How to convert density from physical unit to comoving unit?

You're asking two different, but somewhat related questions: One has to do with a practical way of describing the expansion of the Universe; the other has to do with a way of dealing with our ignorance of the exact expansion rate of the Universe.
I will answer your specific question in the last paragraph.
Comoving and physical coordinates
A comoving coordinate system is one in which particles that don't move through space have fixed coordinates, even is that space is expanding (or contracting, or otherwise warping).
In contrast, the particles' physical coordinates are the actual distances you would measure between them if you froze space and started laying out measuring rods.
Expansion, scale factor, and redshift
Cosmological distances are often measured in megaparsec (Mpc), i.e. one million parsec, or roughly 3.3 million lightyears.
Our Universe expands, and we describe this expansion not by its size (because we don't know how large it is, or even if its finite), but by a scale factor $a$, which is defined to be equal to $1$ today. Thus, in the past, when everything was half the distance from each other compared to today, we had $a=0.5$. Because light moving through an expanding space is redshifted, light from a certain time when $a$ had a certain value, is redshifted ("stretched") by a certain factor. It turns out that if we observe some light to have been redshifted by a particular value of $z$, it must have been emitted at a time when the scale factor was $a = 1/(1+z)$.
Definition of comoving coordinates
Comoving coordinates are defined to coincide with the physical coordinates today. That is, if the physical distance to galaxy "CLASH 2882" is 3.4 Gpc, so is the comoving distance, always was, and always will be (save for a very small movement through space, but this can be neglected). The light we receive from CLASH 2882 is redshifted to twice its emitted wavelength, so $z=1$, and $a=1/(1+1)=0.5$. At earlier times, when $a$ was, say, $0.75$, $0.5$, and $0.1$, the physical distance to CLASH 2882 was 2.5 Gpc, 1.7 Gpc, and 340 Mpc, but its comoving distance was always 3.3 Gpc.
In other word, a comoving meter is only equal to a real, physical meter today, but was smaller in the past and will be larger in the future. To be explicit, we sometimes prepend a unit with a "c" or a "p", like this:
$$
1\,\mathrm{pMpc} \equiv \frac{1\,\mathrm{cMpc}}{1+z}.
$$
The use of comoving coordinates
Comoving coordinates are practical for instance when comparing properties of something that may or may not evolve with time, but also change with the size of space, e.g. densities. One example is the number density of galaxies. If galaxies had just been around forever, without evolving, then their number density would decrease with time proportionally to $1/a^3$. To study how galaxies form, evolve, merge, etc. it is helpful to factor this out, i.e. to use comoving coordinates. If the number density in comoving coordinates change, it must be due to some "real", astrophysical effect, not just because of cosmological expansion.
Another example is the density of neutral hydrogen, in $\mathrm{g}\,\mathrm{cm}^{-1}$. In physical units, if nothing happened this density would decrease as $1/a^3$, whereas in comoving units it would be constant. Nevertheless, when $a$ was about $0.1$ the global, comoving density dropped several orders of magnitudes, evident of a physically significant process referred to as the Epoch of Reionization.
Converting between physical and comoving units
So, if you want to convert between physical and comoving units, just remember that the comoving coordinates factor out the expansion of the Universe, and you have, for distances,
$$
d_\mathrm{physical} = a d_\mathrm{comoving} = \frac{d_\mathrm{comoving}}{1+z},
$$
and consequently, for densities,
$$
n_\mathrm{physical} = a^{-3} n_\mathrm{comoving} = n_\mathrm{comoving}(1+z)^3,
$$
Little $h$
The little $h$ commonly seen in length units (and units derived hereof) is a way of factoring out the exact magnitude of the expansion rate $H_0$ of the Universe. It is defined by
$$
H_0 = 100h\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}.
$$
That is, if $H_0=70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, then $h=0.7$, and if $H_0=67.81\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, then $h=0.6781$. The use of $h$ is something of a reminiscence from the times when we only knew that $H_0$ was of the order of $50$–$100\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, and factoring out the exact value allowed people assuming different values to more easily compare their results. It may be argued — and has in fact been (Croton 2013) — that today, where $H_0$ is known rather precisely to be around $70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, $h$ only acts to confuse. It does, however, seem to prevail, especially in numerical, cosmological simulations.
$h$ notation
The way you express your question, namely that you

want to express length in co-moving kilo-parsec per hubble constant

shows, in my opinion, a common misinterpretation of the use of $h$: The factor $h$ is not a part of the unit of some number; rather it should just be thought of as a number that is multiplied on the quantity. So, although the result is the same, I think it's better to write, say,
$$
d = 2300h^{-1}\,\mathrm{Mpc}
$$
than
$$
d = 2300\,\mathrm{Mpc}/h.
$$
$h$ in quantities
$h$ appears whenever a derived quantity depends on $H_0$. Because some quantities can be derived in different ways, you may actually encounter the same quantity multiplied by different powers of $h$, depending on how it was derived. For instance, deriving a galaxy's mass from its luminosity introduces two factors of $H_0$, so if its "true" (stellar) mass was, say, $M_\star=10^{10}\,M_\odot$,
then if you had assumed $h=0.7$ you'd write
$M_\star=4.9\times10^9h^{-2}\,M_\odot$.
But if the same galaxy's mass was derived from dynamics, you'd only have one factor of $H_0$, so you'd write
$M_\star=7\times10^9h^{-1}\,M_\odot$.
Converting to "$h$-units"
In general, to convert a number to "$h$-units" (bearing in mind that $h$ is not a part of the unit), you need to know how it was calculated. Derived distances, for example, usually scale inversely with $H_0$ (to the power of one), so it makes sense to measure in $h^{-1}\,\mathrm{Mpc}$, that is to divide the quantity by $h$. Thus, assuming $67.8\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ we may write the distance to CLASH 2882 as
$$
\begin{array}{rcl}
d & = & 3.3\,\mathrm{Gpc}\\
  & = & 2.3h^{-1}\,\mathrm{Gpc}.
\end{array}
$$
On the other hand, since derived times scale proportionally to $H_0$, they may be measured, e.g., in $h\,\mathrm{years}$ and thus, in velocities these two $h$ factors cancel, such that velocities are just measured in, say, $\mathrm{km}\,\mathrm{s}^{-1}$.
The answer to your question
We are now ready to answer your specific question of converting from cm to comoving kpc with $h$ factored out.
You need to know two things: The assumed value of $H_0$, and the redshift of the object you're considering. For the sake of this calculation, let's assume that $h=0.7$ and $z=0.1$, and that we want to write the diameter of a galaxy whose physical diameter is, say, $10^{23}\,\mathrm{cm}$ (a Milky Way-sized galaxy). Then
$$
\begin{array}{rcl}
R_\mathrm{phys.} & = & 10^{23}\,\mathrm{cm}\\
                 & = & \frac{3\times10^{-22}\,\mathrm{kpc}}{\mathrm{cm}} 10^{23}\,\mathrm{cm}\\
                 & = & 32.4\,\mathrm{kpc}\\
                 & = & 32.4 \times h \, \times \, h^{-1}\,\mathrm{kpc}\\
                 & = & 22.7 h^{-1}\,\mathrm{kpc}.
\end{array}
$$
and
$$
\begin{array}{rcl}
R_\mathrm{com.} & = & (1+z) \, R_\mathrm{phys.}\\
                & = & (1+0.1) \, 22.7\, h^{-1}\,\mathrm{kpc}\\
                & = & 25 h^{-1}\,\mathrm{kpc}.
\end{array}
$$
Similarly, a density of $n_\mathrm{phys.}=1\,\mathrm{g}\,\mathrm{cm}^{-3}$ could be expressed as$^\dagger$ $n_\mathrm{com.}=3.9 h^3\,\mathrm{g}\,\mathrm{cm}^{-3}$.

