What if a global killer hit Mars? Compared to the effects an impact of a global killer would have on Earth, what if such an asteroid, with a diameter around 7 miles (11.2 km), impacted the red planet with its different atmosphere, size and gravity? Would any volcanoes erupt? And what effects would such impact have on the probes and rovers currently on Mars and on its exploration?

Imagine a similar asteroid as Chicxulub, the one that likely killed the dinosaurs here on earth. It is thought to have impacted at about 20 km/s.
Initial atmospheric effects: When the asteroid entered the Earth's atmosphere, it pressurized the atmosphere and caused faster than 1000 km/hr super-heated winds - https://www.lpi.usra.edu/science/kring/Chicxulub/regional-effects/ that blasted out in a 1500km radius.  Since Mars's atmosphere at the surface is less than 1% the density of earth's, the heat generated by pressurization would have been correspondingly at 1% given the ideal gas law tells us that heat and pressure have an inverse relationship. The super-heated winds, glowing heat release on atmo-entry, and asteroid ablation would be disappointing on Mars compared to Earth.
Impact and initial ejecta: On Earth, the Chixculub impactor hit in the shallow ocean near the Yucatan, causing a huge tsunami that covered most of what is now the South East United States.  The water landing softened the blow, but enough energy was released to vaporize the meteor and blow out enormous amounts of ejecta.  The big ejecta (a few centimeters to 300 meters in diameter and bigger) landed hundreds of kilometers away in just a few minutes.  Much of the smaller ejecta floated down thousands of kilometers away.  A convection reaction dust plume at the impact site pumped soot and dust into the upper atmosphere and beyond. This ejecta (which was mostly vapor) sped up as it rose. As much as 12% of this ejecta reached escape velocity and never came back down to earth! https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2001JE001532
On Mars, there wouldn't be any tsunamis of course, since there is no ocean.  Hence the impact would kick up more debris.  Since Mars's surface gravity is about 3/8 of Earth's, and since there is much less atmospheric resistance for objects moving in the atmosphere, the big ejecta would have traveled thousands of kilometers instead of hundreds.  A lot of this ejecta (especially the vapors and liquids) would have gone into orbit or reached escape velocity!  However, the convection dust plume at the impact site would have been smaller than on Earth, with less atmosphere to feed the plume.
Debris layering: Since the area of a sphere is $4\pi r^2$ and the radius of Mars isn't much larger than half that of Earth, we can estimate that the surface area of Mars isn't much more than 1/4th that of Earth.  So the debris layer would be almost 4 times thicker (really much more because there would be more debris in an impact with no water).  On Mars, the debris would be centimeters to meters thick (nearer the impact site) rather than on Earth, were it is only mm thick opposite the impact strike. Also, due to the lower atmospheric density, particles would settle out much more rapidly than the decades long global cooling Earth experienced.
Volcanism and Earthquakes: I agree with the other answer, an asteroid strike of this magnitude likely caused volcanism on Earth and would likely do so on Mars.  This would be to a lesser extent since Mars has a thicker crust, is cooler, and has less current volcanic activity. The Chicxulub strike was thought to be equivalent to an earthquake of magnitude 10 on the Richter scale. Enormous shockwaves would move across the entire planet and meet at the antipodal point, likely causing eruptions or even raising hills or mountains.
Rovers and Probes: The big effect an asteroid strike like this would have on rovers and probes is to destroy them if they were near the impact site, damage them with earthquake effects if they were somewhat distant, and envelope them with debris wherever they were on the planet's surface!


Beamsize in radio/microwave observations In radio/microwave observations, I understand the beamsize is the response of the telescope to a point source - effectively, the telescope's resolution.

Now, the Planck satellite (par exemple) combines different frequency channels in a variance-weighted map. These different channels in general will have different sensitivities, and different beamsizes.

Suppose you coadd two (or any number really) of these channels. What will be the effective resolution of the resulting map?
For instance, suppose we take channel 1 with a 5 arcminute beamsize (FWHM), and channel 2 with a 12 arcminute beamsize (and let's suppose they have the same sensitivity).
What will be the resolution of the inverse-variance weighted map? How will this change if the sensitivities differ?

This is a fundamental problem in modern multi-color/frequency astronomy. You can not simply add two spectrally significant different channels because they contain information about different solid angles on the sky.

If you have two maps with different inherent beam sizes one common approach is to resample the map with the higher spatial resolution to the lower resolution of the other map by convolving with a Gaussian beam of size:

$\theta_{Gauss-kernel}=\sqrt{\theta_{low}^2-\theta_{high}^2}$

For example, if you have two maps with resolution (FWHM) 5" and 15" you need to smooth the 5" map with a Guassian kernel of size

$\theta_{Gauss-kernel}=\sqrt{15^2-5^2}=14.14"$

to obtain a comparable map.

Of course the beams are never true Gaussians and it depends whether you observe point sources or extended (w.r.t. the beam) source.

As an example you can take a look at the SPIRE Observers manual. SPIRE was an instrument onboard the Herschel Space Observatory. They discuss the involved problems in somewhat more detail.


Why are gas giants colored the way they are? As I understand it, Jupiter, Saturn, Uranus, and Neptune are all made primarily from varying proportions of hydrogen and helium.  Despite this, Jupiter is very red, Saturn is yellow, and Uranus and Neptune are blue.  Why the difference in colors for planets with similar chemical makeups?  Or do they not have similar chemical makeups?

If such an explanation is available, I would like specific descriptions of why each planet is colored the way it is.

The simple answer is that the composition in terms of elements doesn't matter for the colour, since that is determined by the molecular composition as well as state (pressure and temperature) of the visual layers (which contain a small but significant fraction of other elements than H and He). Given the differences in the giant planets temperature and pressure, this molecular composition is most likely different.


Does cosmological horizon grow or decrease as it radiates? Black holes decrease as they evaporate and their radius decreases as well.

So what is with a cosmological horizon?

If cosmological horizon is just a black hole centered at the opposite side of the universe, we should see the radius of the external space groving as the radius of the BH decreases.

But if the radiation decreases the horizon entropy we should see the area of the horizon decreasing.

What is the correct conclusion?

I think there's a misinterpretation here that's causing your confusion. There's a difference between the event horizon of a black hole and the cosmological horizon.

The event horizon of a black hole is the place inside which nothing can escape from the black hole. The cosmological horizon, on the other hand, is the place beyond which an observer cannot "retrieve information." The difference between the two is basically that the event horizon of a black hole is in the same place for observers in different locations, while the cosmological horizon differs for observers in different locations.

Also, if the cosmological horizon was the edge of a black hole, the black hole would have to be centered right where we are - which is highly unlikely!


How are neutrinos produced in blazar jets? I was watching the press release about measuring neutrinos and gamma rays from a distant blazar. One of the presenters mentioned that the neutrinos are associated with very high energy protons caught up in the jet produced by the galactic nucleus. What I don't understand is what interation(s) is producing the neutrinos in this scenario. What sort of particles must the protons be interacting with to produce these neutrinos?

Neutrinos are typically produced in AGN jets through what we refer to as hadronic processes. Protons are accelerated to relativistic speeds and interact with nearby photons. Depending on the particular type of AGN, there are a number of possible sources of these photons:

Ambient light within the jet
Emission from the accretion disk
Photons from the broad line region surrounding the AGN
Emission from the nearby dust torus

As I understand it, contributions from the latter two sources are less prominent in blazars, although they may play an important role in the radio-loud flat spectrum radio quasars (FSRQs). Murase et al. 2014 have an excellent diagram showing the possible sources of this radiation:

The photons interact with the protons in the jet and produce other hadrons, typically pions. Neutral pions decay to a pair of photons, one key mechanism of gamma ray production in AGN jets:1
$$\pi^0\to\gamma+\gamma$$
However, it is quite possible for the proton-photon interactions to produce charged pions ($\pi^{\pm}$), which then decay to muons and muon neutrinos (Atoyan & Dermer 2002). The muons decay to electrons, while the muon neutrinos may be observed here on Earth. This appears to have been the source of the neutrino detected by IceCube in 2018, originating from the blazar TXS 0506+056.

1 The other main set of gamma ray production pathways in the jets are the leptonic processes, where electrons and positrons transfer energy to ambient photons via the inverse Compton effect; see Degrange & Fontaine 2016, which I found quite helpful when learning about those mechanisms.


Is there a limit to how hot an object an get? Temperature is a measure of the average kinetic energy of particles in a given place (correct me if I am wrong) so there must be a definite limit to heating like there is an absolute zero to cooling. And what process could achieve this temperature?

First, let's get a bit of background. "Absolute hot" is the concept that a maximum temperature exists. It describes your question exactly. For a long time, no one was able to figure out whether an absolute hot exists, but in the 20th century major revolutions in theoretical physics gave us answers.

"Absolute hot" isn't the name of the hottest temperature, but it's the concept that one exists. So what is our answer — is there a highest possible temperature that fits the concept of an "absolute hot"? Well, yes. But first we must talk about heat and temperature.

Photons act both like particles and waves. As we add more energy, the photon's wavelength decreases, and its frequency increases. Thus, the wavelength of the light emitted by hot objects generally depends on the temperature of the object. Hotter objects will emit more energetic photons with shorter wavelengths.

According to modern physics, a Planck length is the smallest meaningful length in the Universe (the reason for that is an entirely different question). If we keep raising the temperature of our object, nonstop, the wavelength of the photons it emits will decrease until it reaches a Planck length. Our object would reach a temperature called Planck temperature. That's about $1.417×10^{32} \; \mathrm{K}$.

Planck temperature is theoretically the hottest possible temperature. At that point, the object could not theoretically get any hotter, as the photon's wavelength cannot decrease beyond a Planck length.

But what happens if we were to add more energy to the object? Would it go beyond Planck temperature? Well...God knows. At that point, we wouldn't even call it temperature anymore. Nobody knows what exactly would happen, but it certainly wouldn't match the definition of temperature.


Why is the difference from a perfect sphere the same for the Earth and the Moon? Wikipedia gives the following physical dimensions for the Moon:

  Physical characteristics

Mean radius        1737.1 km  (0.273 of Earth's)
Equatorial radius  1738.1 km  (0.273 of Earth's)
Polar radius       1736.0 km  (0.273 of Earth's)

In all cases, the proportion of Earth's equivalent measurement is 0.273. If these figures are accurate, why aren't these numbers (0.273) different from each other at this scale? Obviously, if measured to enough decimal places, differences will appear.

