How large can a ball of water be without fusion starting? How large can a ball of water be without fusion starting?

Peculiar question: some explanation might be necessary. My young son is into ‘space’ and astronomy. One of his posters says that Saturn could float, if a sufficiently large ocean could be found. Obviously that wouldn’t work: Saturn’s atmosphere would peel off and join or become the atmosphere of the larger body, and then Saturn’s dense core would sink.

But could such an ocean even exist without fusion starting?

You really need a full-blown stellar evolution model to answer this precisely and I'm not sure anyone would ever have done this with an oxygen-dominated star.
To zeroth order the answer will be the similar to a metal-rich star - i.e. about 0.075 times the mass of the Sun. Any less than this and the brown dwarf (for that is what we call a star that never gets hot enough at its centre to initiate significant fusion) can be supported by electron degeneracy pressure.
A star/brown dwarf with the composition you suggest would be a bit different.  The composition would be thoroughly and homogeneously mixed by convection. Note that other than a thin layer near the surface, the water would be completely dissociated and the hydrogen and oxygen atoms completely ionised. Hence the density of protons in the core would be lower for the same mass density than in a "normal star".  However, the temperature dependence is so steep I think this would be a minor factor and nuclear fusion would be significant at a similar temperature.
Of much greater importance is that there would be fewer electrons and fewer particles at the same density. This decreases both the electron degeneracy pressure and normal gas pressure at a given mass density.
The star is therefore able to contract to much smaller radii before degeneracy pressure becomes important and can thus reach higher temperatures for the same mass as a result.
For that reason I think that the minimum mass for hydrogen fusion of a "water star" would be smaller than for a star made mainly of hydrogen.
But how much smaller? Back-of-the-envelope time!
Use the virial theorem to get a relationship between perfect gas pressure and the temperature, mass and radius of a star. Let gravitational potential energy be $\Omega$, then the virial theorem says
$$ \Omega = -3 \int P \ dV$$
If we only have a perfect gas then $P = \rho kT/\mu m_u$, where $T$ is the temperature, $\rho$ the mass density, $m_u$ an atomic mass unit and $\mu$ the average number of mass units per particle in the gas.
Assuming a constant density star (back of the envelope) then $dV = dM/\rho$, where $dM$ is a mass shell and $\Omega = -3GM^2/5R$, where $R$ is the "stellar" radius. Thus
$$\frac{GM^2}{5R} = \frac{kT}{\mu m_u} \int dM$$
$$ T = \frac{GM \mu m_u}{5k R}$$
and so the central temperature $T \propto \mu MR^{-1}$.
Now what we do is say that the star contracts until at this temperature, the phase space occupied by its electrons is $\sim h^3$ and electron degeneracy becomes important.
A standard treatment of this is to say that the physical volume occupied by an electron is $1/n_e$, where $n_e$ is the electron number density and that the momentum volume occupied is $\sim (6m_e kT)^{3/2}$. The electron number density is related to the mass density by $n_e = \rho /\mu_e m_u$, where $\mu_e$ is the number of mass units per electron. For ionised hydrogen $\mu_e=1$, but for oxygen $\mu_e=2$ (all the gas would be ionised near the temperatures for nuclear fusion). The average density $\rho = 3M/4\pi R^3$.
Putting these things together we get
$$h^3 = \frac{ (6m_e kT)^{3/2}}{n_e} = \frac{4\pi \mu_e}{3}\left(\frac{6 \mu}{5}\right)^{3/2} (Gm_e R)^{3/2} m_u^{5/2} M^{1/2}$$
Thus the radius to which the star contracts in order for degeneracy pressure to be important is
$$ R \propto \mu_e^{-2/3} \mu^{-1} M^{-1/3}$$
If we now substitute this into the expression for central temperature, we find
$$ T \propto \mu M \mu_e^{2/3} \mu M^{1/3} \propto \mu^2 \mu_e^{2/3} M^{4/3}$$
Finally, if we argue that the temperature for fusion is the same in a "normal" star and our "water star", then the mass at which fusion will occur is given by the proportionality
$$ M \propto \mu^{-3/2} \mu_e^{-1/2}$$ .
For a normal star with a hydrogen/helium mass ratio of 75:25, then $\mu \simeq 16/27$ and $\mu_e \simeq 8/7$.  For a "water star",  $\mu = 18/11$ and $\mu_e= 9/5$. Thus if the former set of parameters leads to a minimum mass for fusion of $0.075 M_{\odot}$, then by increasing $\mu$ and $\mu_e$ this becomes smaller by the appropriate factor $(18\times 27/11\times 16)^{-3/2} (9\times 7/5\times 8)^{-1/2} = 0.173$.
Thus a water star would undergo H fusion at $0.013 M_{\odot}$ or about 13 times the mass of Jupiter!
NB This only deals with hydrogen fusion. The small amount of deuterium would fuse at lower temperatures. A similar analysis would give a minimum mass for this to occur of about 3 Jupiter masses.


Why can't different regions of the universe be the same as they have the same origin? The horizon problem states that because of the speed of information can't be faster than light and the distances between far regions are too big there is a problem because they look quite the same. Therefor, the inflation is introduced.
But what I don't understand is why those regions can't be the same without inflation or something like that? They have the same origin and the universe was in the beginning too empty to make a difference for the direction you went. So in my simple understanding, it is rather expected that they should be the same but why are cosmologists think different about that?

We don't really know what happened at the time $t=0$. It doesn't really make sense to say that everything "touched each other" at $t=0$, particularly if the Universe is infinite. But we can calculate the distance that a photon — and hence the maximum distance that any information — can travel in the expanding Universe in a given time. This calculations depends only on the Hubble parameter, and the densities of the various components of the Universe, and whether you start your calculation at $t=0$ or $t=10^{-12}\,\mathrm{s}$ makes little difference after a second.

The answer is related to your previous question about the cosmic event horizon which was also about the particle horizon, i.e. the distance that light has been able to travel since the Big Bang. If you were present when the CMB was emitted, you could also calculate your particle horizon. It would of course be much smaller than today, since light at this time had only traveled for 380,000 years. If the Universe hadn't been expanding, this horizon would of course be 380,000 lightyears, but due to the expansion, it is quite a lot larger, roughly 850,000 lightyears.

This means that, at the time of the CMB, regions farther apart than 850 kly had not had the chance to be in causal contact.

We can also calculate how large an angle 850 kly at the CMB would suspend if observed today. It turns out to be 1.7º. That is, if the Universe had just been expanding like we observe its expansion today — i.e. depending only on the densities of the known components — patches on the CMB map separated by an angle $\theta > 1.7^\circ$ shouldn't look the same.

But they do.

Inflation solves this problem by saying that the Universe initially went through a much, much faster expansion, such that regions much, much farther apart than the 850 kly have been in causal contact. Still, even with inflation, the particle horizon is not infinite. So on scales much, much larger than the observable Universe, it may be inhomogeneous.


Transform asteroid rotation to heliocentric ecliptic coordinates I'm working with data from the DAMIT database of asteroid shape models.  I'm adding them to a visualization in which the sun is at [0, 0, 0] and the X, Y axes constitute the ecliptic plane of the solar system.

Each asteroid model comes with some attributes that define its orientation and spin:

λ (ecliptic longitude),
β (ecliptic latitude)
P (sidereal rotation period)
φ0 (initial rotation angle)
JD0 (initial date).

I've applied the matrix formulas suggested by the folks at DAMIT:

in which

This is where I get confused.  My understanding is that my visualization uses a heliocentric ecliptic coordinate system.

I set r_ast to the XYZ location of a vertex in my visualization.  I'm not sure this is correct.

Then I compute r_ecl from the equation above.  But I think variables λ and β are angles in a geocentric ecliptic coordinate system and I need to do some transformation into heliocentric ecliptic coordinates.

Is my approach correct and if so, how can I convert the asteroid's rotation to my visualization's heliocentric ecliptic coordinate system?

r_ast is a vector that goes from the center of the asteroid to each vertex of the shape model as if you had stuck a set of rods in the center of a potato and pointing along the long axis, to the pole on top and 90 degrees to these to give X,Y,Z directions.

The matrices above that you give then transform these r_ast vectors for the co-ordinates of the surface of the asteroid into the ecliptic plane as a function of time as the asteroid rotates, giving you r_ecl. It says in the Kaasalainen paper (I think) that r_ecl has an origin shifted to the center of the asteroid. If you are wanting a visualization of what the asteroid looks like relative to the Earth or Sun etc, then you will need to calculate the Sun->asteroid vector and the Sun->Earth vector using e.g. JPL HORIZONS and using the Vector Table ephemeris type or a programmatic wrapper to this (e.g. astroquery) or the NAIF SPICE toolkit and getting or generating a SPK kernel file for the asteroid (the normal e.g. de430.bsp JPL planetary ephemeris will handle the Sun->Earth part)


What are the conditions for an ionosphere to form? What are the conditions for an ionosphere to form on a given planet?

Factors I could think of are

Class and age of the star the planet orbits
Distance to said star
Density and composition of the planet's atmosphere

According to the paper "Simultaneous ionospheric variability on Earth and Mars" (Mendillo et al. 2003), the overall structure of any ionised region of a planetary atmosphere depends on

  a blend of in situ production and loss
  processes, plus effects of transport of ionization into or out
  of the local region of interest

Specifically, according to Mendillo et al., the only significant influence is from the photon flux from the parent star - so, the further the planet is from the star, the less the incoming photon flux.  The other mechanisms are suggested to be planetary-based, including (from the article):

planetary rotation rate
orbital obliquity

which both also affect the photon flux reaching that particular part of the planetary atmosphere. Other mechanisms are to do with the planetary atmospheres themselves, with factors including:

the thermal structure, chemistry and dynamics of the planetary 'neutral' atmosphere itself.
by how much energetic particles (from the stellar wind and/or magnetosphere) affects the planetary atmosphere.
diffusion and electrodynamics from coupling from above the layer; and tides, waves and electrodynamics from coupling with below the layer.