$^\dagger$ Note however, that if instead of particles per cubic centimeter you were to measure Solar masses per cubic Mpc, then depending on how you had measured the mass, there could be up to two factors of $h$ canceling some of the $h^{-1}$ in this expression.


What makes scientists already so certain that Comet C/2019 Q4 (Borisov) is a comet and not an asteroid?

It's the appearance that is mainly used to distinguish comets from asteroids. As noted in the discovery Minor Planet Electronic Circular (MPEC), numerous observers observed extended diffuse emission (i.e it looks "fuzzy" with a Full Width Half Maximum (FWHM) greater than the field stars, or shows a tail) typical of a comet. In the case of 1I/'Oumuamua, deep searches with 8-m telescopes failed to record any evidence for any extended emission, indicating it was most likely a bare rock. The much brighter magnitude of C/2019 Q4 does play a role in this, making a search for extended emission easier.

The distinction has become more difficult in recent years with the discovery of activated asteroids such as (6478) Gault, Main Belt Comets but also objects like 1I/'Oumuamua on very long period, comet-like orbits which don't seem to show any sign of emission.


Temperature of a substellar point on an airless tidally-locked planet

The planet would reach an equilibrium where the amount of heat absorbed is the same as the amount of heat radiated. If there is no way to transfer heat on the planet (no conduction, no atmosphere), then that condition must apply locally.

The flux radiated from a blackbody surface (in W/m$^2$) is given by $\sigma T^4$, where $\sigma$ is Stefan's constant and $T$ is the temperature.

If the substellar point is a distance $d$ from the star, and the luminosity of the star is given by $L \simeq 4\pi R^2 \sigma T_{*}^4$ (assuming it too is a blackbody with radius $R$ and temperature $T_*$) and assuming $d \gg R$ to avoid some unnecessary geometric unpleasantness, then the flux absorbed at the substellar point is $L/4\pi d^2$. The flux is all absorbed, since you wish to assume a blackbody.

The temperature of the substellar point at equilibrium is therefore given by
$$\sigma T^4 = \frac{4\pi R^2 \sigma T_*^{4}}{4\pi d^2},$$
$$ T = T_* \left(\frac{R}{d}\right)^{1/2}$$

Since we assume that $d \gg R$ then clearly $T < T_{*}$.

At other points on the planet's surface it will receive a reduced flux from the star, simply becasue the flux from the star is incident at an angle to the exposed surface, so the equilibrium temperature will be lower.

On the unlit side of the planet there is no illumination from the star, but an almost isotropic flux from the cosmic microwave background equal to $\sigma T_{\rm CMB}^4$ over the entre surface. Therefore in the absence of any other source of heat, then that side will assume the temperature of the CMB at equilibrium.


Would exoplanets like those of Kepler-70 possess magnetic fields?

Although it's not proven, I've just read a research paper which provides evidence for the planets having magnetic fields. Assuming that they exist, of course.

The paper is "A tidally destructed massive planet as the progenitor of the two light planets around the SDB star KIC 05807616."

The authors state:

  Considering that the planets have suffered a recent strong tidal deformation with continuous strong tidal force from the central EHB star, and that they are heated by the central star radiation, they are expected to be hot and liquid, and possess a differential rotation. It is very probable that a strong dynamo is operating in each planet, leading to the required magnetic field.

Bear and Soker also calculate the expected time of evaporation for the closer of the two planets, resulting from the star's intense UV radiation. They arrive at a figure of $\approx 10,000,000$ years. However, Kepler-70 has been a B-subdwarf for 18.4 million years (source: Charpinet et al.) and its exoplanets would therefore be at least that old.

They state:

  the expression is highly uncertain, and we might overestimate the evaporation rate.

However, their calculations show that a planetary magnetic field about 10 times as strong as that of Earth is plausible, and would suffice to protect the exoplanet from evaporation. They find this more likely than uncertainty in the aforementioned calculation.

  Our preferred explanation for a low evaporation rate is the existence of a magnetic field ...  More likely we find the possibility that a planetary magnetic field, about 10 times as strong as that of Earth ... will substantially reduce the evaporation rate by holding the ionized gas.


Can stellar black holes cancel each other?

Your logic isn't unsound. By symmetry there would be a gravitational neutral point between them.

Of course, the black holes could not then be described by the Schwarzschild metric, since that only applies to a spherically symmetric situation.

As you brought the black holes closer together, the isopotentials would become distinctly aspherical, the event horizons would be complicated shapes and ultimately would merge - although as the black holes would presumably be in orbit and rotating, the spacetime geometry would be even more complex than that.


How does the Sky in Google Earth Pro work?

Stellarium is an open-source software that has similar features. The code is available on github, and the developer's documentation is here.

Some information about Stellarium scripting can be found in Stellarium simulation and “Real time” Stellarium Scripting?


About more than 15 stars moving in straight line through the sky

These will be SpaceX's second batch of 60 Starlink communications satellites. They were launched on 11 November 2019, and are still in roughly the same orbit, although they are slowly dispersing. Enjoy the show. There should be plenty more. SpaceX plans to launch several thousand of these satellites. Unfortunately, this is not good news for astronomers, since they may interfere with ground-based observations.


Why is there a supermassive black hole at the center of every galaxy?

As some of the comments pointed out, there isn't a definite answer since this is still ongoing research. However, there are some things that might explain it:

Galaxy formation: Current theory says that dark matter is non-uniformly distributed in the early universe. Once radiation and baryonic matter decouple, matter flows towards the dark matter overdensities. We know that stars (Pop III) must form very early during this, because their light reionizes the universe. These early stars are thought to have a mass of up to 1000 solar masses. It's possible that some of the infalling gas collapses straight into a black hole.
Density profile: A galaxies' density rises sharply towards the centre. So, during the formation of the galaxy, the gas rushing inward might have formed a black hole without going through the process of star and supernova. Also, this means there's enough to feed the black hole once it's formed.
Dynamic friction: Imagine a self-gravitating system of heavy and light objects (like a star cluster or, indeed, a galaxy). A heavy object passing through an ensemble of lighter ones will lose some of it's kinetic energy to them, which means it will "sink" towards the centre of the gravitational well. This is well observed in globular clusters. Problem: The timescale for this to happen is far too long to matter for large galaxies.
There seems to be a correlation between the mass of the central black hole and the mass of the elliptical galaxy or the bulge of the spiral galaxy. So, some process makes sure that the central black hole gets to grow along with its galaxy.

Coming back to your question: There are two parts to it, one: why a black hole at all, two, why at the centre. The second part is simpler, it would get there eventually, and probably (let's say in a galaxy merger) quite quickly. For the first part, there is no definite answer but the points above are some indications as to why it's not too surprising that there should be one.

As a last point, black holes at the centres of other galaxies are not at all easy to find. Unless the black hole is active (quasar), there are at best very indirect detection methods (velocity dispersion in the central part etc.).


B-modes in the raw CMB data can be caused by primordial gravitational waves, as well as by cosmic dust. After subtracting the cosmic dust polarization from the BCEP2 data, the residual signal is too weak with respect to noise to be statistically sufficiently significant to claim a discovery.

Or stated in a different way: The observed B-modes can be explained away by cosmic dust.

This doesn't prove the absence of primordial gravitational waves. But it doesn't provide sufficient evidence for their presence either.


