Why black holes are extremely cold? "The most massive black holes in the Universe, the supermassive black
holes with millions of times the math [sic] of the Sun will have a
temperature of 1.4 x $10^{-14}$ Kelvin. That's low. Almost absolute zero,
but not quite. A solar mass black hole might have a temperature of
only 0.00000006 Kelvin."
September 5, 2016 by Fraser Cain, Universe Today

Black holes absorb every form of energy, even light. Absorption of energy should raise its temperature but still it is extremely cold, why?

Under General Relativity (GR) alone, a Black Hole's (BH's) event horizon is a point of no return -- anything that passes through the event horizon is lost and gone forever, and nothing comes out. Hence, under GR alone, BHs are utterly black and don't have a temperature at all.

This is why the absorption of radiation (or anything else) by a BH doesn't raise its temperature -- it just gets swallowed up and lost. (It's mass, angular momentum and charge do remain, but that's all -- see the No Hair Theorem.)

(Note: The accretion disk that surrounds a BH can be very hot indeed, but that's another thing entirely.)

Stephen Hawking discovered that applying quantum mechanics to BHs showed that BHs would emit a random spray of radiation, and that that radiation was precisely what a black body would emit -- black body radiation.  This is called Hawking radiation.

Black body radiation is simply the thermal emission of a perfect absorber of radiation, and leads to the inescapable conclusion that a BH does have a non-zero temperature.  Interestingly, Hawking's analysis showed that the effective temperature of the BH is inversely proportional to its mass and that solar-mass BHs (which are the smallest for which we have actual evidence) would have a temperature of about 0.00000006 K. Kinda cold, but still not zero.

Note that, unintuitively, a solar mass BH get colder as it absorbs radiation. Because any radiation (or anything else) it absorbs increases its mass, and since higher mass BHs are colder, the more energy you dump into one, the colder it gets!


Calculating the Size of Earth in the Apollo 8 Earthrise Photo I was debating with someone online (I know, great way to get nowhere fast) regarding the size of Earth as it appears in the famous 'Earthrise' photos from Apollo 8.
Below is an accurate (almost pixel-perfect) calculation of the apparent size Earth should appear. In this instance it checks out, however if there is a better approach in calculating this and similar images I would be interested to know :

The photograph was shot on 70mm film using a 250mm lens. We can calculate the what the Field of View should be for the image, using the following formula :

         FOV (rectilinear) = 2 * arctan (frame size/(focal length * 2))

i.e.
             FOV (rectilinear) = 2 * arctan (70/(250 * 2))
             = a FOV of 15.93 degrees.

From our point of view on Earth, the average angular diameter of the moon is 0.5 degrees. From the Moon, the Earth has an angular diameter of about 1.9 .

From this it can be calculated how many pixels in size the Earth should appear in the photograph :

Diameter of Earth in Pixels = Photo's Diagonal Resolution * (Earth's Angular Diameter / FOV)

i.e.

Diameter of Earth in Pixels = 3841.87 * (1.9 / 15.93)

= 458.23 pixels

For reference, The full, un-cropped version of the 'Earthrise' photo that the above calculation is basing the image resolution upon can be found at this location :

https://www.nasa.gov/images/content/297755main_GPN-2001-000009_full.jpg

I imagine that this could be further refined by deriving the correct angular diameter of Earth from the exact distance of the moon at the time the photo was taken, as well as the altitude of the Apollo module above the moon.

What is your actual question? Are you asking others to confirm your process/formulas?
Correct, specifically whether the diagonal fov is always the correct method?

No, an incorrect approximation is never "always the correct method".
Even with the basic pinhole camera model (the basis for almost all of the better approximations) position on the focal plane is never linear with incident angle. Instead, you draw similar triangles.
Here is a simple 2D analogy, usually people use 3D, Projective Geometry and Homography and the matrix math it requires.
$$\frac{w}{d}=\frac{x}{f}$$
$$x=f\frac{w}{d}$$
where $d$ and $f$ are the distances along the optical axis from the source and image plane to the lens pupil, and $w$ and $x$ are distances from the axis to the points on those planes which are perpendicular to the axis.

Most higher order methods that incorporate lens distortions star with the pinhole camera approximation, and quantify the imaging system's deviations from it.
no arctangents were harmed in the making of the projective plane.


Why is Saturn invisible in this radar image of its rings? The image below is from Radar imaging of Saturn’s rings Nicholson, P. D. et al., Icarus 177 (2005) 32–62, doi:10.1016/j.icarus.2005.03.023 and discussed further in this answer to How did Arecibo detect methane lakes on Titan, and image Saturn's rings?
I believe I can see Saturn's radar "shadow" blocking the furthest parts of the ring near the top of the images, but I can not see any reflections from Saturn itself.
The paper addresses a possible signal from the planet itself in Section 3.2. Ring images:

Echoes at low Doppler shifts might be expected to arise either from near the subradar point on the planet itself or from ring material far outside the main rings. Our images show no evidence for any echo from the subradar point on Saturn, which would appear near ν = 0 and τ = −2RS/c = −402 ms.

The same answer describes radar measurements of Saturn's moon Titan as well.
It might be possible to make a hand-waving argument that Saturn itself is invisible because it's "just gas" but according to ESA's Saturn's atmosphere

The top visible cloud deck, made of ammonia clouds, is found at about 100 kilometres below the top of the troposhere (tropopause), where the temperature is about -250°C.
The second cloud deck, made of ammonium hydrosulphide clouds, is found at about 170 kilometres below the tropopause, where the temperature is -70°C.
The lowest cloud deck, made of water clouds, is found at about 130 kilometres below the tropopause, where the temperature is about 0°C (freezing point of water).

So I'm thinking that the various clouds of different colors and compositions will contain droplets or particles, rather than just be regions of different gas composition.
Why then does there seem to be nowhere near as much radar return from the planet as there is from the rings?

The main issue is that there is relatively little material in Saturn's atmosphere that can efficiently scatter radar waves, so the radar basically just gets absorbed.

The key point is that it's much harder to get a radar return from the very small objects (aerosol droplets or tiny ice particles) that would make up clouds in Saturn's upper atmosphere than it is to detect the meter-sized ice chunks in the rings.

The figure below (from here) shows $\sigma / (\pi r^{2})$, which is the "effective" scattering cross section relative to the geometric cross section of a scattering object, plotted versus the size of the object $r$ relative to the wavelength of light $\lambda$ (e.g., the radio waves used in radar). When the size of the object is roughly the same size as wavelength of light, or larger, it efficiently scatters them: its effective cross section is roughly the same as its geometric cross section (ignoring other effects such as chemical composition, surface roughness, etc.). Sometimes it can even be several times larger!

But when the objects become smaller than the wavelength, you transition to the Rayleigh scattering regime, where the scattering depends on the size of the object to the sixth power.

Imagine a sphere 10 cm in radius, roughly the same as typical radar wavelengths (so $2 \pi r / \lambda \sim 1$). It will have an effective cross section similar to its geometric cross section: about 300 cm$^2$. Now imagine subdividing that object into spheres one-tenth the radius. This would mean about 1000 objects 1 cm in radius (so that the total volume is the same), with a total geometric cross section of about 3000 cm$^2$ (realistically a little less, because of shadowing). But the plot shows that the effective cross section goes down by a factor of 1000, so the total radar cross section of all the small objects -- and thus the backscattered energy of the radar waves -- would be ten times smaller than it was for the single big object of the same total volume. And that's just for a reduction in size of ten; in reality, we're talking about going from centimeter-to-meter-sized objects (the ring) to sub-mm and micron-sized cloud particles in Saturn's atmosphere.

This is why weather radar on Earth can show you precipitation (rain, snow, hail) -- because the raindrops/etc. are large (a few mm in size), and can scatter radar waves with some efficiency -- but usually can't show you clouds -- because the water droplets making up clouds are so small. (And obviously radar wouldn't be nearly as useful to the military if it was easily scattered by clouds.)

So -- absent floating chunks of centimeter- or large-sized ice in the upper atmosphere of Saturn -- you're not going to get much of a radar return from Saturn itself. The original discovery of radar returns from the rings surprised people, because the assumption had been that the rings were made up of very small (e.g., micron-sized) bits of ice, which wouldn't efficiently reflect radar waves either.


Why is Mercury's Density So Low? I know the title sounds odd.  You might be thinking "Doesn't Mercury have the highest uncompressed density of any terrestrial planet?  Much higher than a planet its size normally should have?" Here's the thing, though: Iron has an uncompressed density of $7.874\ \mathrm{g/cm^3}$ and silicon has an uncompressed density of $2.329\ \mathrm{g/cm^3}$. I know the distinction for planet composition is usually metals-silicates, not iron-silicon, but iron and silicon comprise the vast majority of those two categories.
Mercury is said to be 70% metals 30% silicates with an uncompressed density of $5.4\ \mathrm{g/cm^3}$.1 However, if you run the numbers, 70 % iron and 30 % silicon gives you an uncompressed density of $6.21\ \mathrm{g/cm^3}$, or over $0.8\ \mathrm{g/cm^3}$ higher. My question is, why is Mercury's actual density lower than my proposed figure? I know that there are other materials in Mercury, like nickel, sulfur, etc., but can these things really make up the difference?

1Wikipedia and NASA

The actual density depends on the mineralogy, we don't have a crystalline iron core and silicon crust. You do have a lot of oxygen available, too when you look at the overall elementary abundance. So the abundant materials are Fayalite (${\rm Fe_2SiO_4}$), Olivine (${\rm (Fe,Mg)_2SiO_4}$), Fosterite (${\rm Mg_2SiO_4}$) etc. which make up most of the crust. These have a bulk density between ${\rm 3g/cm^3}$ and ${\rm 4.5g/cm^3}$ at normal pressure and of course a somewhat higher density under pressure.
In the core we are talking about some iron nickel sulfide alloys which have a density of less than pure iron, too (typically ${\rm FeS}$ is around ${\rm 4.8g/cm^3}$ at norm pressure). So conversely knowing the total mass from celestial mechanics and satellite date, the volume from imaging and the surface composition from spectroscopy we can actually make density estimates on the core's composition (with some further guesstimates on the equations of state on the alloys in question).
Add to these figures that we will have in the interior some high-pressure phases which have somewhat higher density, the estimated mean density of ${\rm 5.4g/cm^3}$ seems to fit quite well.
Also mind, whether a source talks about weight% (thus referencing the weight contribution of single elements. One iron atom weighs nearly 4x as much as one oxygen and twice that of magnesium) or whether the source talks about number fractions (thus counting actual atoms, disregarding weight). For one single molecule of Olivine we have ${\rm Fe:Mg:Si:O}$ in the weight ratio 56:24:28:64 while we have the number ratio of 1:1:1:4.