The Moon has less gravity and rotates at a different speed from the Earth. Why don't these cause variability in the different radius measurements? Do they cancel out? It is a coincidence? Is there something about the Earth-Moon dynamics that means over millions of years these proportions gradually equalise?

Why is the difference from a perfect sphere the same for the Earth and the Moon?

It isn't, as soon as you look one more digit further.

From NASA's Earth Fact Sheet and Moon Fact Sheet which I found here

                             Moon (km)   Earth (km)      Ratio     (rounded)
Mean radius (volumetric)     1737.4       6371.000      0.27270      0.273
Equatorial radius            1738.1       6378.137      0.27251      0.273
Polar radius                 1736.0       6356.752      0.27310      0.273

If you would like to compare the Moon and Earth parameters further, see the Moon Fact Sheet:


What exactly is interplanetary scintillation; what was the Interplanetary Scintillation Array looking for? Did it successfully observe any? The Interplanetary Scintillation Array is the radioastronomy observatory (i.e. big antenna) where the first pulsar was discovered by then graduate student Jocelyn Bell Burnell through careful and painstaking review of chart recorder data.

Question: What exactly is interplanetary scintillation; what was the Interplanetary Scintillation Array looking for? Did it successfully observe any?

For observing ionizing radiation such as cosmic rays and photons, a scintillator is used to convert energy to lower energy photons (usually visible light) but I don't know if the term scintillation in radio astronomy is related to that process in any way or not.

update: I've just asked in Aviation SE: What is “scintillation” and are “qualified pilots” aware of it? which was quickly answered and points out that in that context scintillation can also be called "twinkling". If a radio source can twinkle as well, is this an atmospheric effect? If so is it neutrals or ions that generate it. If not, is it turbulence in the ionized interstellar medium?

To fully answer your questions, let me introduce scintillation before interplanetary scintillation.
Atmospheric scintillation
The imaging of an astronomical source is affected by a collection of effects that goes under the name of astronomical seeing, the main ones being smearing, motion and scintillation of the image. All these effects are caused by the deformation of the light wavefront due to random inhomogeneities in the refractive index of the atmosphere.
Let's visualize the basic idea behind the seeing. Suppose to have a light source located at infinite distance, such that it's ideally a point source and its light reaches Earth in plane waves. When entering the atmosphere these waves face changes in the refractive index, and in geometrical optics approximation the wavefronts are deformed according to Snell's law. In the following image you have two simplified cases: on the left, it is shown that a vertical gradient in the refractive index produces a simple tilt of the wavefront; on the right, it is shown that a horizontal gradient produces a deformation of the wavefront. Here $\textrm{WF}_1$ is the incoming unperturbed wavefront, $\textrm{WF}_2$ is the same wavefront after entering the atmosphere and $n_i$ are the refractive indices.
$\hskip2in$
Let's now approach the real case. It is known that variations in the refractive index are related to variations in density (e.g. via Gladstone-Dale relation). Since our atmosphere has an overall more-or-less stationary distribution of temperatures, densities, and pressures, the most dramatic variations of the refractive index are found only in the turbulent layers. Here each turbulent eddy can have different density, temperature and so on, causing local variations in the refractive index. These layers are located in the first km of atmosphere (the planetary boundary layer) and at $\sim$10 km (near the tropopause). In the following image, it's depicted how a plane wavefront is deformed after crossing a turbulent layer and all of its eddies with different $n_i$. The lenght $r_0$ is the Fried parameter, which can be interpreted as the characteristic lenght of the turbulent eddies.
$\hskip2in$
What is atmospheric scintillation then? Well, as you see from the previous image, the curvature of the wavefront leads to a convergence (or a divergence) of the light rays (i.e. the lines perpendicular to the wavefront). This means that when you take an image by collecting photons with your sensor (i.e. when you collect a portion of the wavefront) the image brightness can rise or decrease depending on whether light rays are converging or diverging. Furthermore, this brightness variation is time-dependent, since the eddies evolve with time and the turbulent layer has an horizontal drift velocity. This is scintillation.
To see scintillation you need that (i) the source has small angular dimension, (ii) the turbulent layer is far from the telescope, (iii) the characteristic length of the collected wavefront's portion (e.g. the diameter of the telescope) is comparable to $r_0$, (iv) the exposure time is less than the lifetime of the deformation. If (i) and (ii) are not satisfied you might see the motion of the source instead of scintillation, while if (iii) and (iv) are not satisfied you will see speckles or a smeared image.
Scintillation in the radio domain
In the radio domain, scintillation is not limited to atmospheric scintillation.
For wavelengths between millimeters and centimeters we still have atmospheric effects, mainly due to the water vapor vertical gradient near the ground. Indeed, for this wavelengths the refractive index of air deviates from unity less than few houndreds of ppm. Since the turbulent layers are closer to the ground and usually the radiotelescope apertures are bigger than the characteristic length of the turbulence, seeing is not dominated by scintillation.
For wavelengths beyonds centimeters, the frequency of the waves gets closer to the plasma frequency in the ionosphere, hence the wavefront deformation provoked by the passage through the electron clouds in the ionosphere becomes relevant. Since their great distance from the ground, this will result in scintillation.
The timescale of ionospheric scintillation is between minutes and tens of minutes, but Hewish (1955) started to notice that there was another scintillation with a timescale of few seconds, and that its intensity was greater for sources near the Sun.  Following this hint Hewish et al. (1964) have shown that in fact this faster scintillation was produced by plasma clouds in the interplanetary medium, delivered by the solar wind.
Going further, Sieber (1982) have shown that there's also a scintillation with timescales between days and months due to plasma clouds in the interstellar medium.
Conclusions

What exactly is interplanetary scintillation?

Scintillation is a time-depending change in the intensity of a light signal, and it is caused by deformations in the wavefront due to random variations of the refractive index. In the case of interplanetary scintillation, the refractive index variations are caused by plasma clouds traveling in the interplanetary medium and delivered by the solar wind.

What was the Interplanetary Scintillation Array looking for?

Precisely interplanetary scintillation. From it you can study the properties of both the interplanetary plasma and the scintillating radio sources.

Did it successfully observe any?

I guess yes!


What particles does a black hole emit when it evaporates itself? Due to Hawking Radiation, after a very long time, a black hole will eventually disappear.

But where does this "evaporation" go? What is this made of?

Although black holes are widely believed to emit Hawking radiation, it should be stressed that it has not actually been observed (yet?). The Hawking radiation should consist of electromagnetic radiation/waves that have a near-perfect black body spectrum, which is at a temperature that is inversely proportional to the mass of the black hole - the smaller the black hole, the higher the temperature.

The radiation is caused because "vacuum" is not empty when considered in quantum mechanics. Particle/anti=particle pairs are created for brief moments of time, before they annihilate again. Close to the event horizon of a black hole, it is possible that one particles from a virtual particle pair travels within the event horizon of the black hole and cannot escape to annihilate with its anti-particle. In principle, these particle/anti-particle pairs could be any types of particle, but in practice they are more likely to be the lightest particles. The lightest charged particles are electron/positron pairs, since these need to "borrow" less energy from the vacuum to be created. I do not think it is essential for one of the pair to disappear, since "Unruh radiation", a close relation of Hawking radiation should also be seen whenever there is acceleration with respect to the vacuum.

To stay out of the black hole event horizon the particles must be accelerating. Accelerating charged particles locally emit electromagnetic radiation, which is then gravitationally redshifted when seen by an observer a long way from the black hole. It turns out that to be in a thermal equilibrium, the radiation must have a blackbody spectrum form.

The temperatures and amount of radiation emitted are very small for stellar-sized black holes. The temperature of a non-rotating Schwarzschild black hole is given by
$$ T = \frac{6.2\times 10^{-8}}{M}\ K,$$
where the mass $M$ is in solar maases.

The power emitted by a black hole as Hawking radiation is
$$ P = \frac{9\times 10^{-29}}{M^2}\ W$$


Why don't planets give off their own light? Why don't the planets glow like stars?

All matter radiates (except if it's at absolute zero temperature), regardless of its composition (you got that of Mercury badly wrong). The most important form of radiation is the black-body radiation which only depends on the temperature of the material, but line emission and absorption may also be important (but depends on the composition and ionisation state of the material) and other emission and absorption processes.

Stars are hot enough (the Sun has surface temperator $\sim5700$K) for the black-body radiation to peak in the visible part of the electro-magnetic spectrum. As a star shines, it looses energy, i.e. it cools, reducing the gas pressure that stabelizes it against gravitational collapse. In stars, this energy loss is balanced by the energy production from thermonuclear fusion in the core (requiring temperatures $\sim10^9$K). The transport of this energy to the surface makes stars non-trivial.

In planets and brown dwarves, there is (by definition) no thermonuclear energy source. Therefore, these objects must shrink, which generates energy from gravity. However, they cannot shrink indefinitely, as ultimately quantum mechanics becomes important: Pauli's exclusion principle demands that the electrons cannot be arbitrarily closely packed. Thus, for brown dwarves and giant gas planets (but also white dwarves) further shrinking is halted at a radius comparable to that of Jupiter (Jupiter itself is still shrinking at a very small rate). These as well as all smaller objects then merely cool down very much like a piece of glowing coal.

The situation is a often more complicated by sources of energy. Planets, for example are
irradiated by their host star, which may dominate the energy gains at their surface (in addition, the Earth gains energy from nuclear fission in its core). The balance between this energy gain and the loss by black-body radiation determines the temperature of a planet.

The Earth, for example, radiates in the infrared. However, the radiation losses are also regulated by line absorption of that infrared in the higher atomsphere by so-called green-house gases, in particular CO$_2$.


Why is there a density dropoff in the stellar winds of magnetic O-type stars? I recently went to a talk on X-ray emissions from stellar winds of O-type stars, and a paper was explored that in turn referenced ud-Doula et al. (2014).

Essentially, O-type stars may show strong x-ray emission due to radial shocks in their stellar winds. In general, the winds are isotropic, and so emission is assumed to be spherically symmetric neglecting rotational effects. However, certain O-type stars (e.g. NGC 1624-2) have strong magnetic fields; this means that outflows are channeled into rather narrow beams within a few stellar radii, which eventually spread out.