What is tilt anisoplanatism for Laser-guide-star usage in adaptive optic systems?

Isoplanatism  commonly refers to a region of angles over which a ground-based telescope observes effectively the same atmospheric turbulence (e.g. an "isoplanatic patch"), such that a laser guide star provides effective correction of atmospheric seeing.

an-isoplanatism refers to a lack of isoplanatism, or a way in which  the science target and the reference laser guide star differ.

A single laser guide star does not provide knowledge of tip or tilt errors since the up-going beam experiences tilts in addition the down-going signal; these two sets of tilts are entangled and in the case where the transmitter and receiver are identical the laser guide star appears stationary. See a detailed overview and history of this problem the introduction of Ragazzoni
 1996, http://adsabs.harvard.edu/full/1996ApJ...465L..73R. This is commonly solved by observing a "natural guide star" position, since its light only passes through the atmosphere once, and using that signal to maintain the position of the target star by changing the angle a small ("tip-tilt") mirror.


Why are nearby stars like Proxima Centauri and Barnard's star not visible to the naked eye?

John's answer is correct.   For a few more details:

Stars brightness works out to roughly the 4th power of their relative mass.  This falls off a bit for very large stars, but for smaller stars it's in the ballpark.   (Red-giant phases not included.)

Proxima Centauri is about 12% the mass of our sun, and about 14% the sun's diameter.   At 12% its mass, its brightness, using the 4th power rule, is about 0.02% or 1/5,000th as bright.   That's total light.  Because it's smaller and colder, much of that light is in the infra-red range that we can't see.   In terms of visible light or visual magnitude, it's about 1/20,000th as bright as our sun.   That means, for it to be as bright as our sun from Earth in visible light, it would need to be about 1 million KM away.   If it was as far away as Pluto, some 40 AU on average, it wouldn't look very different than Venus at Venus' peak brightness, with a distinct red color.

Larger stars can have a brightness over 100,000 times brighter than our Sun.   So, because the mass of the star affects the brightness to about the 4th power, what it really comes down to is size matters.

Distance matters too, but only to the square of the distance, so mass of the star matters more.   Here's an article with a chart of the visually brightest stars in the sky.  Negative numbers and smaller numbers correspond to greater brightness.

It might go without saying, but virtually none of the stars that can be seen with the naked eye are red dwarf stars even though they're the most common type of star.  They're too dim for us to see at multiple light-years distance.


Explanation of resonance stability and instability

Mercury and Spin-orbital resonance is pretty straight forward.  Planets and Moons are gravitationally lumpy and large bodies are somewhat fluid, even rocky bodies.   Both aspects are prone to tidal forces and that can lead to spin-orbital resonance if the tidal forces are strong enough.   Mercury's somewhat high eccentric orbit balances out with a 3:2 spin-orbital resonance.

Resonance in 3 body orbital system is much more complicated and the math is above my pay-grade.  A proper answer to this question would have lots of math where orbital stability can be measured over long time periods and thousands of orbits.  Something precision is probably lost in an intuitive explanation.

The standard, 2 paragraph answer on Kirkwood gaps is that, if the Kirkwood object is in resonance with Jupiter, then Jupiter gives it a tug whenever it passes closest to the planet and those tugs over time destabilize the 2:1, 5:2, 7:3, 3:1 and less populated 4:1 orbital resonances.

See this summary.

  The Kirkwood gaps are regions in the main asteroid belt that have been
  cleared of asteroids by the perturbing effects of Jupiter, named for
  Daniel Kirkwood who discovered them. The Kirkwood gaps are due to
  resonances with Jupiter's orbital period. For example, an asteroid
  with a semimajor axis of 3.3 AU makes two circuits around the Sun in
  the time it takes Jupiter to make one and is thus said to be in a 2:1
  resonance orbit with Jupiter. Once every two orbits, Jupiter and such
  an asteroid would be in the same relative positions, so that the
  asteroid would experience a force in a fixed direction. Repeated
  applications of this force would eventually change the semimajor axes
  of asteroids in such orbits, creating gaps at that distance. Gaps
  occur at 4:1, 7:2, 3:1, 5:2, 7:3, and 2:1 resonances

This is only a partial answer to your question, because 3:2 (Hildas) and the much less populated 4:3 (Thules) collect around the resonance, they don't move away from it, and as you noted, Pluto and other Resonant Trans Neptunian Objects (TNOs) or Plutionos collect around the resonance as well.   They also get tiny tugs from the planet each orbit, but those tugs tend to keep them around the resonance rather than pulled away from it.

As a perhaps obvious sidebar, Resonant objects, while they orbit the sun, the need to be "herded" so to speak, by a single planet.   That's why there are relatively few resonant objects between the gas giants, and the vast, vast majority of them are between Mars and Jupiter, because Mars is small enough to not have much gravitational influence, or outside Neptune, because there are no large planetary bodies in the vicinity outside Neptune.   That's why Kirkwoods and Hildas and Resonant TNOs dominate the known orbital resonance objects.

Cis-Jovian, for inside Jupiter's orbit or Trans-Neptunian, for outside Neptune's.   In time, we might discover Planet Nine and it might have a collection of resonant Trans and Cis-Ninian objects, but that remains to be seen, and I only mention it because I wanted to be say the word Cis-Ninian.  Obviously that won't stick if/when the planet is found and given an official name, but for now, there are theoretical resonant Trans and Cis-Ninian objects.   In fact, it's that theoretical resonance that lead to the hypothesis that Planet Nine was out there and gave a clue where to look.

So . . . meat of the answer:

Jupiter's 3/2 Hilda's, for the most part, have their aphelions passing close to one of Jupiter's Lagrange points, L3, L4 or L5.  That's why the Hilda's form a kind of triangle.   They don't orbit in a triangle, they orbit in slightly perturbed elipses, but together, they form a triangle because there are 3 main categories of Hildas, those with aphelions in Jupiter's L3, L4 and L5.

This is explained in more detail here, under Dynamics

  The Hildas taken together constitute a dynamic triangular figure with
  slightly convex sides and trimmed apices in the triangular libration
  points of Jupiter—the "Hildas Triangle".2 The "asteroidal stream"
  within the sides of the triangle is about 1 AU wide, and in the apexes
  this value is 20-40% greater. Figure 1 shows the positions of the
  Hildas (black) against a background of all known asteroids (gray) up
  to Jupiter's orbit at January 1, 2005.7

  Each of the Hilda objects moves along its own elliptic orbit. However,
  at any moment the Hildas together constitute this triangular
  configuration, and all the orbits together form a predictable ring.
  Figure 2 illustrates this with the Hildas positions (black) against a
  background of their orbits (gray). For the majority of these
  asteroids, their position in orbit may be arbitrary, except for the
  external parts of the apexes (the objects near aphelion) and the
  middles of the sides (the objects near perihelion). The Hildas
  Triangle has proven to be dynamically stable over a long time span.

This resonance where the Aphelion is about 60 degrees or 180 degrees away from Jupiter keeps the Hilda objects from getting too close to Jupiter, which assists their long term stability.  Too close for a resonant object is generally bad.

Pluto, may have a similar resonant pattern with Neptune (note, I need to check my sources, I misread the linked answer) - will try to follow up.
 some details in the accepted answer to this question.

The Kirkwood objects are mostly too close to the sun to be gravitationally tied to one of the 3 Trojan points.   Passing close to Jupiter's orbit would make a 2:1 resonance impossible.   As a result, there isn't a stable area for them to settle in, and they are pushed by Jupiter away from resonance, unlike the 3:2 resonant Hilda's which orbit through one of the Trojan regions.  That's the best I can work this out at least.