What is a White Hole? White holes are purely hypothetical at this point and are considered only because it is assumed something like it could exist as the balance to black-holes.

Reading this article seems mostly writers taking liberties with the information but poses an interesting question.

What is a White Hole?

Well according to the article "Have we seen a white hole?" (Byrd, 2011) there are some claims that scientists (specifically "The Revival of White Holes as Small Bangs" (Retter and Heller)) may have detected a white hole; however, this claim seems to be shrouded in controversy, as this observation may actually have a more stellar explanation.

However, the article from Retter and Heller offer a hypothesis as to what white holes are, from the article linked before:

  We thus suggest that the emergence of a white hole, which we name a ‘Small Bang’, is spontaneous – all the matter is ejected at a single pulse. Unlike black holes, white holes cannot be continuously observed rather their effect can only be detected around the event itself.

Which, if the authors are correct, would explain why we don't see/detect them and white holes (unlike black holes) are unique because they supposedly spew matter, rather than extract.

One major caveat is that it is not clear as to where this matter that spews out comes from.


Can an observer on Earth only see half of the sky? Is this the following statement true?

  An observer on earth only may see half of sky from the northern or southern hemisphere, and even if the observer stands on the equator on top of a very high mountain, he will hardly be able to see all the stars in the sky in given period!

Do most globe-like objects in space have this property?

I'd appreciate any more remarks on the subject.

At any given instant of time, in any place on Earth, if the sky is clear and the horizon is low and flat, you see half of the celestial sphere - at that very instant.

But as the Earth keeps turning, you may end up seeing more, depending on where you are.

If you're at the North or South Pole, you see exactly half of the sky no matter how long you wait. That's all you'll ever see from there.

If you move closer to the Equator, you'll end up seeing more than half, if you're willing to wait.

At the Equator, you see pretty much all the sky, if you wait as the Earth keeps spinning around, revealing all the sky to you eventually.


What is the upper and lower limit of temperatures found on stars? What are the most extreme temperatures (both hot and cold) stars have been detected  at?  Is there an upper and lower limit for the detected temperature of stars?

The answer depends on what you'd want to consider as a "star." If you're just thinking about stars on the main sequence, then you can just refer to the classical stellar type letters, "OBAFGKM" (which has relatively recently been extended to accommodate the coolest brown dwarfs with the letters "LTY"), where O-stars are the hottest stars (~30,000 K) and Y-stars are the coldest, so-called "room-temperature" stars (~300 K).

Self-gravitating, gaseous objects are incapable of fusing deuterium below about 13 Jupiter masses, and thus simply collapse and cool perpetually (as is the case for all the giant planets in our solar system). These objects can be colder than 300 K but are not technically stars as they do not undergo nuclear fusion.

For stars that leave the main sequence, two possible outcomes are a white dwarf star or a neutron star, both of which are born extremely hot: White dwarfs are born with surface temperatures of ~10^9 K, whereas neutron stars are born with surface temperatures of ~10^12 K. However, both white dwarfs and neutron stars cool as they age, with the coldest known white dwarfs being ~3,000 K, and neutron stars cooling to ~10^6 K.

So to answer the first part of your question: The coldest known stars are Y-stars (i.e. brown dwarfs) and the hottest known stars are either O-stars or young neutron stars, depending on whether you consider objects that have left the main sequence or not.

And as for strict lower and upper limits, the coldest stars possible are likely black dwarfs, which are what white dwarfs become after cooling for a very long time (>10^15 years). The hottest stars are likely the newly-born neutron stars I previously mentioned, it is very difficult to get much hotter than 10^12 K because any excess energy is carried away via neutrinos.


Do great spot like features favour appear in south hemisphere? As we know, both Jupiter and Neptune have great spot like features in South hemisphere, and Saturn has dragon storm in south hemisphere, is it true that great spot like features appear in South hemisphere more frequently? If so, what is the reason?

Given that we have very few examples, we could not, statistically, draw any significance from any apparent bias.  It's also hard to see any theoretical basis for expecting a bias, so my gut feeling would be there probably is no bias.

It's also worth remembering that human observations of these things are essentially a very brief snapshot in a huge period during which we have no idea about what may have been happening.  So again the statistical significance of what happens at the moment needs to be carefully considered.


Is it plausible for a planet that is positioned in the habitable area of a solar system to have little extractable metals? I am big fan of Asimov's Foundation and I am wondering about the plausibility of some of Foundation's imagined words.

According to this article Terminus was a special world with little metals:

  It was the sole planet orbiting its isolated star and had almost no
  metals. The nearest planet was Anacreon, 8 parsecs (26 light years)
  away. Being on the fringe of the galaxy, there are almost no stars in
  the sky.

Since the planet hosted the first Foundation it is clearly habitable (i.e. atmosphere, no extreme temperatures). However, I am wondering how could it have so little metals if it seems so similar to Earth (habitability-wise) which has lots of metals. Is this possible in our Universe?

I am interested in answers that use current discovered exoplanets information or some articles that deal with planets come to be.

Question: Is it plausible for a planet that is positioned in the habitable area of a solar system to have little extractable metals?

Interesting question, though probably a better fit for  world-building.

Firstly "metals" is often used by astronomers to refer to any element other than hydrogen or helium. That is clearly not the intended use here. Terminus has a solid surface at temperatures that humans can stand, and a breathable atmosphere (perhaps after terraforming). So, what can it be made of?

The most obvious option is silicate rock (mainly aluminium and magnesium silicates) similar to what most of the outer layers of the Earth are made of. It is certainly imaginable that a planet might be more thoroughly differentiated than Earth, so that all the iron etc. ended up in the core, and there were no, or almost no, useful ore bodies in the crust.

On the one hand, this clearly fails, since magnesium and aluminium are metals. On the other hand, it is quite possible in context that what is meant is "easily extractable metals" and/or "the metals required for a specific range of technologies", which might include iron, copper, etc. (given that Asimov was writing in the '30s that is probably what he was thinking of).

In fact, with enough energy you can extract aluminium or magnesium from these rocks. We don't do this, because it's cheaper to find rare compounds of those elements from which the metal is more easily extracted, but the Foundation could certainly have had aluminium and magnesium in this scenario assuming they had enough power.

A stranger possibility would be carbon. Several gas giants are believed to have carbon-rich cores. Suppose one of those lost its outer layers in some cataclysm, the resulting smaller planet could differentiate to have a thick diamondoid crust over a rocky mantle and iron core. Although if diamonds were literally as common as dirt on Terminus, I can't help feeling someone would have remarked on it.

A similar but slightly less exotic option would be silica -- so a planet with a mainly or entirely quartz crust. I can't see a scenario for that to form -- one where you'd have silicon and oxygen and not much else around when a planet was forming.

Now we're definitely into worldbuilding territory, but I wonder is ice is a contender? If you found a world with an icy crust a few hundred km thick, is there a way to terraform it without melting all (or even most) of the ice? YOu could imagine covering the ice in a layer of vacuum-filled diamond honeycomb and then dumping a few meters of imported or manufactured soil over it, creating an atmosphere somehow and then putting a heavy greenhouse layer (CO2 and SF6 and so on) to warm the surface. Could the first-empire scientists have done something like that?   Taking this back to the original question -- an ice world could exist at the outer edge of the habitable zone (or just beyond it, if you like) and then be made habitable by adding a greenhouse layer and changing its albedo. If you can avoid melting the ice, that would meet the requirements.


Redshifting restframe SED I have a template spectral energy distribution (SED) at z=0. I want to shift the template SED to an arbitrary redshift (z>0), how to do this in python?

This is exactly what is done for template fitting. The shifted SED is one of the intermediate data products of template fitting codes such as Le Phare or Phosphoros.

For instance, with Phosphoros, you can shift your SEDs to any number of redshifts, then find the shifted SEDs (also called model grids) in the IntermediateProducts folder in the Phosphoros directory.

Based on your comment, you are also applying a filter to the shifted SEDs. This can also be done with Phosphoros, and you can directly get the flux through each filter for the shifted SEDs.

Phosphoros can be used from the command line, so you use the Python library subprocess to make calls to Phosphoros from your Python script.


Was the Earth-Sun distance ever calculated based on the time difference between night and day? If we define sunrise and sunset as the time when the center of the Sun goes over the horizon then day time is a little bit shorter than night time because from a single point one cannot see a complete hemisphere. Computing backwards from a known Earth radius of around 6700km and a known Earth-Sun distance of around 150 million kilometers one can compute that the angle of the Sun shining on the Earth is 13,4/150.000 which comes to the day being roughly 5 seconds shorter than the night.

The radius of the Earth was already known to the ancient Greeks. You also need a clock that is accurate to at less than a second per day and a spot to observe the Sun rise and and Sun set at exact times. If you are not at the equator there will be differences in day time due to the tilted Earth axis but one could either measure directly at the equator or measure over a whole year and average it out.

Question: was this method ever used to estimate the distance between the Earth and the Sun?

