these stars, then these events occur every 800-5000 years (for events of energy 1e34 - 1e35 ergs) on a slow-rotating G-star. (Though you need to read carefully - it is probably a function of temperature and rotation). The authors note that superflares may be associated with the formation of very large starspots (or starspot groups). So that may be our clue to duck and cover.


How likely are planets to form after neutron star collisions? It is well known that planetary collisions can create moons orbiting the result of the merger if they happen in the correct way, and this is how the Earth's moon is believed to have been formed. See the animations on this Durham University page to get an idea of how the mechanism works http://icc.dur.ac.uk/giant_impacts/.
It seems to me that it should be at least theoretically possible for the same process to happen when neutron stars collide, which would produce bizarre extremely-high-metallicity (or rather high-average-atomic-mass) planets. However, I also know that the physics is very different in some ways: the colliding objects are much denser; the collision is much higher-energy; radioactive decay creates a burst of extra energy from any matter thrown off the objects; the gravity and velocity are high enough that relativity matters a lot; they probably are in very circular orbits spiraling toward each other rather than hitting each other from the angles that protoplanets do; etc.
It's also possible that most of the mass of the neutron star might be thrown away and leave a low-mass remnant that might expand into an high-atomic-weight planet or white-dwarf, or that some bit of ejected matter might be thrown out at similar enough velocities (speed AND direction) to eventually coalesce into a rogue planet.
I'm just wondering whether anyone has looked into this before, or if anyone has any input as to whether this would be more or less likely than moons forming from planetary collisions, or if anyone knows how to test this with simulations.
EDIT: I've just realized the reason why it is probably impossible for a planet to form in the same way the Moon formed around the Earth: The outward force is way stronger than gravity except for close objects, which would be inside the Roche limit of the resultant black-hole or neutron-star and thus form an accretion disc or ring rather than a planet (due to the fact that any potential planet would be ripped apart by tidal forces). I haven't done any math on this, and this is just my impression, though. In addition, this doesn't mean a planet couldn't form from the ejecta in other ways; for instance, the disc of matter close enough to be held in orbit after the initial explosion might be pushed out to include a planet-forming region outside the Roche limit during a later phase of the event.
EDIT 2: I've had an idea for how this might happen, but I think this might really be a different question. The idea is that, if another star was in the same system as a kilonova (collision between stellar remnants that ejects matter and radiation), the kilonova might leave enough of the star to stay in the system, or perhaps leave enough matter for the other star to capture it somehow. One thing about this scenario, though, is that the idea of another star being in the same system as a compact binary merger rather implies that this third star has already been hit by at least one supernova, possibly multiple and maybe several novae, depending on whether a parasitic binary was formed. (This wouldn't apply if the third star was captured into the system after both of the other stars had already died, though.) Supernovae are stronger than kilonovae in terms of energy that gets thrown out, so the previous supernovae would already have had a stronger affect on the object. I believe that kilonovae are thought to produce heavier elements than any type of supernova, so stars hit by kilonovae would be different in composition than ones hit by supernovae, but it's still basically the same question: What kind of remnants can survive from stars hit by supernovae/kilonovae/novae at close range. I think it's pretty obvious that this could form some kind of remnant, possibly depending on the distance to the third star, so that already answers my question, though I don't know what compositions are possible or what masses are likely, but I think this is really a different question that should probably be asked separately if I or anyone else want it answered on Stack Exchange.

There do appear to have been some studies on the properties of potential fallback discs formed after neutron star mergers, for example:

Rosswog (2007) "Fallback accretion in the aftermath of a compact binary merger
"
Cannizzo et al. (2011) "Fall-back Disks in Long and Short Gamma-Ray Bursts
"
Lyutikov (2013) "The Electromagnetic Model of Short GRBs, the Nature of Prompt Tails, Supernova-less Long GRBs, and Highly Efficient Episodic Accretion
"

These studies focus on explaining X-ray flaring in the aftermath of gamma-ray bursts rather than the potential to form exotic planets in these environments. It does seem fairly likely that some kind of disc does form around the remnant of a neutron star merger, but it's going to be extremely hot and likely so close to the remnant that it will be unable to form planets.
As noted in Menou et al. (2001) "Stability and Evolution of Supernova Fallback Disks", planet formation in fallback discs depends on the timescales for the disc cooling and how long it takes to spread beyond the Roche limit: if the disc becomes neutral before it spreads beyond the Roche limit, spreading becomes reliant on interactions within the remaining disc of rocks. While they consider the case of merging white dwarfs (noting that this scenario leads to a more favourable environment for planets than post-supernova fallback discs around black holes or neutron stars), they do not study the case of merging neutron stars.


How can we avoid needing a leap year/second? Given the Earth's current speed around the sun and current rate & axis of rotation, what is the best way to keep time to avoid a leap year? How many hours should we have in the day and days in a year would keep things balanced to not need to add or remove days from the year? Further, how many minutes per hour and seconds per minute should we have to avoid a leap second?

Leap years exist for two reasons:

There are not an integer number of days in a year.
People perceive a need to keep the seasons where they are on the calendar.

Given the above, there is no way to avoid leap years, or something similar. Defining the calendar year as being a fixed number of days (e.g., 365 days) would result in the seasons shifting by one day per four years.

Leap seconds exist for two reasons:

The length of a day as measured by an atomic clock is not constant.
People perceive a need to keep midnight at midnight, noon at noon.

Given the above, there is no way to avoid leap seconds, or something similar. Defining the day as being a fixed number of atomic clock seconds (e.g., 86400) would result in your clock and the Sun disagreeing on mean local noon, but by a very small amount.