A further study in the paper "Ionospheric photoelectrons: Comparing Venus, Earth, Mars and Titan" (Coates et al. 2011) compared the sunlit portion of planetary ionospheres, which is sustained by photoionization of primarily neutral atmospheric constituents by solar EUV (Extreme UV), a process originally identified for Earth, but recently:

  the Mars Express, Venus Express and Cassini-Huygens missions have revealed the importance of this process at Mars, Venus and Titan, respectively.

It should be noted, that for planets without magnetic fields (Venus and Mars for example), the Max Planck Institute for Solar System Research page "Research about the planetary plasma environment",

  The ionosphere is assumed to display two distinct states, "magnetized" and "unmagnetized", depending on the penetration depth of the solar wind magnetic field.

One more factor is suggested - dust, theorised from observations reported in the paper "On the role of dust in the lunar ionosphere" (Stubbs et al. 2011). According to the NASA page summarising the paper "Mystery of the Lunar Ionosphere", ubiquitous grains of dust from the Lunar surface may be the source of the Lunar ionosphere.  The research from Stubbs et al. is summarised as:

  UV rays from the sun hit the grains and ionize them.   According to their calculations, this process produces enough charge (positive grains surrounded by negative electrons) to create the observed ionosphere.

The process with the lunar ionosphere is illustrated below (from the NASA page linked):


does a minority of all planetary systems show transits? do most or only a minority of all planetary systems to show transits?

I feel like its minority due to the random orientation of their orbits?

But the thing is I seem to find conflicting information.

Could someone explain which answer is correct because now I am confused?

Consider a circular orbit and approximate the planet as a point.

The distance from Earth to a star is so large compared to the diameter of the star, which we'll call d, that we approximate the lines joining the two poles of the star with the observer as parallel to each other and to the x-axis that we choose. Thanks to symmetry, we choose our coordinates so that the plane of the planet's orbit passes through the tangent plane of the celestial sphere at the star in two places on the y-axis. By symmetry again we ignore half the cases for rotation and also ignore the direction of the orbit and only have to consider how the orbit is inclined at between 0 and pi/2 radians from the x-axis (Edit for clarity: in the XZ plane). Call that angle a.

When the radius r of the planet's orbit is such that d/r = sin(a), the planet will just barely touch the limb of the star when it is closest to the Earth. If sin(a) > d/r it will never happen and if sin(a) < d/r there will be a transit of finite length. Since there is no preference for the value of a, we conclude that the probability of the planet transiting must be $$P(sin(a) < d/r) = P{ \left( a < {arcsin \left( { d \over r } \right)  } \right) } = { 2 arcsin \left( { d \over r } \right) \over \pi } $$

You will notice that for a star-grazing planet, the probability of a transit approaches certainty. Overall probability will depend on the probability distribution of semi-major axes and possibly eccentricity which I ignored here.


How do stars change in size over their main sequence lifetime? I know that main sequence stars increase in temperature and luminosity as they age. However, I do not know about how they change in size. Do they expand? If so, why do they expand? Do they grow very much? I’ve tried to search up the answer, but details were limited, so I would appreciate an answer here.

Very roughly.
The star is in hydrostatic equilibrium so
$$ \frac{dP}{dr} = -\rho g$$.
Replacing $dP/dr$ by $P_c/R$, where $P_c$ is the central pressure and $R$ the radius, and letting $\rho \propto M/R^3$ and $g \propto M/R^2$, then we get
$$ P_c\propto \frac{M^2}{R^4}$$
But the central pressure is proportional to $\rho T/\mu$, where $T$ is the interior temperature and $\mu$ is the mean molecular mass. So
$$ \frac{\rho T}{\mu} \propto \frac{M^2}{R^4}$$
$$ \frac{MT}{R^3 \mu} \propto \frac{M^2}{R^4}$$
$$ R \propto \frac{\mu M}{T}$$
Now during the main sequence lifetime, the mass stays (roughly) constant and $T$ stays (roughly) constant because the temperature sensitivity of hydrogen burning is high.
However, $\mu$ increases because 4 hydrogens (4 protons + 4 electrons) are getting turned into 1 helium (1 nucleus + 2 electrons). So $\mu$ increases (in the central regions) from 0.5 to 4/3. Hence the radius increases.
The radius increase is indeed about a factor of 2 over the whole main sequence lifetime.
A different way to think about it, more appropriate for solar mass stars and above, is in terms of the radiative energy transport. The increase in $\mu$ is also accompanied by a decrease in the number of free electrons per mass unit. This reduces the opacity in the central regions and decreases the temperature gradient. This means the star is bigger for roughly the same central temperature.


Who is considered as the father of modern astronomy? Googling is giving me both Galileo Galilei and Nicolas Copernicus as the answer. I am sorry and will remove this question if it is an opinion based question.

I would say Copernicus. He did his work maybe 100 years earlier and formulated the heliocentric view of the universe.


Do massive OB stars accrete mass the same way as the black holes they turn into do? It is often said, in spite of sci fi often having illustrated the opposite, that stellar mass black holes do not gravitationally attract material in their environment into themselves anymore than a star with similar mass does. But how and why do blue stars of about 8-10 or more Solar masses which turn into black holes, change how frequently and observably they accrete material from their environment when they transform into BH? Maybe because the supernova ejects material, some of which doesn't get far and is re-accreted when the stellar wind ceases?

BH's are visible when they have active accretion disks. Do massive O and B stars also have such accretion disks as frequently?

Many, if not all, stars will, in their birth phase, have an accretion disk, or circumstellar disk, around them, formed of the material from which the star forms. This disk dissipates in a few million years, both due to the material accreting onto the star, due to material being blown away by radiation pressure, possibly assisted by dust grains, and due to depletion of the material to rocks and planets. Friction in the disk causes it to heat up and be observable in the infrared. After this phase, the star usually won't have an accretion disk. That is, until they die.

As you suggest, a star that later on goes supernova can accrete matter from its "reverse shock", i.e. matter that is falling back due to the shockwave hitting the surrounding interstellar medium. More often, though (I think), will the accretion disk be made of material from a companion star: If one star of a binary system turns into a black hole, and then the other star starts to evolve into a red giant, threspassing the Roche limit, it will transfer matter to the black hole.

As you say, the black hole's mass isn't stronger than the original star's mass, and thus the gravity at a given distance is also not stronger (in fact it's smaller, since it has blown out most of its mass in the explosion). However, due to its much smaller size, the accreting matter can come much closer before being sucked in. Coming much closer means being in a much steeper gravitational potential and reaching much higher speeds, i.e. being heated to much higher temperatures. Hence, black hole accretion disks can be detected in X-rays


How do they keep lights from guided lasers in adaptive optics from interfering with the image? From amateur searching of the info, I can find no mention of the configuration of the lasers when they work. From these pictures, I imagine that those 4 lasers are just outside of the 4 corners of the telescope field of view. Since the lasers leave such strong reflections on the equipment, how do they keep those lights from messing with the result images? Also I think the artificial stars may somehow affect the final picture too, even if they're not in the FoV, simply because they're so bright.
Maybe the scientists do it by simply filtering out the wavelength of the laser in their results? But that way they won't be able to see any sodium lights in the faraway stars, either.

The laser guide stars are indeed outside the science field of view. In the setup for the Wide Field Mode of the MUSE instrument, there is a mirror with a large hole in the middle: the light for the instrument goes through the hole, while the light from the adjacent part of the sky, where the laser guide stars appear (and also the natural star that's used as the "tip-tilt" reference in the Wide Field Mode) is reflected off to the side to the adaptive-optics system ("GALACSI").
Here's a figure (from the MUSE Instrument Description web page) showing the possible positions of laser-guide stars (LGS), and the tip-tilt star (TTS) for MUSE's Wide Field Mode (the TTS can be any reasonably bright star within the GALACSI field; it doesn't have to be in the upper right). The green square is the science field of view; the gray annulus is the part that gets reflected to the adaptive-optics system:

The alternate Narrow Field Mode allows for proper AO correction down to (in principle) the diffraction limit of the telescope. In this case, all the light goes through the central aperture, and then a special dichroic mirror reflects the laser wavelengths to part of GALACSI, while another dichroic reflects all the near-infrared light to a different part of GALACSI, where the TTS star's light (here called the on-axis guide star = OGS) is analyzed; the parts of the optical light not reflected by the sodium-wavelengths dichroic go to the actual science instrument.

(See here for more details.)
But, yes, scattered light from the laser guide stars is a concern. I believe that the Narrow Field Mode deals with thus using the aforementioned sodium-wavelength dichroic; for the Wide Field Mode, there is a narrow-band filter in the science instrument to block light from the laser. Both approaches do exclude light from astronomical sources with the same wavelengths, so you have to plan your observing accordingly. (E.g., if you really want to observe those wavelengths, then you can't use the laser guide stars.)
The figure below (from the MUSE user manual) shows the instrument throughput in the Wide Field Mode as a function of wavelength. The blue and red curves show sub-modes using the laser guide stars ("AO" = adaptive optics), and the fact that they go to zero for wavelengths of $\sim 5700-6000$ Angstroms is due to the filter being used for those modes.


Fitting PSF to an image to get total flux (Python) I have an image of a single unsaturated source (asteroid) that I'd like to fit guesses for various model PSFs to, so that I can then compare the total flux in each case. I've only recently begun studying PSFs, so I am a bit confused.

I know that astropy offers a package called photutils.psf which does the PSF fitting for you, but I would like to better understand what's going on behind the fitting, and how the flux of the source is resolved from it. Furthermore, when I tried using this package, I wasn't getting any outputs anyway (it said a source was found, but that it didn't meet the sharpness or roundness criteria, despite me changing those parameters around quite a bit). I have tried something simpler below (where model_data is the 2d numpy array representation of the PSF, and img is the asteroid image data):

import numpy as np
from astropy.modeling import fitting
from photutils import FittableImageModel

y, x = np.mgrid[:np.shape(img)[0], :np.shape(img)[1]]

fitter = fitting.LevMarLSQFitter()
model2d = FittableImageModel(model_data)
result = fitter(model2d, x, y, img)
print(result)
print(np.median(img), np.max(img), np.sum(img))

The output I get is then:

flux       x_0       y_0

-----     -----     -----

9282.784  466.44    531.69

2705.82 25301.56 980812434.98

However, these are not the correct centroid coordinates of the source in the image. The correct coordinates are around (300, 300). Also, to get this result, the PSF data was normalized so that the max intensity point is 1. But when I change this normalization so that the total flux (sum of all the pixels) is 1, the result for the flux after fitting I get is way higher.