I'm not sure I explained that as well as I should have, but that's the gist of it, as I understand it.  The requirements for resonance are very specific and the Kirkwood objects don't have a stable zone to gather around, so the periodic tugs of Jupiter sheppard them between the resonances rather than hold them to the resonances.


How long can afterglow last after a fireball?

This is an persistent ionization trail. As the meteoroid goes through the upper atmosphere at high speed (sometimes more than 45km/s) electrons are stripped from atoms. As the electrons recombine, the gas glows.

Persistent trails are formed by brighter meteors, and can be visible to the naked eye. They may be visible for up to about 45 minutes, but that depends on the meteor. Generally brighter meteors have brighter trails.

The movement and streaking in the trail can reveal the presence of upper atmosphere winds. As radio waves bounce off the ionised trails, they can be used for long distance radio communictions.

Persistent trails were discussed by the "Bad Astronomer"

Image taken during the 1966 Leonid Meteor storm, produced by a Mag -6 meteor


How did we differentiate lights from stars?

If two lights are shining in your room, one red and one blue, how do you know which one is red and which is blue?

If you look at the two lights, the lens in your eye will focus the red light at a different place from the blue light, so you can see two lights in different places. Even though the light that reaches your eye is mixed, the lens can separate the light.

Telescopes have lenses (or curved mirrors that do the same thing). The light from different stars is focused by the telescope, to form an image, and so the light from different stars is separated. The light from individual stars can then be split into colours. This is the star's spectrum.

The job of the telescope is two-fold: collect a lot of light and then focus it so that individual objects can be seen and analysed individually.

There are many ways of classifying a spectrum, Stars have continuous spectra crossed by dark lines. The temperature of the star changes the spectrum, the stars are classified in order O B A F G K M from hottest to coolest. Nebula sometimes have emmission spectra which looks mostly dark, crossed by bright lines.


How does an absolute horizon form before the apparent horizon?

I guess you are talking about black hole formation here. If we take the Oppenheimer-Snyder model for the spherically symmetric collapse of a star, then an event horizon forms first followed later by an apparent horizon that is at, or interior to, the event horizon.

The event horizon is the surface behind which light rays will never reach an infinite distance from the black hole. The apparent horizon is the surface behind which outwardly directed light rays will move inward.

The two will coincide for a static black hole, but for a forming black hole with a growing mass, it is possible for an outwardly directed photon to be outside the apparent horizon, but inside the event horizon, because the black hole mass is inevitably larger in the future if some of its future mass is still beyond its final Schwarzschild radius. As the event horizon forms at the centre of the star and moves outwards, then it is clear that the apparent horizon, which is always interior to the event horizon, must form later.

In the very simplified Oppenheimer-Snyder model, featuring an initially uniform star comprising of pressureless "dust", the apparent horizon first forms just as the surface of the collapsing star coincides with the event horizon and the apparent horizon is therefore always coincident with the event horizon. In more realistic models the apparent horizon forms a little earlier and then moves outwards to coincide with the event horizon as the surface of the collapsing star moves inside the event horizon.

NB: Note that whilst the apparent horizon can be defined at any point in time, the "event horizon" referred to above is a theoretical ideal, since we cannot preclude that the black hole might accrete some more mass in the future!


What is a good way to start learning Astrophysics

This question is very general/broad and there is no clear answer possible (neither is there a single good way to start learning Astrophysics) so you will get a lot of subjective answers. I imagine that it may not be the typical Q&A style on this platform. But anyway, here's my subjective answer:

The first book that comes to my mind is:

The Anthropic Cosmological Principle by John D. Barrow and Frank J. Tipler

This is the highest level astronomy book that I read during middle/high school and made me decide to study astronomy. It is relatively easy to follow the physics and it has a nice story in the background (the anthropic principle) which makes it a page turner

My introductionary freshman's University book

An Introduction to Modern Astrophysics by Bradley W. Carroll and Dale A. Ostlie

This contains the basic stuff. It is easy to read and covers a broad range of topics.

What helped me understand relativity, but I imagine it is a great source for much other science topics:

Feynman Lectures by Richard Feynman

When I did not understand my course note the story from Feynman helped me through it. I actually only read the special relativity part when I needed to pass my examy, but it covers a much larger range (it is a shame I have still not read the rest while that chapter about relativity had such great impact).

Understanding quantum mechanics was much more difficult for me than relativity. I had to pass the barrier of accepting the principle rather than keep challenging it.

Introduction to Quantum Mechanics by David J. Griffiths

This is a very standard introductory text. Clearly written with simple mathematics.
Inleiding tot de golvings- en de quantum-mechanica by Adriaan D. Fokker

I like reading old-fashioned stuff. It keeps me stick to the hard facts and potentionally makes me understand better why, but espescially how, we got to construct the theoretic ideas (This particular one is hard to find; another more available good classic is Ernst Mach's the science of mechanics).

This work is an early description of the application of wave-mechanics. You can sort of see the birth of the idea of quantum mechanics

Nice works to read:

From quantum fluctuations to cosmological structures: Proceedings of the First Moroccan School of Astrophysics

There must be as many books as stars? This is one of those books. I got it from some second-hand store. This sort of stuff is what keeps me connected to astronomy after quitting my studies in astronomy.

I like to read in it since it covers a wide range of topics and is not too much in-depth. It is pretty old and a bunch of articles, but it is nice to read from this moment in time (I hate it that I can not set back the clock on the internet and get an old snapshot from it. We have the wayback archive but it isn't working sufficiently to generate an old snapshot).
Introduction to cosmology J.V. Narlikar

.... to be continued

Stuff on relativity

Space - Time - Matter Paul S. Wesson
Spacetime and Singularities an Introduction Gregory L. Naber

.... to be continued

Some other stuff that worked on me

Hydrodynamic and Hydromagnetic Stability by Subrahmanyan Chandrasekhar

This was my reference work for my Bsc thesis
an amateur radio telescope by George W. Swenson

When I was a teenager I though I could easily make a radio telescope (not).


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.


Would Hubble Space Telescope improve black hole image observed by EHT if it joined array of telesopes?

No, it would not, because it operates in the visible spectrum and the EHT is an array of radio telescopes. For the "very long baseline interferometry" technique to work, all the telescopes have to be operating at the same wavelength, because combining the signals involves measuring exactly how well the peaks and troughs of the radio waves from the different telescopes line up.
You can do VLBI in the visible spectrum, but you have to match up the waves even more precisely since light has shorter wavelengths than radio waves. The EHT collected all the data first, and then spent a lot of time combining it by computer, but that required using very precise clocks to sync the data. We don't have clocks precise enough to do that for light, so a direct optical connection is required between the multiple telescopes. So there isn't a good way to do a planetwide VLBI optical telescope yet.
Addendum inspired by the comments, especially from TazAstroSpacial: To get the same resolution with optical light, you can use a much smaller telescope array. I didn't mention that fact because I was thinking about keeping the size of the array the same and getting better resolution. But in any case the problem is that the challenges of doing the interferometry at smaller wavelengths more than make up for the advantages of needing a smaller aperture. At least at the current state of the art.


Current delay for radio signals from the Mars InSight mission.

The delay at the time of the landing was 8.07 minutes (8 minutes 4+ seconds), as Mars was at the time at a relative distance from Earth of 0.9703 Au.
Current delay as at 8th December 2018, we're at a distance of 8.85 (8 minutes 51 seconds) light minutes or 1.064 Au.

Using WolframAlpha's tool here, you should be able to satisfy your curiosity without effort.