That said, there are serious proposals to eliminate leap seconds. Some people such as those who use UTC to timestamp financial transactions do not like them. So far, those proposals have been rejected. The standard response is that it's not UTC that's broken; it's using of UTC in a context where it shouldn't be used that is broken. If you need a monotonically increasing time scale, use TAI or GPS time instead.


Why do gas giants have clearly delineated surfaces, whereas the Earth's atmosphere fades into space? I've just seen this Forbes article.

Why do gas giants appear to have clearly delineated surfaces, whereas the Earth's atmosphere fades into space?

Is it just a matter of scale?  Or is there some form of "surface-tension" for the hydrogen gas?

In an isothermal atmosphere, the exponential scale height of the atmosphere is
$$ h \sim \frac{k_\mathrm B T}{\mu g},$$
where $g$ is the gravitational field, $\mu$ is the mean mass of a particle and $T$ is the temperature (in kelvin).

i.e. The pressure/density of the atmosphere falls exponentially, with an e-folding height given by the above expression.

I suppose what matters when you look at a photo, is how this height compares with the radius of the planet.
$$ \frac{h}{R} \sim \frac{k_\mathrm B T}{\mu g R}$$

Jupiter is half the temperature, 11 times the radius and with 3 times the gravity of Earth. However $\mu$ is about ten times smaller (hydrogen vs nitrogen/oxygen). Overall that means $h/R$ for Jupiter is of order 5–10 times smaller than for Earth and so it will appear "sharper".

EDIT: If you put some reasonable numbers in for Jupiter ($T \sim 130$ K, $\mu=2$, $R=7\times 10^7$ m), then $h/R \sim 3 \times 10^{-4}$. This means even if Jupiter fills a photo that is 3000 pixels across, the atmosphere will be about 1 pixel high.


Declination change due to precession I have this problem: I want to calculate the declination of the star in the year 5000. And I know RA (right ascension) and Dec (declination) of the star right now.

How can I calculate it using spherical trigonometry.

  I know that the declination changes over time due to the precession
  and somewhere I read that I should use spherical triangle, but I am not
  really sure how to use it.

Thank you for your ideas.

I'm not absolutely sure if it is correct, but I use information from this site: http://star-www.st-and.ac.uk/~fv/webnotes/chapt9a.htm

And I convert equatorial coordinates (RA = α, Dec = δ) to ecliptic coordinates (λ, β)  using this formula.

  sin(β) = sin(δ) cos(ε) - cos(δ) sin(ε) sin(α)

  cos(λ) cos(β) = cos(α) cos(δ)

  ε = Axial tilt (23,5°)

I know that due to precession only λ changes (β remains same for ecliptic coordinates). So I add the angle of precession to the λ. Then I convert the new λ and β back to the equatorial coordinates.

  sin(δ) = sin(β) cos(ε) + cos(β) sin(ε) sin(λ)

  cos(λ) cos(β) = cos(α) cos(δ)


Where can I find the positions of the planets, stars, moons, artificial satellites, etc. and visualize them? What resources are available to find the positions of planets, stars,
moons, artificial satellites, asteroids, and other heavenly bodies?

There are many resources online, so this is a community wiki
answer. Please feel free to add to it!

If you want to visualize the stars/planets/etc (as viewed from Earth
or another location), you are looking for planetarium software:
https://en.wikipedia.org/wiki/Planetarium_software

If you want accurate positions for stars/planets/etc, you are looking
for horizons:

https://en.wikipedia.org/wiki/JPL_Horizons_On-Line_Ephemeris_System
http://ssd.jpl.nasa.gov/?horizons

If you want to compute star/planet positions yourself, you have
several options:

SPICE (http://naif.jpl.nasa.gov/naif/tutorials.html) will give you
results that match HORIZONS very closely. You can also use some of
SPICE's functionality online at
http://wgc.jpl.nasa.gov:8080/webgeocalc/#NewCalculation

SPK (Spice kernel) files are available at https://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/ -- while these are primarily intended for use with CSPICE, the format is documented and you can use it directly. It's described at https://github.com/skyfielders/python-skyfield/issues/19 and implemented in Python at https://github.com/brandon-rhodes/python-jplephem/tree/master/jplephem

If you want to do the SPICE computations yourself (numerically solving the differential equations), see https://astronomy.stackexchange.com/a/13491/21

You might also want to use an n-body simulator to do the computations yourself: https://physics.stackexchange.com/questions/25241/what-open-source-n-body-codes-are-available-and-what-are-their-features

IAU SOFA (http://www.iausofa.org/) will give you the International
Astronomical Union's "official" libraries to compute positions.

VSOP theory (https://en.wikipedia.org/wiki/VSOP_%28planets%29) is
another option.

PyEphem (https://en.wikipedia.org/wiki/PyEphem), which derives
from MIT's libastro library
(https://stuff.mit.edu/afs/athena/project/xephem/src/xephem-3.5.2/libastro/)

Skyfield: https://github.com/skyfielders/python-skyfield

libnova (http://libnova.sourceforge.net/)

If you want to treat planetary orbits as simple ellipses and ignore perturbations, you can find orbital elements at https://ssd.jpl.nasa.gov/txt/p_elem_t1.txt but you'll probably want to visit https://ssd.jpl.nasa.gov/?bodies#elem first

Note that these computational libraries don't always agree with each
other or with planetarium software:
https://stackoverflow.com/questions/16293146

If you want a catalog of stars/etc, NOMAD
(http://www.usno.navy.mil/USNO/astrometry/optical-IR-prod/nomad) and GAIA (http://gea.esac.esa.int/archive/) are
the largest such catalogs (about 1 billion entries each, lots of overlap). The NOMAD link also provides references to smaller catalogs.