I'm not sure what's happening behind the scenes here (I got lost in the documentation), and I'm not sure if I can trust any of these flux values, considering the x_0 and y_0 are always off. So I'd like to better understand -- how exactly does one find the flux of an object by fitting it to a PSF? And how should that PSF be normalized?

You can't really get a PSF for an extended object.  The Point-Spread-Function is a parameter of your imaging system which describes the shape of the image of a putative point source.

I don't have any knowledge of what your python package is intended to do, so I can't comment directly on the results.  I will point (sorry) out that the geometric centroid of the image is not generally the same as the energy(brightness) centroid of the image.

Further, the total flux is just the sum of all the pixels covered by the asteroid's image (less background noise and whatnot).  It's not clear why you need the PSF if you are just doing energy/power calculations.


How rare are earth-like solar eclipses? Solar eclipses on the earth are so spectacular because the moon has roughly the same size as the sun when viewed from the earth's surface. This is an incredible coincidence, and my guess is that it is a very uncommon phenomenon in the universe.

Is it possible to say anything sensible about the likelihood of a planet having a moon and a sun with same apparent size? Is it possible to make an educated guess about how many places in the universe you would be able to see an earth-like solar eclipse?

We can certainly speculate.

There are dozens of moons in our own Solar System. I've done some preliminary calculations of their apparent size as seen from the planet vs. the apparent size of the Sun at that distance. The large moons (Jupiter's 4 Galilean satellites, Saturn's Titan, Neptune's Triton) are all substantially larger in the sky than the Sun is. I think there are some satellites that are fairly close to the apparent size of the Sun, but I haven't done all the calculations.

The apparent size of our Moon has changed over time, as the Moon has gradually moved farther away. We are coincidentally in a period of history in which it happens to be very nearly the same apparent size as the Sun.

The criterion for an exoplanet to have solar eclipses like the spectacular ones we have here on Earth is a moon that happens to have an angular size large enough to cover the photosphere, but not so large that it also hides the corona.  For a larger moon, you'd have eclipses in which the corona is visible, but not all the way around the moon. For a smaller moon, you'd only have annular eclipses.  (It could also get interesting if you consider eclipses as seen from other moons.)

So given the distribution of sizes and distances of moons in our Solar System, we can guess that situations where the apparent size of a moon and sun closely match is fairly rare, but since there are moons whose apparent size is smaller than the Sun and others whose apparent size is larger than the Sun, it probably happens sometimes.

The one piece of the puzzle that we're still missing, I think, is the typical sizes and distances of satellites of Earth-like planets. Moons the size of ours orbiting habitable planets might be common or very rare. If they're rare (say, if Mars-like moon systems are far more common), then Earth/Moon style eclipses might be very rare for habitable planets.


Do we have knowledge of the geology of the moon? With reference to the Ebb and Flow lunar satellites, was there any data about its geology? Is there a central core, or is it homogenous?

We know a considerable amount about the moon, but the picture is not complete.  There are considerable resources available and so much data available that it would take too long to list every aspect of it, so instead the following are examples of resources (with annotations):

Our knowledge of lunar geology comes from the results of direct sampling (from rocks brought back) and orbiting satellite observations.  Also, with inferences and comparisons with the Earth.

NASA's Lunar Geology page, this page approaches lunar geology by focussing on the 2 main sub-branches, the geochemistry and geophysics of the moon.

The geochemistry focuses on the crustal composition, volcanic petrology and the existence of polar ice, hydrogen and helium.  The geophysics focuses on the crustal magnetism, moonquakes, outgassing, crustal features.

Both of which help scientists develop a picture of the geological history and the geological structure of the moon.  Exploration of the moon's geological features in comparisons with Earth, are described as described in the online chapter Geology of the Moon as a geological 'Rosetta Stone'.

In terms of the internal structure of the moon, observations from direct sampling, sensors on the lunar surface and from satellite data have built up a reasonable picture of what the internal geology of the moon looks like.

According to the NASA page NASA Research Team Reveals Moon Has Earth-Like Core, re-analysis of Apollo mission seismic data has revealed that there are several similarities between the lunar and Earth internal structure (as seen below):

Caption from NASA: An artist's rendering of the lunar core as identified in new findings by a NASA-led research team. (NASA/MSFC/Renee Weber)

As can be seen in the diagram (and explained in all the links), the moon's internal structure is thermochemically layered, broadly similar to Earth (there are considerable differences).  But the main point, in answer to your main question, the interior of the moon is not homogenous, but layered with a central core.


Could the largest SMBH swallow an entire galaxy? It is unlikely that the largest supermassive black hole in the universe has already been discovered, there may be much larger ones.

Are there, or could there be, supermassive black holes large enough to swallow whole galaxies as opposed to the occasional star?

Are there other factors beyond size that would also affect a SMBH's ability to swallow it's own galaxy?

This question might be related to Is there a maximum size for a black hole?
Typical SMBHs do not contribute much to the overall mass of a galaxy, much less than the sun is of the total mass of our solar system. And as we can see, even such mass centered quite nearby does not swallow all objects in the solar system very easily.
While there is no theoretical maximum, in practice - based on simulations - maximum mass for a black hole seems to be around 50 billion solar masses†, which will give a radius of 0.01 light years. This is very small compared to the size of the galaxy, so it will not affect the situation much. Thus the main effect of huge black hole vs galaxy would be via plain old gravitation.
The main impediment for a black hole to swallow anything is angular momentum. Any object a black hole is going to consume must shed its angular momentum in order to get closer to event horizon. This is not very easy thing to do, and the main reason black holes are surrounded by discs of extremely hot gas - friction will turn angular momentum into heat, allowing matter to descend towards black hole.
†practical maximum of 50 billion suns is due to fact that accretion disc becomes unstable due to new stars forming from it. See for example Andrew King's 2015 paper How Big Can a Black Hole Grow?

Abstract:
I show that there is a physical limit to the mass of a black hole, above which it cannot grow through luminous accretion of gas,  and so cannot appear as a quasar or active galactic nucleus. The limit is $M_{max} ≃ 5 \times 10^{10}$ M⊙ for typical parameters, but can reach $M_{max} ≃ 2.7 \times 10^{11}$ M⊙ in extreme cases (e.g. maximal prograde spin). The largest black hole masses so far found are close to but below the limit. The Eddington luminosity $≃ 6.5 \times 10^{48} \text{erg s}^{-1}$ corresponding to $M_{max}$ is remarkably close to the largest AGN bolometric luminosity so far observed. The mass and luminosity limits both rely on a reasonable but currently untestable hypothesis about AGN disc formation, so future observations of extreme SMBH masses can therefore probe fundamental disc physics. Black holes can in principle grow their masses above $M_{max}$  by non–luminous means such as mergers with other holes, but cannot become luminous accretors again. They might nevertheless be detectable in other ways, for example through gravitational lensing. I show further that black holes with masses ∼ $M_{max}$  can probably grow above the values specified by the black–hole – host–galaxy scaling relations, in agreement with observation.


How can I find out the maximum magnification of my telescope? I have a Meade DS-2080AT-LNT. It's an inexpensive scope that I've had for several years. It has a couple of eyepieces with it, but nothing that will let you get a really good look at planets. I doubt I can see any Messier objects as more than a blip.

But it's more than good enough to take up and leave at my family's cabin in the far North and I think all of the kids that visit will enjoy it. But to get more out of the scope I was thinking about getting a Barlow lens. I see lenses of 2x and even 5x for sale. I don't know enough about telescopes to know what "x" of Barlow I should get for this scope.

The specs are as follows:

80mm (3.1") Aperture: Plenty of brightness to reveal planets, clusters,
nebulas and more.
Sturdy Fork Mount: Lightweight and solid aluminum mount and
tripod for stable views.
494 AutoStar® Controller: Automatically locates more than 1400 objects
and points the telescope toward them for you at the push of a button.
Series 4000 Super Plössl 1.25" Eyepieces: (26mm and 9.7mm) Enjoy low
and high power viewing with crisp, wide fields of view.
Altazimuth Set-up: Easy-to-use mount moves up/down, left/right.
SmartFinder™/Red Dot Viewfinder™: Makes stars and other objects
easy to find. Electronic level sensor, north sensor, and precision
internal clock help get your scope aligned with the heavens quickly.
AutoStar Suite™ DVD: Amazing planetarium software and instructional
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I appreciate any help!

The magnification is given by (focal length of telescope)/(focal length of eyepiece).  The spec I saw said the telescope focal length is 800mm, and the maximum magnification is 160x, so to get that would need a 5mm eyepiece.

Barlow lenses have the effect of increasing the telescope focal length: the 2x barlow would therefore give an effective focal length of 1600mm, and using the 9.7 mm eyepiece would give magnification of 160 (approx).  However be aware that the occasions when 160 magnification gives really clear views will be quite rare, as it is at the limit of what an 80mm scope can achieve.  The rule of thumb for maximum useful magnification for a good scope is 50x the diameter of the objective lens (in inches).


Pulses from a pulsar What is/was the motivation behind attributing the pulses received from a pulsar to the rotation of an object (neutron star) and emission from the magnetic poles (of the said neutron star), given that fact that pulsars have not been resolved yet in various wavelengths?

Could the pulses have been a product of a non-rotating but radially vibrating radio emitting star?

The key is to compare the range of pulsar periods and ther behaviour with the typical dynamical timescales of stars. Pulsar periods range from just less than $10^{-3}$ s to $\sim 10$ s. And $\dot{P}$ is positive - the periods are getting longer for most pulsars (and all of them that aren't in binary systems).

The dynamical timescale is $\tau \simeq (G\bar{\rho})^{-1/2}$, where $\bar{\rho}$ is an average density for the object.

For a star like the Sun $\bar{\rho} \sim 10^{3}$ kg/m$^3$ and $\tau \sim 1$ hour.

For a white dwarf star $\bar{\rho} \sim 10^{10}$ kg/m$^3$ and $\tau \sim 1$ second.

