What makes small interferometers useful? Like NIRISS on JWST NIRISS is an instrument on the James Webb Space Telescope. It has a "non-redundant aperture mask" which obviously covers most of the area of the sensor. It seems to be advantageous for high contrast imaging (like finding an exoplanet next to a star) and an alternative to coronagraphs. But however does that work? Why is it good to cover most of a sensor?

I have associated interferometers with creating as large as possible baselines for higher resolution, like the Very Large Baseline Array and the Spectr-R radio space telescope which gives up to a 390,000 km long baseline. So what is the magic with sacrificing sensor area to turn a single small telescope into an interferometer? Aren't all photons welcome? Would such an instrument do as well with a correspondingly smaller main mirror (maybe in separate fragments)?

A supplemental answer to probably someone's answer:

Why this can indeed be called interferometry:

Once one thinks in terms of physical optics (e.g. $\text{exp}(j(\omega t - \mathbf{k} \cdot \mathbf{r} ))$ ) instead of ray optics, imaging is always an interference problem, and the math behind correlating signals from an array of radio telescopes to produce an image is not so much different than the math behind calculating an intensity pattern at the focal plane of an optical telescope.

From the Aperture Mask Interferometry section on page 5 of the JWST Pocket Guide (see below) you can see that they have no problem using terms like "interferometry" and "non-redundant baselines" as one would when laying out a sparse pattern of dish antennas in an array. See for example a Google Maps image of the Meerkat Array Core below, which has already produced images with only 16 sites occupied with active dishes.

While the mask shown in the question is a pupil mask, the pupil plane is conjugate to the telescope's aperture. So it's very similar to having seven large hexagonal holes in front of the JWST, or just using seven of the telescopes primary mirror's hexagonal mirror elements (with additional masking, these hexagons to not cover the full size of the individual elements, as shown in the illustration).

An important distinction though is that these seven small apertures are of the same order in size as their separation, while in a large radio telescope array, the spacing between receivers is usually somewhere between much larger and much much larger than the aperture of individual elements. So the analogy breaks down at this point.

What Aperture Masking Interferometry is for:

The purpose of using the mask is to enhance the system's resolution by narrowing the central peak of the point spread function compared to what you would get from the full aperture.

By loosing contrast in the full field due to diffraction artifacts, as well as loosing ~90% of the transmitted intensity, you can enhance contrast near the center of the field by narrowing the central diffractive peak, in this case in the range of 70 to 400 milli-arc seconds, according to the JAM Team's (JWST Aperture Masking Team) slides linked below.

From the first link:

  Light admitted by 7 apertures in an otherwise opaque pupil mask interfere to produce an interferogram on the detector. This interfogram has a sharper core than is provided by normal "direct" imaging. The advantage is significant: while the ability to separate closely spaced objects with normal imaging is given by the familiar Rayleigh criterion (separation  δθ=1.22λ/Dδθ=1.22λ/D , where λ is the wavelength of light and D is the diameter of the telescope),  interferometry can resolve objects as close as δθ=0.5λ/Dδθ=0.5λ/D (the Michelson criterion).  AMI allows planetary or stellar companions that are up to ~9 magnitudes fainter than their host star and separated by 70–400 mas to be detected and characterized.  AMI  can also be used to reconstruct high-resolution maps of extended sources, such as active galactic nuclei.

You can read more about it in:

JWST's NIRISS Aperture Masking Interferometry
ArXiv NIRISS aperture masking interferometry: an overview of science opportunities
The JAM Team's: Non-Redundant Aperture Masking Interferometry (AMI) and Segment Phasing with JWST-NIRISS

Below: MeerKAT array core from Google Maps at (30.7136109S, 21.4399576E).


Can expansion of space blueshift galaxies with respect to one another? Suppose we have two galaxies in an otherwise devoid space, and suppose the galaxies are gravitating. Suddenly, dark energy kicks in, and accelerated expansion of space begins ...

(1D force diagram. Ruddy rectangles represent the two galaxies, violet ellipses represent dark energy.)

Because the galaxies are gravitating, dark energy between them expands space less, while "behind" the galaxies, where their mutual gravities don't overlap as much, dark energy exhibits a greater effect. The arrows above the galaxies compare force vectors. The black arrows are forces of gravitational attraction, the violet arrows are forces of expansion. Eyeballing the vectors' sum for each respective galaxy, it seems that the galaxies get an extra kick of acceleration from the regions of greater expansion behind them.

If we play the movie in our heads with these assumptions: the galaxies start off by accelerating toward one another under their own mutual gravitational attraction (each observes a blueshift), then dark energy kicks in and the galaxies accelerate with more energy than can be accounted for by acceleration due to their own gravities alone (blueshift observed much greater than it should be).

I have always read and heard of redshifted galaxies, not blueshifted ones (well, not so much, anyway), so something must be wrong here.

Question: Can the expansion of the universe drive two galaxies together like this? Does the net effect of expansion behind and between the galaxies cancel out, resulting in no greater acceleration? Have I got this all wrong?

Expansion will still redshift the light from these galaxies.

The universal expansion, at least by my understanding, isn't so much a force that pushes these galaxies further apart, but instead creates additional space everywhere, constantly. This includes the space between the galaxies that are approaching one another.

The light travelling from one galaxy to another will be affected by this expansion, since the space that light and matter occupy is also expanding. The result is that light "stretches" as it travels, lengthening its wavelength, because the space between one wave's crest and the next is expanding. Since that light continues to travel at a constant speed, these crests will never speed up or slow down to "catch up" with their original wavelength, so the lightwave gets stretched into a longer wavelength with lower energy - that is, it gets redshifted.

The light will still be blue-shifted by the galaxies' movement toward one another, but the Universe's expansion will not push them together faster. Rather, the Universe's expansion will universally (heh) increase the distance between all points in space, including the space between those galaxies. This expansion rate may not stop these galaxies from reaching each other, but it will always serve to increase the distance that they must travel to do so. Because of this, the Universe's expansion will always redshift light, just to a greater or lower degree based on initial distance.

For further clarification, the fact that there is a larger amount of expansion happening "behind" these galaxies has no effect on the expansion of the space between these galaxies. It's not so much a force that pushes all matter away from each other, but the literal creation of more space at all points in space all the time.

Creation of vs expansion of space

Now, there seems to be semantical ambiguity on this subject (or I may simply have an insufficient understanding of the subject - a very likely scenario) concerning the idea of dark energy as a force. Many interpret it (I think) as being a force that pushes matter away from other matter, but I don't think that's what it is. From my understanding, it seems more like a force that pushes space itself away from the surrounding space.

That is, if you were to somehow identify two "adjacent" points in space, their dark energy would push them away from each other. This of course means that those imaginary adjacent points are no longer adjacent, as there is now space between them. We can now identify new points in space between the two formerly adjacent points (note that in practicality, it is impossible for two points to be adjacent in this sense, as we can always divide the space between them further. This example is just a thought experiment).

Following this thought experiment, we find that, counterintuitively, this dark energy remains at a constant density. That is to say, these points are pushing the surrounding space away at the same rate as before, in spite of now being more spread out, and the "new" points between them are also now pushing the space surrounding them away at the same rate. Now, if the density of dark energy remains constant while the space it occupies is expanding, that can only mean that there is now more dark energy between the two original points than there was before.

This is why I like to say that space is "created" rather than simply expanded. It makes it easier for me to conceptualize. If space is thought of as the thing that creates dark energy (whatever it is), and that energy doesn't lose its density when space expands, then it makes sense to me to say that there is now more space there than there was before. To be honest though, I don't actually know what terminology is primarily used among those who regularly study this, just that this way of thinking makes it easier for me to conceptualize.

Disclosure: I am not an expert or even particularly knowledgeable on this subject. If I have made mistakes here, others should feel free to offer corrections. My answer is based on my own understanding gained from my own limited studies of this subject, and I won't pretend to even begin to understand the mathematics behind any of it.


How large must an object be to be seen through a telescope? I've studied this in a few optics classes before, but I'm drawing a blank at the moment.  If I know the aperture size and focal length of a telescope, as well as how far away the object I want to view is, can I calculate how large/wide/big that object must be in order to be seen?

A few clarifications.

Telescopes in general operate at very large distances - "at infinity" is the term used in optics parlance.

A bright enough object can be seen from any distance, no matter what its size is. All that matters is that:

It's bright enough to produce an impression on whatever sensor
you're using (or your eye)
The background is dark enough to produce sufficient contrast

But then it would be just a bright but tiny spot.

I believe what you're really asking for is: what is the combination of factors that shows the object as bigger than a simple dot? In that case, it's two factors: aperture of the telescope, and angular size of the object.

Assuming a flawless telescope, its aperture is what determines its resolving power. The resolving power is the angle at which two dots can be separated by the telescope. The formula is:

resolving power = 1 / (10 * aperture)

where resolving power is in arcseconds, and aperture is in meters. Examples:

aperture         resolving power
10 cm            1 arcsec
20 cm            0.5 arcsec
1 m              0.1 arcsec

As long as the object's angular size is bigger than the resolving power, it will appear bigger than a dot.

That's all. In astronomy, we don't speak of an object's absolute size, we only speak of the angular size. But that should be enough. As soon as you have the angular size, and say the distance, then you could deduce the absolute size, it's a simple matter of trigonometry.

absolute size = distance * tangent(angular size)

E.g., this is the size of an object of 1 arcsec angular size, situated at 384,000 km (the orbit of the Moon):

http://www.wolframalpha.com/input/?i=384000+km+*+tangent%281+arcsec%29

It's 1.8 km (in case the link above is unavailable).

In other words, that's the minimum absolute distance resolved by a telescope 10 cm in aperture, for objects on the Moon. Any two dots closer together than 1.8 km, placed on the Moon, are seen as one dot in a 10 cm telescope.


Is gravity a source of infinite energy at a cosmological scale As I understand it, gravity cannot be attenuated by any medium (in the way that EM radiation can be, for instance).

Does this, then, not make it a source (theoretically, I am not talking of practicalities) of infinite energy - if we assume the universe itself is infinite and looks the same everywhere (ie there are objects with mass everywhere)?

Or is it more correct to say that gravity has a net contribution of nothing to the universe's energy density because the attractional energy is balanced by a negative potential energy?