Times taken are at midnight on those dates and vary minimally throughout the day (eg. the delay on the day following launch at midnight was approx four seconds longer).

This does not take into account the time delay between the signal being sent to the satellite and it being decoded by your box and displayed by your tv or the time it may have taken to be distributed over the net if you watched it that way. I'm not able to determine if any Broadcast-Delay was deliberately introduced, which is a common practice and commonly is about 7 seconds.


Gaia Gbp - Grp: why does it get larger as star gets redder?

But you demonstrate that you know the answer to this question! Fainter objects have larger magnitudes. So as the white dwarf becomes cooler and redder, its blue magnitude $(G_{BP})$ grows by more than its red magnitude $(G_{RP})$ and the colour-index $(G_{BP} - G_{RP})$ becomes larger. This is very similar to the commonly used $B-V$ colour, which is larger for cooler stars.


Could rocks from Earth have reached the Kuiper belt, or Neptune at least? If so, how?

(I have tracked down the reference that I made in my comment on the question.)
Presumably earth rocks once blasted into space (by volcanism or meteorite impact) gain or lose energy primarily by orbital interactions with planets or moons. Since these interactions will occasionally give enough energy to escape the solar system entirely, presumably anything short of that (Neptune's orbit, the Kuiper belt) is also possible. The questions is if this will happen often enough to create a reasonable expectation that an Earth rock could have collided with any given solar system body.
Various attempts to identify the odds have been made.  In one, researchers simulated the path of fragments from the Chixculub impact, and found that many may have made it to the outer Solar System within 10 million years.
This is the paper in Astrobiology


Without using absolute magnitudes or isochrones, how might we tell a star's age and evolutionary status?

The spectra of a red giant and a red dwarf are completely different, so there isn't really too much to say about this and distinguishing giants and dwarfs is simple. For example, alkali lines are almost non-existent in red giants, but strong in red dwarfs. The theory as to why this happens is to do with the surface gravity and pressure broadening; it is the stuff of a standard graduate/undergraduate course on stellar atmospheres, not an SE answer.

The fact is that a R=50,000 spectrum with decent signal to noise ratio will quite easily
give you the temperature (to 100K), surface gravity (to 0.1 dex) and metallicity (to 0.05 dex), plus a host of other elemental abundances (including Li) to precisions of about 0.1 dex.

What can you do with this:

You can plot the star in the log g vs Teff plane and compare it with theoretical isochrones appropriate for the star's metallicity. This is the best way to estimate the age of a solar-type (or more massive) star, even if you don't have a distance and is the most-used method. How well this works and how unambiguously depends on the star's evolutionary stage. For stars like the Sun, you get an age precision of maybe 2 Gyr. For lower mass stars, well they hardly move whilst on the main sequence in 10Gyr, so you can't estimate the age like this unless you know the object is a pre-main sequence star (see below).

You can look at the Li abundance. Li abundance falls with age for solar-mass stars and below. This would work quite well for sun-like stars from ages of 0.3-2Gyr and for K-type stars from 0.1-0.5 Gyr and for M-dwarfs between 0.02-0.1 Gyr - i.e. in the range from where Li  starts to be depleted in the photosphere until the age where it is all gone. Typical precision might be a factor of two. A high Li abundance in K and M dwarfs usually indicates a pre main sequence status.

Gyrochronology is not much help - that requires a rotation period. However you can use the relationship between rotation rate (measured in your spectrum as projected rotation velocity) and age. Again, the applicability varies with mass, but in the opposite way to Li. M-dwarfs maintain fast rotation for longer than G-dwarfs. Of course you have the problem of uncertain inclination angle.

That brings us to activity-age relations. You can measure the levels of chromospheric magnetic activity in the spectrum. Then combine this with empirical relationships between activity and age (e.g. Mamajek & Hillenbrand 2008). This can give you the age to a factor of two for stars older than a few hundred Myr. Its poorly calibrated for stars less massive than the Sun though. But in general a more active M-dwarf is likely to be younger than a less active M dwarf. It should certainly distinguish between a 2Gyr and 8Gyr M dwarf.

If you measure the line of sight velocity from your spectrum, this can give you at least a probabilistic idea of what stellar population the star belongs to. Higher velocities would tend to indicate an older star. This would work better if you had the proper motion (and preferably the distance too, roll on the Gaia results).

Similarly, in a probabilistic sense, low metallicity stars are older than high metallicity stars. If you were talking about stars as old as 8Gyr, these would be quite likely to have low metallicity.

In summary. If you are talking about G-dwarfs you can ages to precisions of about 20% using log g and Teff from the spectrum. For M dwarfs, unless you are fortunate enough to be looking at a young PMS object with Li, then your precision is going to be a few Gyr at best for an individual object, though combining probabilistic estimates from activity, metallicity and kinematics simultaneously might narrow this a bit.

As an add-on I'll also mention radio-isotope dating. If you can measure the abundances of isotopes of U and Th with long half lives and then make some guess at their initial abundances using other r-process elements as a guide then you get an age estimate - "nucleocosmochronology". Currently, these are very inaccurate - factors of 2 differences for the same star depending on what methods you adopt.

Read Soderblom (2013); Jeffries (2014).

EDIT: Since I wrote this answer, there is at least one more promising method that has emerged. It turns out that the abundance of certain s-process elements (e.g. barium, yttrium) are enriched quite slowly during the lifetime of the Galaxy (by the winds of dying asymptotic giant branch stars), and more slowly than enrichment with iron and much more slowly than alpha elements like Mg and Si. Thus a measurement of the relative fractions of these elements, like [Y/Mg] can give the age to precisions of a billion years or so (e.g. Tucci Maia et al. 2016; Jofre et al. 2020). This method is probably the best for solar-type stars older than a Gyr, but remains unexplored/uncalibrated for lower mass stars.


According to the blog How old is the Sun in Galactic years? (Siegel, 2008), a Galactic year is about 223 million Earth years.

The age of the Sun, according to the Stanford Solar Centre's page How old is the Sun? is approximately 4.57 billion years.

This would put the sun at approximately 20.5 galactic years old.


What stabilizes rings or accretion disks?

I believe the two biggest things that determine whether or not a ring system forms into a moon depends on are the parent planet's Roche Limit and whether or not there are other moons that can disrupt the ring.

The roche limit of a planet is how close a large body can orbit the planet without being torn apart by the planet's gravitational pull. The article I've referenced can show how this is calculated. Small moons like Jupiter's Metis can orbit within the roche limit as they are so small, the difference between the tidal forces acting on their near and far sides is not large enough to tug the moon apart.

The presence of other moons - or in the case of stellar disks, other planets - will play a large role. Like this article explains about our asteroid belt, the gravitational perturbations of Jupiter prevented a planet forming between it and Mars. Smaller bodies cannot easily gather together when a beast like Jupiter swings by to scatter them all!

This also means though that most asteroids within the belt are confined between Mars and Jupiter. Around Saturn's F ring, there are two small moons that slow down / speed up debris within the ring so that it remains gathered in a small space. This does keep rings and disks stable, but it also stops larger bodies forming.

Long story short; if the disk if far enough away from its parent, and nothing else large is orbiting nearby, you should get a moon / planet. If it's too close to the parent and being shepherded by another large body, it's more likely to remain a stable disk.


Buoyancy, following Archimedes' law, depends on the volume the body and the density of displaced material. The force is
$$ F_{buoy} = \rho_{gas} \cdot V_{body}\cdot g$$
which has to act against the bodies' own weight $F_{weight} = -m_{body} \cdot g$.
Here $\rho$ is density, $V$ is Volume, $g$ the local gravity.