For a neutron star $\bar{\rho} \sim 10^{18}$ kg/m$^3$ and $\tau \sim 10^{-4}$ seconds.

Whatever mechanism you choose to cause the pulsar phenomenon, it cannot happen any faster than a few times this dynamical timescale if it is something involving the whole star.
This basically rules out pulsations from normal stars and white dwarfs because they could not have pulsation periods that are short enough to explain most pulsars.

Next one might consider binarity, but again, the minimum orbital period, where the components are touching, is a few times the dynamical timescale and some sort of phenomenon involving binarity is ruled out by the long dynamical timescales of normal stars and white dwarfs.

Next one considers rotation; again, simple Newtonian physics can be used to show that an object will rip itself apart if it spins any faster than the dynamical timescale, so again this rules out normal stars and white dwarfs.

So we are left demanding that whatever object causes the pulsar phenomenon is highly dense. In addition there is microstructure in the pulses which indicates the entire source is varying on timescales shorter than a ms, which means the entire emitting region must be much more compact than about 300 km (i.e. $c\times 10^{-3}$s).

So why a rotating neutron star? I think the argument here is that if this were some kind of oscillatory motion you might expect the period to be stable. For some kind of binary phenomenon you would expect the period to get shorter, due to the emission of gravitational waves. The rotation of a magnetised neutron star and the loss of power by magnetic dipole radiation can explain why pulsar periods get longer.


Simulating Noise in Astronomical Images I want to add realistic noise to a simulated image but I am a little confused about the process. I want to have some Gaussian random noise representing the readout noise and to also add Poisson noise correlated with the signal in the image.

I understand how to add the Gaussian random noise; I draw samples from a Gaussian with fixed width and add the samples onto the pixel values.

I am confused about the Poisson noise though. Is this noise only correlated to the noiseless signal?

i.e I should draw samples from the Poisson distribution with $k = \mathrm{pixel  value}$ (before adding the random noise) and add it to the noiseless image.

Yes, one can superpose these two noise sources: add a Gaussian readout noise (with a spatially constant sigma) to a Poisson noise (with spatially variable number count k).

In practice, as the number counts are high for optical CCD images, one often approximates the Poisson noise by a spatially-variable Gaussian noise, using the square root of the photon number counts as sigma. If your images have some significant sky background level (e.g., hundreds of photons), this is fine even for faint sources that add only few photons.  A Poisson distribution gets very close to a Gaussian already for k > 20. Note that despite this common approximation by a Gaussian, such a simulated noise contribution is still often called "Poisson noise", to stress that it originates from the spatially variable signal.

One warning: using the square root of the signal as sigma of a Gaussian to approximate Poisson noise only holds for a signal in units of photons (or almost equivalently, in units of electrons). If an image is given in ADU (analog to digital unit), the gain is required to convert the pixel values into electrons. The same holds when using a true Poisson distribution: one has to be careful that the counts have not been rescaled.


Precession of the Equinox I just learned about precession of the equinox, i.e. the Earth's axis tilts one degree every 72 years.

And it takes 72 years/degree x 360 degrees =  25,920 (years in one cycle)

There are many theories to this, i.e.

Sun's gravity is affecting the tilt of Earth's axis
Sun itself is revolving around another star, and it takes 25,920 years, hence the slow tilt of Earth's axis

I googled whether the Sun revolves around something else, and according to NASA.gov, Sun revolves around center of Milky Way it it takes 230 million years.

Now, I am confused between precession of the equinox and Sun revolving around another star.

Does precession of the equinox have anything to do with the Sun revolving around another star? Or is it simply the Sun's gravity acting on the Earth?

What is the explanation according to modern science? I understand there were astronomers in the past who deduced that the Sun revolved around another star and it took 25,920 years.

The sun does not orbit another star. A star, even a very small brown dwarf, would have been seen, if not in visible light then in the infra-red by the WISE survey.

However there is no mystery about precession. The Earth is non-spherical, it is wider around the equator, and because the Earth is tilted with respect to the ecliptic, there is an asymmetry in the force of the sun, and the moon, on the Earth. This force pulls the axis of the Earth's rotation in a circle. The basic maths of this is not so hard, and is in the Wikipedia page on Axial precession. This explanation of axial precession was known to Newton. There is nothing controversial about it.

The sun does orbit the galaxy it takes about 230000 years to complete one orbit. This has nothing to do with axial precession.

The page you link to is worthless. It offers an "alternate explanation" of a phenomenon that is very well understood, and pretends that there is some mystery, where there is none. It the uses this to justify a lot of astrological nonsense.

To repeat.

There are not "many theories" explaining Axial precession.
Axial precession is a well understood phenomenon, due to the Earth not being a sphere and the axis of the Earth not being perpendicular to its orbit
The sun does not orbit another star. We know this because we could have seen it.
No astrology based on the presence of another star is valid.


Incidence angles of meteoroids I've been reading the book Planetary Geology. In chapter 7, section 7.6, the author states that

The angle of incidence at which asteroids or comets encounter the surface of
a planetary body follows a Gaussian probability distribution with vertical and
horizontal collisions being extremely unlikely and the highest probability existing
for 45°

Why is this the case? Why doesn't the angle of incidence for meteoroids follow a uniform law? Isn't there just as much chance for an asteroid or a comet to arrive from any direction?

Horizontal atmospheric entries are unlikely because the object would have to almost miss the Earth and brush against it rather than impact it (and probably would disintegrate in the atmosphere if it's a small one, due to the long flight through it in such case).
Vertical entries are unlikely for the opposite reason: the object would have to fly at precisely 90 degrees relative to the Earth, meaning the Earth is in the very center of the object's orbital path.
But if an asteroid or comet comes at another angle its path may be changed by the Earth's gravity towards about 45 degrees. This is an approximation, it is about 40-50 degrees, but 45 degrees is the most likely one out of specific angles.


Is it correct to say vertical conjunction? This question is related to this one: Are we seeing a conjunction these mornings?

Last Feb 27, the a reporter said that, the day before, at sunset, Venus and the Moon were in vertical conjunction. Is that correct?

Sorry, but I cannot understand the meaning of near conjunction.

On the evening of the 27th, Venus and the moon were not in conjunction. They didn't have the same right ascension.  However, from Europe they did have about the same azimuth, so would have appeared with Venus above a crescent moon in the sky.

This is, presumably, what the reporter meant by "vertical conjunction", but that is not a technical term.  And while this arrangement is pretty, it isn't very significant, and happens every month, when Venus is visible.


Jupiter appearing odd through a 6" reflector Recently I have been trying to get a good look at Jupiter through my 6" Orion reflector, but the image seems to always come out as an oblong white streak. I've tried changing the lenses I'm using, but that only changes the size of the oblong streak. I looked at the moon with the same lenses to see if it was a general problem with the reflector, but the image of the moon came out clear as usual. What sort of problems could be causing this?

Most likely would be the the main mirror needs collimating. On big things like the moon you won't notice bad collimation as much as something small like a planet or star.


Why hasn't the "9th Planet" been detected already? In the comments to this question, there was considerable doubt placed as to the subject of if the so called "9th planet" really exists. That wasn't really the intent of the previous question, so I ask this. Why hasn't this "9th planet" been detected before now, if it even exists?

Brown and Batygin, the authors of the paper on the possible planet, have a webpage addressing this.

A few reasons not already covered:

It moves quite slowly - the authors estimate 0.2-0.6 arc seconds per hour - so standard surveys may not notice the movement and fail to recognize it as a solar system object.

  Eris, which is the most distant confirmed object still known in the
  solar system, moves at a speed of 1.5 arcseconds per hour, which is so
  slow that it was missed the first time around. Most surveys of the
  outer solar system would not be able to find Planet Nine, even if it
  were quite bright, as they would just think it is a stationary star.

If the planet is near aphelion, it might be an order of magnitude further away than any major or minor planet we've found so far (excluding exoplanets, which are found by methods that don't apply in this case). The authors suggest an aphelion between 500 and 1200 AU. For comparison, Pluto is at 30-50 AU, while Eris at around 100 AU wasn't discovered until 2005. The potential 9th planet would be far larger than Eris, but is also likely to be much further away, and thereby fainter.
The WISE survey eliminated Saturn-sized planets within 10,000 AU, and Jupiter-sized planets within 26,000 AU. But the potential 9th planet is far smaller than those. WISE has also done a more sensitive search, which would pick up Neptune-sized objects, but that search has so far covered only a limited part of the sky.
The planet will be far harder to spot if it has the Milky Way in the background - there are too many stars potentially drowning out a faint object.

Here's the authors' summary:

Estimated orbit for the putative 9th planet. The horizontal axis is the right ascension. The colored segments are regions where it should have been found by existing surveys.
Illustration by Brown and Batygin, assuming fair use applies.

  The biggest unexplored territory is where, statistically, it is most
  likely to be: near aphelion. Sadly, aphelion is also very close to the
  Milky Way galaxy. Ugh.

  So where is it? Probably distant. 500 AU+. Probably fainter than 22nd
  magnitude. Very possibly in the middle of the Milky Way galaxy.

  Now go find planet nine.

More details on the authors' webpage: http://www.findplanetnine.com/p/blog-page.html

Finally, the gravitational dominance of the Sun reaches halfway to the nearest star. There's still plenty of unexplored territory for planets smaller than Saturn to hide in.

Note the log axis. We have a good map for the inner 50 AU, and are starting to find objects around 100 AU, but solar system objects might exist all the way to the outer edges of the Oort cloud.
Illustration from wikipedia.


Moon's revolution around the Earth and it's rotation around it's axis? Does Moon rotate around itself while it makes one revolution around Earth or not?

Yes. Exactly once. This is how we get the light and dark side of the moon. The reason for it being exactly once is due to something called tidal forces. Wikipedia explains this quite well: https://en.wikipedia.org/wiki/Tidal_locking.


Could information be transmitted from inside a black hole using gravitational waves? Could you use gravitational waves to communicate from inside the event horizon of a black hole to someone who was outside?