You have a good idea. You mentioned 4 effects already (clock accuracy, latitude, time of year, and clear horizon), but there is another effect that is larger than those: atmospheric refraction. Refraction causes a rising object to appear to be half a degree higher than reality. Refraction depends on atmospheric pressure and temperature, so it may be harder to compensate for it.
This article on Sky & Telescope indicates that the amount of refraction is not well understood: We Don’t Really Know When the Sun Rises

But Michigan Tech dissertation work by Teresa Wilson (now at the U.S. Naval Observatory) suggests that our estimates are often off by 1 to 5 minutes.
She found that, overall, predicted times varied in accuracy by location and season, with sunrise times over land generally being early in the summer months and late in the winter. Summer showed the largest discrepancies, probably due to the pronounced refractive effect that the large temperature difference in the atmosphere has during those months. Mirage effects due to cold air topped by warm over water horizons also exacerbated lags in sunset times throughout the year, sometimes by up to 5 minutes.


Is Earth the most inclined or not? I'm trying to understand orbital inclination by looking at the planetary orbit inclination table on https://en.wikipedia.org/wiki/Orbital_inclination (there are orbital inclinations for terrestrial and gas giant planets listed there).

By inclinations to ecliptic or Sun's equator, it looks like the Earth is the most (or least) inclined from all the listed objects having the smallest (or biggest) inclination value. However, the inclination to invariable plane seems to be bigger for some planets than for the Earth, and smaller for others.

At first, I thought that displaying the absolute value (dropping the minus sign) in case of a negative angle is the case. But if those three columns were just three different reference planes, the relative difference between planet inclinations should at least be the same in all cases. But it's not the case as, for example, Mercury's and Venus' orbital planes have 3.62° between themselves as per the inclination to ecliptic, but they have 0.48° as per the inclination to the Sun's equator.

So, how should one read those numbers?

The confusing thing is that inclination is not the only thing defining an orbital plane; it also matters where the ascending node is. Only if all ascending nodes would be in the same direction, you could do the trick with the absolute values you mentioned.

I can't draw a picture right now, but imagine you have a table before you now, and use it as a reference plane (the Sun's equator). Put a sheet of paper on the table, and lift the right end a bit so that it inclines 7.155° w.r.t. the table. This is Earth's orbit, and it looks a bit like the picture in the Wikipedia article.

Now take another sheet of paper, and lift the far end a bit so that it inclines 7.155° w.r.t. the table. This is supposed to be the orbital plane of a hypothetical planet with the same inclination to the Sun's equator, but it is also inclined to the ecliptic (i.e. Earth's orbital plane).


Distribution of Stars in Milky Way and globular cluster analogy Just a quick (and hopefully simple) question on the distribution of stars in the Milky Way and in globular clusters. I was wondering why it IS NOT possible to describe the distributions of stars in both these systems by analogy to the distribution of air molecules in a room.

Are we talking of the more random motion of air molecules as well as the relatively large number of collisions between air molecules...something which we wouldn't see in a galaxy. I'm assuming also that air molecules do not have the same central gravtational potential...and therefore do not move relative to some fixed point. That is to say, that it is much more random? Any advise on this point would be greatly appreciated.

Particles in a gas approximate to point-like objects that interact roughly elastically through short-range forces when they collide, but are otherwise non-interacting.

Stars interact gravitationally over long ranges, occasionally with each other, but always with the overall gravitational potential of the system.

Sometimes people do talk thermodynamically about star clusters. You can discuss the "temperature" of stars when you are referring to the velocity dispersion in a cluster. The concept of heating or cooling a cluster also has some merit.


Binocular highlights from Tasmania next few weeks? My friends are visiting southern Australia and Tasmania for the next few weeks.

I am lobbying them to bring my Nikon 8x42 binoculars so they can appreciate the Milky Way and possibly see a few other things.

What I need is:

What parts of the Milky way will be most spectacular in binoculars at night at about 40° South latitude around this time of year?
Anything else? I've never stargazed in the Southern hemisphere so I have no idea what to point them towards.

I can use online star maps (e.g. in-the-sky.org) so I can give them directions how to find things if it's not hard, but I don't have any idea what "things" there are.

My favorite resource for this sort of thing is the Skymaps.com evening sky map.
The southern hemisphere edition is drawn for 35°S latitude.
The November issue doesn't show the Milky Way, which hugs the horizon at that time.
However, both the Large and Small Magellanic Clouds are well placed in the evening, and the back page lists several other objects "easily seen with binoculars."
For example:

47 Tucanae, a globular cluster visually near the SMC
NGC 2516, an open cluster in Carina
M42, the famous nebula in Orion


Which are the best American universities/colleges for observational astronomy? I'm a French student and I'm planning to go to the United States for a graduate program in astronomy.

When I will arrive in the U.S. I will begin by the first year of graduate. Then I will pursue with a Ph.D. and some postdoctoral research.

I would like to know which are the best American universities/colleges for what I want to do.

Concerning what division of astronomy I am interested in:
I'm not interested in the conception of telescopes, instruments and other systems. I'm not interested in engineering. And I'm not interested in extremely theoretical astronomy either.
I'm interested in observational astronomy.

Regarding what subfields of astronomy I am interested in:
I'm not interested in extragalactic astronomy or cosmology.
I'm interested in planetary astronomy, exoplanetology, exobiology, SETI, stellar astronomy in the solar neighborhood.
I'm not interested in things that are less than 2 Astronomical Units or more than 70 light years away from us.

Out of curiosity: where does this limit of 70 lightyears come from? Exoplanets can be detected to way beyond that. This strikes me as a strangely restrictive limit.

Be that as it may. There are several great astronomy departments in the US, and it really depends on what you want to do exactly. For exoplanets, Harvard would be a great choice. Professor David Charbonneau is a heavy weight in the field.

Similarly, many of the other renowned Universities have good astronomy departments. Yale, for instance, or Berkeley (although since the Geoff Marcy incident I imagine the exoplanet department has taken a hit).

So at this point it is not really a matter of choosing 'the best' University. I suggest you have a look at some papers by the professors at different Universities, to get a sense of what you'd like to work on.

And don't forget: for your research to be fruitful the environment is also very relevant. Just have a look and see what city you'd like to live in.


Are astronomers continuously monitoring exoplanetary systems? I'm reading many of the Wikipedia pages about exoplanets and the different methods they are using to detect them.  But I wonder, it seems that the emphasis is on detecting and finding new exoplanets, but are astronomers continuously monitoring the exoplanets that have already been discovered?  For instance, if there is a collision between a detected exoplanet and a previously unknown planet, this could cause a change in some of the orbital parameters, and could be potentially detected.  If this happens, would we learn of this change, or will it have to wait until astronomers again take another look at the system?

Exoplanets is an almost over populated field of research now, so yes they are monitored. But maybe not too much beyond what is needed to confirm their existence.

I think you have to humble yourself in the enormity of time. There are some transient events in astronomy, but mostly, watching the sky is like watching a rock. Nothing much changes. Phoebes is crashing into Mars. But not until several million years. It will be a very slow process of it breaking up and parts of it landing one by one. Even in the midst of that event human intuition would think of it as a frozen constant non-event. Like the rings of Saturn. Look at a mountain and see how fast it grows or shrinks. Astronomers cannot see exoplanets crashing into each other in the way I think you mean.


How do they measure sub-centimeter wave height on a moon of Saturn? Michael Stevens from V-Sauce has a video called The Mandelbrot Set on a different channel. At 12:56 he says:

  There’s a moon of Saturn that has liquid hydrocarbon lakes that literally have waves that are only like a fraction of a centimeter large, but, you could surf that, right?

While Michael occasionally does indeed say things that appear to be wrong I'm guessing he got this correct.

How are waves a fraction of a centimeter tall measured on a moon of Saturn? How is the existence of these waves either measured or deduced?

According to the Wikipedia page on Titan, that data was deduced from Cassini.

  On December 21, 2008, Cassini passed directly over Ontario Lacus and observed specular reflection in radar. The strength of the reflection saturated the probe's receiver, indicating that the lake level did not vary by more than 3 mm (implying either that surface winds were minimal, or the lake's hydrocarbon fluid is viscous).

That claim cites two sources:

newscientist.com (definitely not a primary source)

Harvard ADSABS which does appear to be a primary source analysis of Cassini data. However, I don't have time right now to see what peer-review made of this analysis.

  Cassini RADAR altimetry data collected on the 49th flyby of Titan (2008 December 21) over Ontario Lacus in Titan's south polar region provides strong evidence for an extremely smooth surface, with less than 3 mm rms surface height variation over the 100m-wide Fresnel zone. Histograms of the raw radar echoes imply a mirror-like specular reflection of the transmitted signal. Such an echo is possible only if the surface is extremely flat relative to our 2.2-cm wavelength. The 3 mm upper bound follows from analyzing the strength of the specular return, which declines exponentially with increasing surface height variance. In this experiment, the strength of the echo was larger than expected, severely saturating the receiver. We developed a method to partially correct the echoes for the distortion incurred. While the implied mm-scale smoothness is not proof that the surface is liquid, it is unlikely that a solid surface is so smooth.

From the abstract, it seems they are deriving this conclusion indirectly, but I'm not educated enough in this area to personally vouch for the methods.


apparent and absolute magnitude of meteors and space debris I have a question regarding the apparent and absolute magnitude of space debris objects or meteors reentering the earth's atmosphere. I already understand the difference between apparent and absolute magnitude. The absolute magnitude is the brightness of the object if the object is at a distance of 100km and at zenith. And the apparent magnitude is how bright the objects appears to the observer.
But how can I determine the apparent magnitude of an objects observed from an observer? The objects has the distance $d$, the zenith distance $Z$ and the height $h$. On the internet I found the following equation:
$$
M = m +5  \log(\cos(Z))
$$
$M$ = absolute magnitude, and $m$ = apparent magnitude. I posted a picture below.
But how is this equation derived? I do not understand the equation. What can I do, if the height $h$ of the objects does not equal 100 km?
The book where i found this is "The Observer's guide to Astronomy" (Volume 2) pages 659 to 660!
https://books.google.de/books?id=TXc54LfKsSQC&pg=PA659&lpg=PA659&dq=zenithal+magnitude&source=bl&ots=fPeMVD1sOq&sig=ACfU3U26N79Xm-qsiC4ocoMHHcCw9Lc3jw&hl=de&sa=X&ved=2ahUKEwiP2Lf3v-bpAhVk-ioKHQ0oD5wQ6AEwAXoECAsQAQ#v=onepage&q=zenithal%20magnitude&f=false

The formula appears to compare the squares of distances, not taking into account the difference in extinction (which is small for small zenith distances, and for different altitudes provided they're high enough), this means $M - m = 5log(\frac{h}{cos(Z)})-5log(h_1)=-5log(cos z) - 5log(\frac{h_1}{100})$. Also, I believe the formula should say -5 cos z ... the meteor from the picture provided is farther away from the one at zenith and at the same altitude and should have higher (numerical value) magnitude.