X-ray emission is characterized by a parameter $X_{T_x}$
$$X_{T_x}\equiv\rho^2\exp(-T_x/T)$$
for density $\rho$ and temperature $T$, with a threshold temperature of $T_x$. Simulations place most of the emissions in a rather narrow spot near the star, because while there are indeed high temperatures experienced all along a beam extending outward from the star, the density quickly falls off, as shown in Figure 4:

From left to right: Logarithmic density, logarithmic temperature, and x-ray emission.

Why does the density drop off so quickly? I would expect that the channeling of more of the wind into a small beam would lead to a rather steady radial density profile along it, as is the case with the temperature graph.

This is unrelated to the question, but I'll add that the temperature isn't really constant. 1D and 2D simulations of spherically symmetric show varying behavior due to repeated shocks inside the wind, and I would expect the same to be true here.

Conservation of mass?

In a steady-state wind, the mass loss rate through each shell is the same.
$$ \frac{dM}{dt} = - 4\pi r^2 \rho(r) v(r),$$
where $\rho$ and $v$ would be the density and velocity of a spherically symmetric wind, with a fixed mass-loss rate $\dot{M}$.

Thus for a fixed wind speed you expect the density to fall as the square of the radius in spherically symmetric wind - and it looks to me that a little outside the closed field loops then spherical symmetry is approximately recovered.

This alone, would not account for the sharp drop off in density seen, but the other factor is that a radiation driven wind accelerates with radius when close to the star, which exacerbates the decrease in density. Typically, $v$ would increase by a factor of a few between say 1.5 and 5 times the radius. Multiplying this by a $r^{-2}$ factor would give a factor $\sim 100$ decrease in density overall.

The behaviour of the temperature is much more complex, governed by heating and cooling processes and the expansion of the gas.


Is Pluto a "proto-planet"? My question is this : could the Charon-Pluto system becomes in the coming few millions years a full fledged planet, as in the IAU classification ?

I ask this because there are two facts that bother me :

The system orbit isn't on the ecliptic
The system hasn't cleared its neighborhood

So, is it possible that that system isn't stabilized yet, and that 1) its orbit is very slowly changing (for lack of a better word) to the ecliptic, and 2) that the system will slowly collapse, and bring all its moons with it, to clear its neighborhood ?

In other words, is Pluto a "proto-planet" (as I would personally name it), or definitely (and for the next millions years) a dwarf planet ?

Edit : I disagree on the fact that this question is a duplicate of How long will it take Pluto to grow to planet size? as this one ask not for the size of Pluto, but rather the other characteristics, the orbit and the clearing the neighborhood.
Moreover, the question was asked in a broader canvas, to know if it was possible that some elements of the solar system (Pluto, but not only) could be instable enough to reach the level of planet sometime in the futur.

Pluto will never be a planet.

There are a number of technical papers that give more precise meaning to the concept of "clearing the neighborhood". It's not just now, it's can the object in question clear the neighborhood of its path while the Sun is still a star.

In the case of Pluto, Ceres, Eris, and a host of other not-quite-planet objects, that will not happen. Ever. Or at least so long as the Sun shines. After that, does it matter?


If Jupiter and Saturn were considered stars, which of their moons would qualify as planets by 'clearing the neighbourhood'? Jupiter has four and Saturn seven spherical satellites. Jupiter's moon Ganymede and Saturn's Titan are larger than Mercury and they're the most massive moons at 0.0248 M♁ (Ganymede) and 0.0225 M♁ (Titan). Do these moons 'shepherd' the other moons' and asteroidal satellites' orbits? Ganymede is three times the mass of Europa and quite close to it, so does it dominate Europa's orbit? Callisto on the other hand is twice as far as Ganymede and the 3rd most large and massive known moon while quite lonely in its orbit. Would it qualify for a planet too? If Jupiter was a star, I think Ganymede and Callisto would qualify as planets, Europa would be a 'dwarf planet' (because of being dominated by Ganymede's and perhaps Io's orbit) and Io I don't know, it probably would be more controversial.
Saturn's moon Titan is more massive than the other six spherical moons altogether, so I think it would clearly be a planet. The other moons (except Iapetus) are quite close to each other so I think they'd be considered 'dwarf planets' except Iapetus which is very far from Saturn and also quite lonely, so it might qualify as a planet (there might be an issue concerning its shape but my question is about the 'clearing the neighbourhood'/planetary discriminant point). Mimas would near-certainly be a 'dwarf planet' because it's about within the rings which alone are about 2/5 the mass of Mimas. So I think Titan and Iapetus would be considered 'planets', Mimas a 'dwarf planet' and the other spherical moons of Saturn probably too.
What would you say? Which Jovian and Saturnian moons would be considered planets and which ones 'dwarf planets' as per the proposed planetary discriminants if Jupiter and Saturn were stars?

I've noted before that the IAU naming critera are guidelines rather than laws of nature.  If applied to Jupiter's moons the four Galilean satellites would probably be "planets" (they don't share their orbits with anything else of comparable size and are large enought to be in hydrostatic equilibrium)  There is a mean-motion resonance 4:2:1 between them but this is a more or less symmetric relationship without one body dominating another.
The other moons are tiny by comparison.  There are 4 major bodies orbiting Jupiter.
For Saturn, Titan certainly dominates its orbit (Hyperion is much smaller.)  Dione and Rhea also seem to have a value of mu that are over 100 (though determining the value is tricky because not all moons and moonlets have been discovered) Enceladus and Tethys and Iapetus also seem to qualify,  As you note, Mimas is too small and borderline spherical. The other moons are just so much shrapnel by comparison.
So there would probably be the six pre-1800 moons, minus Mimas.
It is possible that the rules might be bent, the Rings are no permanent features, so they might be ignored for the sake of planetary determination. Remember the rules exist to serve us.  Humans make the rules. Some can be bent, some broken.


Cooling of astrophotography camera sensor chip For astrophotography using 'medium size' land telescope at good location,

How many time longer exposure, in view of boosting the system's ability to photograph even darker objects, can one gets from cooling down digital camera sensor chip?
Is there a 'sweet spot' temperature where most thermal related noise is suppressed and further lowering of temperature yields litter extra benefit.
What is the ball park temperature for low (amateur), medium and high (professional) grade cooled camera?

Web search shows that Peltier electric cooled camera works at around "minus a few tens degrees centigrade". Does the really professional one cool to more extreme level, somewhere close to liquid nitrogen temperature?

Does CCD, CMOS or other sensor have different 'sweet spots' temperature?

The practical length of the time exposure will depend on your equipment. Cameras used in astrophotography are generally able to be used with exposure times ranging from milliseconds to hours. As your question stands, cooling the chip will have no effect on the length of the exposure time. What it can affect is the amount of noise in the image. Cooling the sensor generally reduces the amount of noise. Cooling beyond -20°C may have limited value, particularly if you are not using research-grade instruments. While you may further reduce the noise, you may still be at the mercy of the quality of the optics, drive mechanisms, etc. and you may have less noise in the image but the limiting factor is not your image noise but the system limit for drift and distortion.

Some professional instruments use liquid nitrogen cooling while others use thermoelectric cooling, depending on the manufacturer and the intended use. The instruments that use sub-zero cooling also usually have a high quality sensor with a pixel well depth that supports longer exposures, without bleeding effects from adjacent pixels. It all depends on the sensor, camera, and other parts of the system, so your mileage may vary.


How probably is it that galaxies will extinguish? In the nebulaes of galaxies are stars born, but the gasses and dust they are made of comes from exploded stars (fe supernovas). So it looks like these processes can go on continuesly. But is that really the case, or does according to entropy(?) galaxies extinguish because fe the galaxie went too big so gravity has no opportunity to form other stars any more. And before colliding with other galaxies it is passed away. So is there any sign that galaxies are getting smaller or less active?

Galaxies are gradually being extinguished. Most star formation activity occurs near the start of a galaxy's life, or in response to merger activity with other galaxies.

The star formation rate of the universe peaked at redshifts of around 3, corresponding to a look-back time of around 9 billion years. Since then the star formation rate has declined as the universe expands; mergers are less frequent, gas is driven out of galaxies by supernovae and active galactic nuclei.

However, most of the stars that have been formed are of lower mass (K- and M-dwarfs) than the Sun and will live on for tens or hundreds of billions of years. They are however much fainter than the Sun. So although high-mass, luminous O- and B stars live their short lives and are not replaced at the same rate, the low-mass stars continue to shine. This means that galaxies will get fainter on average as the remaining stellar populations increase in average age and decrease in average mass.

It is a slow process though. High mass stars are still being formed in our galaxy after 12 billion years, and most spiral galaxies have ongoing star formation. However, star formation has more-or-less ceased in gas-poor elliptical galaxies.


Can rogue/nomad planets and other starless planetary bodies orbit each other in a stable system? Nomad planets (also called planemos or rogues planets) which drift in space without  a star or substellar object to orbit. The majority of such planets we have discovered are gas planets with thick atmospheres and no landmass, although some of them could be frozen deserts with rocky surfaces like Earth's.

Most nomad planets we have found are alone/solitary, and do not orbit any other companion objects. We have only discovered 1 (as in 1.0) binary planemo called 2MASS J11193254–1137466.

What I'm asking here is that can a large system of 6 or 7 rogue planets and other planetary bodies orbit each other and remain stable; basically like a tiny open cluster or huge multiple star system, just comprised of planets instead of stars?

One of the main hypotheses for the formation of rogue planets is ejection from a preexisting planetary system. This is an unlikely scenario for the formation of a cluster of rogue planets, because it doesn't seem feasible for a whole bunch of bodies to be ejected together and remain in a stable system. If multiple planets were ejected from a planetary system, I would expect them to exit randomly and separately.

That said, a couple sub-brown dwarfs have been observed to host what may be protoplanetary disks (although that term may not be applicable in these cases). The important objects here are Cha 110913-773444 ($\sim8M_J$) and OTS 44 ($\sim11M_J$). These objects have not been classified yet; they could be considered rogue planets or sub-brown dwarfs (or even brown dwarfs, honestly). Assuming the normal processes active in protoplanetary disks take place - collisions, accretion and the like - then in the future, additional bodies could form.

OTS 44 is thought to have a disk of around 30 Earth masses (Joergens et al. 2013); depending on the efficiency of planet formation, we might see bodies similar in mass to terrestrial planets (note that this also depends on disk composition; much of it is likely to be hydrogen and helium, not silicates). Giant planets seem unlikely, as the disks mass is comparable to those of Uranus and Neptune, within a factor of two, and it is highly unlikely that the entire disk would coalesce into one planet.