A black hole is defined as a region of spacetime where the gravitational effects are so strong that the escape velocity within the event horizon is greater than the speed of light.  According to relativity, nothing can propagate faster than the speed of light, so at most gravitational waves would propagate at this speed.  According to particle physics, mediation of gravity would be controlled by massless particles called gravitons, which, by virtue of being massless, would travel at the same speed as photons, i.e. the speed of light.  Suffice to say gravitational waves (analogous to electromagnetic waves) would not be able to escape from inside the event horizon.

The question of information, in regards to black holes is open however. The No-Hair Theorem would seem to imply the destruction of information as the black hole consumes matter.  However, the holographic principle, if correct would imply that information regarding the matter that formed the black hole would be encoded on the 2-dimensional boundary of the event horizon.  This information could very well be exhausted by Hawking radiation or gravitational waves.  Since this information was outside of the event horizon this would be fully consistent with the formulation of black holes.


How was the hypothetical ninth planet kicked so far out of the Solar System? I was reading this article on science news and ran into the following statement:

  Such a remote super-Earth probably originated closer to the sun only to be kicked out by the other giant planets during the solar system’s formative years, Batygin and Brown suggest.

I was wondering, how was this planet kicked so far out of the Solar System?
Why only one planet?

It's hard to say much about this planet, given that most of its properties are unknown. It hasn't been directly observed; instead, its effects on Trans-Neptunian Objects (TNOs) have been simulated and match observations. That said, its mass can be estimated, which is why it is conjectured to be the core of a giant planet.

One of the papers that Batygin & Brown cite is Morbidelli et al. (2012). Morbidelli has done prior work on the evolution of the Solar System if a 5th gas giant formed early in its history, which led to the Jumping Jupiter scenario. This is all described in a modification of the Nice Model.

Jumping Jupiter models with the four gas giants lead to one getting ejected from its stable orbit via interactions with the others (see
 Nesvorný (2011)). This is one of the ice giants - Uranus or Neptune. However, if we postulate the existence of a 5th planet - a third ice giant - then it can be ejected from its orbit, leaving the other four in a stable arrangement.

Batygin & Brown are implying that the object perturbing the TNOs may be the core of this 5th gas giant.

Since writing the first draft of this answer, I've found out that Batygin has made some estimates that this planet would have been ejected long before the Late Heavy Bombardment - thought to be the result of a Jumping Jupiter scenario - occurred, meaning that it could not be that ejected ice giant, unless we accept a different cause of the LHB. This makes things more interesting, as different initial setups are possible.


If water vapor is always blown away into space, how is it able to create chemical compounds on Venus? This is the follow-up of this chem.SE question.

According to Wikipedia, water vapor on Venus is present in trace amount (20 ppm). There are multiple reasons why Venus has very low water content:

The rich carbon dioxide atmosphere generates the strongest runaway greenhouse effect in the Solar System reaching temperatures at least 735 K, hot enough to boil most of the water vapors.
Free hydrogen and oxygen atoms  has been swept into interplanetary
space by the solar wind because of the lack of a planetary magnetic
field
Water vapors are continuously blown away by the solar wind through the induced magnetotail.
UV and photodissociation of water vapor creating hydrogen and oxygen atoms and radicals
High D/H ratio in Venus due to hydrogen atoms escaping into space since it is light thus unable to form water
The atmosphere is rich in sulfuric acid clouds which is a powerful dehydrating agents. Most of the water reacts with H2SO4 to form hydronium (H3O+) and bisulfate (HSO4-) ions. As a result, the concentrations of “free” H2O in the acid solution and in the vapor over the acid are extremely low.

So, the question is "since water vapor is always blown away or reacting in some way and there is only a trace amount of stationery and unreacted water vapor available, why/how is it able to create chemical compounds like phosphoric acid and sulfuric acid?"

Background information
A 1986 paper1 concluded that P4O6 is the main phosphorus bearing gas on Venus from the Vega mission results. It was also reported that some phosphorus bearing particles were present which was found to be phosphoric acid, H3PO4.
I am pretty sure phosphoric acid/phosphorous acid is the result of reaction between phosphorous anhydride and water vapor in sulfuric acid environment (which is later speculated to be the pathway of phosphine formation). But "is water vapor present on Venus sufficient enough to form phosphoric acid? What is the concentration of phosphoric acid on Venusian atmosphere (which would later contribute to formation of phosphine)?"
Reference

Krasnopol'skii, V A. Vega Mission results and chemical composition of Venusian clouds. United States: N. p., 1989. Web. doi:10.1016/0019-1035(89)90168-1.
https://www.vanderbilt.edu/AnS/physics/astrocourses/AST101/readings/water_on_venus.html

Think about the water constantly being both added to and removed from the atmosphere.
Adding water:

Water migrating from deeper layers of the planet crust and mantle into the atmosphere.
Water being formed from sun wind hydrogen and the available oxygen in the atmosphere (the high D/H ratio hints that the primordial water dominates on Venus)
Water being formed from hydrogen and oxygen produced by decomposition of other substances in the atmosphere.

Removing water:

Photolysis - water decomposed to hydrogen and oxygen, both of them engaged in other chemical processes
Escaping hydrogen and possibly whole water molecules into the space.

There is always some equilibrium between the available water and hydrogen and also other hydrogen containing substances like sulphuric acid (that's what Venus clouds are made of) and the widely-discussed today phosphine.

On Earth, we have stratospheric and mesospheric clouds made of water, sulphuric acid and nitric acid in atmospheric layers that are otherwise notoriously dry.  On Venus, the atmosphere has little free oxygen so it is tolerant to much more of a chemical diversity. (E.g. phosphine would self-ignite in Earth's oxygen-rich atmosphere)


How much mass does our sun need for exploding like supernova? Beside that, is that possible our sun became pulsar? And if not (and i know likely) but how much more mass does it need to become one ?

The smallest stars that could explode in supernova have a mass of at least about 8 times the mass of the sun. So it is impossible for the sun to become a pulsar at the end of its life. It will form a red giant which will then lose its outer layers and form a white dwarf.


How will the recent appearance of a supernova, the closest in 27 years, help reduce the uncertainties in our measurements of dark energy? A supernova has been recently spotted on M82, also known as the cigar galaxy. Being the closest to earth in 27 years and given the advances of technology this article proposes that:

  ...as the Cigar Galaxy has been studied in detail thanks to images from the Hubble Space Telescope. "It is likely that the star that blew up has been directly imaged"...

  If astronomers do find the dying star in past images and are able to make accurate readings of its brightness before and after the explosion, we might be able to reduce the uncertainties in our measurements of dark energy by a factor of two.

How will it help reduce the uncertainty?. Furthermore, how would this help us better understand the way the universe is going to end?.

The current supernova is a supernova of type Ia. Supernovae of type Ia are used as standard candles for distance estimates, especially used to determine the Hubble constant. Hence by a better calibration of this kind of supernovae, more about the reliability and accuracy of distance estimates can be learned.

The expansion rate (in relation to the distance) of the universe, expressed by the Hubble constant, is an important incredient for the estimation of the fate of the universe.


App Stellarium and hourly wobbling motion of Neptune In Stellarium when I zoom in on Neptune to be able to see your daily rotation. I select Neptune, center it (space bar) and then make increments of one solar hour (command +).  I notice that the axis of rotation seems to wobble (just focus on the big dark spot to notice that), but the precession period of Neptune's axis of rotation is hundreds of years, not hours. How is this possible?. I think it may be due to the combination of the daily rotational movement of us (the Earth) and Neptune, but I cannot get a clear geometric representation of this. Does anyone know why this phenomenon happens?
Thank you.
I have: Stellarium 0.20.4 for Mac OS.

This appears to me as just the consequence of having an altazimuthally mounted “camera” (the simulation). Objects are tilted towards the left when they rise, towards the right when they. set. You can verify that easily by looking at the Moon when it rises and sets from your own place. This is due to the horizon being “tilted” with respect to the Earth’s plane of rotation (unless you are exactly at the pole).
I’m not super familiar with Stellarium, but as far as I know, there’s a way for the observer to not be located on the Earth, or at least, you can change the latitude of the simulation. Try that, and if you set yourself as being at the Earth’s (north or south) pole, you shouldn’t see that anymore.


Are all the planets differentiated? The Earth is differentiated, as it has a distinguishable core, mantle and crust with different compositions and densities. Are all the other planets differentiated as well?

They are - somewhat by definition.
One criterion for a planet is that it is large enough that it is in hydrostatic equilibrium so that it takes on spherical shape. Any body large enough for that to occur will to a certain degree be a differentiated body as hydrostatic equilibrium means that masses need to be able to move, thus the heavier elements will sediment towards the center. Due to the speed of geological processes, differentiation (=de-mixing) may initially be faster, but is never really finished, unless a body is completly solid and cold (which no planet is).
The existence or non-existence of hydrostatic equilibrium is one of the differences between dwarf planets and so-called small solar system bodies.


Would a tidally locked planet be always increasing in temperature? Okay, say we have a tidally locked planet that has one side facing its star. On the side that always faces the star, my question is will that side facing the star always be increasing in temperature? If the side is always facing the star, it doesn't have time to cool.

Hot objects radiate heat. The hotter something is, the faster it radiates heat.

The side of the planet that faces the sun would heat up until the rate at which it is absorbing heat from the star is equal to the rate at which it is radiating heat. Once it reaches this point it will be stable, and neither heat up more or cool down.

As a simple analogy, consider standing next to a heater or a fire. You will heat up, but you won't heat up forever. You will reach thermal equilibrium.


Why the torque exerted by the moon upon the Earth makes the Moon to increase his orbit? For what I know, this torque exists because of the misalignment of the tidal bulge with the apsidal line Earth-Moon a certain angle $\alpha$, which makes the earth rotation to slow down a bit. But, how that pulls away the moon? I thought this "loss" of rotational momentum transferred into tidal heating. I fail to see how this energy is transferred into more orbital momentum, because I thought the conservation of angular momentum didn't have to do with the rotation of the objects. I mean, isn't true that $c$ remains constant in the two-body problem (being $c = \vec{r}\times \dot{\vec{r}}$)? How the rotation plays into that?

because I thought the conservation of angular momentum didn't have to do with the rotation of the objects.