Gravity travels at light speed (or less, possibly), so even in an infinite non-expanding universe of finite age you'd only be gravitationally interacting with a finite mass in a finite volume. Our universe is observed to be expanding, further inhibiting us from coming into contact with new objects.

Furthermore, a common assumption of cosmological models is that the universe is homogeneous and isotropic on large scales. Meaning it is pretty much identical in all directions. Observations support this assumption. One net effect of this is that the gravitational potential energy you experience from objects outside (roughly) your supercluster is zero (and really close to it from things outside your own galaxy, actually). They just balance each other out.


Metal rich or metal poor? What does the image in the link mean?  Are the stars in the halo high in metallicity or low?  How about at the bottom by the disk?

http://science.psu.edu/alert/images/SDSSmetals.jpg/image_view_fullscreen

The ones in the halo  have blue colors, which is the "old" part of the spectrum.  Each type of metallicity is usually measured using a form of logarithmic scale, since the flat percentage of metals (especially iron peak elements) of any star will be very small.  In this case, we have
$$[Fe/H] = \log(N_{\text{Fe}}/N_H)_{\text{star}} - \log(N_{\text{Fe}}/N_H)_{\text{sun}},$$
So we take the ratio of iron to hydrogen in the target star, measure it as well in our own sun, then take the ratio of those two and take the logarithm of that value.  Since the $\log$ function is monotonic and has $\log(1)=0$, a negative value of $[Fe/H]$ means our Sun has more iron in it (relative to the amount of hydrogen).  A value of $[Fe/H]=-1$ means our sun has an iron to hydrogen ratio that is 10 times greater than the given star's, for example.  The more negative this value, the more metal-poor the star is (or, at least, iron poor) than our own sun, which is considered of a fairly recent generation.  Lower metallicities are typical of earlier generations of stars, and so may be described as "old".  Note, however, that they need not necessarily be old stars.  In principle it is possible for a recently formed star to have formed in an unusually metal-poor region, or otherwise been unable to accrete the normal quantity of metals.


What does observational evidence say about the number of protoplanets believed to exist in the early solar system? Currently, the solar system has 8 major planets, a handful of dwarf planets and a plethora of asteroids, comets and natural debris, and of course, a rather locally dominant star (the Sun).  There are many references that describe the early solar system as a chaotic environment with many more protoplanets.

Based on observational evidence, how many protoplanets are believed to have existed in the early solar system?

Earth and moon are thought to be formed by the collision of two planets. So we have at least these two protoplants. The cores probably ended mainly in the core of Earth; that's inferred by the much higher mean density of Earth in relation to Moon.

Iron-nickel meteorites strongly indicate at least one pre-existing planet large enough to form an iron-nickel core.

Comets of the Kuiper belt are hypothized to be a collision product.

As you go further back in time the building blocks of planets get ever smaller and more numerous till the interstellar dust and supernova remnants our solar system is formed of.

Some of the early material will most likely have been ejected from the solar system; this may include protoplanets.

By the chaotic nature of multi-body systems like our solar system it's getting more and more difficult to track single bodies back into the past.

The NICE models try to describe the early solar system.


When black holes such as those detected by LIGO merge, in what ways does the Schwarzschild Radius warp? I was reading the paper about LIGO's detection of gravitational waves, and I wanted to know what the Schwarzschild Radius did. For example, I feel like the simulation here is oversimplified, and that reality wouldn't actually have that simple bump together of the two radii.

This is a better simulation SXS did for the LIGO detection (still frame below), the two event horizons (Schwarzschild Radii as you say) just simply come together and merge quite effortlessly like two bubbles in a bath coming together and merging into one. I don't know if you were expecting something more dramatic than this but there is no reason to suggest its a violent event.


Actual meteor sightings formula I'm trying to realistically approximate the value of meteor sightings during a meteor shower because I don't want to have high expectations going in. I have a couple questions about this formula I found from this site

Number of meteors=ZHR×sin(θ)/(F×r^6.5−m)

Question 1
I would like to know for the calculation of F, where

F=1/(1−k)
where k is the fraction of the sky which is which you cannot see, because it is obscured by cloud or other obstructions such as trees.

Assuming I am on a flat plain with no trees/building whatsoever and I have a clear view from 0-180 degrees (cross section view of the earths crust). Will k be 0.5 based on the fact that I cannot see whats below ground (180-360 degrees).
Also will the limitation of the human eye affect this value? I.e If I am facing north my peripheral vision can only reach maybe north east to north west hence further increasing k value above?
Question 2
What value should I have for m in the above formula if I am at this location? I currently assume a value of 6 according to this wiki. Is this right?

The factor k, the fraction of sky you cannot see, is not adjusted because of the ground or your peripheral vision. When on a flat plain (a "sea-level" horizon) and none of the sky is covered by clouds, buildings, or trees, then k=0. (Therefore, F=1, and the Zenithal Hourly Rate is not diminished.)

I do not know how the light pollution map was created and whether it accounts for the humidity. The humidity can have a significant effect on the limiting magnitude. In other words, if it is a dry day and the air is clear, you can see fainter stars than if it is a humid day and the sky is murky. If your location is clear (so you have blue skies during the day and black skies at night), then assuming m=6.5 is realistic. (Note that when you click on a location on the light pollution map, it will give the Bortle classification. Clicking the link for the Bortle class will give the expected limiting magnitude.) Of course, the limiting magnitude also depends on your eyesight. Younger people can see fainter stars than older people. (I just tried the map for my location, and it predicts 5.5 to 6.0. I can usually only see stars in the range of 5.0 to 5.5 because of the light pollution and humidity.)


What do we mean by space is expanding? I know this question had often been asked on Physics.SE as well as this site but I don't get them. What do mean when we say that space is expanding? I mean that space is no physical entity that can expand. Space is just (apparently) "nothingness". Please solve my confusion.

Indeed space is not a physical entity (as far as we know). Saying space expands is another way to say that galaxies recede away from each other at a rate proportional to their distance.

The "expanding space" picture is unfortunately a source of endless confusion and misconceptions, as can be seen on the page of this very question: mpv mentioned that "The real expansion of space would manifest itself in the following way: you have 2 rockets in empty space, stationary with respect to each other. None of them starts its engines, but yet, they start moving away from each other."

This is plain wrong, the "expansion of space" is not some magical force that pulls objects away from each other. If you mean empty space as in no dark energy, no dark matter and no visible matter then the two rockets would move towards each other due to their mutual gravitational attraction (if you neglect their mass then they would not move at all). In the general case what will dictate the motion of the rockets is their gravitational attraction, the matter nearby (visible and dark) and dark energy (whatever that might be), the combination of which might pull them closer or further away from each other. The expansion of space is not an additional force. Galaxies don't keep moving away from each other because some undetected substance is created between them, but because they have an initial velocity which they acquired at the time of the big bang: without anything to stop them they keep moving away.

Joan.bdm said: "As I said in a large scale all objects recede from us, if it wasn't for the space expansion it would mean we are at the center of the universe."

The second part of the above sentence is wrong as well. As long as all galaxies recede from one point at a rate proportional to their distance, it follows that from ANY point all galaxies will be seen to recede at a rate proportional to their distance. This is easily seen with a little drawing (each * corresponds to a galaxy):

Before

After

If all galaxies are seen to recede at a rate proportional to their distance from the point of view of the galaxy at the center, then from any galaxy all other galaxies will be also seen to recede at a rate proportional to their distance.

I don't think we have any satisfying explanation as to why galaxies move away from each other at a rate proportional to their distance in the first place. The "expansion of space" is not an explanation, it is an analogy. A related and interesting read: Expanding Space: the Root of all Evil?


Is there a chance that UFOs could be messages from other planets? Description

Lots of people have thought to of seen UFOs (Unidentified Flying Objects), in fact, more than 83,000 have been reported this year alone. (and it's only February!) But most of the time it's just a shooting star or an airplane. But some of the sightings don't seem to be any of the things a UFO is commonly mistook for, but there isn't any evidence supporting if it is really a UFO. I might not be alone on this theory, but I believe the sightings that potentially could be UFOs are messages from the cosmos.

Question

There may not be much to back up my theory, but I just wish to learn some background knowledge:

  Are UFOs (and I guess all the UFO sightings reported, too) messages from another planet?

I also believe that if they are messages, they could possibly be responses to NASA's Voyager spacecrafts, for their purposes were to inform any forms of life out in space what earth is like with the 'Sounds of Earth' music disk attached to it.

No.

The purpose of sending a message is to have someone else read it. It defies logic to send a message over light years of space and then have it do anything but broadcast its presence very clearly to anyone who might possibly want to investigate it. If the goal of these UFOs was to tell us something, they wouldn't be such elusive specimens that only appear to specific people at specific times and places, and then disappear without a trace. If UFOs are in fact meant to communicate with humanity, they are indeed rather poorly designed, because they haven't accomplished that goal very well at all!


Does our current understanding of the mass of black hole only allow for it to be located at the singularity? Does the distribution of mass or density of mass inside the event horizon matter for a black hole to exist as we currently theorize? Could the mass inside the event horizon be distributed in any way at all, or have any density so long as it is dense enough to locate 100% of the mass inside the Schwarzschild radius?

As a star or neutron star collapses to form a blackhole there will, of course, be a moment when the matter is distributed everywhere within the event horizon.  So, there may not be a singularity within a full fledged blackhole, at first.  But, the inward forces within the event horizon are irresistible (considering that escape velocity is greater than the speed of light) and so everything within the event horizon must fall into a singularity.  How long this takes is complicated by the issue of which reference frame you chose to time it.


How does a black hole spin? This seems like a pretty basic question, but I couldn't find a duplicate upon quick inspection. From what I understand about black holes they actually contain no matter. All of the mass that maintains the black hole (ie. keeps space warped) is found in the energy bound in the space itself. I may not be stating this quite right, but this is basically what Kip Thorne states in "The Science of Interstellar". Therefore, it seems that to maintain angular momentum during black hole formation that space itself must "spin". This would imply that space is actually "spinning" even in the vicinity of the Earth (to a much lesser degree of course). Is this true?

As an aside, I don't take everything that Kip says as gospel truth. He states several times that Io is the innermost moon of Saturn in Chapter 7.

What we call a black hole is a solution of Einstein's equations of gravity.