What effect does the Moon have on the near Earth asteroid population? Does the big (0.012 Earth masses) Moon of Earth clear away NEAs, Earth orbit crossing asteroids, in a significant way? Venus and Mars don't have large moons, do they therefore have larger or smaller population of near asteroids than Earth would have if it were in the same orbit?

The outer planets have large moons, but also lots of captured asteroids, trojans and centaurs. Does a large moon even help gathering such objects, rather than ejecting them? Moonless Mercury and Venus seem to be pretty clean.

I think you answered your own question with your second paragraph.

According to Newton's law of gravity

the Gravitational force is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.
So the biggest body in the neighborhood has the most pulling power provided it agrees with the equation.
As a consequence Jupiter and the outer planets should win out over Earth and keep them from hitting Earth or at least divert the collision course. Especially since the NEA come from their part of the cosmos.
I suppose the same logic would also work with the comets.

Please see the following link for more:
http://earthsky.org/space/is-it-true-that-jupiter-protects-earth


A camera and time dilation? If I travelled near a black hole, my time would progress slower relative to someone on Earth. This is clear enough. However, what if we sent a probe with a camera to a black hole? When we watch the screen, would we see time through the camera's perspective — that is, would the Universe appear to progress faster as the probe got closer and closer to the black hole?

If I travelled near a black hole, my time would progress slower relative to someone on Earth. This is clear enough.

Yes, no problem with the gravitational time dilation.

  However, what if we sent a probe with a camera to a black hole? When we watch the screen, would we see time through the camera's perspective — that is, would the Universe appear to progress faster as the probe got closer and closer to the black hole?

No. We'd see the universe progressing normally, because we aren't subject to that gravitational time dilation. (I presume we're at a safe distance). Let's suppose it was a TV camera that took a picture 25 times a second and sent it back to us, adequately catering for redshift. The camera starts off taking 25 pictures a second as measured by us. But after a while we notice we're only getting 24 pictures a second, then 23, and so on. We see events in the wider universe progressing at their normal rate, but eventually the movie starts getting jerky as the frame-rate reduces. In the end the frame-rate reduces to zero, and that's the end of the show.


Why isn't most hydrogen in the universe molecular (diatomic), instead of atomic (monoatomic)? Similar questions have been asked before; but, why?
Is the monoatomic hydrogen left over from the Big Bang?  And hasn't had the opportunity to collide with other hydrogen atoms yet?
Or are hydrogen molecules that do form split up after absorbing ultraviolet radiation or X-rays?  Which occurs more frequently than molecule-forming random collisions?
What is the 'activation energy' needed to form diatomic hydrogen from atomic?  Is it too much for the randomly-colliding hydrogen atoms to form molecules in the depths of space?
I mean, if the loose protons and electrons shot out by stars can (usually) find each other to form atoms, which is why neutral atomic hydrogen is more common between the stars and planets than ionized hydrogen, why can't the lonely atoms find each other to form ${\rm H_2}$?

Yes, the atomic hydrogen is probably mostly left over from the Big Bang. [Edited to add: Not sure how much that is true and how much present-day atomic hydrogen is the result of recombination.] And, yes, ${\rm H}_{2}$ does get dissociated by high-energy photons -- and also by cosmic rays, which can penetrate dense, dusty clouds that block most of the high-energy photons.
The real issue, as I understand it, is that's actually very difficult to make ${\rm H}_{2}$ by colliding two H atoms in the gas phase. This is because when two atoms collide, the resulting (temporary) molecule almost always has too much energy to remain stable and will very rapidly dissociate back into the two atoms -- unless you can somehow get rid of the excess energy before it does so.
Now, you can do this if the reaction actually involved producing extra components (like a more typical molecular reaction -- e.g., $AB + C \rightarrow AC + B$), since then the excess energy can be put into the kinetic energy of the products (while still preserving momentum). But in a reaction like ${\rm H} + {\rm H} \rightarrow {\rm H}_{2}$, there's no second body to allow that.
You could also do this if the "extra component" is some third body that happens to hit the other two at the same time (without becoming part of the molecule), since it can rebound with excess kinetic energy. But while this can work in dense gasses (e.g., in a laboratory on Earth), it will happen much too rarely in interstellar or intergalactic space.
You can also do this if the temporary molecule can emit a photon carrying away the excess energy. However, such transitions are strongly forbidden, especially in the case of a symmetric molecule like ${\rm H}_{2}$, which means that on average they will take too long to occur.
Finally, you can get rid of the energy if the reaction actually takes place on a surface that acts as a catalyst by absorbing the excess energy. This is thought to be the primary way ${\rm H}_2$ is formed in the interstellar medium: two H atoms that are adsorbed onto a dust grain combine on the grain's surface.
I mean, if the loose protons and electrons shot out by stars can (usually) find each other to form atoms, which is why neutral atomic hydrogen is more common between the stars and planets than ionized hydrogen, why can't the lonely atoms find each other to form $H_{2}$?
In addition to the fundamental problem outlined above, it's also easier (as Alchemista pointed out) for protons and electrons to "find each other" because they are attracted by their opposite electrical charges, something which doesn't happen in the case of neutral atoms.


Which binary star system has a greatest angular distance? Visible from the Earth, which binary star system has greatest angular distance ?

Wide binary stars it turns out is an active research topic, with groups using statistical searches of catalogs (Tycho Gaia ...) to find in 3D space pairs of stars that are moving together taking into account possible orbital motion. Here people are looking at the physical separation not the angular separation. On the theory side it is suggested that nearly all stars were formed as part of a binary. So somewhere out there is sister of our Sun. One paper looked at nearby stars. They find several very wide binary pairs visible to the eye plus more fainter. In particular they mention Capella and 50 Peg which have an angular separation of 20 deg. This they class as an escaped binary : so no longer in orbit (thanks Connor for pointing this out).
There is however Fomalhaut; a faint (mag 12.7) 3rd member of the system was discovered in 2013. It is 5.7 deg away from the primary. The other companion is  2deg away in another direction. The whole system spans 8 deg. The Sky and Telescope article claims this as the widest (on sky) binary, beating Alpha AB Proxima system.


Is there a difference between the solar elevation angle and sun declination? I need to use the sunrise equation but one of the variables is the sun declination. On the other hand, I have the values for solar elevation angle that I need. Are they the same thing?

Elevation and Declination are from different co-ordinate systems, so solar elevation is given in Alt/Az co-ordinates and refers to the elevation above the local horizon. Sun declination is measured in RA/Dec co-ordinates (which is an equatorial co-ordinate system) and measures the Sun's inclination above or below the equator on the celestial sphere.


How can we focus radio telescopes on a star when the earth is spinning? Reading about the Star KIC 8462852, it has been said that the SETI project turned its radio telescopes towards the star to search for extra terrestrial radio signals as the star had strange fluctuations in light. How can we point from earth a radio telescope towards a star which is 1480 light years away whilst the earth is spinning at 1675 km/h and keep it focused or, in the case of a radio telescope, aligned in order to try and receive radio waves??

Part of the answer that I suspect the original questioner needs is that although the Earth is indeed spinning very fast, the amount the surface of the Earth moves relative to an astronomical object is tiny.

So you put motors in the base of the telescope so that it slowly turns to look at the same patch of sky.   You don't need to refocus because telescopes are looking at objects so far away that focus doesn't matter.  You don't need to do anything else because the Earth's movement is smooth and continuous, and it's not about how fast you're moving, it's about how quickly you're turning.  In our case, one complete circle every 24 hours which is pretty slow.

Focus at infinity just means that you set the focus of the telescope so that an object which is infinitely far away would be in perfect focus.  It depends on the quality of the telescope, but the practical difference between focus at infinity and focus at the actual distance disappears after a few miles or so.  At the distance of stars, there is essentially no difference at all.


Moving from mathematics to astronomy/astrophysics I'm currently studying for a PhD in pure mathematics (the topic is the cohomology of differential graded categories, just to give you an idea of how pure I mean), although the first year of my undergraduate degree was in mathematical physics before I transferred to mathematics, so I have a little (but unfortunately not that much) background in physics as well.
I'm seriously thinking about whether I want to pursue pure mathematics as a career, and since astronomy and astrophysics were, one could say, my first love when it comes to science-y type things, I've been wondering if it would be possible for me to switch fields, and what would be necessary to do so. Although I'm aware that astronomy/astrophysics is a wide area and could mean lots of different things in terms of a profession.
To give you a little more background, my current level of astrophysics knowledge is propped up mostly by my short time studying physics at university and a continuing interest and reading various non-technical books and low level textbooks of various topics. The topics I've found most interesting are things like supernovae, and the formations of stars and star systems, although I haven't studied the mathematics or technical details of these things in any serious depth. I should say that I split off from applied mathematics some time ago, so topics such as PDEs and statistical analysis are not areas I'm adept in (although I have no doubt that I would have relatively little trouble learning them at this stage). I've also spent a short amount of time (~ 1 year) working as a software developer, so I have some proficiency with programming. And actually I thoroughly enjoyed my time as a developer, and I know that software development/programming is an important aspect of work and research in this field for analysing images, data sets, creating simulations and that sort of thing. So something of that sort is of great interest to me, especially if it could directly involve mathematics. So perhaps working with simulations would be a good option for me?
Is a change of this kind possible for me? And if so what would you recommend I do to make it happen?
Apologies if this question is off-topic by the way. Please do let me know if there are more appropriate places to find this information.