The angular momentum of the system is conserved.  As rotation can contribute to angular momentum, it does affect the equation.  You can ignore it for point-like objects.  In the case of the earth-moon system, the earth's rotation is a significant portion of the total momentum.

A not-too-wrong version would be to assume an axis centered on the earth.  Then the total angular momentum of the system could be found by summing the individual momenta for:

the earth's rotation
the moon's rotation
the moon's orbit

Assuming the moon is tidally locked to the earth and has a nearly circular orbit, this gives you only two variables.  You can then calculate that, if you slow the earth's rotation, the only way to keep angular momentum of the system constant is to increase the angular momentum of the moon by an equal amount.  This would correspond to a larger orbit.

Further if you calculate the KE from both configurations, you find the second one has less energy.  So there must be an energy loss to transition between them.


How does large body gravity affect planet formation? If Jupiter's gravity is what kept the asteroid belt from forming a planet, then why didn't the sun's gravity keep the inner planets from forming?

why didn't the sun's gravity keep the inner planets from forming?

The short answer is, the asteroid belt orbits the sun, it doesn't orbit Jupiter, and you shouldn't expect the same effect with 2 different relations.

The sun is the gravitational object that the Asteroid belt orbits.  Jupiter is not and it has a stabilizing effect on the material that orbits around it.  Jupiter has a (mostly) destabilizing effect on nearby material that orbits around the sun, though there is some stability with orbital resonance and trojan points.   The reason why that is gets a little bit mathy, but I'll cover the basics.

When a solar system forms, probably out of a combination of strong solar wind from a supernova and a dense enough gas cloud, most of the solar-system material forms the young sun and maybe 2%-4% remains in orbit and over time, spirals and flattens out into a disk.  (see cool video)

The sun's gravity stabilizes that disk.  There's no reason for it to disburse it, though the young and usually strong solar wind can blast it and clear out the closer regions of smaller particles and any ices.

Jupiter is entirely different.  Jupiter and other planets which form inteh disk, tends to eat everything in their path.  That's one of the definitions of a planet.  It clears out it's orbital area.  These planets on their own would remain in orbit around the sun, but they can interact over time with each other.   Pretty much anything in Jupiter's path, once Jupiter gets large enough, either gets eaten or gets thrown into a very different orbit, either further our or closer in.

The asteroid belt is (as Suhrid Mulay points out) a relatively low density region of space that doesn't have enough mass to coalesce into it's own planet, but it might not have always been that low density and that's probably only part of the reason.  The other part of the reason is the proximity of Jupiter which tends to disrupt anything that orbits too close to it and Jupiter may have at one point, moved in quite a bit closer to Mars before moving back out.  Tossing much of the material in that region far away.

Planets don't do well forming too close to other planets because they will gravitationally disturb each other.  A planet the size of Jupiter had pretty significant reach when it comes to disturbing other planets and planet formation.


Can earth escape sun's gravity with the help of a black hole heading towards our solar system? To escape the Sun's gravity, the Earth would need sufficient amount of energy. Let's say a stellar black hole has entered our solar system and at a certain distance from earth, the black hole exerts same and equal gravitational force on the Earth as the sun is exerting. Earth feels a zero net force. Will it help earth to fly away?

Assuming the black hole had an empty stomach before eating our solar system.

There is no need to hypothesise a black hole. A black hole's gravity is indistinguishable from that of a similarly sized star, except in the region closely surrounding the event horizon.

Black holes can't be only 1 solar mass, there is no known mechanism that could cause a single solar mass object to undergo complete collapse. Black holes have at least three times the mass of the sun.

So could a massive star disrupt the orbit of Earth? Yes it could. If the star were to pass by the Earth, it could cause a slingshot effect, transferring some of it's momentum to the Earth, which could cause Earth to leave solar orbit.

In the precise situation that you describe, Earth would initially be travelling in a straight line, but then would follow a complex and possibly chaotic path, as the sun and the black hole pass each other. It is quite possible that the Earth could be ejected, or transferred to an orbit of the black hole in that case.

Fortunately the gaps between stars are so vast (and black holes even more widely spaced) that in the nearly 5 billion years of the Earth, not one has come close enough to cause major disruption to Earth (we know because we are still here).


What is the difference between the two terms named "Eccentricity" and "Ellipticity"? I don't get the difference between the two terms named "Eccentricity" and "Ellipticity", especially, in astronomy. I understand eccentricity as a measure of the curvature of a orbit but what is ellipticity?

Both ellipticity $f$ (also called flattening) and eccentricity $e$ are measures of how elongated an ellipse is, based on the semi-major axis $a$ and the semi-minor axis $b$  (figure from wikipedia).

They are defined respectively as
$$f=\frac{a-b}{a}$$
and
$$e=\sqrt{1-\frac{b^2}{a^2}}$$
For a circle, $a=b$, which implies that $f=e=0$.  In modern orbital mechanics, $e$ is used rather than $f$.  This is due to convention as $e$ is one of the six orbital elements Kepler defined.
Notes:

We can extend the definition of eccentricity to parabolic and hyperbolic orbits: $$e=\sqrt{1+\frac{2\epsilon h^2}{\mu^2}}$$ where $h$ is the angular momentum of the orbiting body relative to the more massive body, $\epsilon$ is the specific orbital energy, and $\mu$ is the standard gravitational parameter.  Then $e=1$ for a parabolic orbit, and $e>1$ for a hyperbolic orbit.

While $e$ is conventionally used to describe orbits, $f$ is still used in astronomy as a factor expressing the oblateness of nearly circular bodies.

Here is a table and plot that illustrates the difference in values by fixing the semi-major axis $a=1$ and varying the semi-minor axis $b$ between 0 and 1.  Eccentricity and ellipticity are only equal when the orbit is either a circle or a line segment (when the orbit has no tangential component of velocity, in free fall towards the more massive body).


How can I estimate the initial luminosity of a SNIa given initial mass of Ni-56? The initial portion (and peak) of a SNIa's light curve is powered by the $\beta$-decay reaction:
$$ ^{56}Ni \rightarrow ^{56}Co + e^+ + \nu_e + \gamma $$
Supposing we know how much Ni-56 is created in the supernova (i.e., an initial mass of Ni-56), how can I go about estimating the luminosity peak? The half-life of Ni-56 is 6.1 days, if that's relevant.
Does it make sense if I just look up/calculate the energy released in this reaction, multiply it by the number of Ni-56 atoms required to comprise the given initial mass, and call that the luminosity? Any other better ideas?

Nickel 56  decays to Cobalt 56 via electron capture decay, with a half-life of 6.1 days and a decay constant of $\lambda = 1.31\times 10^{-6}$ s$^{-1}$.
About 1.75 MeV of energy is lost as gamma rays and a further 0.41 MeV in the form of an electron neutrino (Nadyozhin 1994)
Let's assume that we are talking about the period of time after the initial detonation, where the fusion energy of carbon and oxygen is released and is sufficient to unbind the white dwarf.
Let's assume that apart from the very first few seconds of the supernova, that the envelope is transparent to neutrinos, so that energy is lost.
Let's further assume that the gamma rays are able to thermalise rapidy in the envelope, that the energy is able to diffuse to the "photosphere" of the expanding fireball in $< 6 $ days and that the work done in expanding any ejecta is also negligible. This latter may not be justified in a type Ia supernova.
The decay equation $N= N_0 \exp(-\lambda t)$, means the rate of gamma ray energy deposition will be
$$ \frac{dE}{dt} = 1.75\ \lambda N_0 \exp(-\lambda t) \ {\rm MeV/s}, $$
where $N_0$ is the number of nickel nuclei you begin with.
Let's assume about $0.5M_{\odot}$ of Nickel is produced (see Childlress et al. 2015, incidentally, this paper illustrates how complex this question really is). This means $N_{0} \simeq 1.1\times 10^{55}$.
Thus I make $dE/dt$ (assumed to result in the emergent luminosity), when $t=0$, to be
$2.5\times10^{49}$ MeV/s, or $10^{10} L_{\odot}$.
This value is about a factor of two higher than the average type Ia supernova peak bolometric luminosity measured by Scalzo et al. (2014), which were well modelled with nickel masses of around $0.5 M_{\odot}$. So I would suggest that one or more of the assumptions above (probably the one about no work being done on the ejecta) is a bit flaky. However, I also notice that the peak luminosity occurs about 2 weeks after the initial rise, so I would suggest that the "instant" reprocessing assumption is also incorrect and that the released energy is smoothed out somewhat so that a lot of the nickel has already decayed by the time you get to peak luminosity.


What is the composition of the Universe in terms of Types Of Stars? There are

Main Sequence Stars
Red Giants
White Dwarfs
Red Dwarfs
Neutron Stars
and other types of Stars.

How many % of stars are Main Sequence? How many are Red Giants? and so on.

I searched around and found some info, but it's confusing me. For example,
I found that 90% stars are Main Sequence Stars, but at the same time, Red Dwarfs are the most common stars in the universe with 70%. What is true?

First of all, there is a misconception in your question that red dwarfs and main sequence stars are two completely different kinds of stars.

Red dwarfs are a type of main sequence star. If a star is in the main sequence that just means that it is at a certain stage of its evolution.
On the main sequence, stars can be further categorized based on their color into brown dwarfs, red dwarfs, sun-like stars, blue stars,...

In a comment, you state that the local distribution of stars is enough for your purpose.

To get the local distribution of star types, you can use the Gaia data, which gives the distance, luminosity and color of many stars in our galaxy. From this, you can plot an HR diagram, where the position of a star indicates its type..

This diagram isn't quite correct, because it uses a cut in magnitude to select stars. This means that fainter stars are under-sampled. But if you only took stars up to a certain distance, you could get something more representative.

In this article, you have a breakdown of stars based on their intrinsic luminosity and their spectral class. Most stars (76%) are in the main sequence, 24% are white dwarfs, and the rest (0.6%) are on the giant branch.

The stellar distribution of our Milky Way seems to be pretty typical for this type of galaxy, but there are other types of galaxies where the stellar distribution is rather different. When a galaxy is rather young, there is a lot of gas available for new stars to form, so you get a lot of big bright blue stars. But these big stars are short-lived, so in elliptical galaxies where the stellar population is older, you get more small red stars.