We know several exact solutions: Empty and flat space corresponds to special relativity. Other known solutions include that of a point mass, a charged point mass, and a solution, known as the Kerr solution in which spacetime around a point mass is being "dragged"

This last solution can be interpreted as corresponding to a black hole that would form if a rotating mass were to collapse. As you note, a black hole contains no matter: it is a vacuum solution of Einstein's equations. However a black hole does have properties like mass and angular momentum.

The shape of space around the Earth can be approximated the Kerr solution of a spinning black hole, with a radius of about 1cm. We would therefore expect some dragging of spacetime around the Earth. In 2008 the Gravity B probe had measured the gravity around the Earth and confirmed that the Frame Dragging effect predicted was present and its results were consistent with  General relativity. Frame dragging around the Earth is fantastically small and the Gravity B probe was working right at its limit of sensitivity. It causes a gyroscope to deviate from its spin by 0.00001 degrees per year.


Did I see a meteor or a re-entry? Last night (January 21, 2015, possibly 5:30 UTC), from my house in northern California, I saw a moving point of light in the sky, going west to east, crossing about 120 degrees of sky in 5 seconds or so.
It didn't look like my experience of meteors - thicker and slower, and quite bright. I think it survived until it was out of sight entirely, rather than burning out.
Aerospace.org's reentry predictions doesn't show anything for that time.
I've seen ISS and other satellites; this was a lot larger and brighter, and my favorite satellite pass schedule also shows nothing.
Was this more likely a rocket body re-entry, or a large meteor?

Partial answer only.

120 degrees of sky in 5 seconds or so.

120 degrees in 5 seconds is about 0.42 radians per second. At a distance of 70 km (roughly the distance where an overhead meteor would start to glow visibly) that would be roughly 29 km/sec.
That's a nice number for a meteor.
But it's quite a challenging number for an originally earth-launched object to achieve during reentry unless it's done some planetary flyby's to change it's orbital energy relative to the Sun.
I know it's been almost 6 years now, but if there's a chance that the apparent rate was a lot less than 0.4 radians per second, it would open up the possibility of a reentering artificial object.
If not, I think it is more likely to be a meteor, though reentry of a deep-space bit of space junk is not impossible.


How fast does a spacecraft have to be to enter a primordial black hole without being torn apart? If there really is a primordial black hole beyond the Kuiper belt, we can send a probe to the black hole and into it. But how fast must the probe be in order to enter the black hole without being ripped apart by spaghettification? The black hole in question has an event horizon diameter of 2-3 inches and 5-15 Earth masses so it is a tiny probe (tiny enough cameras do exist) we would have to launch. Let's say the probe is 1.5 inches broad and has a length of 2.5 inches. How fast must it be in order to not be torn apart and successfully enter the black hole?

Let's try a crude back of the envelope calculation to get orders of magntitude:

The tidal force near the event horizon is of the order of $G M/r^3$ which is something like $10^{18} g/m$. So a probe with a mass of 1 gram and diameter of 1cm would experience about $10^{12} N$ of force trying to "spaghettify" it by accelerating ends of it at about $10^{16} g$ relative to the centre. If it experienced that force for a time $t$ the ends would likely move about $10^{17} t^2 m$ relative to the centre, assuming $t$ is short enough that relavity does not complicate things, so if we want to limit that to say 1mm we need $t < 10^{-10}$. So basically we want the probe to go from a few Schwarzchild radii down to 1 in about 100ps in its internal time frame, so that the tidal forces do not have time to tear it apart. Using the Newtonian approximation this is a few times faster than the speed of light, so what we learn is that the probe must be going at relativistic velocities, to have a chance of surviving. From its perspective, that flattens out the gravitational field around the black hole in the direction of travel, so that the tidal force seems to even briefer in duration.


Is a perfectly circular orbit possible? Let's assume that there is a perfectly spherical planet and there is a moon which is also perfectly spherical. Let's assume that there is no atmospheric drag and no other gravitational pull. If the moon is put in a perfectly circular orbit around the planet somehow, would the moon eventually "fall" towards the planet and form an elliptical orbit or would it continue to follow the perfectly circular orbit?

Edit: What i actually wanted to ask is that would the gravity of the planet cause the moon to "fall" towards the planet or would the gravity allow the moon to continue its orbit without bending its path any further towards the planet. I know that no planet can be a true sphere or a cube due to the shapes of the particles.

"Perfectly" is a funny word.

Perfect circles are a mathematical abstraction. Real objects are not "perfect".  So supposing a "perfectly spherical planet" is to suppose something that does not and could not exist. All real planets are made of atoms and anything made of little clumps of matter cannot be perfectly spherical.  Even if you built a planet that was as spherical as possible, it would be distorted by its rotation and the tides. So there are no perfectly spherical planets.

Now you say "put in a perfectly circular orbit". This is like drawing a line that is exactly $\pi$cm long. Again you are supposing something that does not, and could not exist.

What we can do is consider a mathematical model of gravity. If you model the sun and the planet as "particles" (ie point masses) and you model gravity with Newton's law of universal gravity, and if you give the model the system with the exact amount of energy to give a perfect circle, then the system will remain in a perfect circle, it will never become elliptical.

If you use general relativity to model gravity, then the release of gravitational radiation will mean that no circular orbits are possible, all orbits will spiral inwards, however it would not become elliptical. Something similar will happen with quantum models of gravity.

So your question can only be answered in the context of a mathematical model of gravity.


Electron Degeneracy Pressure The Z/A term for calculations of degeneracy pressure for white dwarfs is given as 0.5. But I can't find an explanation as to why. For my assignment, I am trying to find the degeneracy pressure for the core of the sun. I am given R = 0.1Rsun and M = 0.1Msun. I am also told that the core contains 50% hydrogen and 50% helium.

I need to find Z/A for this scenario in order to find the degeneracy pressure, but I don't know how to approach it.

Z is the atomic number, A is the atomic mass.

A white dwarf is typically made of carbon (A=12, Z=6) or Oxygen (A=16, Z=8). So Z/A = 0.5.


Why does Neptune only have one major moon? Neptune is profoundly anomalous when it comes to moons among the gas giants.  Jupiter has its four massive moons, Saturn has a several large spherical bodies and Uranus, a world similar in size to Neptune has five or so major moons.

Yet Neptune, the ice giant which sports one of the largest hill sphere radii in the solar system only has one major moon, of which is likely to be ripped to shreds in the future.

I find it peculiar because Uranus has five major moons yet Neptune only has one major moon.  What explains this disparity?  Did some disruption happen to earlier moons Neptune may have had?  What's known about this?

It seems possible that Neptune once had a system of moons similar to other outer planets. However Triton, a Kuiper belt object was captured by Neptune, and this disrupted the existing moons. It would have initially had an eccentric orbit, intersecting with the orbits of other moons, setting them into unstable orbits and ultimately either colliding with each other, hitting the planet or being ejected.

The minor moons formed from the debris of these collisions after Triton had entered a more stable orbit. So the disruption was the capture of Triton.


Why Free-Free emission is regarded as Thermal emission? I read that thermal emission/radiation are the ones whose spectra is similar to that of a Black body radiation. Also thermal radiation depends solely on the temperature of the object. How can we consider Free-Free emission and spectral line emission as thermal emission processes? kindly explain

Thermal radiation $\neq$ blackbody radiation.

Thermal radiation is radiation that comes from a system where an equilibrium has been reached, where the various energy states are occupied according to the Boltzmann distribution and the particle velocity distributions are Maxwellian at some given temperature.

That does not necessarily imply that the radiation field is in equilibrium with the matter at the same temperature. It is this latter equilibrium that is required in order for the radiation field to approximate to a blackbody.

So the following statements are true:

Blackbody radiation is thermal radiation. Thermal radiation is not necessarily blackbody radiation.

An example of thermal, but non-blackbody, radiation would be  thermal bremsstrahlung. The particle velocities in the gas follow a Maxwellian distribution at a characteristic temperature and this determines the spectrum of the emitted free-free radiation. However, the bremsstrahlung radiation is able to escape from the gas without further interaction - in other words the gas is "optically thin". The result is a thermal bremsstrahlung spectrum that is not at all like a blackbody spectrum, especially at frequencies $<kT/h$.

On the other hand, the bremsstrahlung process is purely caused by the interaction of free electrons and ions. If you allow there to be atoms and partially ionised atoms in the gas, then it is feasible for those to be in equilibrium at the same temperature. The population of the various energy states in the atoms and ions is determined by this temperature, and transitions between these levels produces spectral emission lines in a completely predictable, temperature-dependent, way.


nodal precession: earth's orbit of sun Does the Earth's orbit around the sun exhibit nodal precession, as does the moon's around the Earth?  If it does, that must mean that the ecliptic is in a periodic wobble with respect to the celestial sphere?

The ecliptic is in periodic wobble wrt the celestial sphere, but that's because of nutation, not precession. Precession (nodal precession) changes where the Sun is during the equinoxes and solstices. The Spring Equinox, for example, has moved from Aries (it's still called "the first point of Aries") to near the Pisces-Aquarius border


What would the sun look like from aphelion of The Goblin's orbit? I was amazed to find out about the discovery of The Goblin 2015 TG387 with an orbital period of more than 32 thousand years and a aphelion of more than 1955 AU! That is an astonishing distance. What would the sun look like from aphelion on The Goblin? Could a naked eye observer distinguish it from the rest of the sky?

Yes, easily.

1955 AU is a long way, but it is only 0.03 of a light year. The sun would still be less than 1/100 of the distance to the next star.

The brightness of the sun would be much less: about 4000000 times less bright, but that is about 16 magnitudes. The sun from Earth is -27 magnitude, so the sun would appear as a magnitude -10 star. It would be much brighter than any other star, and would easily outshine Venus as seen from Earth. In fact it would be 100 times brighter than Venus. It would be as bright as a half-moon (strong enough to cast shadows, but not to read by)


Is it possible to get the distance to a star in the IPHAS DR2 catalog? I'm looking at the IPHAS DR2 catalog. It has the Right Ascension and Declination but I'm not seeing any distance information (lightyears or parsecs). I'm wondering, given the columns in the dataset, if the distance can be calculated.