If you're referring to this video, then the aerogel that was created in the 1980's is essentially a high-tech balloon.

There are a few points that need discussion here.

1. Balloons the size of continents don't occur naturally anywhere.
I think we can take this as given.

2. The filling gas
acts to decrease $F_{weight}$ and its pressure must be enough to hold up the body volume. Else the body will collapse and the buoyancy decreases to zero.

A floating continent would need some appropriate filling gas in large quantities, that has lower density than the gas giant atmosphere.
Impossible is a strong word, but this seems very impossible to me. Also for the continent then be able to uphold its structure under its own weight would be very difficult.

3. Particles either float or experience drag.
If they float, they're coupled to the gas flows. Once they would grow larger, they start experiencing gas drag. When this happens, they loose their momentum and are dragged down into the depths of the planet.
They then will contribute to the structure of the deep planet, but are lost to any continent-building process.

So even if your floating continents seem impossible, I think is not a stupid question at all. I might just throw this on our students and test their understanding of buoyancy!


Why are there uneven bright areas in this photo of black hole?

No, you aren't seeing the shape of the accretion disk. Although its plane is almost that of the picture it is far larger and fainter than the ring that is seen. The reason for this asymmetry is almost entirely due to Doppler beaming and boosting of radiation arising in matter travelling at relativistic speeds very close to the black hole. This in turn is almost entirely controlled by the orientation of the black hole spin. The black hole sweeps up material and magnetic fields almost irrespective of the orientation of any accretion disk.

The pictures below from the fifth event horizon telescope paper makes things clear.

The black arrow indicates the direction of black hole spin. The blue arrow indicates the initial rotation of the accretion flow. The jet of M87 is more or less East-West (projected onto the page), but the right hand side is pointing towards the Earth. It is assumed that the spin vector of the black hole is aligned (or anti-aligned) with this.

The two left hand plots show agreement with the observations. What they have in common is that the black hole spin vector is mostly into the page (anti-aligned with the jet). Gas is forced to rotate in the same way and results in projected relativistic motion towards us south of the black hole and away from us north of the black hole. Doppler boosting and beaming does the rest.

As the paper says:

  the location of the peak flux in the ring is controlled by the black hole spin: it always lies roughly 90 degrees counterclockwise from the projection of the spin vector on the sky.


In general, solids are made from from atoms that originate naturally in stellar nucleosynthesis,
see for example the composition of our sun's atmosphere from an older paper of
Asplund et al (2005):

Here you see for example, that silicon (Si), an important rock-forming element, has an abundance relative to hydrogen of $10^{-3}$, or for every 1000 hydrogen atoms there is one silicon atom. Also we find lots of oxygen, carbon, magnesium, iron,... all known rock-forming elements.

The above work has been performed using spectroscopy, and in a similar way we can detect solids in space mainly through three means:

Spectroscopy (again): Small particles of $\mu m$-sizes exist in vast numbers in the spectrae of red giants (I'll comment on the formation later), white Dwarves, in cold molecular clouds and even in far away Quasars. Those numbers are so huge that we can detect prominent features like the 10 $\mu m$-feature of the $Si-O$ bond that we find in many earthly minerals. See also the following example of a clear detection of silicate minerals in the envelope of a young star (the broad trough at 10 $\mu m$) (c)Uni Hawaii:

As mentioned, when solids grow, their numbers diminish. Thus big solids around the sizes of mm become invisible in spectrae, but sometimes can be detected through scattered light, as they undergo Mie-scattering with visible or infrared light when being very near to a stellar source.
Then one can take pictures like the one from Benisty et al. 2015 of the young stellar system MWC758, where we see dust spirals possibly caused by the existence of young planets:

Reflection of solar light: As solids grow to cm, meter, kilometer-sizes they become invisible for the above methods. Only at sizes of tens to hundreds of kilometers and only in our solar system can we then detect them again through the reflection of solar light.

Now that we have established that solids exist in space, let us review the theoretical perspective of growth mechanisms:

dust of up to $\mu m$ condenses out in the relatively cool atmospheres of red giants (AGB stars), where the ionized environment helps to grow solids through charge separation. Details e.e. in here.
this dust is blown away in the strong AGB star winds into the interstellar medium, where it helps to create new star systems (see also the second image I linked).
All successive growth to larger sizes has to happen in the interaction with gas, thus it usually happens in the relatively dense environment around young stars
The dust has now to overcome the bouncing barrier in order to grow into cm to meter-sized particles that we call 'pebbles'. How particles overcome this barrier is unclear in general, but the article I've linked mentions a few solutions.
Pebbles are defined by their property that they interact with the orbiting gas in protoplanetary discs through friction. This interaction allows them to trigger the streaming instability, a very efficient mechanism that is able to compress clumps of pebbles directly into km-sized solids.
Now small comets grow through direct collisions, up to planetary sizes. Only in this stage geological mechanisms can play a role, like the igneous separation of minerals that you've mentioned.

That's it for now. I think with this description it is clear that the whole problem is not chicken-and-egg, but if you think otherwise then please comment.


How to interpret error of absolute magnitude of a star given data from wikipedia?

Spica (α Vir) is a binary star.
The Wikipedia article cites Herbison-Evans et al. 1971, whose Table III says:

  Absolute magnitude of primary (Mv1)   -3.5±0.1
  Absolute magnitude of secondary (Mv2)   -1.5±0.2

So the confusing "-3.55 (-3.5 / -1.5)" can be interpreted as "combined (primary / secondary)."


How to calculate when the moon is highest in the sky to the earthbound observer?

The Moon moves round the ecliptic. At full Moon it is opposite the Sun. Hence when the Sun is furthest South (in the Northern Winter), that full Moon is that which will be furthest North in that year and so on. But plainly, the Moon moves round the ecliptic every 28 days, so every 28 days or so it is furthest North etc.

In other words there is no answer based on phase or time of year alone.


It has been observed 1977, and in 1994.
It should be near its perihelion this year (2014) around august 7 (using wikipedia data, perihelion at March 6, 2005 and period 4.71 years), give or take a few days.


Do asteroids originally come from debris of star remnants?

The asteroids (which are mostly found in the asteroid belt between Mars and Jupiter) and more distant, icy bodies found in the Kuiper belt and the Oort cloud were formed at the same time as the rest of the solar system.

The sun and the planets formed from a molecular cloud, containing mostly hydrogen and helium, but also some other elements that came from the remains of stars, including supernovae. This gas cloud collapsed under gravitation to form a star, surrounded by a disk of dust and gas. The matter in the disc formed the planets, and the Kuiper belt. More distant matter in an extended halo around the star, formed the the Oort cloud.

Thus the sun neither captures nor produces asteroids and other solar system bodies. They were formed at the same time as the Sun.

Note "Kuiper" not "Kepler" belt, and the Oort cloud is the source of comets, not asteroids. Asteroids are rocky; Oort cloud bodies are made of ices.


This is a known issue affecting multiple locations in Google Sky, including a region in Orion and this area in Pegasus. If you look closely at the interface, you can see that Google Sky gets its data from preexisting surveys, including the Digitized Sky Survey and the Sloan Digital Sky Survey - Google doesn't operate its own fleet of telescopes (at least not yet), so it uses existing archival data. You can query the DSS online.

Here's what I see when I look at the region in question on Google Sky:

Now here's what I see when I look at the same region with DSS-1 data:

The blank spot is now filled with the field of stars, as expected.