There are fewer placements than qualified applicants so competition is strong.  Moreover, suppose this was the other way round:  A person doing a PhD in (say) planetary nebulae comes to the maths department saying that they are really interested in maths, they're good at equations and calculus and really like all that topology stuff (you know, rubber sheets and so on) so can they start doing cohomology of differential graded categories?
It's not impossible, and a strong foundation in maths is a prerequisite for progress in Astronomy but the question you should ask is why should a university hire you, when there are 5 others with PhDs in astrophysics?
Talk to your supervisor and careers advisor before burning any bridges.


How close are we to be able to detect and measure gravitational lensing inside the milky way? Gravitational lensing takes advantage of statistically uniform distributions of background objects in order to infer a credible map of dark matter concentrations that isn't too noisy. If there are too many background objects and is not clear there is an uniform pattern of distribution that one can compensate numerically, then the noise of the measurement makes the reading essentially worthless

So, I'm curious if there is any research attempt to try to pull this kind of local measurement of lensing in the "backyard". If one takes the Milky Way's galactic rotation curve, and compares it on what's expected given baryonic matter, one would be able to have a radial distribution of expected dark matter in order to compare with the results of lensing

But is that kind of lensing measurement something that is possible today?

Well, there are microlensing experiments, the results of which have largely removed MACHOs as a significant source of Dark Matter.  The background field of stars is usually a nearby galaxy (a Magellanic cloud one or Andromeda, say), or the galactic bulge, and the experiment looks for lensing objects between us and there.  In particular, they're pretty much always looking for the microlensing effect within our own Milky Way or its halo.  These experiments have detected a number of microlensing events (and many, many more candidates that were due to other things), so, yes, this is not only possible, it is already done.

The wiki page I started with linking has this to say, in particular:

The first two microlensing events in the direction of the Large Magellanic Cloud that might be caused by dark matter were reported in back to back Nature papers by MACHO and EROS in 1993, and in the following years, events continued to be detected. The MACHO collaboration ended in 1999. Their data refuted the hypothesis that 100% of the dark halo comprises MACHOs, but they found a significant unexplained excess of roughly 20% of the halo mass, which might be due to MACHOs or to lenses within the Large Magellanic Cloud itself. EROS subsequently published even stronger upper limits on MACHOs, and it is currently uncertain as to whether there is any halo microlensing excess that could be due to dark matter at all. The SuperMACHO project currently underway seeks to locate the lenses responsible for MACHO's results.

Despite not solving the dark matter problem, microlensing has been shown to be a useful tool for many applications. Hundreds of microlensing events are detected per year toward the Galactic bulge, where the microlensing optical depth (due to stars in the Galactic disk) is about 20 times greater than through the Galactic halo. In 2007, the OGLE project identified 611 event candidates, and the MOA project (a Japan-New Zealand collaboration) identified 488 (although not all candidates turn out to be microlensing events, and there is a significant overlap between the two projects). In addition to these surveys, follow-up projects are underway to study in detail potentially interesting events in progress, primarily with the aim of detecting extrasolar planets. These include MiNDSTEp, RoboNet, MicroFUN and PLANET.


What is the proper interpretation of a negative parallax? The Gaia data release 2 dataset has some parallaxes listed as negative. What is the proper interpretation of this? Should they be discarded as bad data? Use their absolute value to calculate the distance? Use them as-is and treat them as negative distances along the right ascension/declination vector?

Did you read this section of the documentation?  It suggests there are ways to deal with it, but I have not examined the paper it refers to.

  • For closely aligned sources (separated by 0.2–0.3 arcsec), which are only occasionally resolved in the Gaia observations, confusion in the observation-to-source matching can lead to spurious parallax values which are either very large or have a negative value very far away from zero in terms of the formal parallax uncertainty quoted in the catalogue. These sources tend to be faint and located in crowded regions and are also associated with unreliable (large) proper motions (Gaia Collaboration et al. 2018b). Guidance on how to clean samples from spurious parallax values is provided in Lindegren et al. (2018).

It also says there's a small bias, but it shouldn't be universally removed.

  • The systematic errors in the parallaxes are estimated to be below the 0.1 mas level (Lindegren et al. 2018) but the following systematics remain. There is an overall parallax zeropoint which, from an examination of QSO parallaxes, is estimated to be around −0.03 mas (in the sense of the Gaia DR2 parallaxes being too small). The estimated parallax zeropoint depends on the sample of sources examined (Arenou et al. 2018) and the value above should not be used to ’correct’ the catalogue parallax values....


How impermeable are the surfaces of the moon? In the event that we want to dig into the surface of either face of the moon do we know yet if the surfaces of the moon are particularly difficult or particularly easy to dig in?

The moon's surface is made up of many layers.

The top few centimeters of the moon are generally composed of a layer of thin powder.  This is easy to dig into, and the Apollo astronauts easily scooped up this stuff.

Wikipedia - Lunar Soil

Below that layer is the bulk Regolith,  which contains a mixture of powder, grains, pebbles, and rocks - the remnants of billions of years of meteor collisions which have thoroughly mixed and distributed the material from collisions all over the surface.  The regolith ranges in depth around the Moon based on the age of the surface - the old highland and farside surface contains regolith up to 15 meters deep, while the mare regolith averages about 4 meters deep.

The regolith is not solid rock,  but it's very compacted and the grains are rough and billions of years of vibrations from impacts and quakes has caused the particles to interlock.  Therefore, it can be extremely difficult to dig into.  The Apollo 14 astronauts only managed to drive a heat probe 70cm into the regolith, using about 20 hard hammer blows to do so.  For later missions they had to use electric drills to get deeper, about 3 meters.

Lunar and Planetary Institute - Lunar Sourcebook, chapter 7

Apollo 14 - Soil Mechanics Investigation

Below the regolith is rock.  On the Mare, the rock is a very hard basalt (8 out of 10 on the hardness scale).  Getting through this is like cutting granite.  On the older highland regions, the most common material under the regolith will be anorthosite, which is less dense but it's still drilling through rock.

The mare basalts range in depth from 100m to about 3-4 km, with an average depth of .74 km.  Below that is old anorthositic crust.  The crust itself ranges from 10-70 km thick, with an average of about 35 km.  The crust on the near side is significantly thinner than on the far side.

Planetary Society - GRAIL maps of crustal thickness

If you want to drill deeper,  you're into the mantle, which is only partially melted near the core, and made up of roughly the same olivine basalt that makes up the floor of the mare, and goes down to a depth of about 1100 km.

The Moon - Internal Structure and Magma Ocean

Hope that helps.


How are all the intermediate images between the "lensed" and "unlensed" endpoints of this video generated? Phys.org's ALMA sees most distant Milky Way look-alike describes the numerical image reconstruction of a strongly and very nicely lensed z = 4.2 galaxy by a by a foreground galaxy at z = 0.263 and links to a short video SPT0418-47: lensed view to reconstructed view found in YouTube and also at ESO.
The caption says only:

This animation gives the viewer an idea of how astronomers reconstructed the true shape of the SPT0418-47 galaxy, from ALMA observations of the gravitationally lensed view of the galaxy.
SPT0418-47 is gravitationally lensed by a nearby galaxy, appearing in the sky as a near-perfect ring of light, shown at the start of the video. At the end, we see the image of SPT0418-47 that astronomers obtained when they used a new computer modelling technique to reconstruct its real shape and motion of gas.
Credit: ALMA (NRAO/ESO/NAOJ)/Martin Kornmesser (ESO)

Question: So the video begins with the currently observed image and ends with the reconstructed image while keeping the size unchanged. But what do all the intermediate images between these two endpoints represent? Are they intermediate steps in the un-lensing procedure, or are they simply lensed images while computationally ramping down the mass of the foreground lensing galaxy just for fun?

For me, it looks indeed very much like an interpolation, both from what I see and how the caption is formulated. Also: If it would be scientific simulation results, I would have expected whatever the value of the variable displayed on each frame - much like most journals would not accept a 2D plot without axis-labels and units.
Another hint: Martin Kornmesser is a graphic designer for public outreach not a scientist. I guess we would only know if we asked him - but I did not figure out his email yet.


Semi major axis of an hyperbolic orbit The semi major axis (a) is not immediately visible with an hyperbolic trajectory but can be constructed as it is the distance from periapsis to the point where the two asymptotes cross. Usually, by convention, it is negative, to keep various equations consistent with elliptical orbits.

I was reading about hyperbolic orbit and found this in the wikipedia, what does it mean?
Can we just make a distance negative just to keep various equations consistent?
I am a high schooler beginning my prepartions for Astrophysics Olympiad so I don't have an in-depth understanding of central forces. Sorry if this question comes as if I didn't do enough research on the topic.

A semi-major axis of an ellipse defines the half diameter of the ellipse where it is longest. That's a convenient measure to compare ellipses and express its properties. Mathematically you can describe an ellipse with semi major axis $a$ (how long is it) and semi minor axis $b$ (how thick is it) as $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
Often, the eccentricity is used which is defined as $$ e = \sqrt{1 - \frac{b^2}{a^2}} $$
So for a circle with $a = b$, the eccentricity is $0$, and for the edge case, a very long ellipse with $a \rightarrow \infty$, we get a parabola which hence has an eccentricity of 1.
Now, what would a trajectory look like with an eccentricity larger than 1? We get an eccentricity larger than one, if the radix of the definition of eccentricity gets larger than 1, thus $-b^2 / a^2$ needs to be positive. We achieve that if we introduce an imaginary semi major axis (not a negative! See the linked wiki article for a brief introduction to the concept of those numbers) $a$ such that $a^2 < 0$.
With that extension we can express hyperbolic orbits with the same formulation as elliptic and parabolic ones, with a single number, the eccentricity:
$$ a \rightarrow a_r + ia_i\\ b \rightarrow b_r + ib_i $$
and thus we get a single number to describe any kind of orbit, closed and not-closed ones:
$$ e = \sqrt{1 - \frac{b^2}{a^2}} = 0 \cdots <1: \qquad\mathrm{ellipse}\qquad a = a_r; a_i = 0\\
 $$
$$ e = \sqrt{1 - \frac{b^2}{a^2}} = 1: \qquad\mathrm{parabola}\qquad a = \infty\\
 $$
$$ e = \sqrt{1 - \frac{b^2}{a^2}} > 1: \qquad\mathrm{hyperbola}\qquad a = i\cdot a_i; a_r = 0\\
 $$
Energetically, objects in elliptical orbits don't have enough energy to escape, objects on a parabolic orbit had no kinetic energy at infinity and thus just escape and it only needs a slight change in orbital energy to capture them. Objects  on hyperbolic orbits were already in motion at infinity, thus have excess orbital energy and would need to loose much more energy to capture them (and in a solar system context are very likely to be of extra-solar origin).