If a rogue planet formed in situ, rather than being ejected from a system, we do have a different picture to consider. In this case, it seems more likely for binary rogue planets to form. At this point, of course, the label "sub-brown dwarf" is more appropriate than "rogue planet".


Quantum death of stars This comes from a comment on this question, to quote:

  The death of (large mass) stars is also based on quantum events with probabilities technically not 1 (and is very fast), so it is technically possible for a star not to die in a way it really ought to by sheer chance.

The question is what quantum events are involved in the death of a star (likely and unlikely)? If many of these very unlikely quantum events happened at the same time all over the star, how could the death of a star differ form the typical nova or simply cooling down?

Stellar fusion and supernovae are governed by quantum particle interactions (on massive scales).  In general there are many possible ways for particles to interact, decay, etc. at various probabilities.  To understand the physics of a system correctly you must take account of all possibilities (up to an error tolerance in actual practice).

One often sees phrases of the form "situation X favors reaction Y".  This means the conditions (X) make the the probability of Y greater than any other possible reaction, but it doesn't equate to Y having probability 1.  Some may prefer to use this phrasing for when Y's probability is, in some sense, much larger than any other event.

The proton-proton chain that governs the main sequence stage of a star is a prime example of why you cannot discount the reactions which aren't favored.  If you did, you would conclude that our own sun is incapable of fusion.  The 'favored' reaction for two interacting protons in a star like our sun is to simply bounce off of each other thanks to Coulomb repulsion.  The protons simply don't have enough energy to overcome this repulsion in order to undergo fusion.  Technically this statement is also not probability 1, but rather very close to it.  But an improbable quantum tunneling event is possible to bypass this potential barrier, allowing the interacting protons to fuse when they collide just right and the dice favor them.  But a diproton is unstable.  The heavily favored reaction in this state is for the diproton to immediately decay into two protons.  But there is a small chance that one of the protons will undergo a beta decay into a neutron, instead, at which point you have a stable helium nucleus.  That these things are so unlikely is a major factor in why our sun needs billions of years to exhaust enough of its hydrogen (aka protons) in order to initiate helium fusion and exit the main stage.  Even very large stars still need millions of years because of this.

So while the proton-proton reaction is rather unlikely, stars are simply so immense that it is not only a virtual certainty that some proton-proton reaction will yield helium, but in fact that it is a virtual certainty that the reactions will occur in large numbers and continuously until the available reactants dwindle.

At the other extreme, we have supernovae.  There are quite a lot of varieties of supernovae, but the commonly understood types--the outer layers collapse and then violently rebound off the core--feature a sudden loss of radiation pressure due to certain endothermic reactions becoming favorable.  Iron production from fusion is the culprit here: it takes more energy to fuse it than is released.  The final split second of this sort of star's life is when conditions favor the production of iron via fusion.  Ridiculously favors it, as I really do mean it is the final split second.  There is an inconceivably tiny chance that this iron production would fail to occur for some chosen period of time, and if it happens to stall long enough then the star would likely undergo some other event.  See also David's answer for another wildly improbable, but technically possible, example of random chance altering a star's death.


Black hole darkness a result of gravity or temporal distortion? Please correct me if I am wrong as I may have made some incorrect assumptions.

Okay so we know that at some stage of "nearness to a black hole", light is no longer reflected back at us from the black hole itself. Some scientists have proposed the speed of light and it's relationship to relativity is natures way of adjusting the rate at which time passes seemingly to disallow "weirdness that could result" .. for example the fast moving object would become more massive.

We also know that the mass of the black hole causes a nearby observer to behave as if he/she were moving at incredible speeds and subsequently slows time for the observer "relatively speaking of course".

I realize that from the perspective of the observer light is traveling to and away from him/her at it's normal rate even though effectively the observer is traveling very fast. Imagine however what would happen to the frequency of the light. The red/blue shift would undoubtedly be obscene.

Since we here on earth are "relatively speaking" experiencing time passing at our normal "earth rate". I can't help but imagine that immense gravity near a black hole is distorting time and its darkness is a result of either extreme red/blue shift or the light is somehow phased or altered by the temporal distortion. Yes gravity is the origin of the darkness but is the direct cause or an indirect relationship where the rate of time passing is the culprit?

As an object falls towards a black hole, light emitted by the object will become red-shifted by the gravity. As photons rise out of the gravity well of the black hole they lose energy, they can't slow down but their wavelength is stretched. The object will go dark as it approaches the event horizon due to gravitational redshifting.

Photons starting from the event horizon would be redshifted to zero energy, and no photons can travel from inside the event horizon.

We also will observe gravitational time dilation, the last second before the object reaches the event horizon will be stretched out indefinitely. The two effects are all part of the same relativistic theory of gravity. You can't unpick the two effects.


Exact ecliptic position of Earth and Moon from Mars as photographed by Curiosity Rover? I've been studying the idea of travelling through space and navigation - in particular: Apollo sextant (used by astronauts to plot the position of their spacecraft relative to stars). And, in order to appreciate the fundamental characteristic of the solar system, in terms of planes and orbits, I've also been reading about the route of the recently discovered interstellar Comet 2I/Borisov, which passed through the ecliptic plane from above and exited below the solar system.
The goal is to visualise how planes (planets) might appear from different parts of the solar system (in terms of Line Of Sight) - parallel to the ecliptic plane and from different compass points... I find the eyes.nasa.gov/apps/ website for Mars really useful for moving through and around the solar system.
Finally, considering that the Moon has a 5° tilt to the Earth, and Mars is on the same ecliptic plane as Earth - where exactly was Mars and Earth and the Moon when this photo was taken by Curiosity Rover as detailed on this space.com article?
I guess Mars was positioned near Earth September Equinox, and Earth positioned just after the Perihelion, and full Moon.
I can't quite understand how the moon seems to be much lower than Earth.
EDIT: answering the general position of the Earth and Moon, which is the first step towards determining the exact.
Interestingly, the light seen from Mars is that reflecting from the surface of the Pacific ocean.
On the question of "exact", there is a difference, obviously, between knowing the exact and general position. And the first knowing is most suited for the purposes of science, navigation, research - whereas the second form of knowing (generally) for the sake of natural curiosity. But, I think there is a great deal to be gained from appreciating both the exact and general position (us and the Earth) in space - I think this can broaden our perspective.
This is (in general) the position of the Earth and Moon as seen from Mars...
Firstly, the mountain range in the background is the same as the range visible in the 2nd image below. We can compare and notice the same peak and shape of the summits.

The Dingo Gap is situated at Gale Crater.
Mars, Earth, Moon, in space...
In general... January 31, 2014 - 10:30pm. This broad view looking downwards at the plane.

And again...

And from beneath the plane... showing us just how far ahead Mars was relative to Earth.

Finally, considering the exact position of the Moon in order to answer the question for why the Moon appears to be positioned beneath Earth.
In fact, the Moon, at the time, was midway through its 5° tilt orbit around Earth. And, from certain angles and from such distance from Mars, could have appeared beneath Earth, as shown in the photo taken by Curiosity.

Because our galaxy is at an angle, it is difficult to determine where planets are relative to one another - consequently, from one perspective, as shown in the above image, Earth appears to be ahead of Mars. The only absolute way to determine the general position (visually) of planets is from beneath or above the ecliptic plane(?).

NASA Photojournal item PIA17936 says:

Researchers used the left eye camera of Curiosity's Mast Camera (Mastcam) to capture this scene about 80 minutes after sunset on the 529th Martian day, or sol, of the rover's work on Mars (Jan. 31, 2014).

Between sols 528 and 532, Where is Curiosity? shows it in a place called Dingo Gap at Martian longitude 137.41°, latitude -4.63°.
At that location on Mars in Stellarium, sunset on 2014-01-31 occurs around 21:00 UT, so the picture was taken around 22:20 UT.
JPL HORIZONS gives these J2000 heliocentric positions at that time:

lon. (l)
lat. (b)
dist. (r)

Mars
168.784°
1.613°
1.662 au

Earth
131.767°
-0.001°
0.985 au

Moon
131.732°
0.010°
0.983 au

and these Mars-centric positions:

lon. (λ)
lat. (β)
dist. (Δ)

Earth
22.935°
-2.535°
1.058 au

Moon
22.826°
-2.522°
1.059 au

In other words, the Moon should appear about 0.11° west of Earth in the Martian sky, or to the right if Mars's north celestial pole is up:

But looking toward the western horizon (zenith up) from a tropical latitude, celestial north is to the right, so the Moon appears below Earth in the photo.


wavelength-based IFU data cubes VS velocity-based IFU data cubes I have 2 simulated IFU data cubes of the same observation, let's say A and B. The 3rd dimension of cube A is in wavelengths. In order to create cube B I just rebinned cube A in log-wavelengths and then converted that to velocity (km/s). The rebinning to log-wavelength was necessary in order to convert the 3rd dimension of cube B in km/s. For an explanation see this:

Converting ångström spectral dimension to galaxy speed (km/s)

Now I want to apply instrumental broadening on both cubes. In order to do that, for each cube, I have to create a Gaussian profile with FWHM equal to the resolution of the instrument (or Line Spread Function, or LSF) and then convolve that with every spectrum of the cube. Of course the LSF of cube A will be in wavelengths while the LSF of cube B in km/s. Here is my question now:

Since all pixel shifts in cube A do not correspond to the same km/s value, one of the two LSFs won't be a perfect Gaussian, right? Shall I assume that would be the velocity LSF?
What is the correct way of creating the velocity LSF? Do I just create a wavelength LSF and then rebin it to km/s, the same way I rebinned cube A to cube B?
Does it worth the effort though? Will it make a difference? I have seen peer reviewed papers using velocity LSFs which are perfect Gaussians. Is this because the difference is negligible or my whole thinking/question is totally wrong?

Your logic seems entirely correct. Whatever you do is an approximation, the difference between a Gaussian profile in cube B and what it "should" look like will be small for any sensible resolution. By sensible, I mean if your wavelength/velocity axis covers many resolution elements.

However, there is a "correct" way. Generate a Gaussian in wavelength space $f(\lambda)$ (if you think that's what your LSD is, where $\lambda$ here is short hand for $\lambda - \lambda_0$, the separation from the central wavelength) and then convert to velocity space in a flux-conserving manner:
$$f(\lambda)\ d\lambda = g(\log \lambda)\ d\log \lambda$$
$$ g(\log \lambda) = \lambda f(\lambda)$$

Maybe a better way to do this is to get some data, like an arc calibration and see how unresolved lines look in your observational space?