For high-resolution astrophotography try ALADIN (http://aladin.u-strasbg.fr/#AladinLite)

If you want to find artificial satellites and spacecraft, start at
https://en.wikipedia.org/wiki/Two-line_element_set

Transits of Mercury and Venus across the Sun? http://eclipse.gsfc.nasa.gov/transit/transit.html

Planetary conjunctions?
http://laps.noaa.gov/albers/ast/conjun/conjun.html

Occultations of stars by asteroids? The International Occultation
Timing Association (IOTA) at http://occultations.org/

Create your own Sun-Earth-Moon model by using a ~17 inches (~44 cm) beach ball or box as the Sun, 4-5 mm object like a pencil eraser as the Earth and a 1 mm sized drill bit or other object to represent the Moon. Then, find a 150-foot string or tape measure. The sun and earth are ~154 feet (~46m) apart. The earth and moon are ~4.7 inches (12 cm) apart.


Finding the Uncertainty in a Spectral Line Currently, I am doing a lab in school on atomic spectroscopy. After taking readings of strong emission lines of various elements, I have fit the data using a Voigt Profile function from lmfit. The function minimizes the reduced chi-squared  and returns the best fit along with fitting parameter values (center, amplitude, sigma, gamma). I am mainly interested in the center of the line profile. But, what is the uncertainty in this reported center? Or how would I report the measurement and its uncertainty.

If you are doing chi-squared fitting, then a value for the uncertainty (or an estimate) should have been returned by the software. If not, you could perhaps manually estimate by fixing the line centre at various values, allowing the other parameters to be fitted, and seeing where the minimum chi-squared increases by 1 over the global minimum.

Failing that, output your spectrum as an ASCII file and fit it with something that does provide uncertainties like gnuplot.

Failing that, here's a rule of thumb. The line centre is determined with a precision given by the FWHM divided by the signal to noise ratio per pixel.

Note that none of the above take any account of uncertainties in the wavelength calibration.

EDIT: It appears that lmfit does estimate uncertainties. https://lmfit.github.io/lmfit-py/confidence.html


The sun's SOLAR WIND VS MAGNETAR As we know magnetars have strong magnetic field but what if we swap earth with a magnetar and the sun will released a solar wind. Is the magnetar affected by the solar wind or not ?

The solar wind is just a stream of charged particles, mostly protons and electrons, some alpha particles and a tiny trace amount of other stuff.

If the solar wind approached an uncharged Neutron star, then the particles would simply fall onto the neutron star or circle around it in a hyperbolic orbit.  The Neutron star's gravity would add considerable velocity to the particles due to it's strong gravity.   Any particle that falls within about 2 radii of the neutron star would inevitably fall towards it due to the space-time properties of relativistic orbits, so there would be some spiraling in and probably some kind of light-show in the process.  The impacts of particles onto the surface would likely be in the x-ray or gamma ray spectrums and might not be that visible.   Ultimately it wouldn't be that interesting, but perhaps there would be some kind of light-show.

If we give the Neutron star a strong magnetic field, then the solar wind would simply get redirected, similar to how it gets redirected by Earth's magnetic field, except in this case, the magnetic field is much stronger and the central object, much smaller.  Probably very few of the solar wind particles would hit the Neutron star and most would just circle around it's magnetic field, which would produce (I think) less of a light show.   There wouldn't be much effect that I can think of.  The Magnetar star is very massive with a very strong magnetic field.  The solar wind is comparatively, especially after spreading out for 93 million miles, quite diffuse and weak.

If a person got hit directly by the solar wind, it might be deadly.   We're not very tough, but a magnetar is one of the most indestructible objects in the universe.   The effect would be negligible.

It might be more interesting if you moved the Magnetar much closer to the star.   Magnetic field lines can't cross each other, but they can push each other and flatten out, and in the case of solar flares they can get very twisted (but still never cross).

A magnetar is several orders of magnitude smaller than a sunspot.   The interaction between the two magnetic fields, if the magnetar passed close to a sunspot would be interesting to see. (though the gravitational effect would still probably be the more impressive of the two)  But if you move a Magnetar as distant as the Earth, the solar wind would have lost much of it's strength and it would just be ejected material from a solar magnetic storm.  It wouldn't have much effect on the magnetar.

That's my hobbyist understanding anyway.   If I missed any key points, please clarify your question or ask a new question.


Icy moons: why hasn't the ice sublimated away? Let's take Europa for example:
The predominant model suggests that heat from tidal flexing causes the ocean to remain liquid and drives ice movement similar to plate tectonics.
The heat generated from the tidal flexing has to go somewhere. Is it all carried away by the water vapor plumes we recently detected? Is more water from the ice cover melting away? Is it just radiated away?
I would expect it to cause sublimation as well: There is hardly an atmosphere (molecular oxygen at 0.1 micro Pascals, or 10-12 times that of the Earth), and a tenuous ionosphere.
And that same article suggest that the surface ice is also broken down by radiation:

the oxygen [...] is formed through the process of radiolysis, where ultraviolet radiation from the Jovian magnetosphere collides with the icy surface, splitting water into oxygen and hydrogen. The same radiation also creates collisional ejections of these products from the surface, and the balance of these two processes forms an atmosphere.
The hydrogen, meanwhile, lack the mass needed to be retained as part of the atmosphere and most is lost to space. This escaped hydrogen, along with portions of atomic and molecular oxygen that are ejected, forms a gas torus in the vicinity of Europa's orbit around Jupiter.

Or do we have reason to believe that any losses are compensated by ice being deposited from external sources?