Only by knowing what kind of a star it is, or by inferring that information from the available colours. From there you use a photometric or spectroscopic parallax calibration. ie. There is a relationship between the absolute magnitude and colours of a main sequence star. Of course there is no guarantee you are looking at a main sequence star.

You can get more idea by cross correlating with the 2MASS JHK catalogue, as the J-H vs H-K colours of giants are usually distinctive.

More specific help needs a more specific question.


What will happen to the shape of a galaxy when a super massive black hole lying in its center dies(evaporates out)? What's at the Center of the Milky Way?
 In this article it is said that a supermassive black hole lies in the center of milky way galaxy.

  At its center, surrounded by 200-400 billion stars and undetectable to the human eye and by direct measurements, lies a supermassive black hole called Sagittarius A*, or Sgr A* for short.
  The Milky Way has the shape of a spiral and rotates around its center, with long curling arms surrounding a slightly bulging disk. It's on one of these arms close to the center that the sun and Earth are located. Scientists estimate that the galactic center and Sgr A* are around 25,000 to 28,000 light-years away from us. The entire galaxy is around 100,000 light-years across.

We revolve around the center every 250 million years.Presumbably we rotate beacuse of the BH.

When the black hole dies in our galaxy will we be thrown out of the revolving orbit?

The shape of the galaxy is expected to change right?It will be some irregular shape not spherical?

Presumbably we rotate beacuse of the BH.

No. The galaxy is being held in one piece due to its own total gravity. The black hole is only a small fraction of that. Basically, the BH doesn't matter.

  When the black hole dies in our galaxy

The BH will probably be the last thing left of our galaxy at the end. And even then it will take some incredibly long time for it to evaporate. BH evaporation for very large BHs is basically the slowest process you could imagine.

  It will be some irregular shape not spherical?

The galaxy is not spherical. Its shape is rather more like a round disk (with some irregularities and some features like arms, etc).


How many new galaxies enter the observable universe each day/year/decade? Each day light has more time to reach our eyes from distant galaxies. In one day, light travels 2.59×10^10 km.

So our observable universe (assuming my simple math skills apply here and there isn't something about relativity to take into account), expands that much every day in all directions. So at what rate are new galaxies being encompassed by the expanding observable universe?

What we see at the outer reaches of the observable universe is a surface of the cosmic microwave background (CMB), which is at the time when the universe, still quite homogeneous (perturbations were a few parts in $10^5$), cooled enough for the electrons to combine with the protons and make hydrogen atoms.  Each day, we can see a light-day farther and so we slowly see a different (larger) surface of the CMB (but there were no galaxies at this epoch).  Therefore, as we observe the CMB over many years, we will always see the same time epoch, but at different spherical shells.  Will we see new galaxies?  Yes. Galaxies are forming continuously in view at high redshifts (2-20?).  In a few million years, the small perturbations that we see today in the CMB will develop into young galaxies within our view.  One cannot provide numbers for how many we might be seeing because small galaxies merge into bigger ones and those merge again.  At high redshifts there is a lot of merging going, so the rate is particularly ill defined there.

How is this changed by the acceleration from the cosmological constant (aka dark energy)? From the Friedmann equation one sees that the Hubble constant reaches a final constant value of $c\sqrt{\Lambda/3}$ or about 40% higher than its present value.  Eventually, in billions of years, the CMB will shift from $3^oK$ to much lower and then the CMB will fade (redshift) away.  After that, those new distant galaxies will also fade away.  And then more and more of the universe will go dim.


what is the dimension of Black Hole? Black hole is the densest .... in universe. We know that dimension of universe is 4D . So, what is the dimensions of black hole? Is string theory explains what is it's dimension?

The singularity itself is, by definition, zero dimensional (well...you could argue one dimensional as it exists over time, but it has zero spacelike dimensions).

The event horizon is a two dimensional sphere, just like the surface of a beach ball - whether it can be said to have any thickness is unknown, there might be a thickness down at the Planck scale, but that's currently unknown.


Life planets orbiting black-holes. Can/Do they really exist? So, I watched Interstellar and if you watched it too you know that there's a planet orbiting a black-hole, they call it Miller's Planet. According to the movie, every hour on Miller's Planet is equivalent to 7 years on Earth due to gravitational pull from the black-hole.

Question: Assuming there are other life forms in the universe, is it really possible for it to be near a black-hole? Is it possible that they came to existence thousands (or even millions) of years before us, but are not as advanced as us because time on our planet is way quicker than that of theirs? is a lot more for us? If they have an assignment due tomorrow, we have, say, 100 years more time to do it (assume what else we can do in that 100 years). Or they actually are more advanced than us, but from Earth, they somehow live in the past?

Well, first things first.   It's not likely to have a planet orbiting near a black hole and in significant time dilation because the tidal effects would likely tear anything that close apart.   Certainly a planet orbiting a stellar mass black hole would need to be quite far away so as to not be torn apart, so any time dilation would be pretty small.

Around a super-massive black hole, the tidal effects are smaller and a nearish orbit with some measure of time dilation is possible.  (see link below for more specifics),

https://physics.stackexchange.com/questions/110044/time-dilation-factor-for-the-circular-orbit-at-3-2-schwarzschild-radius

But a stable planetary orbit, you probably max out at about 20% time dilation and only around a super-massive black hole, where there's only 1 per galaxy.    The idea of 1 day to 100 years isn't practical, it's 80 days to 100 days if you're talking about a stable planetary orbit.

and, I'm not sure you'd want to be that close to the black hole in the center of the galaxy, not because the orbit isn't stable, but because stars are in that orbit too.  It might not be a safe place to be.

So, in reality, you'd want a stellar mass black hole and a distant orbit, where the time dilation would be quite small, and in that scenario, yes, life is possible because of tidal energy, so a planet could have liquid surface water and an atmosphere, even if the black hole gave off very little light and heat.

Such a planet in a tidal-energy orbit would probably be tidally locked which would protect the far side from any gamma rays the black hole spits out when it eats, so it would in theory be a good place for life.   No significant light source, unless it was a binary system, so, plants would have it harder, but there would be heat.

There's another problem.   The creation of black holes tends to blow everything appart in a huge explosion.  It's unclear that a planet would survive a black hole's birth, so you might need a captured planet.

Finally, intelligent life . . . we really don't know enough about how common intelligent life is on other planets.   Life might be fairly enough, but intelligent life is far less clear and there's other factors than just time.

Today, we simply don't know enough to predict if there's intelligent life out there or not.   There probably is life elsewhere in the universe, though even that isn't 100% certain, but regarding intelligent life, there's far too many unknowns in that equation.   I think a black hole might not be optimal for the formation intelligent life because of the lack of light so, much less photosynthesis, so, slower oxygen formation (if it follows the same pattern the earth does) and the unlikeliness that a planet would survive the black hole's creation.


What is the name of the cycle whereby the lunar azimuth and altitude recur on a given day? Like the Metonic cycle, except for position in the sky, not phase.

Original question: If I look up at the moon tonight, how long before it is in the same apparent position, and same phase, again?

I realized I can calculate this myself, if only I know two piece of information. Unfortunately I do not know the name of the second.

My question should be answered by the product of:

the Metonic cycle, and
the cycle where the moon's azimuth and altitude recur on a given day

What is the name of this second cycle?

You are looking for the Metonic cycle.  This is almost exactly 19 years, and represents 235 synodic months.

Now you have edited the detailed question so that it is out of sync with the original Title.  The other period you need is the Draconic month, which is the period between the moon crossing the ecliptic (in the same direction).


About the formula for taking in count of the observer altitude in sunrise/sunset formula I'm new here and searched for solution but did not found one. I have found algorithms in order to calculate sunrise and sunset times for an observer on the surface of Earth with explanations of derivations of these formulae, but I want to improve it by including the altitude of the observer. I found an article in Wikipedia.

Could one explain to me, from where is this "2.076 degrees" in the formula $-2.076^{\mathrm{o}} \sqrt{\mathrm{elevation\:of\:observer\:in\:meters}}/60^{\mathrm{o}}$ derived and calculated?

Thanks in advance!

It's simple geometry, you are calculating the downward angle you need to look relative to the horizontal, from a given elevation, so see the horizon. To lowest order, that angle simply gets added to the angle of rotation of the Earth to make the limb of the Sun reach the horizon.

To calculate that angle, let's measure all distances relative to Earth radius, and say you are at altitude $x$ above the surface. Draw a straight line to the horizon, where that line makes a right angle to the radius of the Earth (it is tangent to the horizon at that point).

So now you have a right triangle with hypotenuse $1+x$, and a side of length $1$, so the angle between the direction you are looking, and the direction to the center of Earth, has a sine of $1/(1+x)$. That means the downward angle we want is however much less that angle is from a right angle. Let's call the downward angle we want $y$, so that $y \ll 1$ if measured in radians, and then what we have from the above is $\sin(\pi/2 - y) = 1/(1+x)$.

Using a trigonometric identity then says $\cos(y) = 1/(1+x)$. But since $x$ and $y$ are $\ll 1$, $\cos(y)$ is close to $1 - y^2/2$ and $1/(1+x)$ is close to $1-x$, so we can say $y$ is close to the $\sqrt{2x}$. That's where the formula comes from, the rest is just unit conversion from radians to degrees and from the radius of the Earth to the altitude in meters.


Gravitational lensing of quasars Are there any cases in which the gravitational lensing of a quasar has been observed in both visible light and in radio waves, and if so, is the radio-wave resolution good enough to show that the lensing at these two different wavelengths is the same?

Yes, there are many. One of the first, if not the first was the radio lensed quasar B1938+666, in which a near-IR Einstein ring was reported by King et al. (1998).

The answer to your real question though is not as straightforward as that. Although lensing should be wavelength independent (according to GR), it is not independent of the size and position of the source. The radio waves and optical emission arise from different source sizes and locations. In addition, if you are interested in testing whether the surface brightness magnification is the same, well that relies on knowing the intrinsic source properties; and the radio to optical brightness of quasars varies wildly.


If the earth lost 1/81th of its mass how would it orbit change? If a planet becomes lighter, then I believe its orbit will change.

For example, the moon is 1/81 of the mass of the earth. So, if the moon disappeared, then the earth-moon system would lose 1/81th of its mass. How would the orbit of the earth change in that circumstance?