Finding the distribution of star types in these different types of galaxies is the matter of stellar population synthesis. You start out with an initial mass function (IMF) that tells you the number of stars there are with any given mass in one galaxy. This gives you a snapshot of the galaxy at time t=0. Then you use stellar evolution models to see how that initial population evolves over time.

De Marchi et al gives the IMF for a cluster of galaxies (rather than for individual galaxies). This might be as close to a "typical" distribution as you can get.

Further reading


Is sky-glow affected by height? I'm an architecture student and my graduation project will consist of building a research center for outer space, my question is whether or not height would affect sky-glow or not ? Thanks in advance, any help is appreciated.

Sky-glow is a generic term for the diffuse light not due to the sun, moon or stars (or aurorae, I suppose). Much of it is light pollution from the ground, some is the zodiacal light from dust in space, and the coolest part may arguably be the air-glow from chemical reactions in the upper atmosphere.
So, depending on height the different components would change strength. Urban light pollution apparently declines as $\propto d^{-2.5}$ ("Walker's law") due to spreading and absorption, so it would decline with altitude. The zodiacal light would get stronger with less absorption in the atmosphere. The air glow is altitude-dependent, and would presumably intensify at first, until one is above it (like the ISS) is.


What is the lifetime of an x-ray binary? What would be the typical lifetime of an x-ray binary? I have trouble finding a reference.

I am not an expert on the topic but probably you have to specify in your search which mass regime are you looking for (low or high mass x-ray binaries). Accretion time must be important and depends on the mass. High-mass stars are young (very short life) < $10^7$ years. This reference is quite old but probably helps a bit:

The Average X-Ray Lifetime of Massive X-Ray Binaries
Observations of accretion disks in X-ray binaries
X-Ray Binaries


Can concentration of gas in exoplanet atmosphere be found out from wavelength and absorbing radius from spectral data? I was working on exoplanet spectral data from which I need to infer the concentration of gases. However, the exoplanet spectroscopy data contains only absorption wavelength and absorption radius. Is there any way with which I can derive the concentration of the elements in the exoplanet's atmosphere?

Sure, but it is not straightforward. There's a lot of degeneracies involved, such as many molecules sharing similar absorption bands, the presence of clouds, scattering, absorption happening at many different pressure ranges, solving the chemical equilibirum equations of the species and the temperature-profile modifying the overall spectrum as well.
There's a lot of "Forward" models available online, which "solve" for the atmosphere (Give you the theoretical spectrum) of a planet given for some initial conditions, such as metallicty, Carbon to oyxgen ratio, or concentration of molecules.  What you want in your case is to do the inverse, i.e. start from your data and go to the mixing ratios of the different species. You can do that with "Retrieval" models, which you should again find information about online.
Assuming that you are working with transit data (which is what I understood from your post), you can often assume a constant temperature for the entire atmosphere at the equilibrium temperature of the planet, since transmission spectrum are usually less affected by the temperature profile, as they act more like passive absorbers.


Could a black hole pass quiescently through the Oort cloud? Imagine a stellar-mass black hole passing through the Oort cloud, perhaps a quarter  to half a light-year from the Sun. How black would the black hole be? That is, any matter that fell into the black hole would release energy on the way in, and we know there are comets out there, and presumably smaller objects down to microscopic dust. But we also know they are very few and far between. Has anyone done a serious analysis of whether there would be enough infalling matter for the resulting radiation to be detectable?

As two commenters already recognized, this really is about radiation from infalling matter, and not the broader question of how we could detect such a black hole. Answers to that broader question in the comments are interesting, though.

An approach would be to use the Bondi-Hoyle (B-H) accretion rate and then assume that some large fraction of the gravitational potential energy was radiated from the innermost stable circular orbit (ISCO) as the material spiralled into the black hole via an accretion disc.

To calculate the B-H accretion rate we need how fast the black hole traverses the Oort cloud and the density of matter there.
$$\dot{M} \simeq \lambda\frac{4\pi G^2 M^2 \rho}{v^3},$$
where $M$ is the black hole mass, and $\lambda$ is an accretion efficiency parameter that is judged to be $\sim 0.01$ on the basis of observed compact objects (see Fender et al. 2013).
The luminosity will be
$$ L \simeq \eta \frac{GM \dot{M}}{R_{\rm ISCO}},$$
where $R_{\rm ISCO}= 6GM/c^2$ for a Schwarzschild black hole. Thus
$$ L \sim \eta \lambda \frac{4\pi G^2 M^2 \rho c^2 }{6v^3}.$$

Getting all my information from wikipedia, I assume there are 5 Earth masses in the Oort cloud in a spherical shell from 20,000 to 50,000 au. This gives an average density $\rho \sim 2\times 10^{-23}$ kg/m$^3$.

This is actually lower by a factor of a thousand or so than the density of the interstellar medium. Thus accretion from the ISM is far more important on average. Putting the numbers in for a 1 solar mass black hole and $\lambda=0.01$, $\eta =1$:
$$L = 3\times 10^{23}\left(\frac{\rho}{10^{-20}\ {\rm kg m}^{-3}}\right)\left(\frac{v}{10\ {\rm km s}^{-1}}\right)^{-3}\ {\rm W.}$$

This is quite an appreciable luminosity and much would emerge at UV and X-ray wavelengths. It would be readily detectable. The flux at a distance of 50,000 au would be $5\times 10^{-10}$ W/m$^2$, which is 4 orders of magnitude above the faint flux limit of the ROSAT all sky X-ray survey.


How to find the shape of an orbit What is the minimum information I need to be able to draw the shape of an orbit of a body that orbits the sun? With this information how can I figure out the shape of its orbit?

If you want just the "shape" you need one number: the eccentricty.
The shape of an ellipse is determined by the eccentricty. It is a measure of how "uncircular" an ellipse is.  Conventionally this is written as $e$
If the ellipse is centred at the origin, and its long axis ends at (1,0), then the focus of the ellipse is at (e,0)
A very simple geogebra construction that does just this: https://www.geogebra.org/classic/ydp9cthr
If you want the "shape" and the "size" of the ellipse, you need a second number, and conventionally this is half the length of the long axis of the ellipse, this is written as $a$.
If the half length of the short axis is $b$, then these are related by the formula
$$e^2 = \frac{a^2-b^2}{a^2}$$
So if you have two, you can work out the third, and plot the ellipse at a given size using parametric equations
$$\begin{align}
x &= a \cos(t)\\
y &= b \sin(t)\\
\end{align}$$
Now, this draws the ellipse in 2d space and oriented along the x-axis.  If you want the ellipse drawn in 3d space and at an arbitrary alignment, you need three more pieces of information, usually these are the "argument of periapsis" (the angle of long axis of the eclipse) "inclination" (the angle of the plane containing the eclipse to the x-y plane) and the "longitude of the ascending node" (the bearing of the point at which the object crosses the x-y plane)
A sixth number is required to define the position of the planet on the ellipse at a given time. (usually given as the "mean anomaly" or "true anomaly")
These six numbers are called the orbital elements
All this assumes Kepler's laws.  Real orbits are perturbed by the gravity of other planets (and slightly deviate from Kepler due to relativity). The real paths of the planets aren't actually periodic, but Kepler's laws are a very good model.


What will the universe be like in a googolplex years time? It will be cold and dark in 10^(10^100) years.

What else can be said of the state of things in the far distant future? Will everything have fallen into supermassive black holes and then radiated back into space so there is no matter left?

Will the universe revert to an exotic state as it was in primordial times, or will it slow and stay unchanged indefinitely? What kind of structure will it have?

That is nearly long enough to reach heat-death, which is estimated as about $10^{10^{120}}$. What that means is rather speculative, since it depends on various events that we have never observed, such as the spontaneous formation of black holes by quantum tunnelling. Such events are utterly rare, but are predicted to occur at very long timescales.

Any matter, will have at some time over a googolplex years quantum tunnelled into having enough mass to collapse as a black hole. So no matter will exist

Any black hole will have evaporated in Hawking radiation.  So no black holes exist.

Redshift will reduce the energy of photons to the point where their wavelengths are comparable to the scale of the universe.

Nothing is happening in the universe.

See https://en.wikipedia.org/wiki/Timeline_of_the_far_future

Now there is a: immovable object/unstoppable force paradox here, for in a universe that is spatially big enough, there are pockets of order among the chaos of heat death.


Can lunar atmospheric density, as it exists today, increase naturally? Are there any mechanisms which could increase the density of the moon's atmosphere.

The moon's existing atmosphere is exceeding rare, with about 1 000 000 molecules of gas per cubic centimeter at the surface, ten trillion times less than the Earth.

The source of this gas is the impact of solar wind on the moon's surface, and some outgassing from the interior of the moon.

Can this density vary significantly? Could it reach a density comparable to Earth's? Are there natural processes which could cause the atmospheric density to increase, either in the short term, or over a longer time span.

There's not enough gassible (is gassible a word?   I'm guessing not), OK, there's not enough of the right elements to form enough gas on or in the moon to create any kind of atmosphere that would measurably slow down falling objects (say, enough to land a probe with a parachute, like they do on Mars).   Atmospheres are heavy.  They may not seem heavy but they are.   Mars' atmosphere has a mass of about 25 trillion metric tonnes and that's a very thin atmosphere compared to Earth.

The Moon, with roughly 1/4th Mars' surface area, would need about 1/4 of that mass in it's atmosphere to have an atmosphere with similar air resistance to Mars, somewhere in the 6-7 trillion metric tonne range.

There's some ice on the moon in craters that never get direct sunlight, but not nearly enough ice to make that kind of an atmosphere.   The Moon also has very infrequent volcanic eruptions, but while some gas is likely released, again it wouldn't be nearly enough.

One possible scenario would be a large asteroid or comet impact that could both release some trapped gas from below the surface and release gas from the impactor.   The problem with such a scenario is that the heat from the impact would likely send much of the gas away from the planet at above escape velocity shortly after impact.    The escape velocity of the Moon is 2.38 km/s and at the top of this theoretical atmosphere it would be a little bit less than that.  A large impact from a comet is the best way I can think of, but that wouldn't be very efficient.