If you want, you can do the same thing with the other DSS and SDSS datasets. I wasn't able to find any blank spots in the raw images, meaning that this is presumably an issue on Google's end, not an indicator of the presence of some object that needed to be hidden.


Delay between neutrino pulse and visible flash of supernova

The neutrino pulse associated with the core-collapse supernova SN1987A in the Large Magellanic Cloud, were estimated to have arrived at the Earth about 3 hours before the rise of the supernova light curve (e.g. the review by Beall 2006).

If neutrinos are massless or have small masses (which they do), then this is your answer.

If they have masses (and they do), then the answer could depend on the neutrino mass and how far away you are from the supernova, since particles with mass will travel slower than the speed of light. Therefore the "delayed", but faster, optical emission will catch-up with the neutrinos.

The neutrino masses are probably of the order of 0.1 eV, whilst the energies of neutrinos from core-collapse supernovae are around 30 MeV in energy. i.e. The neutrinos travel with a Lorentz factor $\gamma = (1-v^2/c^2)^{-1/2} \sim 3\times 10^{7}$. Thus
$$\frac{1}{1 - v^2/c^2} \sim 9\times 10^{14}$$
and so $v/c \sim 1 - 1/1.8\times 10^{15}$. That is, the neutrinos probably travel at about 1 part in $10^{15}$ slower than the speed of light.

At the distance of the Sun, this amounts to a delay of $5 \times 10^{-13}$ s, at the distance of the Andromeda galaxy about  0.006s, but for a galaxy at a distance of a Gpc, the delay would be about 100 s.

So actually, unless you are detecting the supernova on the other side of the observable universe, it makes no odds and certainly makes no difference at the distance of the Large Magellanic Cloud. So I say your answer is about 3 hours.


Is it correct to say vertical 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.


What limits the use of the H-R diagram to measure distance (main sequence fitting), what distances is it useful for?

It depends how precise you need to be. Main sequence fitting assumes that the star(s) in question is(are) on the main sequence. If you have a cluster of stars (at the same age) then defining what is on the main sequence and what isn't becomes much, much easier and of course you have lots of stars with which to beat down the statistical uncertainty.

In fact, for coeval groups of stars what you term "main sequence fitting" is rarely done. The process is the fitting of an "isochrone" (a line that links points in the HR diagram at a single age), so that both the distance, age (also extinction and metallicity) are possible free parameters.

The subtle distinction here is that what is termed "the main sequence" does not really exist in practice, or at least not as a uniquely defined locus in the HR diagram. Star begin their main sequence lives on the "zero age main sequence" (ZAMS) and end their main sequence careers at the "terminal age main sequence" (TAMS), gradually changing their luminosity and temperature as they do so.

Here is a diagram from Martignoni et al. (2014) showing the ZAMS and TAMS for stars of different masses. there is typically a factor 2-3 in luminosity between them (a larger difference at larger masses). That means whether you use the ZAMS, the TAMS, or something in between to determine the distance from a vertical displacement in the HR diagram, you could vary your answer for the distance by $\sqrt{2}$ to $\sqrt{3}$. In other words you need to know the age of a main sequence star before "main-sequence fitting" can give you an accurate distance.

Of course, lower mass stars are longer lived. Anything of say 0.7 solar masses or below has hardly moved from the ZAMS in the age of the Galaxy, so there would be little error in assuming a ZAMS locus. Conversely, the effects of age are far more rapid and therefore far more important on the main sequence at higher masses.

If you were to try to use main-sequence fitting to estimate the distance to individual stars then there are several hazards. For one, it can be nearly impossible to estimate the age of an individual star. Therefore if it has a mass greater than 0.7 solar masses then there will be an uncertainty in its position in the absolute HR diagram that leads to an inevitable uncertainty in estimated distance. Further, the intrinsic position in the HR diagram depends on the star's chemical composition. Such ancillary information might be available, but it might not, in which case that is another source of error. A further source of systematic uncertainty is stellar rotation. Fast rotating stars have extended lifetimes on the main sequence and somewhat different intrinsic positions on the HR diagram; again a source of systematic uncertainty that is especially problematic for high mass stars. Finally, it can be that what you think is an isolated main sequence star is in fact a binary system. A companion can increase the luminosity of the system and make a star appear closer than it actually is (by up to a factor $\sqrt{2}$).


What's the furthest object observable by the naked eye from earth?

This is partly a question of 'limiting magnitude' - the faintest magnitude that is visible using a particular instrument, or in this case, just the eye.

+6 - maybe +6.5  - magnitude is sometimes used as a baseline expectation of what can be seen in dark skies sites.

The most distant star that can easily be seen is Deneb. It is actually quite bright, at brighter than +2 Magnitude, but is (probably) 1425ly away. So though it is at quite some distance, it's luminosity means that actually, it isn't limiting magnitude that is at play.

There are stars much further away that are at the barely perceivable limit, and seeing them will depend on conditions. The +4 mag mu Cephei is probably fairly achievable, at 5900ly, and V762 in Cassiopeia is about +6 mag and maybe 16000ly, which would probably be the record if you can actually see it, but you'll need special conditions.

When it comes to galaxies, Andromeda is 2500000ly away, and +3.4, very visible and much further away than any individual star. If you have good conditions, the +5.7 mag M33 is slightly further away at 2900000ly, but it would be much harder to see.

While the generally accepted limit is maybe +6.5, some people claim to have seen to +8. If it is possible, M83 is 14700000ly away at +8.2, with naked-eye claims reported.


Why is the peak value(s) of the spectral radiance of the CMB different if you calculate with freq. instead of wavelength?

Because $B_\lambda$ is not just $B_\nu$, with $\nu$ replaced by $c/\lambda$.
The relationship between the two functions is that
$$ B_\lambda\ d\lambda = B_\nu\ d\nu$$
since one is defined in terms of flux per unit wavelength, the other as flux per unit frequency.
Thus
$$B_\lambda = B_\nu\ \left|\frac{d\nu}{d\lambda}\right| = \frac{c}{\lambda^2} B_\nu, \tag{1}$$
where you can then substitute $\nu =c/\lambda$ into the expression for $B_\nu$.
To find where the maximum of a function, you differentiate and set to zero. Because of the above, the wavelength where $B_\lambda$ peaks is not $c/\nu_{\rm max}$. We can see that by differentiating both sides of eq (1).
$$\frac{dB_\lambda}{d\lambda} = -\frac{2c}{\lambda^3}B_\nu + \frac{c}{\lambda^2}\frac{dB_\nu}{d\nu}\frac{d\nu}{d\lambda}$$
$$\frac{dB_\lambda}{d\lambda} = -\frac{\nu^3}{c^2}\left(2B_\nu + \nu \frac{dB_\nu}{d\nu}\right).$$
The usefulness of this expression is that you can easily see that if $dB_\nu /d\nu$ is zero at some frequency, then $dB_\lambda/d\lambda$ cannot be zero at that same frequency.


What are the most popular galaxies for which we have images?

Any such list is going to be terribly subjective. Since I'm an astronomer who studies galaxies, I'll go ahead and throw out a subjective list of the more famous, photogenic, and/or scientifically well-studied galaxies. The first six are in the Local Group (LMC and SMC are satellites of the Milky Way, M32 is a satellite of Andromeda).

Milky Way
Andromeda (M31)
Large Magellanic Cloud
Small Magellanic Cloud
Triangulum (M33)
M32
Sombrero (M104)
Pinwheel (M101)
Whirlpool (M51a)
M64 (Black Eye)
M74 (NGC 628)
M81
M82 (Cigar)
M87
M100
NGC 891
NGC 1068 (M77)
NGC 1300
NGC 1365
Centaurus A
Cygnus A


Do any particles in AGN jets escape the galaxy?