What is the cause of the variation from high and low mean obliquity periods of Mars? It is reasonably well known that Mars has a greater obliquity range than Earth, due to Mars lacking a stabilising influence of a large moon.  However, the Martian obliquity seems to have gone through periods of high and low mean obliquity, as shown in the top chart below:

Image source: Mars Climate Modeling Group

What is the current accepted theory as to why there are periodic high and low mean obliquities on Mars?

(Please note, I am not asking about orbital eccentricity)

This figure comes from numerical simulation by Laskar et al. (2002), see Fig. 2. An extended figure from a longer simulation can be found in Laskar et al. (2004), Fig 10.

As you can see more clearly from the extended figure, the simulation doesn't particularly suggest that this is periodic behavior. The question is, then, what happened ~5 million years ago that changed up Mars' obliquity like that?

Laskar mentions the event in the 2002 paper, but doesn't bother to offer any particular physical cause. These are numerical orbital simulations after all, not recorded observations. As unsatisfying as it might be, I think your answer is that evidently the planets aligned, so to speak (albeit probably not in a literal line).

Fig. 8 in the 2004 paper gives a tiny bit of context in terms of uncertainty, it shows that the simulation is only precise for the last 10 million years, and then goes chaotic.

Laskar mentions that his data are free to anyone who requests it, although I can't say how accessible it would be. In theory, someone could go in there and look for correlations (e.g. maybe there was some kind of strong resonance with the Earth and/or Jupiter around that time). I looked through ADS a bit, and found a newer paper by Edvardsson and Karlsson (2008) that pretty much supports Laskar's results in over the past 8 million years, but my brief search didn't turn up anything looking into what exactly caused that anomaly.

Finally, from context it appears that Touma and Wisdom (1993) were the ones to first discuss that obliquity drop, although their paper unfortunately appears to be hiding behind a Science paywall.


Are astronomers waiting to see something in an image from a gravitational lens that they've already seen in an adjacent image? @RobJeffries' answer to the question Does gravitational lensing provide time evolution information? points out that there can be a substantial different in arrival times of light from a given source seen in different images from a gravitational lens.

The linked paper there shows "$\Delta t$" values of the order of 30 days, but it is hard for me to understand what the actual observable is.

What I'm asking for here is (ideally) if there is a well defined event that a lay person could understand, something blinking or disappearing or brightening substantially that has already been seen in one image produced by a gravitational lens that has not yet been seen in one of the other images, and is expected to be seen in the (presumably near) future.

If something like this does not exist, a substitute could be a case where this has happened, and the second sighting of the same event was predicted, waited for, and observed on time.

I have no idea if this happens all the time, or has never happened yet.

What you do is cross-correlate the observational datasets for the multiple sources and look for the "lag" that maximises the cross-correlation function. Generally speaking, the "events" are not really individual flares or dips, but the summation of all the time variability that is seen.

The variability in question usually comes about from the central portions of the "central engine" of a quasar or active galactic nucleus. For a supermassive black hole at the centre of a quasar, the innermost stable circular orbit is at 3 times the Schwarzschild radius ($= 6GM_{\rm BH}/c^2$). This basically defines the inner edge of any accretion disk and if we divide this by $c$ then we get a a timescale for the most rapid variations in luminosity output. So this is very nearly the same formula as presented in the linked question
$$\tau \sim 3\times 10^{-5} \left(\frac{M_{\rm BH}}{M_{\odot}}\right)\ {\rm sec}\, ,$$
except that the supermassive black holes are much less massive than entire foreground lensing galaxies (usually). This the timescale of variation is much shorter than the potential delay time due to gravitational lensing. It is this difference in timescales that means there is plenty of "structure" within the light curves that can be locked onto by the cross-correlation.

There is a notable example however of a type Ia supernova being seen in a multiply lensed image (Goobar et al. (2017), but the predicted delay in the light curves was $<35$ hours and the light curves are not good enough to measure this. This technique is an active area of research and a major bit of science that is exprected to be achieved by the Large Synoptic Survey Telescope (Huber et al. 2019).

Finally, the thing you are really looking for has happened in terms of SN "Refsdal". This was a type II supernova seen to "go off" in a multiply imaged galaxy, seen through/around a galaxy cluster. A prediction was made, based on a model for the cluster gravitational potential, that another image ought to appear within a year or two. This further image was then detected by Kelly et al. (2016) in a paper entitled "Deja vu all over again".

From Kelly et al. (2016) ("Deja vu all over again"). See "SX" in the third panel:

  Figure 1. Coadded WFC3-IR F125W and F160W exposures of the MACS J1149.5+2223 galaxy-cluster field taken with HST. The top panel shows images acquired in 2011 before the SN appeared in S1–S4 or SX. The middle panel displays images taken on 2015 April 20 when the four images forming the Einstein cross are close to maximum brightness, but no flux is evident at the position of SX. The bottom panel shows images taken on 2015 December 11 which reveal the new image SX of SN Refsdal. Images S1–S3 in the Einstein cross configuration remain visible in the 2015 December 11 coadded image (see Kelly et al. 2015a and Rodney et al. 2015b for analysis of the SN light curve).

  Kelly, P. L., Brammer, G., Selsing, J., et al. 2015a, ApJ, submitted
  (arXiv:1512.09093)

  Rodney, S. A., Strolger, L.-G., Kelly, P. L., et al. 2015b, ApJ, in press
  (arXiv:1512.05734)


Why doesn't star size distribution conform to space rock size distribution? The number of space rocks is exponentially related to the size of the rocks. There are more small space rocks than big ones. Stars are most commonly the size of the sun, big and smaller stars are rare, and giant stars are the rarest. Stars that are 0.5 solar masses should also be very bright and visible, but they are less numerous, when they should be more frequent than the sun.
What about planets? There should be many unaccompanied planets wandering in the darkness of space, without a star, and they should be a lot more common than stars themselves, becaues of the size distribution of space rocks and stars should be roughly parallel.
What am I missing?
graph of star size distribtion.
graph of space rock size distribution.

You are comparing distributions in a way that they are not easily comparable and the eye is misled: watch the scaling of your axes of the plots you compare! In order to compare, you want to make sure that you use similar, either log-log for both graphs or linear-linear or something else - but identical in both graphs.
Mind also that the size distribution of the small solar system objects is NOT an exponential, but a power law distribution. A power law distribution is a sloped straight line in a log-log plot and the exponent of the power law is found in the slope in such plot.
The stellar size distribution usually is found under the keyword "initial mass function". And when you look at that (see the plot therein), you will find, that the size distribution of stars also follows a similar power law as does the size distribution of smaller bodies as found in our solar system. This ignores the evolution of stars - but that is what you want to do in order to compare the sizes of objects as they are created.
The slopes of the size distributions course differ as different processes play a role in formation of stars and the objects forming around a star-being-born.


We are in Green Lake, WI, 8/12/18, seeing bright orange “planet”/“star” almost due South, perhaps 20 degrees above horizon, what is it? I have a “sky” app on my ipad and expected it to be Mars, but my app shows Mars just below horizon and closer to East. Oh, I should add that we saw this about 10:15 tonight. Hanging out watching meteor shower, orange object didn’t move much, if at all. I’ve never seen something so orange (or perhaps it is more red).

Not all sky apps get DST right or compute planetary positions accurately.
Stellarium at your location shows Mars 17 degrees above the SSE horizon at 22:15 CDT, or 20 degrees above due south at 23:45.
Skymaps.com northern edition also shows Mars above the horizon before 10 PM in mid-August.
As Mars is unusually close to Earth this summer, it appears brighter than usual--and orange as usual.


What can the average Astronomy Enthusiast do to learn more Maybe this sounds like a silly question (and hopefully is legitimate), but I will elaborate a little more.

I seem to love astronomy, every single thing about it, I find it brilliant, interesting, mysterious, insanely amazing and mind boggling.

There is only so many documentaries you can watch, read articles on the web etc before you start to see repetition.

I just wondered if there was some advice of what I could do to keep that interesting fire burning...because I absolutely love it!!

I recommend that you learn your way around the night sky, so start by buying a book of star charts.  Next try a book that covers interesting objects in each constellation, such as Burnham's Celestial Handbook, or The Constellations by Motz & Nathanson.  This should keep you busy for a while.


Are any bow shocks seen in the visible spectrum? Are any of the images of bow shocks created from visible light, or are they all from infrared light?  While I'm including two of the better known images, my question is about any known bow shock.

LL Orionis

Zeta Ophiuchi

Source of image-Wilipedia

I'm far from an expert in this topic, but I was curious about this question and looked into it myself. Sadly, I was only able to come up with a pretty weak answer.

The very Wikipedia page you got the images from contains this line,

  Bow shocks are also a common feature in Herbig Haro objects, in which a much stronger collimated outflow of gas and dust from the star interacts with the interstellar medium, producing bright bow shocks that are visible at optical wavelengths.

But it is not cited. Perhaps it should have a [citation needed]. That sentence does link to this page about the Herbig Haro Object, which does contain a cited line,

  HH objects associated with very young stars or very massive protostars are often hidden from view at optical wavelengths by the cloud of gas and dust from which they form. The intervening material can diminish the visual magnitude by factors of tens or even hundreds at optical wavelengths. Such deeply embedded objects can only be observed at infrared or radio wavelengths.

So it seems to me the answer to your question is that no, images of bow shocks in visible light are not produced because the visible light from the bow shock cannot reach us due to there being stuff in the way.

I understand that an answer based on 2 Wikipedia pages as sources is pretty weak by Stack Exchange standards, and that these Herbig Haro Objects are not the only thing in the universe that create bow shocks, and therefore the possibility exists that visible bow shocks around other objects may exist. I hope this answer inspires an expert to give an answer deserving of that green check mark.