I'll pick up on your example as it can be used very well for illustrating the main points:

First: The solar intensity in Jupiters orbit is less than 4 percent compared to Earths. So much less energy comes from the Sun which limits possible sublimation significantly
Second: Sure tidal flexing leads to fountains. These fountains blow material into space. But this does not mean that all the material will leave Europas surface forever. It can fall back or it stays in Europas orbit and returns to the surface due to the effects of Jupiters magnetic field or gravitational forces of the other moons.
Third: Sure the heat of the tidal flexing has to go somewhere. And it is largely irratiated away. As Europas surface is quite bright it reflects most sunlight away instead of absorbing it. This makes Europas surface quite cool (50-125K). Compare it with the darker neighboring moon Ganymede (70-152K). As you can see the albedo (which means the reflectivity of the surface) is far more important than the tidal heating.
Fourth: Jep some of the water is split up by the radiation. This leads to an oxygen 'atmosphere' on Europa with 0.1 µPa surface pressure. While this is detectable it still is practically a vacuum. There is very little oxygen there, probably not more than a few hundred or thousand tons on the whole moon. Not enough to threaten the existence of moon with a mass of 4.8 * 10^22 kg

In summary: Yes, icy moons lose some of their material. But for exactly this reason they only appear in the outer solar system where the solar radiation is low enough to make this effect negligble.


How many tsunamis have been caused by meteorites falling? I don't know if this is a better fit for Earth Science (because of the tsunami part) or for Astronomy (because of the meteorite part).
The Wikipedia page on tsunamis says that tsunamis can be caused the the fall of a meteorite. The first reference after that claim mentions that tsunamis can be caused by meteorites, but in an unsubstantiated way. The second reference is behind a paywall so I can't access it.
How many recorded tsunamis have been caused by the fall of a meteorite?

None in historical times.
A tsunami is caused by a displacement of water, so for a meteorite to cause a measurable tsunami it would have to be very large.  The largest known impactor in historic times, in Tunguska, was about 50m diameter could have created waves, but probably not a tsunami.  There is evidence that a much larger impact, he Eltanin impact, 2.1 million years ago in the pacific ocean off Chile (a 1.5km diameter impactor) also did not create a tsunami.  The waves generated were short wavelength turbulent waves and the energy was lost in friction. Tsunami need to be very long wavelength to cross oceans and cause damage remotely.
Much larger impacts would create tsunami, and the size of the tsunami may exceed that created by earthquakes. See my source Are ocean impacts a serious threat."


Can you see the stars when it is trickling? I am writing a book with the following scene in one chapter: It is trickling and my protagonists are outside looking into the night sky, looking to the stars. Reviewing the scene I am wondering: Is this possible? Can you see the stars when it is trickling, or even when it rains? Are there images showing this?

Yes you can. You just need a partly clear sky (to see the stars) and part cloud (to produce the rain).

For an example of suitable conditions, look for pictures of a moonbow.  (I've only seen one once, personally.)


Is it possible to throw a camera attached to a rope within the event-horizon of a black hole and then pull it back in? I have heard that when something passes the event-horizon of a black hole nothing can stop it from getting pulled in. Is that true for objects partially past the event-horizon? Would it be possible to attach a camera to a stick or rope, put it inside the event-horizon, take a picture, and pull it back up and look at the picture?

According to the observer holding the rope, the camera would never make it past the event horizon. It would appear to slow down to a standstill just above the horizon (though would effectively disappear because of redshift). At some point the rope would snap, but the camera would still not appear to cross the horizon.

According to an observer with the camera, they would cross the horizon. But if they were lowered slowly, the rope would snap before they got there, since infinite force is needed to "hover" just above the event horizon.

They would then find themselves inexorably falling inwards in the same way that you and I are inexorably drawn into the future and can't revisit the past.

That different observers in different frames have different versions of events is not unusual in relativity.


Does gravity repel when dark energy is involved? So I am currently reading the book We have no idea by Jorge Cham and Daniel Whiteson.
In one of their chapters they talk about how gravity is a force that only attracts and doesn't repel, most of the time. Dark energy is what makes up 68% of the universe and the only way we know it exists is from its rapid expansion.

The expansion isn't slowing down, however, it is speeding up. Gravity is a force that is said to be a force that only attracts and doesn't repel, unlike electromagnetism, the weak force and the strong force. Gravity should be slowing down that expansion, and I am wondering if anyone can explain this or is this concept out of our reach.

In addition to not knowing, there's other reasons to suspect gravity wouldn't reverse.   There's no known force that reverses over distance (OK, the strong force, kinda/maybe but over very short distances).  Also, the expansion of space explains things that we seem to observe or suspect occurred like galaxies moving away from us faster than the speed of light and the rapid expansion that's suspected to have happened very shortly after the big bang.   Gravity can't do either of those things.

There's also the problem of how would gravity know that things are so far apart that they should start repelling, where nearby galaxies are still too close but more distant galaxies the repulsion kicks in?   It doesn't make a whole lot of sense that gravity would account for both attraction and expansion of space.

That's not to say it's impossible.  We don't know how gravity works or what it is, so, maybe.  Never say never, until something is understood.

@eshaya's answer, "We don't know" is probably the more straight forward answer to this question, but I wanted to point out some problems with repulsive gravity theory as well.

In addition (and I don't understand precisely why), but the gravitational waves discovery has ruled out some theoretical gravity/dark energy proposals.


Behaviour of black holes This question could be strictly linked to my another question related to black holes.

It’s known that the  horizon event is a place where dust,star components are ready to be incorporated to the black hole*.

Actually, my question can be divided in few points:

Might be reasonable to believe that ,if it’s true that in many spiral galaxies there is a black hole in the middle, could these objects have so a powerful gravitational force to consider them the creators of the galaxies?
(*) when the black holes are sucking up dust from the horizon event, are they increasing their mass?

If so, it’s reasonable to believe that their gravitational force will increase in a way to attract farther objects?
Could they eject the dust or anything else if they are “eating” so much?
     This can be in contradiction with above point.