I will answer the specific case of the Moon disappearing since the more general question, is a bit open-ended.

Of course the Moon cannot just "disappear." However, there are very real circumstances where a planet or other stellar object might be disrupted. The best example I can think of is if a sufficiently large asteroid or moon from another planet somehow make their way into Earth's orbit and collide with the Moon. Such collisions between Solar System objects are rare, but occasionally occur in models of the future evolution of the Solar System (see work by Jacques Laskar).

However, one must be careful to first consider whether that mass is actually lost from the system. For instance, in the Giant Impact Scenario a Mars-sized object is theorized to have slammed into the Earth, but the material never left the system. If a similar collision were to disintegrate the Moon and the mass remained as rocks and dust in orbit around the Earth, the effect on Earth's orbit would be minor. This is because it is the Earth-Moon system that orbits around the Sun. Note that there is a small change since the Earth-Moon barycenter has shifted position, but the effect is tiny so let's ignore it for now.

Instead, let's consider a case where the collision disrupts the Moon, and its material leaves Earth's orbit. If this occurs sufficiently quickly, (faster than the orbital period, i.e., 1 year) then from an orbital dynamics standpoint, the mass is effectively lost instantaneously. The Dutch astrophysicist Adriaan Blaauw, first calculated the relevant physics. Earth's orbit will expand, but by an almost miniscule amount. The reason is that the dynamics of the Earth's orbit around the Sun is dictated by the total mass of the system (i.e. Earth's mass plus the Sun's mass). To lowest order, the new orbit's orbital separation ($a$) will be:

$$ a \approx 1 + \frac{\Delta M}{M} {\rm AU} \approx 1 + \frac{1/81 M_{\oplus}}{M_{\odot}} {\rm AU} \approx 1 + 4\times10^{-8} {\rm AU} $$

So, the orbit will expand by roughly 40 parts in a billion. A quick calculation using Kepler's third law ($a^3\sim P^2$) shows that the length of a year would increase by about 2 seconds. A similar change in the Earth's orbital eccentricity will occur ($\Delta e \sim \frac{\Delta M}{M}$), but since this change is smaller than the precision to which we have measured the Earth's orbital eccentricity, the effect is negligible.


Why don't we orbit the center of our galaxy? At the center of the milky way, there must be some giant mass that keeps the galaxy from just floating away from itself. So, obviously, we feel the immense gravitational pull of whatever this is. But, on a smaller scale we can see this with the earth and the sun. The sun keeps us from spinning into oblivion with its gravitational pull. The question I am asking is, why do we orbit the sun, but not the center of the galaxy, which has a much stronger gravitational pull?

There are a number of misconceptions with this question.

First off, the "gravitational pull" (which I'm interpreting as gravitational force) by the Milky Way on the Earth is seven orders of magnitude smaller than that exerted by the Sun.

Secondly, gravitational force is the wrong metric. The Newtonian gravitational force exerted by the Sun on the Moon is more than twice that exerted by the Earth. Yet all but a tiny, tiny minority of professionals who deal with the solar system will say that the Moon does indeed orbit the Earth.

Thirdly, it's not correct to look at "orbit" as being a mutually exclusive concept. That the Moon does indeed orbit the Earth doesn't mean that it doesn't orbit the Sun. It does. The Moon not only orbits the Earth and Sun, but it also orbits the Milky Way, the Local Group, and the Local Supercluster.

So what is the right way of looking at orbits amongst a hierarchy of masses? One issue that needs to be addressed is what the word "orbit" means. Note that people do write and talk about parabolic and hyperbolic "orbits". I'm assuming that "orbit" in the context of this question means a bounded trajectory, which would rule out calling a hyperbolic or parabolic trajectory an "orbit".

That leads to the first of three characteristics of whether object A can be said to be orbiting object B: Object A can be said to be orbiting object B only if object A is gravitationally bound to object B. The appropriate metric here is energy, at least for cosmologically small distances. (Note: The distance between the Milky Way and the Andromeda Galaxy is very small in a cosmological sense.) From the perspective of a frame centered at object B, the total mechanical energy of object A needs to be negative he mechanical energy to be able to say that object A is (at least temporarily) orbiting object B.

Bounded orbits remain bounded forever in the two body point mass problem. That may not be the case when other larger objects are involved. This leads to a second criterion: Object A can be said to be orbiting object B only if the gravitational influences of larger objects on object A are but small perturbative effects that don't result in instabilities. The appropriate metric here is an energy-based sphere of influence. The two most widely used spheres are the Hill sphere and the Laplace sphere. For example, because the Moon is gravitationally bound to the Earth and because the Moon is well inside the Earth's Hill and Laplace spheres with respect to the Sun, it is appropriate to say that the Moon does indeed orbit the Earth.

What about the Sun and the Milky Way (and even larger objects such as the Local Group)? Can the Moon also be said to orbit those? The answer is yes. One way to look at it is to ask what would happen if the Earth suddenly disappeared. Would the Moon's path around the Sun change much? The answer is no. The Moon would still be orbiting the Sun at about one astronomical unit. The same applies to the Earth with respect to the Milky Way. The Earth's trajectory about the galaxy would not change by much at all if the rest of the solar system magically disappeared. This point of view however leads to some troubling issues. Planets (and stars and galaxies) don't magically disappear. Moreover, if Saturn somehow did magically disappear, its rings and all but its very outermost moons would be on hyperbolic trajectories with respect to the Sun. A way around this magic is to say that if object A orbits object B, and object B orbits object C, then object A also orbits object C. "Orbit" is not a mutually exclusive concept.


How high does light pollution reach into the sky? I want to know how high does light pollution typically reach into the sky straight up? Has anyone ever researched this because I wonder if being at 10000ft elevation would reduce the amount of light pollution from nearby cities because of the height assuming that light pollution weakens the higher you go. How high would it go straight up the sky (not to the sides about how far I would have to travel to have no light pollution disturbing the night sky)?

Light pollution occurs because light from the ground refects off atoms in the atmosphere. So you can reduce light pollution either by getting away from  light, or getting above the atmosphere. 50% of the atmosphere lies below 5500m,  if you can get 5500 m high, you half light pollution.  Getting high also improves clarity, and reduces the disturbance caused by atmospheric turbulence, which is why most major telescopes are situated on mountain tops.


What is the "lost light" in this unusual Hubble Deep Sky image? The Daily Galaxy article “The Lost Hubble” –New! Deepest Image of the Universe Ever Taken says:

  To produce the image a group of researchers from the Instituto de Astrofísica de Canarias (IAC) led by Alejandro S. Borlaff used original HUDF images from the Hubble Space Telescope. After improving the process of combining several images the group was able to recover a large quantity of light from the outer zones of the largest galaxies in the HUDF. Recovering this light, emitted by the stars in these outer zones, was equivalent to recovering the light from a complete galaxy (“smeared out” over the whole field) and for some galaxies this missing light shows that they have diameters almost twice as big as previously measured.

The image looks really strange, what is going on? Is there a technical article associated with this work?

Let me see if I can explain the main aim and accomplishment of this work.

First off: the picture you're puzzling over is a "luminance RGB" image, in which the bright regions are represented by color (a sort of pseudo-true color using near-infrared images), with the second faintest parts represented with black and the very faintest parts with white. The latter aren't qite "garbage", as Hobbes suggests in their answer, but they are relatively speaking the noisiest parts of the image, so there's little real information be found there.

This paper (Borlaff et al.; see the link in Hobbes's answer) is about re-processing near-infrared HST images originally taken about ten years ago as part of the Ultra Deep Field. Previous processing of these images (e.g., Koekemoer et al. 2013 ["HUDF12"] and Illingworth et al. 2013 ["XDF"]) was focused on getting information about the smallest, faintest galaxies, which are mostly the really distant, high-redshift galaxies. Because of this, the crucial stage of sky subtraction had some biases: in particular, it tended to treat the faint outer regions of the large, nearer galaxies as part of the sky to be subtracted. This is actually fine for the analysis of the small, distant galaxies, but means that if you do want to analyze the outer regions (outer disks, faint stellar halos, remnants of merger structures, etc.) of the larger, closer galaxies, you have the problem that their outer regions are over-subtracted (hence the "missing light") and thus unmeasurable.

(The "sky" being subtracted is a combination of emission from certain atoms in the tenuous outer atmosphere above HST, sunlight scattered from dust grains in the inner solar system, and the so-called "extragalactic background" = the combined light from unresolved distant galaxies.)

The abstract mentions four improvements the new study implemented when they re-processed the HST images: "1) creation of new absolute sky flat fields, 2) extended persistence models, 3) dedicated sky background subtraction and 4) robust co-adding."

I would suggest that the third item is perhaps the most important: they implement a method which does not subtract off the faint outer regions of the larger galaxies, and thus the resulting images still have information about the outer parts of these galaxies.

The plot below (extracted from Figure 20 of the paper) illustrates the kind of improvement they were after. It shows the surface brightness (in the F105W near-infrared filter) of one of the largest galaxies (a giant elliptical -- I think it's the big, round, yellow galaxy in the lower middle of the color image) as a function of radius (measured in elliptical annuli). The red triangles were measured using the XDF-processed image, the blue squares used the HUDF12-processed image, and the black points use the newly re-processed image produced as part of this paper [ABYSS]. You can see that the XDF points drop off at a radius of about 55 kpc, the HUDF12 points fall off at around 90 kpc -- but the light from this galaxy can be traced out to 140 kpc in the ABYSS-reprocessed image. This also means you can potentially analyze structure in the outer part of the galaxy out to ~ 150 kpc, which wouldn't be possible with the previous versions of the image.

(I should point out that I'm friends with, and have co-authored papers with, a couple of the authors, so I may be a bit biased -- but I think this is really impressive work!)


Why did Venus not lose its atmosphere without a magnetic field? It is often stated that the magnetosphere not only shields the planet from cosmic radiation, but also prevents atmospheric loss. Why then did Venus not lose most of its atmosphere if it doesn't have a strong magnetic field? Is there another mechanism at play, or is the statement about the importance of magnetosphere to atmospheric loss prevention wrong?