Can I calculate the apparent position of comet NEOWISE C/2020 F3 from a topographic position with Skyfield? I'd like to get .altaz(next_several_weeks) of this comet from Skyfield and .separation_from() other planets.
It reads the JPL development ephemerides and can import star catalogs, but is there a way to add some orbital elements for a given comet or asteroid from some standard database and at least approximately propagate them to get an astrometric position and and all the other goodies Skyfield can provide from there?
If so, are these the best places to find the appropriate orbital elements for comet NEOWISE C/2020 F3?

https://minorplanetcenter.net/mpec/K20/K20G05.html

this answer

find_orb (from this answer)

From Horizons:

JPL/HORIZONS                 NEOWISE (C/2020 F3)           2020-Jul-09 02:25:23
Rec #:90004465 (+COV) Soln.date: 2020-Jun-16_22:44:15      # obs: 353 (70 days)

IAU76/J2000 helio. ecliptic osc. elements (au, days, deg., period=Julian yrs):

  EPOCH=  2458958.5 ! 2020-Apr-19.0000000 (TDB)    RMSW= n.a.
   EC= .9991762343449366   QR= .2946766901808353   TP= 2459034.1806295365
   OM= 61.00968684540688   W= 37.27670558969492    IN= 128.9375082086831
   A= 357.7190774701175    MA= 359.9889750746684   ADIST= 715.1434782500542
   PER= 6765.8344565066    N= .000145677           ANGMOM= .01320322
   DAN= .32818             DDN= 2.87461            L= 35.4449114
   B= 28.1060796           MOID= .362501           TP= 2020-Jul-03.6806295365

Comet physical (GM= km^3/s^2; RAD= km):
   GM= n.a.                RAD= n.a.
   M1=  10.9     M2=  14.9     k1=  17.25  k2=  5.      PHCOF=  .030

COMET comments
1: soln ref.= JPL#10, data arc: 2020-03-27 to 2020-06-05
2: k1=17.25, k2=5., phase coef.=0.03;

The Skyfield docs don't show a way to import orbital elements directly, but it can read any BSP format ephemeris you have.
If you request a HORIZONS-generated binary SPK file for the comet, that might be enough to do the job.


How and when did we learn about the universe? I'm 13 - I'd like to know how we found/discovered important details about the Universe. I know we have found many different things about the Universe, but when did we find how amazingly large the Universe is?

Throughout man's history, we have been learning things about the universe we live in.

Probably one of the most interesting discoveries that let us know how amazingly large the universe really is was not really very long ago, only in the 1920s when Edwin Hubble demonstrated that the Milky Way was not in fact the total sum of the universe, but just a galaxy, and there were other galaxies beyond our own, much further away than any of the stars that we see in the night sky with our naked eyes. This fundamentally changed our perspective of our place in the universe.

We have subsequently made many more significant discoveries about the nature of the universe, and many of those are covered in other questions already. If there are specific details you would like to know about, raise a specific question. The whole topic cannot be broached in one answer.


Any new info about solar flares hitting the Moon added by the LADEE mission? This article from 2011 mentions simulations about the sputtering effect caused by a solar flare hitting the Moon:

  "We found that when this massive cloud of plasma strikes the moon, it
  acts like a sandblaster and easily removes volatile material from the
  surface," said William Farrell, DREAM team lead at NASA Goddard. "The
  model predicts 100 to 200 tons of lunar material – the equivalent of
  10 dump truck loads – could be stripped off the lunar surface during
  the typical 2-day passage of a CME."

The researchers said they were waiting for LADEE to confirm these simulations and add new data, but I didn't find any conclusions on this subject on the mission website. I was interested if this event can create a significant static electricity difference between the 2 sides of the Moon, causing some sort of lightnings on the edges or even a mini-version of a dust storm.

You'd probably be most interested in the results of the Lunar Dust EXperiment (LDEX).  A 2015 paper states

  LDEX data show no evidence for an electrostaticallylofted
  dust component at densities greater than a few
  per m3

I am assuming the solar flare stuff didn't pan out, otherwise it would be mentioned in the various LDEX summary papers.  In addition, the NASA DREAM group site doesn't seem to have anything relevant LADEE results, and they're the ones that brought the subject up in the first place.


Interstellar movie: What is the "portal" to the other galaxy? I recently watched Interstellar with some friends and we didn't come to the same conclusion. In my opinion, the portal they use to go to the other galaxy (to visit the three planets) is not a black hole. My friends tell me it is.

(warning, contains spoiler!!)

To me, it's not because of these points:

When they are in front of the portal, they can see the other galaxy, which is not the case when Coop goes to the black hole at the end
When they go to that portal, they "flow" in a tube like channel. Again, that's not the case when Coop goes to the black hole at the end.
When Coop goes to that black hole, he goes through some sort of light dust that hits his ship, which is not the case when they go to the portal.
When the robot (I forgot his name) goes in the black hole, his purpose is to collect some quantic data. If the initial portal was also a black hole, he would have made that collection at that time.
When they goes to the portal, they seems "confident", but when Coop goes to that black hole, it's for killing himself. If the portal was a black hole, they would behave more identical.

So I believe it's not a black-hole, but what is it exactly (maybe not in scientific terms, but in Interstellar's terms)? A Wormhole?

Yes, it is a wormhole indeed. This has been indicated quite clearly in the movie as well, when Dr. Romily explains with a pen and paper to Dr. Cooper. The explanation goes like this :

Imagine a sheet of paper to be 2-D space, then a line joining two points on the sheet of paper is the shortest distance possible to reach that point, but if due to some disturbance, the space is bent ( achieved by folding the sheet of paper), you can pierce a hole in the sheet after aligning the two points together. That is a 2-D wormhole (which is a circle indeed). So what happens when you consider a 3-D wormhole? It becomes a sphere.

Now, you can see the other end of a 2-D wormhole (which is effectively the other hole in the sheet visible from the first hole, and one can see beyond the hole in the other direction as well). Same happens with the 3-D spherical wormhole where you can see to the other side of the wormhole as well.

Now, the means to bend spacetime : Space time can be bent by having a very big mass placed inside the space time (read Einstien's General Relativity ). So, it is safe to assume that the wormhole is a space time disturbance created due to a massive object, but it is not a blackhole.


Is the Sun slightly blue in the center? - Wavelength-dependent limb darkening of the Sun I've slightly modified the title to try to attract some attention. If we call sunlight "white" and limb darkening is a result of seeing deeper at normal incidence and shallower at oblique incidence, then the center end edge of the solar disk viewed from earth should appear to have slightly different effective color temperatures, since the temperature is varying rapidly with depth.

I'd like to have an approximate expression for the wavelength-dependent limb darkening of the sun in the visible spectrum, either a relatively simple analytical expression that I can understand (with the appropriate coefficients for the sun) or just some linear images of the sun in various bands in the visible wavelengths so I can try to make one of my own.

This question received this helpful comment which links to here but honestly I can't make my way through that to a practical expression I can use. The Wikipedia article is not helping me much either, except for the image there. I plotted scans of RGB but by the time an image gets into the internet, things like sRGB and gamma mean it may not be linear.

update: At the Solar Dynamics Observatory (SDO) website, I just found the image sdo.gsfc.nasa.gov/assets/img/latest/latest_1024_HMIIC.jpg. The color gradient of the limb darkening seems very similar to the Wikimedia image below. I've discovered that it is called a "colorized intensitygram" and the color gradient is purely artificial - the data is single channel intensity. The limb darkening is certainly real (compare to the artificially "flattented" display!)

From http://www.solarham.net/latest_imagery/hmi1.htm

I appreciate (the existence of) the complexities and intricacies of photon transport, instrumental effects, and color perception, but I am just starting to do astronomically correct animations, so please for right now, something imperfect, or not absolutely correct is good enough for me.

20 pixel wide averages of horizontal (-) and vertical (--) "scans" across the center of this 600x600 pixel image from here

note: It seems this is a somewhat futile example, as the image is likely to be a monochrome continuum image with false color!

You might find this paper helpful:

http://www.physics.hmc.edu/faculty/esin/a101/limbdarkening.pdf

Look specifically at Eq. 5 for the wavelength dependence.  To get the intensity, use Equation 1.  The value of $u$ can be found in the caption to Figure 2.


Why is the asteroid belt shaped like a triangle? So, in this question, the user JollyJoker posted this image depicting the orbits of the asteroids in the asteroid field in the comments:

[

In this image, you can see that while the individual asteroids follow elliptical orbits, the asteroid belt as a whole is a giant triangle with the Jovian Trojans at two corners, and a third corner directly opposite Jupiter, and the whole triangle orbits in sync with Jupiter.

Is this image accurate? Is the Asteroid Belt actually a giant triangle, rather than the loose circle it is often depicted as? If so, why is this the case?

It's not. The image doesn't show the main asteroid belt. It shows the Jovian Trojans (in green) , and the Hilda Asteroids (in red).

The Hildas are a dynamical group of a few thousand known asteroids in elliptical orbits that are locked in a 3:2 orbital resonance with Jupiter, and reach aphelion coinciding with the regions near Sun-Jupiter Lagrange points L4, L3, and L5 in succession.

"Animation showing Hildas with 3 indicated orbits" from "Asteroid (and Comet) Groups", Petr Scheirich, Retrieved 2020/02/14.

In accordance with Kepler's Second Law, objects in elliptical orbits move slower near aphelion than perihelion, and as a result, when you just look at the Hildas, they bunch up near the Lagrange points, and the pattern they appear to show at any one time resembles a triangle.

A more complete view of all the Asteroids inside of Jupiter looks like this.


What is the mass of hotspots in the Cosmic Microwave Background radiation? A lot of pop science articles (for example, this Space.com article) discuss the cold spots on the WMAP data of Cosmic Microwave Background (CMB) radiation, but have any studies been done on the large mass of hot spots on the right of the image below?

Is this a side-effect of our observation methods or does it represent some actual feature of the early history of the Universe?

I don't remember reading any paper title specifically about the hot spots you refer to, so I guess that they are not as "strange" as the cold spot.

Assuming that the temperature fluctuations are well described by a Gaussian distribution with mean 2.72548 and variance 0.00057 source, the probability of having the cold spot of the observed size and temperature is very low. According to a recent Planck paper this probability is of order of 1% or smaller. The probability of the hot spots is larger (1.5 to 5%), making them more "normal".

This probably explains while the cold spot got more attention that the hot ones.

  Is this a side-effect of our observation methods or does it represent some actual feature of the early history of the Universe?