No, it seems that the formation of a supermassive black hole is a consequence of galaxy formation, not a cause of it. The mass of the black hole is a very small proportion of the mass of the matter and dark matter in the galaxy. The details are not yet worked out but it is the distribution of dark matter that forms the galaxy, and then galaxy mergers contribute to the growth of a large black hole in the centre.
As indicated in the answer to your previous question, "sucked" is entirely the wrong metaphor for how black holes work. Black holes have lots of mass in a small space, so they "pull" gravitationally.

Now gravity gets weird when it is very strong, as it affects the time. From the perspective of a person outside the black hole, a clock on a body falling towards the horizon will slow down, and will appear to stop as the body reaches the black hole, This means that we will never see something fall into the black hole.  However the mass of the body that falls towards the black hole will increase the effective mass of the black hole, and so increase the radius of the event horizon.

For objects further from the black hole, the acceleration towards the black hole is proportional to its mass. So it is not the case that an object will "start to feel the pull", but instead the pull will increase.  Note that mass that is orbiting close to the black hole will act almost like part of the black hole from the point of view of someone further out. This is not a special property of black holes, just ordinary gravity.
Black holes form an accretion disk around themselves, a disk of matter orbiting the black hole at very high speeds and getting very very hot. They also produce high energy jets of particles moving close to the speed of light from their poles. These are produced from the accretion disk, not from the black hole itself. Objects that fall towards the black hole don't come back.


Are X-ray telescopes with glancing angle surfaces basically "funny-looking" Cassegrain telescopes mathematically? In this answer I included the image below of a reflective X-ray telescope. It is made from two elements; the first is concentric shells of glancing (high incidence) angle paraboloidal surfaces, and the second element is the same except that they are hyperboloidal surfaces.
A "Classic" Cassegrain telescope is also a parabolic mirror followed by a hyperbolic mirror, except that the configuration looks very different since for visible light we can use near-normal incidence and are not restricted to high incidence angles.
Question: Despite looking so different, are these basically the same thing? Is the mathematical relationship between the parabolas and the hyperbolas the same in both cases?
Related in Space SE: How does this strangely-shaped horn at Honeysuckle Creek Tracking Station work?

Source

Short answer, yes, the x-ray grazing incidence mirrors set up like this still function like a Cassegrain to minimize off-axis aberration.  However, it is still different in that a classical Cassegrain uses a fast positive paraboloid and a slow negative hyperboloid.
The Cassegrain design combines the paraboloid and hyperboloid to minimize off-axis aberrations effectively. In 1952, Hans Wolter outlined 3 layouts of paraboid-hyperboloid mirrors which could be used in a grazing-incidence for x-ray applications.  The type 2 layout is most equivalent to a Cassegrain, using a positive paraboloid followed by negative hyperboloid, the application used here (in the XMM-Newton observatory) is a Type 1.  As a result, the remaining aberrations are slightly different, though still minimized rather effectively. Also notable, is that the focal lengths of the paraboloid and hyperboloid in the Wolter designs are of similar focal lengths.
Type 1:

Type 2:

https://www.cosmos.esa.int/web/xmm-newton/technical-details-mirrors
https://en.wikipedia.org/wiki/Wolter_telescope


Is there evidence for galactic "seasons"? The Solar Year has huge effects on our climate, and the signs of passing seasons are physically evident on Earth. I recently learned about the Galactic Year. To my understanding, our sun orbits the Milky Way galactic center about every 200-225 million years. Do we know if there are any "seasonal" changes due to our position in the galaxy? One season of the orbital period (50+ million years) feels like a long enough time to leave a mark, if any effect exists:

in the formation of solar system objects?
the bombardment of planets and moons with meteors?
the number, type, distribution, or trajectory of comets?
the shape or density of the Oort Cloud?
in Earth's geologic record?
(at a stretch) mass extinction events?

Obviously you can just say "gravity affects everything, of course the solar system is influenced to some degree by other objects." But I mean specifically, is there evidence for any cyclical pattern which might be related to our position in the galaxy?

Mass extinctions have been linked with the Solar System's oscillations up and down through the galactic plane (the Galactic Cycle). You could also take a look at this question.


What was the first astronomical measurement which demonstrated that "the Earth is surrounded by vacuum"? The question Who was the first to realize that the Earth is surrounded by vacuum? was closed because some users felt it was answered by answers to a different question in an different SE site: Who was the first to postulate that space was a vacuum?

To me the distance between realize and postulate is astronomical!

That aside, I've asked a new question: What was the first astronomical measurement which demonstrated that "the Earth is surrounded by vacuum"?

By demonstrated I'm talking about a measurement that shows it is, or at least is likely to be the case that Earth is surrounded by vacuum. It should be something such that it could be presented as scientifically convincing evidence to other scientists of the time, unencumbered by non-scientific predispositions (e.g. religion, rivalry, philosophy...)

Torricelli, the inventor of the Mercury Barometer (~1644) argued that the height of the column of mercury was governed by atmospheric pressure (the "weight of the atmosphere" as he would have put it).   He asserted that the space above the mercury in his tube was a vacuum, a totally anti-Aristotlean concept at that time.
To test this he enlisted the help of Blaise Pascal and Florin Perier in carrying one of two barometers up a mountain, the Puy de Dome in central France. His prediction was that the pressure at the top of the mountain would be less than at the base because of a lesser weight of air.
The experiment showed a drop in height (ie of pressure) and Torriceli himself concluded that the height of the atmosphere would be about 20km, above which there would be vacuum - the same vacuum as in the top of the barometer tube.
The experiment was IMHO one of the greatest experiments in the  history of physics.  See for example  https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3768090/


Who is Seyfert, and what is "the so-called Seyfert flare"? The BBC's Milky Way's centre exploded 3.5 million years ago says:

A cataclysmic energy flare ripped through our galaxy, the Milky Way, about 3.5 million years ago, a team of astronomers say.
They say the so-called Sifter flare started near the super massive black hole in the centre of the galaxy.
The impact was felt 200,000 light-years away.
[...]The flare created two enormous "ionisation cones" that sliced through the Milky Way.