There is an interesting article on the magnetosphere of Venus on the ESA Science and Technology site. You can find the article here and it will probably answer your question.
The article states, like you did, that some planets, like Earth, Mercury, Jupiter and Saturn, have magnetic fields internally induced by their iron core. These magnetic fields shield the atmosphere from particles coming from solar winds. It also confirms your statement that Venus lacks this intrinsic magnetosphere to shield its atmosphere from the solar winds.
The interesting thing, however, is that spacecraft observations, like the ones made by ESA's Venus Express, have shown that the Venusian ionosphere's direct interaction with the solar winds causes an externally induced magnetic field, which deflects the particles from the solar winds and protects the atmosphere from being blown away from the planet.
However, the article also explains that the Venusian magnetosphere is not as protective as Earth's magnetosphere. Measurements of the Venusian magnetic field show several similarities, such as deflection of the solar winds and the reconnections in the tail of the magnetosphere, causing plasma circulations in the magnetosphere. The differences might explain the fact that some gasses and water are lost from the Venus atmosphere. The magnetic field of Venus is about 10 times smaller than the earth's magnetic field. The shape of the magnetic field is also different. Earth has a more sharp magnetotail facing away from the Sun and Venus has a more comet-shaped magnetotail. During the reconnections most of the plasma is lost in the atmosphere.
The article explains therefore that although Venus does not have an intrinsic magnetic field, the interaction of the thick atmosphere with the solar winds causes an externally induced magnetic field, that deflects the particles of the solar winds. The article suggests, however, that the different magnetic field may mean that lighter gasses are not as protected and therefore are lost into space.
I hope this sufficiently answers the question.


What are the characteristics of solid carbon dioxide on the Martian poles? When carbon dioxide forms on Mars is it like ice or snow? How dense is newly formed solid carbon dioxide? Would more carbon dioxide form inside a pit vs a flat plain?

How thick or deep does solid carbon dioxide accumulate on Mars?

https://space.stackexchange.com/questions/31724/martian-carbon-dioxide-turbine-oxygen-generator

On Earth we know of frozen carbon dioxide as "dry ice." If shaved, it could exist as "dry ice snow." On Mars the dry ice snow could precipitate, then become compressed into dry ice. It would probably be a stiffly packed granular solid.
I had only heard of "water ice" on Mars. Check the article on kottke.org  Link:  https://kottke.org/18/12/a-massive-ice-filled-crater-on-mars


Once a black hole is formed, is there anything other than Hawking radiation which shortens its life? Hawking radiation is supposed to very slowly evaporate a black hole (terms and conditions apply :] ).

Apart from Hawking radiation, is there any mechanism or effect that can make a black hole cease to exist? Or once they are formed are they expected to exist in this form "forever"?

PS: I posted the same question to Physics but I think it applies well to both SEs. So I hope it is okay to try here as wwell.

It can merge with another black hole, if that counts. A simulation of that is illustrated here

That aside, no black hole with decay more quickly, as far as I can find out, than the "normal" decay of a non-rotating, uncharged black hole into which nothing is falling. Let's call this the "standard rate of decay".

"Feeding it" (allowing any matter or energy to fall into it), will reduce the decay below the standard rate, or even reverse it. Since the CMB pervades space and will fall into the black hole, blocking this (for instance by surrounding the hole by an opaque refrigerated shell) will be necessary to allow it to decay at the standard rate. Indeed for a black hole of mass over roughly $10^{22} kg$ the CMB would feed it faster than it decays, if not blocked.

Energy from the spin of the  black hole can be extracted using the Penrose process. Since this answer shows that rotation also reduces the emission of Hawking radiation, you can clearly speed up the decay by extracting energy from as much of the spin as possible, but the fastest decay you get is, once again, the standard rate.


How to approximate the distance at which the thin disk peaks along a given LOS from Earth for a given galactic coordinate per 1 sq deg? The density of stars drops off exponentially from the center of the galaxy and can be approximated as follows using a cylindrical coordinate system:
$$ \rho(R, z) = \rho_0 \exp \frac{-R}{h_R} \ \exp  \frac{-|z|}{h_z} $$
screenshot
The radial scale length and height of the thin disk are ~ 3000 pc and ~ 200 pc respectively. At the moment rho_0 will just be taken as 1. We observe at a given l and b of 90 and 15 respectively using the galactic coordinate system (just look at l and b for now, the rest are annotations I added that I will describe shortly). The galactic coordinate system uses spherical coordinates:

We want to know where the density of stars peak along at given LOS (from Earth) per 1 sq degree at a given l and b. There are 41253 sq degrees in the entire sky, so 1/41253 will be the fraction we want to multiply the density equation by. However, we must also consider how the volume element changes by distance along the given LOS, so we need to also multiply rho and 1/41253 by this change in volume. This volume element can be represented by $4 \pi r^2 dr$ (the surface area of the sphere and the small change in the distance along the LOS). We are almost done but need a way to transform between coordinate systems by relating galactic coordinates to distance r and R (as seen in the diagram above). Mathematically, they are related as follows (Eq. 3 in Where Does the Disk Stop and the Halo Begin? Kinematics in a Rotation Field Morrison, Flynn and Freeman 1990 Ast. J. 100 (4) 1191-1222):
$$ r^2 = R_☉^2 + R^2 - 2 R R_☉ \cos b \cos l $$
screenshot
So, for a given r (distance from star or LOS to sun), l and b, we can find R (the distance from the G.C. to the star or LOS). This is in cylindrical coordinates, so we need to use cos(b)R and sin(b)R to find R and z in the density equation above (note these are different r's). The volume element needs to be the LOS from Earth, which corresponds is the R in the diagram and equation for the paper (for the given l,b, and r). Putting this all together, I have the following code. Please let me know if this attempt seems reasonable, or if not, where I went wrong.
import numpy as np
import matplotlib.pyplot as plt

C=8500 #Distance from center of galaxy to sun (R_sun_

#Galactic Coordinates
b=np.deg2rad(15) #above plane
l=np.deg2rad(90)

#Scale heights for the thin disk
h_R=3000
h_z=200

nsqdeg_in_sky=41253 #square deg in whole sky

#R: Distance from star to sun
R=np.arange(0,15000,1)

#distance from star to G.C.
#r: distance from star to G.C. kpc
r=np.sqrt(C**2+R**2-2*R*C*np.cos(b)*np.cos(l))
print(r)

#Spherical Galactic Coordinates to cylindrical coordinates
#z component of R
R_z=np.sin(b)*r
#r component of R
R_r=np.cos(b)*r

#density equation describing thin disk
rho_thin=np.e**(-R_r/h_R)*np.e**(-np.abs(R_z)/h_z)

#desnity equation combining volume element and portion of sky
N_thin = (1./nsqdeg_in_sky)*(4*np.pi)*R*R*rho_thin

plt.title("Number of stars per 1 square degree")
plt.plot(R,N_thin,color='blue',label='thin')
plt.legend(loc='best')
plt.xlabel(r'$distance$ [pc]')
plt.ylabel(r'$N_\ast$ [pc$^{-1}$]')

edit to add plot for given l and b above:

If you have a ray from an origin point $\mathbf{x}_0$  with unit direction $|\mathbf{v}|=1$,  $\mathbf{x}(s)=[x(s),y(s),z(s)]=\mathbf{x}_0+\mathbf{v}s$ and look at the density times distance $\rho(s)=\rho_0 e^{-R(\mathbf{x})/h_R}e^{-z(\mathbf{x})/h_z}s^2$ the derivative is
$\rho'(s) =\rho_0\left\{ [-e^{-R(\mathbf{x})/h_R}R'(\mathbf{x})/h_R]e^{-z(\mathbf{x})/h_z}s^2 + [-e^{-z(\mathbf{x})/h_z}z'(\mathbf{x})/h_z]e^{-R(\mathbf{x})/h_R}s^2 +[2s]e^{-R(\mathbf{x})/h_R}e^{-z(\mathbf{x})/h_z} \right\}$
$=\rho(s)\left\{[-R'(\mathbf{x})/h_R]+[-z'(\mathbf{x})/h_z]+[2s]/s^2\right\}$.
Now, $R'(s) = (x(s)v_x+y(s)v_y)/R(s)$ and $z'(s)=v_z$ so the criterion for a maximum or minium is $0 = -(x(s)v_x+y(s)v_y)/h_RR(s)-v_z/h_z + 2/s$ or $$\frac{2}{s} = \frac{[x_0v_x+y_0v_y] +[v_x^2 + v_y^2]s}{h_R\sqrt{(x_0+v_xs)^2+(y_0+v_ys)^2}}+\frac{v_z}{h_z}$$
I can't see how to simplify it further.
I think the right approach here is just to do a numeric solution using golden section search rather than do a lot of algebra that anyway ends with a root finding algorithm. Doing that also allows you to use any stellar density model.


What are the maximum and minimum impact speeds for an asteroid that would strike the Earth? I gather that the impact speed depends on the radius of the orbit of the asteroid.

Is the orbit of asteroids in the same direction around the sun, or can they move in the "opposite" direction?

Although most objects orbit the Sun in the same direction — having emerged out of the same rotating gas cloud that spawned the Solar System — some asteroids and other minor planets do move in opposite, or retrograde, orbits (see this Wikipedia article for a list of such objects).

The minimum speed for an asteroid is achieved if it has more or less the same velocity around the Sun as the Earth. In this case the gravitational attraction of Earth will accelerate the object to the escape velocity of Earth, i.e.
$$
v_\mathrm{min} = v_{\mathrm{esc,}\oplus} = \sqrt{\frac{2GM_\oplus}{R_\oplus}} \simeq 11\,\mathrm{km}\,\mathrm{s}^{-1}.
$$
Here, $G$, $M_\oplus$, and $R_\oplus$ are the gravitational constant and the mass and radius of Earth, respectively.

The maximum speed is achieved at a "head-on" collision. Earth's speed around the Sun is
$$
v_{\mathrm{orb,}\oplus} = \sqrt{\frac{GM_\odot}{d}} \simeq 30\,\mathrm{km}\,\mathrm{s}^{-1},
$$
where $M_\odot$ and $d$ are the mass of the Sun and the distance from Sun to Earth (1 AU).