This is tricky. A few scenarios:

The CMB map in the question shows fluctuations of order 10^-4 K. But extracting this signal from observations is very complicated by the fact that on top of the CMB radiation is "polluted" by foreground objects (mostly our own Galaxy, but also extragalactic point and extended sources) that create spurious fluctuations orders of magnitude larger than the CMB ones. This video is great to have an idea of this. If any of the above foreground components is not well understood or some is not accounted for, we could be left with some residuals that we wrongly interpret as CMB anomalies.
The physics we know is the correct one and the cold spot is simply there. But we physicists don't like that in the Universe there are special places
The cold spot is a signal of new/unknown physics or that our standard model is not correct. For instance could be that the assumption that the CMB is Gaussian could be just wrong and we have to re-think our theories about the early universe


Why didn't all the iron and heavier elements find their way to the center of the accretion disc in the early solar system? I guess that's pretty much it. I just heard that that a meteor fragment was found to be solid iron and it just seems inexplicable that it was floating around.

The process of differentiation is how metals like iron and nickel can become separated from less dense substances. It generally occurs in large, partly solid bodies in the early solar system, whose interiors are melted either via accretion energy, radioactivity or some combination of the two. A solid iron meteor is likely a fragment of a differentiated larger body that has been disrupted by a collision.

I think what you are asking is why such a differentiation process does not occur in the primordial disc from which the solar system formed. The answer lies in turbulence and other mixing mechanisms that act against differentiation. If the mixing timescale in the disc is shorter than the differentiation timescale then the differentiation just won't occur. This is for example the same reason that stars like the Sun do not have all their heavy chemical elements at the core.


Could a star revolve around a planet? They actually do, but let's set Relativity aside for a moment. And the whole barycenter thing too. Here's the scenario:

Somewhere in the backwaters of a galaxy there is this huge planet. It is NOT a failed star. It could be considered a gas giant, except most of the gas is nitrogen, with traces of oxygen. The planet has a nucleus with a rocky surface covered in part by water. There's life on the planet, maybe even intelligent life. Orbiting this planet is a dwarf star. It is sufficiently close to the planet to provide energy and warmth, and not close enough to incinerate it. Its orbital speed is sufficient to warm the equatorial parts of the planet without singing them when it passes over them and to come back to them before they freeze over. Or maybe there are two stars of slightly different sizes, traveling along different orbits.

Is this at all possible?

And the whole barycenter thing too.

You're throwing out the definition of orbit at this point, and moving into a different realm. If you throw out barycenter, you have to leave me with my frames of reference.

A star can definitely orbit a planet, if one such as this did in fact exist. You are assuming it is simply a planet-star system, which makes things easier. If you're on the planet, the star is orbiting you, if you're on the star (ouch), the planet is orbiting you.

Once they are of comparable size, the barycenter thing has to be taken into account, you can't just throw out this crucial part of the physics, even in a thought experiment. If you want that part out, the planet must be much bigger than the star.


How does gyrochronology deal with differential rotation and axial tilt? K2 seems to become an even better mission than what a continuation of Kepler would have been, at least to some astronomers. By what means is differential star rotation being dealt with? The rotational period of the Sun is 45% longer at its poles than at its equator. Together with the unknown orientation of the star's rotational axis, can this really be sorted out from Doppler broadening of spectral lines? Is a star's rotational period defined as that of its equator or as some average across its surface?

Gyrochronology uses the rotation periods of stars, caused by rotational modulation by starspots to estimate a stellar age. In the absence of differential rotation, the rotation axis inclination has no effect on this measurement. The doppler broadening of spectral lines plays no role in gyrochronology.

The rotation period of a star is just that. It is the period that is measured from the light curve. So it is some sort of emission-weighted average rotation period that depends on the distribution of starspot latitudes, the stellar inclination and the limb darkening. Sometimes there is evidence of differential rotation because the rotation period of the star changes from epoch to epoch. This can be used to calibrate the uncertainty in an individual rotation period measurement. Differential rotation is a source of uncertainty in gyrochronology, however its effect is limited because (i) at least on the Sun, and maybe other stars too, spots are confined to relatively low latitudes; (ii) spots occur at a range of latitudes over this latitude range. It is likely that younger, faster rotating stars do have spots at higher latitudes, but these stars also appear to have much weaker differential rotation than the Sun ($\Delta P/P =0.05 P^{0.3}$ according to Donahue et al. 1996, where $\Delta P$ characterises the range of measured periods for a typical star of period $P$).

The effect of differential rotation on gyrochronology is discussed at length by Epstein & Pinsonneault (2014) and rather more concisely by Jeffries (2014). It appears that differential rotation probably does set the precision limit of a measured period to be around 10 per cent (though I would argue that it is smaller for younger stars). Because the rotation rate of a star roughly follows the Skumanich-type spin-down law $\Omega \propto t^{-1/2}$, then this leads to an age uncertainty of around 20% in older stars. For younger stars it is the initial spreads in rotation rates which are more important than differential rotation as a source of error.


Is a Wormhole a tunnel? In films such as Interstellar, 2001: A Space Odyssey, and Contact (and also Star Trek Deep Space Nine) wormholes are depicted as tunnels along the edges of which fantastical colours and shapes lie. For example, this is a picture from 2001:

What would passing through a wormhole be really like though?

A wormhole, or Einstein-Rosen bridge, is inherently unstable. It will collapse before anything can traverse it. Keeping it stable enough for even a photon to get through requires negative energy densities. And that requires exotic and suspicious new physics in most cases. Technically there are well-known effects which can produce negative energy densities, such as the Casimir effect. But being able to harness such an effect to stabilize a wormhole is beyond current technology (to say nothing of creating the wormhole in the first place). We have no observational evidence to suggest wormholes exist, especially not on the scales you see in Sci-Fi.

So in the rapidly collapsing case it looks like nothing except possibly a spray of subatomic particles (which you might not even know aren't the usual background ones), and in the stable case it's impossible to say because you'd have to know what the exotic matter or other negative densities look like.

As for what it's like to traverse a stable wormhole, as long as the exotic matter doesn't mess with you then it would be pretty much the same as any other spaceship travel. You could just have interesting time travel effects and beat light rays that went around the wormhole. A number of these are mentioned on the wiki page.


Do stars have "radio photospheres"? Are they different from their optical photospheres? Comments below What is the maximum distance measurable with parallax? discuss challenges associated with parallax measurements of Betelgeuse and link to Wikipedia’s Betelgeuse; Distance measurements which contains the intriguing paragraph:

In 2008, using the Very Large Array (VLA), produced a radio solution of 5.07±1.10 mas, equaling a distance of 197±45 pc or 643±146 ly.83 As the researcher, Harper, points out: "The revised Hipparcos parallax leads to a larger distance (152±20 pc) than the original; however, the astrometric solution still requires a significant cosmic noise of 2.4 mas. Given these results it is clear that the Hipparcos data still contain systematic errors of unknown origin." Although the radio data also have systematic errors, the Harper solution combines the datasets in the hope of mitigating such errors.83 An updated result from further observations with ALMA and e-Merlin gives a parallax of 4.51±0.8 mas and a distance of 222 (+34/−48) pc.[10] Further observations have resulted in a slightly revised parallax of 4.51±0.80.10
83Harper et al. (2008) A New VLA-Hoppocaros Distance to Betelgeuse and its Implications
10Harper et al. (2017)A Updated 2017 Astrometric Solution for Betelgeuse

When radio astrometry is used to measure positions of Betelgeuse in an effort to determine its parallax, I am guessing (see below) that it is the thermal black body radiation from the star's "radio photosphere" rather than maser radiation from a cloud well outside of the star itself.
(This answer to How far have individual stars been seen by radio telescopes? discusses maser radiation from stars for example)
Harper et al. (2017) contains:

For the purpose of this work, the channels containing line emission were excluded from the analysis and a single continuum data set centered at ≃338 GHz with a ∼5.9 GHz bandwidth was used.

and Harper et al. (2008) mentions:

We have used the highest spatial resolutions available with the VLA, i.e., A-configuration with the Pie Town VLBA antenna, and these allow us to obtain positional uncertainties comparable to Hipparcos. Good u–v coverage was obtained for six frequency bands (Q, K, U, X, C, L)7 at five epochs. For each band we used two 50 MHz continuum channels recording full Stokes polarizations.
7These bands have nominal wavelengths: Q = 0.7 cm, K = 1.3 cm, U = 2.0 cm, X = 3.6 cm, C = 6 cm, and L = 20 cm.

Questions:

Would these then be from the star's "radio photosphere"?
Is there such a thing as a "radio photosphere" that differs substantially from a star's optical photosphere?

Stars certainly do have radio emission from their photospheres, since blackbodies emit at all wavelengths. But that’s not usually what is detected, because it’s so faint.  Doing some simple numbers, emission from a star like the Sun peaks at a wavelength of about 500 nm (= $5 \times 10^{-7}$ m), and the long-wavelength side of the Planck function, in the Rayleigh-Jeans approximation, goes as $\lambda^{-4}$, so at a wavelength of 5 cm (= $5 \times 10^{-2}$ m) the emission will be $10^{20}$ times fainter.  With nearby or very luminous stars it is possible to detect this in some cases (e.g. with ALMA which is both very sensitive at operates at somewhat shorter wavelengths), but more typically the radio emission detected is from other processes like synchrotron emission.
Whether the radio photosphere is at a substantially different radius depends on the opacity at that wavelength, but Reid & Menten (1996) Radio Photospheres of Red Giant Stars suggests that for some giants it may be a lot bigger than the optical photosphere.
From the abstract:

[...]These observations suggest that long period variables have a “radio photosphere” at about two stellar radii, where the stellar radius is defined by line-free regions at optical wavelengths. The radio photosphere is just inside the SiO maser shell, and the limited variability of the radio emission suggests that stellar shocks are mostly damped within this region. The density and temperature of the radio photosphere are estimated to be >~ 1012 cm-3 and $\approx$ 1400 K, respectively. For these physical conditions, free electrons, obtained predominantly from the ionization of potassium and sodium atoms, provide the opacity through interactions with neutral H atoms and H2 molecules.


Ground Induced Currents on Mars Since Mars is only weakly magnetized does Mars not suffer from  ground induced currents or are they far worse than on Earth?

There is no net internal magnetic field to perturb, so how could one induce currents?

The effect arises on Earth because some transient event (e.g., coronal mass ejection or CME) hits Earth's magnetosphere and causes a sudden $\tfrac{dB}{dt}$ which, from Faraday's law, results in induced electric fields.  These induced electric fields then drive currents in things like power grids (typically long-distance wires running east-west at high latitudes), oil pipelines, etc.