It mentions "The team - led by Professor Joss Bland-Hawthorn from Australia" and "co-author Magma Guglielmo from the University of Sydney" and says "The findings will be published in the Astrophysical Journal."
Right now I'm just asking the following:
Question: Who is Sifter, and what is "the so-called Sifter flare"?

update: @bertieb's comment notes that the BBC has corrected the passage. The new sentence says:

The so-called Seyfert flare started near the supermassive black hole in the centre of the galaxy, they add.

When first published, the article called it a "so-called Sifter flare" and that's what originally inspired this question.

As has now been corrected, BBC has misunderstood term "Seyfert flare", instead calling it a "Sifter flare".

A "Seyfert flare", is not really a common term, but the authors refer to an energetic outburst from the type of active galaxies called Seyfert galaxies (after Carl Seyfert).

Like a quasar, a Seyfert galaxy is powered by gas accretion onto a central, supermassive black hole, although they're less luminous by roughly two orders of magnitude.

In this case, there is evidence for a highly energetic ($10^{56\text{-}57}\,\mathrm{erg}$) explosion occurring only a few million years ago, resulting a an ionizing, bipolar cone extending outward from the Milky Way's central black hole, Sgr A*.

This powerful flare resulted in huge, 10 kpc-scale, X-ray/gamma-ray-emitting bubbles. The line ratios of $\mathrm{C\,IV}/\mathrm{C\,II}$ and $\mathrm{Si\,IV}/\mathrm{Si\,II}$ points toward ionizing radiation energies of at least 50 eV.

The orientation of these cones is seen in this figure from the paper, which will be on the arXiv tomorrow. The first-author, Joss Bland-Hawthorn, explains in a video here.


How close to Earth were the asteroids moving through this Hubble composite image? Recently Hubble saw several asteroids traveling erratically through a deep field perspective.

What region of the sky is this, and how near were these asteroids to the Earth?

The photo is composed of multiple shots and the asteroids have copies because of Hubble's movement through frames as the photo assemblage shifted.

Was Hubble viewing towards the plane of the solar system and the asteroid belt?
How did they explain that there were five in the same narrow frame?

https://astronomynow.com/2017/11/03/hubble-sees-nearby-asteroids-photobombing-distant-galaxies/

I visited the linked article in Astronomy Now and found the URL of the image contained the string STSCI-H-p1733a. STScI stands for Space Telescope Science Institute.

Searching that string leads to

http://hubblesite.org/images/news/release/2017-33 (four images)
http://hubblesite.org/news_release/news/2017-33
http://hubblesite.org/image/4080/gallery

the last two of which identify the location as Abell 370, at about RA: 2h 40m, Dec: -1.6°. At about RA = 0h the ecliptic and equator coincide, but by 2h 40m the ecliptic has a declination of about +16°, so this photo is looking about 18° below the ecliptic.

The first paragraph of the caption contained in both of those links says:

  Like rude relatives who jump in front your vacation snapshots of landscapes, some of our solar system's asteroids have photobombed deep images of the universe taken by NASA's Hubble Space Telescope. These asteroids reside, on average, only about 160 million miles from Earth — right around the corner in astronomical terms. Yet they've horned their way into this picture of thousands of galaxies scattered across space and time at inconceivably farther distances.

If the distance is about 160 million km, then they are suggesting these asteroids are in the inner part of the main asteroid belt.

The Earth is moving at about 30 km/sec around the Sun, and the Hubble Telescope is "only" moving 7.6 km/sec with respect to the Earth, in roughly the same direction. So much and probably most of the motion is due to the Earth's velocity. I'm suggesting this because if the distance to the asteroids is really roughly 160 million kilometers, then they would be near opposition to the Sun and the Earth's motion would be roughly perpendicular to the viewing direction.

There are a lot of asteroids! According to Wikipedia:

  One hundred asteroids had been located by mid-1868, and in 1891 the introduction of astrophotography by Max Wolf accelerated the rate of discovery still further. A total of 1,000 asteroids had been found by 1921, 10,000 by 1981, and 100,000 by 2000. Modern asteroid survey systems now use automated means to locate new minor planets in ever-increasing quantities.

Asteroids are discovered constantly, and sometimes this happens in images that were not intended to be searches for asteroids, as in this case. While most most asteroids orbit close to the plane of the ecliptic, some of their orbits do stray substantially above and below it, as in this case.

While the date of the News Release is 02-Nov-2017, I don't know the time frame of all of the exposures that contributed to this final composite image.

Here is a screen shot of the Planetarium Mode of Dominic Ford's in-the-sky.org. The marker shows the position of Abell 370 just below the equator, and the Ecliptic is the thicker line above it.


How does light affect the universe? When light is emitted by for example a star, that star loses energy - which causes it to reduce its gravity. Then that energy begins a journey for potentially billions of years, until it reaches some other object.

When that light reaches a surface, such as another star or galaxy, it will give that energy to the destination star in the form of heat. This causes the receiver to increase its energy, in turn restoring a sort of balance. It also causes the receiver to emit a minute amount of more light again, almost like a reflection.

It will also excert pressure on the receiving surface once it reaches its destination, be it a star, a rock or anything else.

But while that light is travelling through space, its energy is "unavailable" to the rest of the universe. Naturally I ask the following question:

Will light cause gravity, while it is traveling?