If the asteroid travels on the same orbit, but in the opposite direction, the impact will then be at 60 km/s. However, if the asteroid comes from far away (e.g. the Oort Cloud), it will be accelerated by the Sun and achieve a velocity equal to the escape velocity from the Sun at the location of Earth. As is seen from the two equations above, the orbital speed and the escape velocity differ by a factor of $\sqrt{2}$. That is, an object falling from infinity toward the Sun, will have a speed equal to $30\,\mathrm{km}\,\mathrm{s}^{-1}\times\sqrt{2}=42\,\mathrm{km}\,\mathrm{s}^{-1}$ when it reaches Earth.

Hence, the maximum impact velocity is
$$
v_\mathrm{max} = 30\,\mathrm{km}\,\mathrm{s}^{-1} + 42\,\mathrm{km}\,\mathrm{s}^{-1} = 72\,\mathrm{km}\,\mathrm{s}^{-1}.
$$


How to make a 65 cm lens with a 20 cm hole in it for a Hamiltonian telescope? This answer to What (the heck) is a Hamiltonian telescope? Is this one? confirms that the telescope in the question linked there is indeed as described and that the first lens is a full 65 cm aperture lens, the second element is a full 65 cm negative meniscus back-silvered, and some corrector lenses are embedded= within a hole in the large primary lens.
Optically I can imagine that it might be possible to let the light transmit again through the primary and still compensate, and mechanically that seems more attractive than polishing a double-sided transmission lens with a hole through the center.
But apparently that's what's been done.
Question: How to make a 65 cm lens with a 20 cm hole in it for a Hamiltonian telescope? I'm thinking about issues including the following:

Is the blank cast with a hole already, or is it drilled?
If drilled, is that before the first side is polished, before one side and after the other, or after both sides?
After drilling does one need to anneal the glass again?

Glass can experience strain-induced birefringence among other things, so I am really interested in finding out how optical surface figures are applied to both sides of this lens with a big hole in it without causing optical problems within the bulk of the glass.

Image from this answer to What exactly is a Hamiltonian telescope? Is this one?

Drill or core the blank
(anneal if needed ... most likely not if annealed to begin with)
use pitch to glue a plug of the same material in the hole
grind, polish and figure as normal then remove the plug.


how to get image coordinates of source points in saoimage ds9? I loaded my fits image to saoimage ds9. Then, Analysis->Catalogs automatically brought up source locations.  how can i get xy image coordinates of this source locations?

OK, it seems to me that what you have is a fits image (from where?) that has some world coordinate system information attached. When you load the 2MASS catalogue, SAOimage is able to use the RA and Dec in the catalogue to calculate the x,y positions of the catalog sources in your image (and marks them as green circles).

Is what you are asking - how do I get the x,y coordinates of these 2MASS sources in my image? Or do you wish to know what the x,y coordinates of the actual sources in your image? If the latter then you will need some other software like Sextractor or DAOphot to do source-searching and parameterisation in your image.

If the former, here's a way of doing it. When you load the catalog you should see a window open entitled "2MASS Point sources" which lists all the sources in your image. From the "File" dialogue, select "Copy to regions". Then, in the main image window open the "Regions" dialogue and select "List regions". Another dialogue box will open asking you to select the format and coordinate system. From this, select "x,y" for the format and "image" for the coordinate system. This will give you a list of x,y coordinates.

If you want to match this up with the 2MASS sources then you have to "list regions" again, but this time select "fk5" as the coordinate system. This will then give you a list of the RA, Dec of the same sources. Use you favourite editor to attach these lists to each other.

Quite probably there is a more elegant solution, but this works.


Description of Henyey tracks on wikipedia incorrect? So, if you search for Henyey tracks on wikipedia (I know, the shame of it!) you will come across this statement:

  The Henyey track is characterized by a slow collapse in near hydrostatic equilibrium. They are approaching the main sequence almost horizontally in the Hertzsprung–Russell diagram (i.e. the luminosity remains almost constant).

Ignoring the bad grammar, I take umbrage with the following:

  the luminosity remains almost constant.

I would say that the apparent horizontal track associated with the Henyey track is more due to the fact that in a typical HR diagram the temperature axis ranges over about one order of magnitude, while the luminosity axis may vary over five or more orders of magnitude.

So, to frame this as a question, would you agree or disagree with the statement on the site or with what I have stated (i.e., that the Henyey track is an apparent horizontal track caused by the nuance in the magnitude scales of luminosity in comparison with the temperature)?

The wording provides a contrast with the Hayashi track phase that immediately precedes it, where the luminosity decreases by orders of magnitude with little change in effective temperature.

You are correct though, the Henyey track for low-mass stars is not horizontal in the HR diagram, however it is a critical point that the luminosity does not change very much (less than a factor of 3) as the effective temperature increases, since it implies that the star slowly shrinks, and by the virial theorem, the core temperature gradually rises and is ultimately able to sustain nuclear fusion.


Intuitive explanation for why the universe is flat I went through Prof. Hitoshi Murayama's online lecture where he describes why the universe is flat. One of the evidences for a flat universe is the evenness of CMB radiation on all directions. My understanding is, that a uniform CMB pattern only indicates that inflation or early expansion of the universe was more or less equal on all directions. Why is this cited for the claim that the universe is flat, or in what sense is the universe 'flat'?

The CMB lets us measure how close to flat the universe is right now.

On the other hand, inflation tries to explain how we got from whatever the early universe was to right now.

The motivation for the latter being that even extremely small deviations from perfect flatness in the early universe should have resulted in very obvious deviations from flatness today.  That we don't see obvious deviations then suggests the universe was amazingly close to perfectly flat early on, and such perfect tuning seems bizarre. Inflation basically lets a (in some sense "very") non-flat early universe get flattened to this precision, thereby resolving the "problem" of having an almost perfectly flat early universe.  It takes the situation from an extremely narrow range of initial conditions for the flatness of the universe and gives us a broad range of equally valid possibilities, instead.

Inflation is also commonly used to explain the remarkable homogeneity of the CMB, which suggests that parts of the universe that cannot ever have received light speed signals from each nevertheless achieved thermal equilibrium (an at-most light speed process).  Inflation throws in a super-luminal expansion to allow these regions to be within light speed communications of each other in the very early universe, with just enough time to achieve thermal equilibrium, and then inflates them beyond light speed communication.


How do K, F, A and B stars die? I know that M (red dwarf) main-sequence stars slowly fade to white dwarf stars, and that G (yellow dwarf) main-sequence stars become red giants and then shed their outer layers in a planetary nebula leaving behind a white dwarf, but do K (orange dwarf) main stars take a similar route to red dwarfs? Or do they expand into red giants like yellow dwarfs?

What about F (yellow-white) main-sequence stars? Do they turn into supergiants and go supernova? Or do they take a similar route to yellow dwarfs? And what about A (white) and B (blue-white) main-sequence stars?

K-type dwarfs behave like a solar mass star, and will leave behind a (slightly) lower mass white dwarf.

F-type dwarfs will behave like a solar mass star and leave behind a (slightly) higher mass white dwarf.

Higher mass (hotter) stars will produce higher mass white dwarfs, up until the initial main sequence star is of about $8M_{\odot}$ (a spectral type of about B3). The corresponding remnant white dwarf mass increase up to about $1.2 M_{\odot}$. All these stars go through a giant phase, He core burning and asymptotic giant branch phases.

Above this then carbon burning will be initiated in the core and it is likely that burning will continue through to iron, followed by core-collapse and a supernova.


What is this gap between hot and cold Jupiters? When plotting mass to semi major axis on exoplanets.org, there seems to be a gap between Hot Jupiters and Cold Jupiters. A lot of Jupiter-mass planets at less than 0.1 AU have been discovered, and a lot at more than 1 AU, but there aren't so many in between.

Since many planets closer and further away from their star have been discovered for the same mass, it seems unlikely that this gap would be due to observational bias.

Giant planets like Jupiter form through accretion in the outer part of the disk, beyond the frost line where the material is cool enough for volatile icy compounds to remain solid. These planets cannot form closer to their host star, since there is there is not enough matter in the protoplanetary disk at smaller radii.

Due to interactions with other planetesimals and the protoplanetary disk itself, the orbit of the young planets changes. This is planetary migration. Interactions with the protoplanetary disk cause planets to migrate inwards. This migration occurs rather fast, so we don't see many Jupiter mass planets at intermediate distance from their star.

The protoplanetary disk is somewhat doughnut shaped. When a migrating planet reaches the inner edge of the doughnut, the sum of forces exerted on the planet change which halts the migration.

This is why there are two clumps in the diagram, for Jupiter-mass planets. One clump (i.e. cold Jupiters) where these planets form, and one clump (i.e hot Jupiters) where they end up after migrating, but nothing in between.


Is a Procyon - brown dwarf - Jupiter - superearth - Mars - Pluto system theoretically possible? I once conducted a thought experiment and with some amateurish calculations based around Hill sphere and Roche limit came up with semi-major axis distances like this:

Procyon-BD:  3.6 AU
BD-Jupiter:  0.7 AU
Jupiter-SE:  0.12 AU
SE-Mars:     0.02 AU
Mars-Pluto:  0.0033 AU

Assuming respective masses of: 496 260 earths, 15 000 earths, 313 earths, 6 earths, 0.107 earths and 0.00218 earths.

Clarification: I use names of specific bodies (Procyon, Jupiter etc) but a Procyon-like star, a Jupiter-like gas giant and so on would be more precise. Think of them only as examples of bodies with such masses.

Could such a system be possible?

Thanks to this answer linked by antispinwards I learned that such system most probably would not be stable, as I calculated it with proximity to upper limit of Hill sphere for each body.

But, if instead, I would have taken distances closer to 1/3-1/2 Hill sphere, it could be theoretically feasible.

  The orbit of a satellite's satellite will be stable if it's deep
  enough inside the Hill sphere, within the so-called true region of
  stability. The limits are a bit fuzzy, but the true region of
  stability is typically the lower 1/3 to 1/2 of the Hill sphere.

On a personal note, if it's so, it surprises me this type of systems are not more pondered on and discussed, with maybe even some probabilities calculated. As it could have a great impact on thinking about abundance of potentially habitable worlds in the universe (imagine how great a number of earth-like satellites could live in such a system in the habitable zone of its host star (it being short-lived on cosmological time-scale notwithstanding)).


Would a comet break up during a close pass of the Earth Would a short-period comet, with a 2-mile nucleus, be able to come within 200,000 miles of Earth without breaking up?