The only changing magnetic field would be that of the solar wind and the small, localized magnetic anomalies on the surface of Mars.  However, these fields are incredibly small so the $\tfrac{dB}{dt}$ would be even smaller.


What does black hole evaporation correspond to in the accelerating universe / black hole analogy? In the same way, as a black hole emits Hawking radiation corresponding to its temperature

$$
T = \frac{\hbar}{r_{BH}}
$$

the accelerating universe can be described as all of the stuff falling towards the cosmic horizon, such that it is analogue to a black hole where the cosmic horizon takes the role of the event horizon with the black hole "interior" being outside.

The corresponding temperature of the accelerating universe is given by

$$
T = \hbar H = \frac{\hbar}{R_{H}}
$$

where H is the Hubble constant. Therefore, the cosmic horizon emits something similar to Hawking radiation too.

The fate of a black hole is to finally evaporate. My question is: What does black hole evaporation correspond to in the accelerating universe when making use of this analogy?

An analogy in the settings of the question would be a deceleration down to an expansion rate, such that eventually light from anywhere in the universe can reach us, meaning the cosmic horizon can vanish. A constant expansion rate of the universe above the speed of light would be the analogy of a black hole of fixed mass.

In an infinite universe this would mean, that its expansion needs to slow down to an asymptotic speed for infinite distance below the speed of light. For a finite 3-spherical universe for a fixed cosmic time a slow-down to below the speed of light would be sufficient. This would correspond to a massive follow-up object of the black hole without event horizon.

For an analogy to a completely evaporating black hole the expansion rate needs to slow down to zero. The slow down between speed of light and zero corresponds to the very last phase of black hole evaporation, when its mass drops below the Planck mass, then without event horizon.

The analogy to Hawking radiation would be radiation or matter emerging from behind the former cosmic horizon.

Thinking in stereographic projections may help for a better understanding: Project infinity of a plane expanding universe to the singularity of a black hole at the north pole of an expanding sphere, as an initial point of further reasoning.


What if Earth and Moon revolved around each other like Pluto and Charon? What would be different for us if Earth and Moon revolved around each other like Pluto and Charon do?

They do, but due to the ratio of masses being vastly different, they seem like they would not to do so as moon seems to rotate just around (the centre of) Earth.

The ratio of Earth and Moon's masses is $\frac{M_{Earth}}{M_{Moon}} = 81.3$ whereas for Pluto and Charon the same ratio is $\frac{M_{Pluto}}{M_{Charon}} = 8.09$.

Because the ratio for Pluto and Charon is relatively small, the center of the system – barycenter, around which the two bodies orbit – is somewhere on a line drawn between the two celestial bodies's mass centers. But for Earth and Moon, because Earth is proportionally much heavier, the system's barycenter does not reach outside of Earth, but instead is about 4,500 kilometers from the center of Earth (see the photo below, too):

  In cases where one of the two objects is considerably more massive than the other (and relatively close), the barycenter will typically be located within the more massive object. Rather than appearing to orbit a common center of mass with the smaller body, the larger will simply be seen to "wobble" slightly. This is the case for the Earth–Moon system, where the barycenter is located on average 4,671 km from the Earth's center, well within the planet's radius of 6,378 km. Source: Wikipedia - Barycenter

The major effect of this co-rotational system is that Earth seems to "wobble" on its orbit, as mentioned in the quote from Wikipedia above.

@JeppeStigNielsen makes a good point about differences in tidal locking in the comments below. In the Earth-Moon system, only Moon is tidally locked (which causes us to see only one face of it, so only roughly a half of it, from Earth), whereas in Pluto-Charon both of the bodies are tidally locked. Earth is not tidally locked because of the higher mass ratio between it and Moon, but the lower mass ratio Pluto-Charon system is, as the lower mass Charon has slowly changed Pluto's rotation to match its orbital movement.


Deriving the first moment of Collisionless Boltzmann Equation in Spherical Polar Coordinates I am following these notes: Dynamics and Astrophysics of Galaxies.
After equation 6.37, we have:
\begin{equation*}
    p_r\,\frac{\partial f}{\partial r} + \frac{p_\theta}{r^2}\,\frac{\partial f}{\partial \theta} +
    \frac{p_\phi}{r^2\,\sin^2\theta}\,\frac{\partial f}{\partial \phi}
    -\left(\frac{\mathrm{d} \Phi}{\mathrm{d} r}-\frac{p_\theta^2}{r^3}-\frac{p_\phi^2}{r^3\,\sin^2\theta}\right)\,\frac{\partial f}{\partial p_r}
    +\frac{p_\phi^2\,\cos\theta}{r^2\,\sin^3\theta}\,\frac{\partial f}{\partial p_\theta} = 0\,.\\
\end{equation*}
This is the Collisionless Boltzmann Equation in Spherical Polar Coordinates.
Then

We now multiply this by $p_r$ and integrate over all $(p_r,p_{\phi},p_{\theta})$ using
that $\mathrm{d}p_r\,\mathrm{d}p_\phi\,\mathrm{d}p_\theta = r^2\,\sin\theta\,\mathrm{d}v_r\, \mathrm{d}v_\phi\,\mathrm{d}v_\theta$ and using partial integration to deal
with the derivatives of f with respect to the momenta

\begin{align}\label{eq-spher-jeans-penult}
    \frac{\partial (r^2\,\sin\theta\,\nu\,\overline{v^2_r})}{\partial r} + \frac{\partial (\sin\theta\,\nu\,\overline{v_r\,v_\theta})}{\partial \theta} & +
    \frac{\partial (\nu\,\overline{v_r\,v_\phi}/\sin\theta)}{\partial \phi}\\
    &
    +r^2\,\sin\theta\,\nu\,\left(\frac{\mathrm{d} \Phi}{\mathrm{d} r}-\frac{\overline{v_\theta^2}}{r}-\frac{\overline{v_\phi^2}}{r}\right) = 0\nonumber\,.
\end{align}
I would like to arrive to this result myself. Multiplying the SPC CBE by $p_r$ & going ahead with the suggested volume element, considering the first term in the above equation, not being worried about the integration limits gives us:
\begin{equation}
\int p_r \frac{\partial f}{\partial r} p_r r^2 \sin\theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi}
=\int v_r^2 \frac{\partial f}{\partial r} r^2 \sin \theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi}
=r^2 \sin \theta \int v_r^2 \frac{\partial f}{\partial r} \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi}
\end{equation}
Using 6.32 from the mentioned notes,
\begin{equation}
\frac{\partial (r^2\,\sin\theta\,\nu\,\overline{v^2_r})}{\partial r} = \frac{\partial}{\partial r} \left( r^2 \sin\theta \int v^2_r f \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi} \right)
\end{equation}
Which is not equal to what I have found just a line above.
What am I doing wrong?

Some question on Math.SE & Physics.SE.

After correspondence with the author of those notes, I arrived at the following conclusion.
When we do $\frac{\partial f}{\partial r}$, we require that all other (other than $r$) variables which f is dependent on, are held constant. That is: $\theta, \phi, p_r, p_{\theta}, p_{\phi}$ are constants.
Note that:
$(p_r, p_{\theta}, p_{\phi})=(\dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta})$
Keep in mind that:

$p_r$, ie $\dot{r}$ is held constant: it does not mean $r$ is held constant (in fact, we are differentiating with respect to $r$).

$p_{\theta}$, ie $r^2\sin^2\theta\dot{\phi}$ is held constant. This does not mean that either $r$, $\theta$ or $\dot{\phi}$ is held constant.

$p_{\phi}$, ie $r^2\dot{\theta}$ is held constant. We are not claiming this about $r$ or $\dot{\theta}$.

Writing the steps emphasizing this statement above:
$$\int d\vec{p} p_r \frac{\partial f}{\partial r}  p_r \Bigr|_{\theta, \phi, p_r, p_{\theta}, p_{\phi}}$$
Writing out explicitly what we mean by $d\vec{p}$:
$$=\int \mathrm{d}(p_r,p_{\theta},p_{\phi}) p_r \frac{\partial f}{\partial r}  p_r \Bigr|_{\theta, \phi, p_r, p_{\theta}, p_{\phi}}$$
We can move out the partial differentiation because all other term than $\frac{\partial f}{\partial r}$ is held constant: [THIS IS THE MAIN STEP]
$$=\frac{\partial}{\partial r}\int \mathrm{d}(p_r,p_{\theta},p_{\phi}) p_r^2 f \Bigr|_{\theta, \phi, p_r, p_{\theta}, p_{\phi}}$$
Changing from the $p$-nomenclature to the variables they represent:
$$\frac{\partial}{\partial r} \int \mathrm{d}(\dot{r}, r^2\dot{\theta}, r^2\sin^2\theta\dot{\phi})  v_r^2 f \Bigr|_{\theta, \phi, \dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta}}$$
ie
$$\frac{\partial}{\partial r} \int \mathrm{d}(v_r, r \sin\theta v_{\phi},  r v_{\theta}) v_r^2 f \Bigr|_{\theta, \phi, \dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta}}$$
This step I am unsure about (ie why I don't have a $dr$ and $d\theta$ term, for example), but we say:
$$\mathrm{d}(v_r, r \sin\theta v_{\phi},  r v_{\theta}) = r^2 \sin\theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi}$$
Rewrite integral
$$\frac{\partial}{\partial r} \int r^2 \sin\theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi} v_r^2 f$$
$$=\frac{\partial}{\partial r} r^2 \sin\theta \int \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi} v_r^2 f$$
$$=\frac{\partial}{\partial r} \left( r^2 \sin\theta \nu \bar{v_r^2} \right)$$
As claimed by the notes.

Same answer on Math.SE & Physics.SE.


Can the Milky Way be seen with the naked eye? Does this apply to any galaxy? If yes, then how and when? I have seen different pictures in a different community to showcasing the Milky Way. I can never tell whether it's edited or it's the real picture. Actually, I can never see any galaxy as such by with my naked eye. But can the Milky Way be seen with the naked eye? If yes, then I would really like to know when and how this is possible.

There are a few nearby galaxies that can be seen with the naked eye, but clear & dark skies are essential. Most notable for northern-hemisphere viewers is the Andromeda Galaxy also known as M31, best viewed in late summer and fall; from the southern hemisphere the Magellanic Clouds, irregular satellites of our own Galaxy, are famous landmarks. These all look like small fuzzy patches, rather unexciting until you consider that they‘re on the order of a million light-years away.