Every single star emits light in every direction, and will eventually reach every other star in the universe. At any single point in the universe, there must be a continous ray of light coming from every single other star in the universe, that has a direct path to that point. Given that all stars on the sky is sending photons that reaches every square centimeter of the earth surface, the amount of pressure should sum up to be quite large.

Is the amount of pressure really neglible, given that every single atom on any surface is receiving light from every single lightsource on the sky?

Based on a calculation found at http://solar-center.stanford.edu/FAQ/Qshrink.html the sun will during its lifetime emit 0.034 % of its total mass as energy. Assuming the sun is average, and that there are about 10^24 stars in the universe, and all of these stars on average are half way through their lifetime, there should be energy amounting to the gravity of about 1.7*10^22 suns distributed throughout the universe.

Yes, light gravitates. The gravitational charge is energy. Well, gravity is a spin-2 force, so you really have momentum and stress as well, but they are analogous to a generalization of electric current.

In general, anything that contributes to the stress-energy tensor will have some gravitational effect, and light does that, having both an energy density and putting a pressure in the direction of propagation.

  But while that light is travelling through space, its energy is "unavailable" to the rest of the universe.

Not quite. It still gravitates. However, the radiation-dominated era was before about 50k years after the Big Bang, but it is long past. Today the gravitational effect of radiation is cosmologically negligible. We live in a transition between matter-dominated and dark-energy-dominated eras.

  Given that all stars on the sky is sending photons that reaches every square centimeter of the earth surface, the amount of pressure should sum up to be quite large.

The light pressure on any surface is proportional to the light energy density incident on it. Thus we can check this line of reasoning directly by observing that the sky is dark at night.

Why it is dark at night is probably deserves its own question (cf. also Olbers' paradox), but it is pretty clear that it is in fact quite small. To be fair, we should check more than the visible range, but even so the sky is pretty dark. Thus on average, light pressure is very small.

We have the privilege of being close to a star, but even during the day, the light pressure due to the Sun is on the order of micropascals.

  ... there should be energy amounting to the gravity of about 1.7*10^22 suns distributed throughout the universe.

And this is a tiny amount. As you just said, this is the equivalent of about 0.034% of the total mass of stars in the universe, which is in turn constitute but a fraction of the matter in the universe. So why you are surprised that its effect is negligible? It's literally thousands of times less than the uncertainty in the measurements of the amount of matter in the universe.


What is the predominant element in the dust of the Sombrero Galaxy? The Sombrero Galaxy contains a rather peculiar ring of dust orbiting it (seen as the dark ring on the outer edge). What is the prevalent element in this dust? Carbon?

I think that first, we have to properly appreciate the size of the Sombrero Galaxy. It is roughly 50,000 light years (15 kilo parsecs) in diameter. That might be only half of the diameter of our own Milky Way galaxy, but still makes each and every pixel on the photograph you're attaching in your question stretch more than 100 light years in distance.

That's a huge distance, so we can probably agree that clouds made of nearly anything and at least as dense as what you'd on average expect the interstellar medium to be comprised of, would obscure even the strongests light sources within that galaxy. This is important to realize, so we somewhat lower your expectations; These areas aren't tightly packed with anything, we're just talking of such vast distances that the light has to travel through, that even almost nothing will be enough to stop it from shining through.

Now to your question what it is comprised of. It is in essence matter from the interstellar medium that stops this light from reaching us, so nothing in particular that you wouldn't expect also in similar areas between stars within our Milky Way, or indeed any other galaxy that is roughly in the same age period of its development. Since the Sombrero Galaxy isn't even all that far away in cosmological sense, merely around 29 million light years, it has an apparent magnitude of ~ 9 and is a rather good target for observations of even hobbyist astronomers with enthusiast telescopes. More importantly though, you aren't looking that far into the past. Again, in the cosmological sense. For comparison, our Sun is roughly 4.5 billion years old, and we're observing the Sombrero Galaxy as it was 29 million years ago.

So what is this interstellar medium made of? On average, it is comprised of mostly gas clouds of hydrogen and helium forming during the primordial nucleosynthesis (read: Big Bang) and at most a few percent of heavier trace elements, some also in gas form and an itsy-bitsy percentage of these would combine into even heavier dust particles, mostly made of carbon, silicon, and oxygen (source for the interplanetary dust composition: NASA APOD). Remember, when I say clouds and dust, they are incredibly thin. So thin, you'd have to be traveling at a significant portion of C to even notice it much, unless you're looking for it with extremely sensitive equipment, like Voyager 1 and 2 do, for example. But even with such low particle density, some regions would be slightly denser, and some slightly thinner. And that is what creates that majestic dust lane of the Sombrero Galaxy.

They are interstellar medium, mostly made of primordial hydrogen and helium, and dust made of carbon, silicon, oxygen and other trace elements, and all of it hundreds of light years in length and obscuring light from the galaxy's stars to penetrate through at varying levels, depending on background light strength, proximity, and local interstellar medium cloud density. The color of the cloud says nothing of its composition. It is mostly black with parts of it shifting towards red hues due to molecules of gas and dust in the interstellar medium scattering blue light from the source of light more than they scatter red light. I.e. the same reason why the Sun or the Moon appear more red when they're low on the horizon.


What is the largest recorded solar flare? For the purposes of this question I would interpret the word "recorded" as loosely as possible meaning that historic records such as ice cores or tree rings would count. Also what effect did such a large flare have on life on Earth at the time?

A spike in Carbon-14 content was recorded from tree rings in Japan, possibly due to a solar flare 20 times more energetic than the 1859 Carrington Event (the largest definitively recorded solar storm). This event has been dated to 774 AD. I think this could be the most energetic (tentatively) recorded solar flare.