It wouldn't break up.

As my Granddad used to say "Close is only good in horseshoes". If the comet doesn't impact the Earth will be able to pass by unchanged. 200000 miles is a long way, about the distance of the moon from Earth. Asteroids have come this close before (eg 2011 XC2).

Comets can be broken up by close encounters with the sun, but the sun is rather hotter than the Earth [citation needed]. 200000 miles is also well outside the Roche limit for the Earth-comet system, and at the speed that comets travel relative to the Earth, tidal effects won't have enough time to act.

So the comet will pass the Earth quite happily. It is a rather small comet, and even the comet's coma won't interact with the Earth. After closest approach it is possible that the Earth could pass through the comet's tail, which would cause a meteor storm.

Even at 100000 miles, there is not much effect on satellites nearly all of which orbit at 22000 miles or less (22000 is geostationary orbit) It is plausible that they could be damaged by meteoroids.

The dust that comes of comets is all quite small. If you are looking for a "near miss that still causes damage" try this. The comet makes a close pass of Jupiter that causes it to split into two parts. The larger part misses Earth by 100000 miles but the smaller part (a 300m chunk that wasn't noticed until too late because story) hits Merchtem 10km outside Brussels


How are the speed, distance and time of India's MOM and USA's Maven Mars missions calculated? How do the scientists calculate the correct time to launch the Mars mission so that the satellite travel time is less? How they are synchronizing the speed of satellite with respect to earth and mars ?
Refer this link for more understanding:
Animation of MOM and Maven launch

There is an extremely accurate model of the solar system and the positions of the planets and so on. So basically you have to lay Newton's equations for Mars, the probe, Earth and so on, you have an equation that models Mars's path, and you have equations for your probe, so you get some x's and y's that eventually will have to match if you want to reach Mars. From there you can then calculate the values for the initial conditions of the probe -such as velocity and direction- that will satisfy you equations.


What is the fifth moon-like object I saw around jupiter through my telescope? Tonight while observing Jupiter I was able to see a five (what looked like) moons. I have observed Jupiter often before and only been able to make out the four galilean moons. It's my understanding that through my 8-inch scope it would be impossible for me to see any others. Switching to a widefield lens I was able to see Jupiter along with some nearby stars however the 5 moon-like points around it all still appeared... well, moon-like in comparison to the bright stars.

I observed Jupiter from about 2016-04-15 05:20 to 05:40 UTC and all five of the moon-like objects were present the whole time. What was I probably looking at?

A seventh magnitude star.  See this link:

http://www.cloudynights.com/topic/533968-jovian-events-414-1516/

Further info: the star in question was probably HIP 54057 (also known as SAO 118636 or HD 95848.) On April 12th it was occulted by Jupiter (that is, Jupiter passed in front of it) for about three hours. So the star would have been in the area of Jupiter for a few days.


What are the aberrations of an SCT? And how can they be eliminated? The Schmidt-Cassegrain telescope is amazing! It is compact, suitable for astrophotography ,and it requires little to no collimation. However, the SCT has some inherent optical issues. What are they? What are their causes? and how can they be eliminated?

The primary aberration present in a Schmidt–Cassegrain telescope is spherical, due to the primary mirror being spherical in shape; that causes light at the edges to have a different path length than light at the center, and thus a different focal point.

The image above, from Wikipedia's "Schmidt corrector plate" webpage, shows an exaggerated corrector; they are actually much flatter.

The secondary aberration is coma, though this is present to a much lesser extent. Coma is a spherical aberration from rays that come in off-axis. Coma shows up as little off-axis comet-shaped blobs that point inwards towards the center of the field and get bigger as you look towards the edge of the field of view.

Persons wanting a lengthy technical and mathematical explanation will enjoy reading sections: "10.2.2. Full-aperture Schmidt corrector: Schmidt camera" and "10.2.2.4. Schmidt-Cassegrain telescope (SCT)" of the https://www.telescope-optics.net/ website.

Recent improvements in the design are offered using aplanatic flat field Schmidt-Cassegrain optics. Here's a screenshot from Celestron's video showing the difference between a standard versus aplanatic Schmidt telescope design:

The competing system from Meade is called Advanced Coma-Free (ACF™) optical design. Claiming also to offer coma-free pinpoint star images and flatter image fields while reducing astigmatism and eliminating diffraction spikes.

The website forum "Starry Nights - Cats & Casses" compares Celestron and Meade.


Maximum distance light can travel? If we assume that there is nothing except the sun in the universe then how far the light can travel ??

Using your scenario, and lets ignore dark energy expansion and red-shifting at greater distances,.  There would be no distance where the sun will disappear.  That might seem counter intuitive but it's true.

On Earth, we receive about 10^17 photons per square CM.  (source).  Using the inverse square rule, to receive 10 photons per second on a pair of human eyes, that distance would be 100 million AU - give or take, or about 1,580 light years.    We probably can't see just 10 photons, but in pitch black, with eyes adjusted to the dark, maybe we could.   That's about the distance where the sun would give of steady but very very dim light.

Further than that, what happens has more to do with the quantum nature of light than the brightness and distance.   As you double the distance you see 1/4th the frequency of photons.   At 5,000 light years you see 1 photon per second, basically a very dim blinking light, but just as bright as it is at 1,580 light years, just appearing less often.     Double the distance again and it flashes once every 4 seconds.    There's no distance where you stop seeing the sun, you just see it less and less often, but it never appears dimmer than 1 photon.


Question about semi-major axis mean separation I am currently doing an assignment but I cannot for the life of me understand what one of terms used means. Notice I am not trying to get help with the assignment itself, but rather to make me understand so I can solve it.
Anyways, in the question (about visual binary stars) I am given their parallax angle etc etc, and I am to find the mass of them. But the sentence I cannot understand is the information given in "The semimajor axis (mean separation) is X.YZ'' and...(continue)", what does it mean? I draw two elliptic orbits and they intersect, I draw the semi-major axes of each ellipse. Is the semi major axis mean separation the angle that is between the semi major axes and the line drawn between each stars center in their elliptic orbits?
Help would be very nice. Thanks in advance!

The semi major axis $a$ of an orbit is the same as the arithmetic average of the minimum and maximum distance $r$ or the object from the central object in sense of $a = 0.5\cdot(r_{min} + r_{max})$; thus semi major axis is sometimes also called mean separation. As such you are given the sum of the semi major axis of two components.
A two-body problem can be reduced / simplified to a mass-less body circling an object of the combined mass - as such you can calculate the combined mass of both stars with the numbers you are given.


Is most hydrogen in the universe in the form of plasma, atomic neutral hydrogen, ionized hydrogen, or molecular? I have had trouble finding an answer somewhere....
Some places say most hydrogen is plasma, such as the stuff stars are made of (mostly) and the 'warm-hot intergalactic plasma'.
Other places say neutral atomic hydrogen (H(I) to astronomers), such as the stuff in the interstellar medium, is the most common...

@HurtHikes, I hope this answers your question,
According to Wikipedia, most Hydrogen is in the form of Atomic and Plasma states.

Throughout the universe, hydrogen is mostly found in the atomic and plasma states, with properties quite distinct from those of molecular hydrogen.

Intuition says that because, as this link says, 99.9% matter in the universe is plasma, most Hydrogen in the universe should also be present in Plasma state.

"99.9 percent of the Universe is made up of plasma," says Dr. Dennis Gallagher, a plasma physicist at NASA's Marshall Space Flight Center. "Very little material in space is made of rock like the Earth."

This link has the calculations to determine the amount of Hydrogen in Plasma in relation to the total amount of Plasma (if anyone is interested...)

Stars are made of relatively simple stuff.  By mass, our Sun is 73% hydrogen, 26% helium, and only 1% of higher Z (atomic number) atoms.

X = mHnH / r = density of hydrogen / total density

Though I am not sure about the actual answer about the same, I will be grateful if anyone can point out to the right answer. And although I have made every effort into the removal and rectification of errors, If there are any errors in my answer, I will be grateful if anyone can point them out.


Could anyone recommend a good book about the basics of astrophysics? I am sorry if a similar question has been asked before. I look at other questions but was not entirely satisfied. So, I am a Psychology student who would like to know more about the basics of astrophysics. Could anyone recommend a good book about this. Also, I am particularly interested in black holes, so introductory literature on that is welcome too.

Answers to these sort of questions tend to be opinion-based which is frowned upon, but Neil DeGrasse Tyson's recent book Astrophysics for People in a Hurry has had very positive reviews. Injecting some facts, it has apparently spent a year on the New York Times bestseller lists and has sold over a million copies. It could be a good place to get you started before starting on a e.g. introductory college/university textbook.


What is the upper bound of number of stable interacting stars in a star system? I remember very well reading in an article that it's impossible to have a stable star system of 4 or more stars. The article justified its claim discussing about the complexity such systems would have in terms of mass, velocity and distance to be stable. Further more I think there was a constriction related to our 3-dimensional universe, unless I confuse it with an other article about why solar systems cannot form in 4 dimensions... I don't remember though where I read the article.
And since then I thought there couldn't be star system with more than 3 stars and that belief was enforced with the movie Contact when Jody Foster traveled through space and encountered a quadruple star system. What an opportunity for a movie to claim the existence of something so exotic, right?
Recently there's been a fuss about the TIC 168789840 six-tuple system. So either I'm very behind in astronomy news or I confuse the terms Star System and Solar System in the article I read...
Here an article about the aforementioned system.

There is no true upper bound. You could argue that a galaxy is a vast multiple star system, with hundreds of billions of stars. Natural galaxies are not perfectly stable (random encounters eject stars over very long timescales), but one can construct stable systems by taking pairs of similar-mass binaries and placing them into a remote circular orbit around each other. This can then be repeated indefinitely, with $2^n$ stars forming a vast system.
In practice such neat systems are unlikely to form naturally (4 levels seem to be the max observed and often involves very unequal masses) and very wide systems will encounter other stars often enough to be destabilized. Eccentricities tend to have a triangular $f(e)=2e$ isothermal distribution, making things less stable. See (Duchêne & Kraus 2013) for more statistics.
(As an aside, Jaron Lanier and Piet Hut have suggested constructing such vast hierarchical systems as a clear indicator that there is intelligent life present.)