Latitude of given altitude of Sun From Question #6 of USAAAO First Round 2018

  (1 point) For the following problem find the range in which the answer lies: on the day of summer solstice, on which geographic latitude is the sun culminating at the angle of +72° 50' north of the equator?

  A. 0° − 15°
  B. 15° − 30°
  C. 30° − 45°
  D. 45° − 60°
  E. 60° − 75°

I have two answers: 41° and 6°, but only one is accepted. No solutions are provided for 2018 and it is an objective question.

It is a strangely worded question I agree.

I would start with the sun appearing directly overhead at the latitude of the tropic of cancer 23.5 degrees. As you go north, vertical shifts north and the Sun appears to be lower towards the horizon due south. To make the Sun appear to be at 73 Degrees from horizontal you would need to be 17 degrees north of the tropic of cancer. 23.5+17=40.5

Local differences in gravitational direction and the eccentricities of the circumference of the Earth due to the equatorial bulge make the answer a bit uncertain hence the range

30 to 45 degrees.


Revolution of the sun It is believed that there is a black hole at the centre of the Milky Way galaxy. And the sun revolves around the centre of the milky way.
So does it mean our sun is revolving around the centre of the galaxy due to this black hole?

The Sun orbits in the entire Galactic gravitational potential, which is very approximately similar to that provided by considering that all the mass interior to its Galactic orbit (about 100 billion solar masses) were concentrated at the Galactic centre.

The black hole at the centre has a mass of 4 million Suns, so contributes less than 0.01% of the mass required to keep the Sun in its present Galactic orbit of about 200 million years. Or to put it another way, if the Sun just orbited the black hole with a period of 200 million years, we would need to be about 800 light years from the Galactic centre, rather than the 30,000 light years we presently are.


Does the expansion of the universe increase with distance? I read something in a book the other day which has gotten me really confused. They were talking about the expansion of the universe, and if I understood it all, they said that the expansion of the universe increases with distance. I’m wondering if I understood this correctly, because I was reading about Hubble’s law and it said that the redshift if galaxies far away (caused by the expansion of the universe) is proportional to the distance (up to a few hundred megaparsecs away). Doesn’t this mean that the expansion is the same but the redshift increases? (I don’t see why the redshift would be proportional and a linear graph if the universe expanded from us proportionally to the distance, or at least more further away). I hope you understand what I mean by all this.

I'm not sure why you say that "expansion is the same" when you also say that "expansion increases with distance", but I wouldn't say the "expansion increases", but rather that "the recession velocity increases".

The recession velocity $v$ of galaxies increases linearly with distance $d$. The constant of proportionality is called the Hubble constant $H_0$. Thus, Hubble's law states that
$$
v = H_0 d,
$$
and currently, $H_0 \simeq 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, but in the past it was higher. That means that, for every Mpc (mega-parsec, $\simeq 3.26\times10^6$ light-years) a galaxy is from us, its speed away from us is roughly $70\,\mathrm{km}\,\mathrm{s}^{-1}$. This is the speed it has right now, but not the speed we see it have, since we see it in the past.

In fact, we don't really see a speed at all. We measure a galaxy's redshift $z$, and $z$ also increases with distance, but not linearly. Thats's because it depends on how much the Universe has expanded since the light was emitted, and the expansion rate changes with time. But for small enough distances, i.e. for galaxies that are not seen that far into the past, the relation is close to linear, so that $v = cz$, where $c$ is the speed of light.

This is indeed approximately true out to a few hundred Mpc. For larger distances, the linear relation becomes inaccurate, and one must instead apply a cosmological model of the expansion history of the Universe, which depends non-trivially on the densities of its various constituents.

Acceleration of the Universe

The term "accelerated expansion rate" refers not to the value of $H$, but to the rate at which the scale factor $a$ — the "size" of the Universe — increases (the Universe may or may not be infinitely large; $a$ is normalized such that, today, $a=1$).

For a given value of $H$, we know the recession velocity of any two points separated by some distance at a given time. At a later time, these two points will have increased their distance, and hence — by Hubble's law — their recession velocity will have increased.

What was higher in the past is the Hubble parameter $H(t)$. That number will, in the future, asymptotically decrease toward a slightly smaller value of
$$
H(t\!\!\rightarrow\!\!\infty) = H_0\sqrt{\Omega_\Lambda} \simeq 56\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}.
$$
The Hubble parameter is defined as $H\equiv\dot{a}/a$, where $\dot{a}\equiv da/dt$ is the Universe's change in size per unit time. If $H$ is constant, and $a$ increases, then $\dot{a}$ also increases. That is, the increase in velocity of any two points accelerates. In fact, we can solve the differential equation and get that, in the future,
$$
a(t\!\!\rightarrow\!\!\infty) \propto e^{Ht},
$$
i.e. the scale factor increases exponentially.

The figure below shows the evolution of $a(t)$ until $t\sim130\,\mathrm{Gyr}$. On the left is a zoom-in of the first $20\,\mathrm{Gyr}$ where you can see how the Universe decelerated the first $\sim10\,\mathrm{Gyr}$, until dark energy started taking over the dynamics.

At the moment the acceleration is actually quite modest, but it will increase without bounds in the future; for instance, in 50 Gyr from now, $a\sim10$, i.e. galaxies will be $10\times$ farther from each other, but in 100 Gyr we have $a\sim150$.


Light that travels eternally because of inflation? What happens with electromagentic waves that are emitted into
intergalactic space, but reach not any object because of inflation? Lets assume that photon A and photon B are emitted. Photon B hits some object, photon A enters a gigantic void or simply has nothing in its path, and heads for the horizon, so to say.

Relativity tells that A and B do not "feel" time passing. From our point of view B travels a finite time, while in the frame of B the object is reached instantly. However, since A will be eternally underway, what is going to happen? We have kind of a zero*infinity travelling time in the frame of A.

Do our models make statements about this? Or is there some explanation

Infinities are a type of singularity. As soon as you introduce an actual infinity into any kind of mathematical model you can't expect it to make useful predictions. Mathematical models can't deal with infinite time. Similarly, using a frame that moves at the speed of light, with infinite rapidity, is singular.

In the particular case you describe, in a regular frame, the photon becomes progressively redshifted by the expansion of spacetime, its energy (to an observer in a inertial frame) becomes lower and lower.


Information about known objects in the Universe I want to retrieve coordinates, photos, and information (size, atmosphere composition, distance from Earth, etc.) about every known object in the universe. Ideally, the information is available via an Application Programming Interface (API) for easy retrieval via a computer program.

What is the best free source of this information? Thank you for your help.

Here's an API: Documentation for Catalog/Image webservices provided by CASG/MAST.

You can also find more search resources in Github, e.g. HackTheUniverse.


Pauli exclusion principle - black holes If a white dwarf compresses to the limit of electron degeneracy, and a neutron star compresses to the limit of neutron degeneracy, what does a black hole compress to the limit of?

In classical General Relativity, there is no limit to the compression in a black hole, hence you get a singularity. However, many astrophysicists feel that's unphysical, and that a theory which unites General Relativity and Quantum Mechanics will impose some kind of limit, perhaps something connected with the quantization of spacetime itself.

We don't have a working theory of quantum gravity, so at this stage we don't exactly know what happens in the core of a black hole. OTOH, we're fairly confident that the core has to be very small, since quantum gravity effects probably don't kick in until a scale much smaller than the size of an atom, and probably smaller than a proton, somewhere around the scale of the Planck length.


Do our sun and moon have names? We seem to have named every moon orbiting other planets. Why haven't we named our own moon? And for that matter, why doesn't our sun have a name since we name or number stars?

What is a name?
A name is a word, that is reasonably unique, that is used to identify a person or thing. When a child is born there is no word that identifies it, and so its parents have to choose a "name".  Similarly when a new astronomical object, such as an asteroid, is discovered, it has no word that identifies it and so it is assigned first a number and later a name.
But there is already a word that identifies our sun and moon:  "the Sun" and "the Moon". There is no need for a new name because they already have a name.  Notice that English (unlike some other languages) gives us a typographical clue that these are names, because the first letter is capitalised.  The Moon is a moon of the Earth and the Earth is a planet orbiting the Sun.
Earth, Moon and Sun are the correct names in English.


Are a black hole's jets caused by the black hole's magnetic field? Do the jets sometimes emitted by a black hole result from its magnetic field?  It would seem Jupiter's magnetic field tends to be concentrated at the poles, and I would imagine this might extend to black holes as well.  I could see the concentrated fields near the poles accelerating particles either into or away from the black hole at high speeds.  I have included an image of Jupiter's magnetic field below:

http://en.wikipedia.org/wiki/Magnetosphere_of_Jupiter#Role_of_Io

It also appears that a black hole's magnetic field is indeed very strong:
http://www.iflscience.com/space/magnetic-fields-can-be-strong-black-holes-gravity

You appear to have a thing about linking the magnetic field generated by Jupiter to that around a black hole (note that a field does not "belong" to an object) - the two things are entirely different, the only connection being that magnetic fields are generated in both cases by currents caused by the motion of charged particles.

In Jupiter's case, the magnetic field is (likely) generated via a dynamo process, involving currents that flow in its metallic hydrogen interior.