Even if you found a way to create that kind of atmosphere on the Moon, it wouldn't last very long.  The Moon is simply too small, with too little gravity and it's too close to the sun to maintain an atmosphere, unless that atmosphere was Xenon or if you want to be a bit more daring, sulfur-dioxide (acidic when it comes in contact with water and stinky) or sulfur-hexaflouride, which isn't toxic in and of itself but becomes very toxic when exposed to UV light from the sun, so, lets not use either of those, but Xenon is safe, so we can use that.

see atmospheric escape velocity.

So the answer to your question is yes, the Moon could be given a thick enough atmosphere to slow falling objects down, but it would probably need to be done artificially, as there's no natural process I can see that would do that, and ideally, you'd use a very heavy and non toxic gas like Xenon.    All the Xenon in the Earth's atmosphere wouldn't be enough.   A rough calculation, Earth's atmospheric Xenon comes to less than a billion metric tons, which is still several thousand times too little mass to create a mars like atmosphere on the Moon.   Now if you harvested all the Xenon from Jupiter and/or Saturn and moved that to the moon, then you might be able to create a somewhat long lasting atmosphere on the moon that would provide some measurable air resistance.   It wouldn't be breathable, but it wouldn't be toxic, which is a plus and with enough Xenon, you might be able to get around on the moon with just an oxygen tank and no pressure suit.  Buildings with Oxygen/Nitrogen would need to be air-tight but not vacuum resistant, which is far easier to build.   Enough Xenon would make living on the moon quite a bit easier.   Still wouldn't be easy, but definitely easier.


Theoretically could the expansion of the universe affect acceleration due to gravity? I was wondering if theoretically the universe expanding affects speed or acceleration due to gravity, compared to if space weren't accelerating.

The way I have it in my head, something is falling towards a planet with gravity G acting upon it. It's been falling for T amount of time and has gone X distance with Y left to go. Currently the universe is expanding, so the distance between its starting and ending point is a very tiny bit longer when it arrives than when it started, either affecting its distance traveled at time T, or its speed at time T.  Compare this to a non-expanding universe where the distance between start and end remains the same, no matter what point in time you choose.

I know that local gravity forces and atomic forces greatly overpower expansion forces so galaxies, solar systems, planets, and watermelons don't just fly apart. I'm just wondering if the value of either gravitational acceleration, or an object's speed could theoretically be changed slightly due to cosmic expansion.

Right now the expansion of space is only observed on scales like that of galaxies and the entire universe. We say that with the fancy term globally. Expansion of space does not take place within galaxies at least for now.
As pela has rightfully noted in the comments, for now the gravitation of galaxies, stars etc. is strong enough to overcome the expansion of the universe. But as General Relativity tells us gravity is local, that is, it only affects objects that are considerably close. So, on large scales expansion continues. (I am no expert, but I think that the reason for expansion not happening within galaxies out smaller systems is that the matter and energy is more dense, resolution by in stronger gravitational fields).
It is hypothesized that the universe may end with a big rip: this means that the expansion eventually would start to affect galaxies. Galaxies, star systems, planets and eventually atoms will be ripped apart due to the expansion of space. When and if that happens, not only gravity but not even electromagnetism or the strong force would work to overcome the mighty expansion.
But this situation is highly hypothetical and maybe it won't even happen. So for now, the answer is no: the expansion of the universe could not affect gravitational acceleration.


Does the Moon's magnetic field affect Earth's magnetic field? I wanted to ask a question; it's simple but I cannot find any possible and perfect solution.

Earth has poles, North and South. By which we can get directions using a compass or a needle compass, but that's not the concern.

My Question:

Would the Moon cause a change in the magnetic field of Earth when they are interpassing or colliding with each other, as the Moon does come between the magnetic path of the Earth.

So would the Moon's magnetic field affect the Earth's magnetic field, just as its gravitational pull affects Earth's gravitational pull for oceans?

So would the Moon's magnetic field affect the Earth's magnetic field, just as its gravitational pull affects Earth's gravitational pull for oceans?

Yes, but only slightly. Firstly, magnetic fields can superimpose, so the field at any point is the sum of the field due to the Earth and the field due to the moon.

However, the moon is rather far away (and has a weak magnetic pole strength), so the magnetic field due to the moon on Earth's surface is nearly negligible (magnetic field also decreases as an inverse-square law)

In addition, the magnetic field of the Moon may bolster or erode the Earth's field as magnets moving relative to each other tend to either lose magnetization or become stronger. But this process has a negligible effect when we take the Moon and Earth.


How can a star be "thrown out of the Milky Way" by Sagittarius A*? A recent publication reports that a hypervelocity star on course to leave the galaxy was found, likely accelerated by the black hole at the center of the Milky Way in a process predicted by Jack G. Hills in 1988. A sentence from that abstract insightfully points out:

  The discovery of even one such hyper-velocity star coming from the Galactic centre would be nearly definitive evidence for a massive black hole.

The process involves a binary system coming close enough to the black hole that one of the stars is "captured" and the other one is flung out.

As far as I can see, the escaping star can at most end up with the kinetic energy of the other star on top of its own, accelerating it by a factor of $\sqrt{2}$ if the stars have the same mass. A "slingshot effect"  is impossible because the central black hole is stationary relative to the galaxy.

Does that mean that this effect necessarily involves binary systems with very different masses so that more energy is available for transfer to the lighter star? Or are there other effects at play?

It's very similar to the question noted in the comments which has a nice mathematical answer.

But some comments in your question are incorrect and worth addressing.

  As far as I can see, the escaping star can at most end up with the
  kinetic energy of the other star on top of its own, accelerating it by
  a factor of 2–√ if the stars have the same mass.

The combined kinetic energy remains unchanged, unless there's a collision or some kind of relativistic interaction, but that's not really important.    It's easier to think of a gravitational slingshot interaction between a larger object (like Jupiter) and a smaller object like a Voyager spacecraft.   The kinetic energy and momentum exchanged is equal between the two, but the acceleration to the larger object is negligible but the acceleration to the smaller object can be up to 2 times the velocity of the larger one plus the initial velocity, or 2U + V, which is more than enough to escape both the central black hole and the entire galaxy when the initial velocities are high enough.

See this question for more math.

And you wrote this:

  A "slingshot effect" is impossible because the central black hole is
  stationary relative to the galaxy.

This isn't relevant, because it's not the central object's velocity that matters but the velocity of the stars orbiting the central black hole, which can be very high.   When two stars orbiting the central black hole pass close to one another, then a gravity assist happens and the two stars exchange velocity and momentum.

This can be a hard thing to picture.   It's often described as bouncing a ball off a moving train, the train's velocity (times 2) can be added to the ball's initial velocity.   The problem with that image is that there's a point of impact where velocity is reflected and that moment never happens in an orbit.    The way I like to think about it, is, imagine you're standing on a station and a train is approaching and the train is carrying a very heavy object - say a neutron star, and you throw a ball towards, but just behind the neutron star.

You throw the ball at 100 kph and the train is moving towards the station at 100 kph.   The neutron star on the train sees the ball approaching it at 200 kph, it then swings around the Neutron star's gravity and flys away from the Neutron star at 200 kph, because orbital speeds are conserved in the hyperbolic orbit.

But if the ball is now flying away from the train at 200 kph and the train is moving towards you at 100 kph, the ball, inevitably is traveling in your direction at 300 kph, or 2U + V.

  Does that mean that this effect necessarily involves binary systems
  with very different masses so that more energy is available for
  transfer to the lighter star? Or are there other effects at play?

This is an interesting question and the math is a bit messy so maybe somebody else could do it.   Two stars of equal mass and opposite directions could accelerate each other and one would fly away from the black hole and the other would fly towards the black hole.    I think it's easier with a larger and smaller object where the smaller objects is accelerated and the larger one only moved a tiny bit, but I see no reason why two equal mass objects couldn't exchange significant velocity to each other, changing both trajectories.   That's a good question whether that would be sufficient to achieve escape velocity for one of them.


How was Mars' very noticeable change in apparent magnitude explained by the Geocentric Model? In the geocentric model, the Sun, the Moon, and the 5 naked-eye planets (Mercury, Venus, Mars, Jupiter, Saturn) all orbited Earth.
Today we know that the planets orbit the Sun, and thus they come closer to Earth at certain times than at others. This means that planets are noticeably brighter at opposition than when they're farther away.

Mercury and Venus are inferior planets and thus their change in apparent magnitude over the course of one orbit is hard to notice.

Jupiter and Saturn's orbital radii are much larger than Earth's and thus their closest and farthest approaches from Earth are not that different. This means that Jupiter and Saturn's apparent magnitudes over the course of one orbit does not change that much and is hard to notice.

However Mars' apparent magnitude very clearly varies  over the course of one orbit. It is much brighter at opposition (reaching a maximum magnitude of -2.94) than even a month later, when it is noticeably dimmer. This is also further exacerbated by the opposition effect.
How did the ancient cultures who believed in the geoentric model rationalize Mars' change in brightness? Surely they must have made the correlation that it is brightest when opposite to the Sun in the sky?

Epicycles: These result in the distance of the planet to Earth changing quite substantially over its orbit (in the geocentric model) and resulting in changes in brightness. With the right choice of epicycles, you can account quite well for the changes in brightness, certainly better than Ptolemy would have been able to measure it.
The less scientifically bent may have assumed it was just "in the nature of a planet" to be brighter at opposition (the god of war, putting out his power making his planet brighter just as it comes to rule the night sky)


Is it possible for a moon to orbit a planet floating free in the galaxy rather than orbiting a star This article got me thinking, can a planet hold a moon in orbit if it is just floating in the galaxy by itself not as part of a star system? Can a celestial body even qualify as a planet if it is floating around by itself?

The answer is Yes.. Planets that don't orbit around a star are known as Rogue Planets. There is nothing preventing a rogue planet from having one or many moons.

Not so long ago, the first candidate for a free-floating exoplanet-exomoon system was presented in this paper. It looks like a gas giant several times larger than Jupiter with a sub-Earth mass moon.

Another study calculated and simulated scenarios where planets where ejected from their orbits around a star and concluded that around five percent of Earth-sized planets who are accompanied by Moon-sized natural satellites would retain them after the event.