Such a solar storm would be far more devastating today than it was back then, in the sense that we rely so much on technology that the EM interference would cost trillions of dollars in damage to power grids and communications.

A flare would need to be disastrously large, I think, to give us on Earth radiation poisoning. Certainly, animals that use Earth's magnetic field for navigation and would get rather confused. But our atmosphere and magnetic field are adequate protection when we're on Earth.

Here is a good write-up of the situation, with a couple of Nature papers if you have access.

If you're going to delve into considering the effects of solar storms on human health (i.e. the effects of magnetic field perturbations), put your tinfoil hat on first. There's a lot of pseudoscience and little substantive research.


What percentage of visual stars are actually binary stars? I know this is such a thing as a binary star where for example two stars rotate around a central point of gravity but to the naked eye are just a star.  What I do not know is how common this is (approximately)?

This is a difficult question, It's been changing through the years and it's difficult to calculate accurately. Now it's know that this percentage changes depending on the star type and ranges from 50% for Sun like stars up to a 80% for type O stars.

Fonts:

http://www.space.com/1995-astronomers-wrong-stars-single.html (2006)

http://www.space.com/22509-binary-stars.html (2013)


How do stars or galaxies get their spin? It is my understanding that when a star, a planetary disk, or a galaxy forms, the rotational momentum of the whole system is conserved.

Due to the smaller size of the resulting object, it will spin with a significantly higher speed than the original nebula.

What I do not understand is where the original rotation comes from. Why should a random dust cloud have an overall spin? Wouldn't the impulses of all particles in the cloud tend to average each other out?

Is there some alternative source of spin or a reason why nebula have an inherent spin?

You could start from the premise that there was no net angular momentum in the universe at all; but it would still be the case that everything of interest was spinning.

On the scales of stars and planets there are (at least) two important mechanisms that result in individual systems having angular momentum. The first is turbulence. If you take a parcel of turbulent gas from a giant molecular cloud it will always possess some angular momentum, even if the total cloud does not. As the parcel collapses to form a star/planets conservation of angular momentum $J$ and dissipative interactions result in an ncrease in spin rate and collapse towards a planar geometry.

Second, stars form in clusters. There is interaction between stellar systems early in their lives. Again, the cluster may have little net J, but groups of stars can, relative to their own centre of mass frame.

On bigger scales (galaxies) the second of these explanations becomes more important. The interaction and accretion of galaxies is what gives individual galaxies a spin, even if the clusters they are born in have much less or even no net angular momentum.

As an example of how turbulent velocity fields lead to gravitational condensations containing angular momentum you could do worse than study the star formation simulation performed by Matthew Bate and collaborators. These simulations start off in clouds with zero net angular momentum, yet produce a host of stars with swirling accretion disks, binary systems of all shape and sizes etc. An example journal paper can be found here: http://adsabs.harvard.edu/abs/2009MNRAS.392..590B  Here is a web page where you can download the animations and study them at length  http://www.astro.ex.ac.uk/people/mbate/Cluster/cluster500RT.html

Turbulent clouds are by their nature random and stochastic in terms of their motions. Often the velocity field is defined in terms of a power law dependence on spatial scale. The formation of vortices is a characteristic of turbulent media. They can be produced in the absence of external forces. The vortices contain angular momentum.

It is also worth noting that not all galaxies have an appreciable spin. Spiral galaxies do, but many elliptical galaxies have little net rotation. See https://physics.stackexchange.com/questions/93830/why-the-galaxies-forms-2d-plane-or-spiral-like-instead-of-3d-ball-or-spherica


Triple Stars v/s Ternary Stars Ternary (or trinary) stars are not the same as triple stars. So, what is the basic underlying difference between these two? What are some popular examples of these two varieties?

Exactly the difference between binary and double stars.

Trinary System

Three stars gravitationally bound to each other.

Here are some examples

Triple Star System

Three stars that appear to be together but any one of them is gravitationally unbound.

Here are some examples.


relative distance and right ascention The redshift-distance relation is $cz=H_0d$. So if we have $z$ we can calculate $d$. But is there a way to calculate or covert or write this $d$ in terms of right ascension (RA) and declination (DEC) ?

Thanks in advance

No. If you only know the distance to your object, that tells you absolutely nothing about where it is. That's like saying, I know the grocery store is 5 miles from my house, how can use this information to figure out exactly where it is? The answer is that you just can't, you need more information. – zephyr

In 3D space, you need three coordinates to locate an object. Ra, Dec are two (position on a sphere), z (or some other radial distance measure for nearby objects) is the third. Astronomy usually tries to make do with just Ra, Dec because that's easy to measure. – Alex

(comments converted to answer under cc licence)


Is it easier to see the young moon when it's closer to the sun? With the naked eye, would it be easier to see the moon when it's close to the sun, because it might reflect some more light (or not, I'm not sure) or farther from the sun because the sun shines brighter, same with how we can't see the stars?
Also, would it be easier to see it a few minutes after the sun sets and the moon is on the verge of setting, maybe at the +5° on the azimuthal grid?

The further the moon is from the sun the easier it is to see.  There are two reasons.  Firstly, when it is close to the sun it gets lost in the glare from the sun.  The sky close to the sun is very bright, and there isn't enough contrast to see the moon against the bright sky.  Secondly, as the moon gets further from the sun, more of the illuminated side of the moon is visible and so the moon becomes brighter. Brighter moon → easier to see.
When I say "close to the sun" remember this only means "appears close in the sky" the actual distance to the sun only changes a little over the year.
The moon is easier to see against a darker sky, so if the sun has set, it is more likely to be visible.  There was a new moon at 2:30 GMT on 12th April 2021 and the moon will become visible in the next few days (which is astrologically significant to Muslims)