In the case of a black hole, as material travels inwards, it forms an accretion disk around the black hole (because angular momentum must be conserved), in which magnetic field can be generated and amplified. The twisting of the magnetic field lines around the rotation axis of the black hole could create a kind of magnetic funnel lined up with the rotation axis, along which charged particles can be accelerated by a strong magnetic field gradient. The field lines would not look like you have shown for Jupiter, they would be twisted into a helical topology along the rotation axis. This is part of the Blandford-Znajek effect, which I do not claim to properly understand.

So the answer to your question is maybe, but its very much a topic of contemporary research.


What's the fastest moving object in the universe? We know that nothing can have proper velocities larger than the speed of light in vacuum. But are there any objects in space that get close to it? Any comets, or other objects thrown by gravity or supernova explosions that were hurled to incredible speeds?

The answer to this is surprising:

We are.

And many (if not all) other galaxies.

And they move faster than light.

See, the universe is expanding, at an accelerating rate. The fabric of spacetime itself stretches out, so that galaxies seem to move away from each other. The interesting thing is that relativity does not forbid these from moving away faster than light. While local space is flat and the local speed of light must be upheld, this need not hold at a global scale, so it is possible to have frames which move away from each other faster than $c$. Indeed, there are some galaxies that are moving away from us faster than light (the only reason we see them is that they used to be closer and moving at a slower speed). Any pair of galaxies that are 4200 Mpc away from each other (that is, with a redshift of 1.4), are moving away from each other faster than light in each other's frames (numbers stolen from the linked page).

Since the only consistent way to talk about motion is relative, one can say that we are moving away from other galaxies faster than light, since the reverse is also true. This can put galaxies in the bucket of the fastest moving objects in the universe. As for which is the fastest, I don't know, we would have to find a pair of galaxies which are the farthest apart (distance measured in the frame of the galaxy, of course), but since the universe is probably more than what we observe1, we can't pinpoint the pair of galaxies for which this is true.

For those who think that it is cheating2 to short-circuit the question with space expansion, there are other objects that go faster than light (they are not the fastest objects in the universe though), and these can be found on good 'ol Earth.

Electrons:

In nuclear reactor cooling pools3, we have a phenomenon known as Cerenkov radiation. Basically, emitted beta particles move faster than the speed of light in water. This creates an effect of similar origin as the sonic boom, where strong light emanates from the medium.

Saywhat? You think I'm cheating again2 by putting everything relative to the speed of light in a medium?

Alright, fine. Here are some fast objects that don't require space expansion to be fast, nor do they involve any trickery of semantics where the medium in which they are being measured is not mentioned. Many have already been mentioned by astromax.

Tachyons:These are particles which go faster than $c$ — this does not violate relativity as long as they never decelerate to subluminal speeds. However, there isn't much (any?) experimental evidence for these. A lot of BSM models do predict their existence, though. So there's still some cheating here, on to bradyonic matter:
Gluons: These are massless, and though they don't occur freely (except possibly in glueballs, though these most probably have mass) they do travel at $c$. But these can't move at any other speed, so again, this is slightly cheating. On to fermionic matter:
Neutrinos: Now these are viable candidates. The electron neutrino is known to have very, very little mass (we have an upper bound for it, which gives ), and as a result it can easily attain very high speeds. Put it in a gravitational field, and it goes even faster. If you want macroscopic objects, however:
Stuff spiraling around spinning black holes: Black holes have a strong gravitational field, and when rotating, they can impart angular momentum (lots of it) to nearby objects like accretion disks. Objects close to a black hole are accelerated to pretty high speeds. In fact, if an object is within the ergosphere, it moves faster than light from the point of view of certain frames of reference.
Stuff falling into black holes: From the faraway frame, an object speeds up and approaches the speed of light as it approaches the horizon of a black hole. Arbitrarily large speeds bounded by $c$ can be attained here.
Black hole plasma jets: Jets being flung out of black holes can get pretty fast relative to each other.

1. Due to cosmic expansion, there can be galaxies that are no longer visible to us. Some galaxies may never have been visible to us, if we start watching from when galaxies started forming.

2. I, for one, agree with you.

3. And other places where you have massive particles being emitted really fast into  a medium


Throughout the year, are all parts of the night sky visible from any given point on Earth? Or are there points on Earth where certain parts of the 360° night sky are never visible throughout the whole year?

Depending on the answer to this question, are there points on Earth where parts of the celestial sphere as such are never in the sky, no matter if it's a day or night sky?

At any one time, only half the sky can be visible, and half is below the horizon.

At the poles the sky appears to rotate about the zenith, and so no new parts of the sky ever become visible.  At the North pole, you can never see any stars in the Southern hemisphere.

On the equator different parts of the sky become visible and over the year all the stars can be seen (but stars very close to the pole will be very hard to see, as they are too close to the horizon.)

For someone who is at (for example) a latitude of 40 degrees North will not see any stars that are more than degrees south of the equator; 90-40=50; and so won't see the southern cross, the Magellanic clouds or other features of the Southern sky. Similarly an observer in Tasmania won't see Polaris, Cassiopeia, or the "big dipper" as these are features of the Northern sky.


Any active solar X-ray imager in orbit? NOAA GOES-14 & 15 went very recently in storage mode. With those also went the solar X-ray imagers (SXI), as GOES-16 & 17 do not have such instruments on board. Hence, my question:

Are there still any active solar X-ray imagers of other space agencies orbiting the Earth?

The Hinode satellite has an active x-ray telescope that is routinely used to image the sun.  And the NuSTAR extreme x-ray telescope can also be used for sun imaging (although its main job is looking for black holes)

Here is the latest XRT image of Hinode

There are also Extreme Ultraviolet telescopes (at 171Å that is less than an octave away from the conventional lower end of X-rays at 100Å) SDO. Stereo (Ahead) and SOHO all have EUV telescopes. The Atmospheric Imaging Assembly on the SDO satellite has sensors for wavelengths as short as 94Å, conventionally classed as X-rays

https://www.nasa.gov/jpl/nustar/searing-sun-seen-in-x-rays


How does the illumination of the sky from the sun compare to that of the moon? I have trouble describing the questions but this is the best I can do.
The sun lights up the sky, until it's about 18 degrees below the horizon. The moon also lights up the sky, and i'm wondering, how does the illumination of a full moon compare to that of the sun? How many degrees does the sun have to be below the horizon to equal the illumination of a full moon?

I hope i'm clear enough. Thanks for any answers

The question is not altogether well defined. One needs to specify what wavelength is being considered and at what position in the sky - since a twilight sky has a very significant gradient of sky brightness.

A paper by Patat et al. (2006) gives the brightness of the sky in the UBVRI photometric bands, at zenith, with the Sun at a variety of altitudes (including below the horizon). Figure 5 from this paper is reproduced below, it shows the zenith sky brightness measured at the Paranal observatory in the V band (approximately 550 nm). Curves for the other bands are qualitatively similar but with different numbers on the y-axis. The vertical dashed lines show where the Sun is 12 or 18 degrees below the horizon (known as nautical or astronomical twilight respectively).

Now to complete an answer to the question we need to know the brightness of the night sky at full Moon. Again, this is not completely defined, since it depends how close you are to the Moon.

The "dark" sky at Paranal is approximately 21.6 mag/arcec$^2$ when the moon is down. That is the horizontal dashed curve in the picture. Another article by Patat (2014) discusses the contribution of the moon as a function of wavelength and angular distance from the Moon. A reasonable number for the full moon at 90 degrees (i.e. the full Moon close to the horizon) is that the sky brightness is increased by about 1.5 mag in the V band (but by as much as 3 mag in the B band). Obviously, if you get closer to the full Moon then the sky becomes even brighter than this.

Using this number and the plot above, we can estimate a night sky brightness of 20.1 mag/arcsec$^2$ (equivalent to a full Moon at 90 degrees) is reached in the V band when the Sun is about 12.5 degrees below the horizon (i.e. very close to nautical twilight).

If you do a similar estimate for other bands, then you arrive at results that are very similar - perhaps 12 degrees below the horizon for the B band and 14 degrees below the horizon for the I band, presumably because moonlight does not have a drastically different spectrum to sunlight (it is a bit redder).

Caveats (at least): The answer will also depend on how far away the Moon is from the Earth, the altitude of your observing site, whether there is any cloud in the sky, what the density of scattering particles are in the atmosphere and the altitude of the full Moon.


Question about space-time In the Big Bang model of the Universe, every observable thing is thought to have expanded very abruptly from a point of infinite density and zero volume. However, the problem with this assertion is that for something to expand it first needs a locality in which to do such, which brings us to my question. How can scientists believe that "everything" was condensed and then suddenly expanded, since such is automatically an oxymoron?

Expansion is a physical property of the universe and yet if "everything" had no where to expand, how could it have expanded? There was "nothing," was there not? Even if we drop the big bang model and talk about "everything's" being in a singularity point, the question still stands.

Where was the singularity point? Asked differently, into what is everything expanding (which is a rhetorical question I ask to implicate space-time as necessarily existing irrespective of any theoretical expansion and likely so vastly that one might as well assume that space-time is infinitely large).

As well, it seems counter-intuitive, as little impressive a word as that is given its context here, that something like... well, "everything" could be condensed in the first place, let alone infinitely condensed and with zero volume. When thinking about the big-bang model, I can't help but conclude that a mere analogy was originally used for stellar birth, evolution and finally explosion. The issue though is still extension.

As far as I understood it, the big bang model states that space itself was also condensed along with everything else in the observable universe to a point of singularity. So, if that's the case, then [where] was the singularity point?

It ends up becoming an infinite regress, of course. Or, if you don't like the issue of "time" presupposed by my question, then look at the question after the universe started expanding. All the matter had to expand somewhere and "somewhere" is not "nothing." - It just seems that the model is somehow overlooking said feature of the actual universe and trying to account for it by mere assertion.

If you want to call it space-time, extension, space, where everything supposedly expanded to and is and is still expanding to, then that's my question. How does the theory allege where everything is yet at the same time assert that prior to the singularity point, there was literally nothing.

In the big bang model of the Universe, every observable thing is thought to have expanded very abruptly from a point of infinite density and zero volume.

Thats not quite true. Have a look at Wikipedia where you can read this:

"Extrapolation of the expansion of the universe backwards in time using general relativity yields an infinite density and temperature at a finite time in the past.[18] This singularity signals the breakdown of general relativity and thus, all the laws of physics. How closely this can be extrapolated toward the singularity is debated—certainly no closer than the end of the Planck epoch. This singularity is sometimes called "the Big Bang",[19] but the term can also refer to the early hot, dense phase itself,[20][notes 1] which can be considered the "birth" of our universe."

You can't necessarily extrapolate back all the way to a point singularity.

  However, the problem with this assertion is that for something to expand it first needs a locality in which to do such, which brings us to my question.

There's no problem with space expanding. Don't think of space as nothing, Einstein didn't, and nor do modern physicists such as Nobel prizewinner Robert B Laughlin.

  How can scientists believe that "everything" was condensed and then suddenly expanded, since such is automatically an oxymoron?

Because there's good scientific evidence that the universe is expanding. And when you "wind back the clock" you find that all the galaxies were in the same place 13.8 billion years ago.

  Expansion is a physical property of the universe and yet if "everything" had no where to expand, how could it have expanded? There was "nothing," was there not?

No. Space expanded, and space isn't nothing. As to what caused the big bang to occur, I don't know. Nor do I know what the universe was like 13.8 billion years ago. But I know of no infinities in nature, and I would challenge the alleged point singularity at the centre of a black hole and refer instead to the frozen star interpretation. In similar vein I would challenge the alleged point-singularity 13.8 billion years ago.

  Even if we drop the big bang model and talk about "everything's" being in a singularity point, the question still stands. Where was the singularity point?

I think you shouldn't drop the big bang model, but instead you should drop the singularity.

  Asked differently, into what is everything expanding?

Nothing. I know this is hard to imagine, but try to. Space is expanding, and there is no space beyond it. So there is no beyond it. As to whether this means you end up coming back round full circle or the universe is some "hall of mirrors" or something else, I don't know.

  (which is a rhetorical question I ask to implicate space-time as necessarily existing irrespective of any theoretical expansion and likely so vastly that one might as well assume that space-time is infinitely large).

No, one might not. Because the universe doesn't expand for nothing. It expands because space has a kind of innate "pressure", as per the energy-pressure diagonal in the stress-energy-momentum tensor. And if the universe was infinite, the pressure would be counterbalanced at all locations. An infinite universe could not expand.

  When thinking about the big-bang model, I can't help but conclude that a mere analogy was originally used

For an analogy, get a stress ball and squeeze it down in your fist. Then let go.

  The issue though is still extension. As far as I understood it, the big bang model states that space itself was also condensed along with everything else in the observable universe to a point of singularity.

See what I said above. Big bang cosmology doesn't actually say everything was condensed to point singularity.

  All the matter had to expand somewhere

There wasn't any matter initially. Remember the wave nature of matter, and pair production. We can quite literally make matter out of electromagnetic waves. These wave propagate through space, and are effectively wave sin space. So we make matter out of waves in space. So start with space, and do somehting to it such that it rings like a bell and is full of waves and starts expanding. As to what, I don't know.

  How does the theory allege where everything is yet at the same time assert that prior to the singularity point, there was literally nothing?

It doesn't. It goes back as far as a very dense universe. The stuff you hear about a point singularity and creation ex-nihilo is popscience, not big bang cosmology.


As a black hole passes through spacetime, does it affect it? Nothing in the universe stands still. It's always on its way from somewhere to somewhere. A black hole follows a trajectory through the universe, moving from one locus in spacetime to another locus.
Deep inside each black hole is a point that we've never been able to characterize.My question is, does the singularity, and possibly the powerful gravitational fields around the singularity have any measurable effect on the area of spacetime it passed through? One imagines an undeformed area of spacetime. After some time, a black hole passes through this spacetime, then moves on. Does the spacetime it passed through simply go back to being everyday, garden variety undeformed spacetime?

No, a black hole, or any other gravitationally attracting object for that matter, does not change spacetime permanently, but only affects spacetime around it. The easiest way to see this, is to view the BH in its own frame of reference. Here, the BH is stationary, and spacetime around it is described by the Schwarzschild metric. If a BH came by you and left a physical imprint on spacetime after it (like a wake), then from the BH's frame that would mean that you would fly by it, and it would start producing a wake in the direction that you moved. You don't have such an effect on BHs.

Wrong scenario of observer and black hole passing each other, as viewed from the observer's (left) and the black hole's (right) reference frames.


Can two singularities combine into one? We know that black holes can collide and merge, but if they each have a singularity at the core, how can the two singularities merge and become one? Wouldn't the mere fact of a second singularity in range of another make it not a singularity?

Hypothetically, Yes. Realistically, No. As James mentioned a singularity isn't a physical object but rather a point in space time. In theory a wormhole is just two singularities which combined.


Does The ISW Effect Predict Net Redshift Over Distance? The Integrated Sachs–Wolfe (ISW) effect is concerned primarily with the study of the Cosmic Background Microwave Radiation (CMBR), but seems to demonstrate that photons passing through voids suffer a net energy loss resulting in redshift, or can be blueshifted passing through high-density regions.

If a photon traveled long enough through alternating high and low density regions of space, can we assume that (depending upon the total ratio of density to void in all the regions traversed) our photon might well accumulate a substantial net gravitational redshift or blueshift after traveling a vast distance?

The (late time) ISW is caused by the evolution of cosmic structures as photons of the cosmic microwave background traverse them on their way to our detectors. It may cause a redshift or blueshift with respect to the redshift predicted for a homogeneous expanding universe.

A bit more detail: If a photon "falls" into a potential well, its frequency and energy increase. But as it exits the potential well it loses the same amount of energy. In a similar way, if a photon crosses a "void", it loses and then gains energy.

However, if the universe contains dark energy (the ISW does not work in a flat universe containing just matter) and is expanding, then the structures the photons traverse are evolving with cosmic time such that the process is asymmetric. A potential well is "smeared out" a little by the expansion, and the photon does not lose as much energy exiting, as it gained on entering. Ditto for a void, but in reverse.

The net effect is small blue- and redshifts (of order 1 part in 100,000 at low spatial frequencies) with respect to the overall redshift you would expect anyway from the cosmic expansion of a homogeneous universe. These shifts are expected to be correlated with the cosmic structures in the same line of sight. The ISW has been detected (albeit with low significance) for the past decade, by cross-correlating CMB spatial variations with catalogues that map out the large scale cosmic structure (e.g. high redshift radio galaxy catalogues). See Nishizawa (2014) for a review.


Approximate formula to find velocity from cosmological redshift From IOAA 2013 (Greece) Theory question no. 15, they stated that the approximate formula to find velocity from cosmological redshift is $$v = c*log_e (1+z)$$ and that it is often used by cosmologists. I did a quick google search but found nothing similar to this formula.
So, where did this equation come from and is it really often used by cosmologists?

This formula is exact if the expansion is linear ($a(t) = H_0 t$) and all peculiar velocities are zero. In that case, the comoving distance to the object is $$\int_{t_\text{then}}^{t_\text{now}} \frac{c\,\mathrm dt}{a(t)} = \frac{c}{H_0} \ln \frac{t_\text{now}}{t_\text{then}} = \frac{c}{H_0} \ln (1{+}z)$$ and the present recessional velocity is $H_0$ times that.
In the real world, the expansion has been not too far from linear, and recessional velocities are not too large, so it's reasonably accurate.


Are these facts about moon true? I don't have much knowledge of Astronomy but today I was googling randomly and found an article, according to which:

  The moon is of the wrong size for its current orbit.
  The moon rings like a bell when hit by an object, which could signal that it's hollow.
  Metal shavings were found in samples taken by the Apollo crew.
  UFO sightings were common on NASA moon missions.
  All these phenomena and more point toward the suggestion that the moon is actually an artificially created spaceship. It's a bit far fetched, but it's odd that astronomers have been discouraged from testing this theory.

Can this theory be true?

Orbital radius depends on speed of the body. Size is largely irrelevant. Imagine a  space craft and much larger space station, able to orbit meters apart at the same speed, with engines switched off. An astronaut going on a spacewalk would be launched away from the station, being much smaller and so on wrong orbit. No. It's as relevant as "Boeing 747 shouldn't be able to fly because it's painted the wrong color." False and based on wrong premise.
For anything to ring you need atmosphere, a medium for sound to travel through. Moon, floating in vacuum, doesn't ring, nor create any other sound for that matter. Even the most powerful meteor impacts are entirely silent. False and based on wrong premise.
Countless iron meteorites hit the surface of the Moon, often shattering on impact. Specks of iron are very common in lunar soil, and I have no doubt it's entirely possible shavings-shaped pieces of lunar meteor might have found their way into the samples. Likely True and nothing odd about it.
...how does that even relate?


Why do stars rotate slower than they're expected to? Due to conservation of angular momentum, I thought most stars would be spinning extremely fast because they have a relatively small diameter. However, it turns out that this is not true and most stars actually spin "slow"? Why is this the case?

There are two phases to this problem.
In order to accrete into stars, a huge amount of angular momentum must be lost to allow so much mass to gather into a small volume.
A second problem is how stars like the Sun end up rotating so slowly, when younger versions of stars similar to the Sun rotate much faster.
The solution to the first problem may be solved by the outward transport of angular momentum in accretion disks. The angular momentum may ultimately be shed by winds originating from the disk surfaces. Another possibility is the loss of angular momentum in jets and bipolar outflows, which are accelerated from the central star-forming core by poorly understood processes. Some of the angular momentum can of course end up in planets.
Stars themselves can then lose angular momentum throughout their lives. Initially this can be through magnetic coupling to the protostellar accretion disk. Later, the principal angular momentum loss process is through stellar winds that become controlled by coronal magnetic fields. The plasma decouples from the co-rotating field at many stellar radii, taking away angular momentum.