Yes, definitely.
While some matter returns to the galaxy as a so-called "galactic fountain" (e.g. Biernacki & Teyssier 2018), some material is ejected at super-escape velocity, becoming part of the intergalactic medium.
This is one of the mechanisms responsible for polluting the intergalactic medium (IGM) with metal-rich gas (i.e. elements heavier than helium). A recent observation of this is presented by Fujimoto et al. (2020).
The IGM itself is too dilute to form stars, and hence metals, itself, but is nevertheless observed to contain quite a lot of metals, usually seen as absorption lines in the spectra of background quasars (e.g. Songaila & Cowie 1996; Aguirre et al. 2008). These must be blown out from the galaxies, either by stellar feedback (through radiation pressure, cosmic rays, and supernova feedback), or by AGN activity (see also Germain et al. 2009).
Note however that although material can reach relativistic velocities, "pushed away at the speed of light" is just a tad too fast. Only massless particles can travel at the speed of light, but if you're massless it isn't really a big achievement — photons do that all the time, which is why we see the galaxies in the first place.


Is Diffraction of light observed in gravitational lensing/microlensing?

I'm not sure how to define the "edges" or obscurations in a case of gravitational lensing.  The image we produce on Earth is the combination of our local telescope lens elements and the "grav lens" a zillion Kessel Runs away from us.  Even if you use the premise that the black hole itself is a central obscuration which should cause a Spot of Arago, I don't think any other diffraction-like effect is observable from such a great distance.  All the wavefronts of  light which passes thru the GravLens region continue to expand, ending up near-parallel from Earth's viewpoint.


Stellar systems: what is the difference between virial, dynamical and thermodynamic equilibrium?

Thermal equilibrium

Thermal equilibrium relies strongly on the idea of equipartition of (kinetic) energy. In a stellar system, this means that the total kinetic energy is divided evenly amongst all the stars. This doesn't imply that the velocities are all the same; they can't, because not all the masses are the same.

Dynamical equilibrium

Dynamical equilibrium means that over dynamical timescales, the system is stable - basically, it will not succumb to core collapse due to a gravothermal instability. Note that it may not be possible for a system to reach thermal equilibrium, even if it is in dynamical equilibrium. In a system with two main types of stars, it must satisfy the Spitzer stability condition (see Fregeau et al. (2001) and these results):
$$\left(\frac{M_2}{M_1}\right)\left(\frac{m_2}{m_1}\right)^{3/2}<0.16$$
where $M_1$ and $M_2$ are the total masses of types 1 and 2.

Virial equilibrium

Virial equilibrium comes about when the system satisfies the virial theorem (see Meylan (2000)), i.e.
$$2\langle T\rangle+\langle V\rangle=0$$
where $T$ and $V$ are kinetic and potential energies.


Why are Delta Aquariids "for the southern hemisphere" while the Perseids are "for the north"?

I suspect that that is simply a statement about where you would have the best visibility of each meteor shower.  The name of each shower tells you where the radiant is, the point in the sky from which the trails of the meteors appear to originate.   Perseus in the northern sky (with the specific radiant point of the Perseids at +58 degrees declination), while Aquarius is farther south (with the radiant of the delta Aquariids at about -16 degrees declination), so a northern observer would have the Perseids radiant higher in the sky and would see more of those meteors, and vice versa for a southern observer and the delta Aquariids.
Since the meteors go in all directions from the radiant, at least some from either shower would be visible for observers over a large range of latitudes, but you’d see the most if the radiant is high in your sky.
Edit: this nice image from spaceweather.com shows clearly why the radiants have their respective locations:

(From spaceweather.com, linked from Sky & Telescope.)
These are reconstructed orbits of fireballs seen as part of these showers.  The light green are the delta Aquariids, and the darker green / teal are the Perseids.  You can see that they have very different inclinations where they intersect Earth, so they appear to originate from different directions in our sky.


I don't have detailed model calculations to hand and I'm not sure they have been done for the small range of metallicity you mention. However, in handwaving terms.

(1) Mass and composition are the most important variables. Helium abundance, in addition to metallicity, could play a role. Rotation could be a very second order effect. Very rapid rotation would lower core temperatures at the same mass.

(2) This is the general assumption if a star plus brown dwarf companion formed in a binary or even if they formed in the same cluster.

(3) Higher metallicity leads to larger opacity in the atmosphere and a larger brown dwarf. This is a small effect in fully convective objects like these, except perhaps when they are very young and have superadiabatic surface layer. However, as this is when D burning takes place, possibly there is a small effect - higher metallicity leads to a larger radius for the same mass and a lower core temperature. So the mass threshold would be slightly higher.

(4) The duration of D burning is mass dependent. The graph below, from Burrows et al. (1997), shows luminosity vs time for brown dwarfs, planets and stars. The D burning phase is the early plateau in luminosity, lasting a few million years for things which are almost stars to maybe one or two hundred million years at 13 Jupiter masses. When D fusion ceases, then cooling recommences and the luminosity will continue to fade (forever), unless the object is massive enough to ignite hydrogen (i.e. a star: the difference between the green lines and blue lines on the plot).


What is the process for vetting the 4,175 candidate planets Kepler has discovered?

Planet "candidates" are Kepler Objects of Interest (KOIs)that have a transit-like light curve and have passed a number of observational tests. They are candidates, because although they do show a transit in the light curve of the star in question, there is no independent confirmation of a planetary mass.

One problem to overcome is that of "false positives". There are other astrophysical phenomena besides planets that can mimic an exoplanetary transit in a Kepler light curve. For instance a grazing incidence stellar binary can produce what looks like a transit in a light curve; so too can a chance alignment of a star with a background eclipsing binary star. There are also a number of instrumental anomalies, such that a signal in one part of a CCD or the Kepler field can produce a "ghost" at another position that might look like a transit.

If you can obtain detailed spectroscopy including accurate radial velocity measurements you can usually rule out most of these false positives, largely by getting a mass constraint for the companion.

Even where you have a very clean transit signal and can estimate a planetary radius, a further problem is that a wide range of masses produce objects with very similar radii. ie. more massive brown dwarfs have very similar radii to exoplanets. Again, only a mass estimate, either through radial velocity measurements or sometimes through "transit timing variations" if the object is in a multiple exoplanet systems.

Now the problem with the KOIs is many of these stars are way too faint ($V>14$) to do the kind of detailed spectroscopy that is easily possible on the much brighter exoplanets around stars found in say the HATNET or WASP ground-based surveys of bright stars.

So, what you can do is tackle the problem statistically, by identifying the kinds of false-positives you might have, quantifying their influence, and throwing away suspect objects (see for instance Batalha 2012). Section 4 of this paper describes in detail some of the tests that are done: e.g. comparing the depths of odd and even numbered transits to look for asymmetries that would indicate grazing incidence binaries; looking between transits for a secondary eclipse that would also indicate a stellar companion; the shape of the transit is diagnostic but cannot easily rule out a planetary candidate; searching for motion of the "photocenter" of the source - if the photocenter moves during a transit it could indicate a background, diluted, eclipsing binary lightcurve.

An early paper by Morton & Johnson (2011) claimed on the basis of a population synthesis approach, that that astrophysical false positives were limited to less than 10%. However, a recent paper by Coughlin et al. (2014) discusses how instrumental effects can be tested for by comparing the transit periods with the periods of other known objects in the Kepler field of view. They claim that around 30% of the KOIs may in fact be false positives. Either way it looks like the big majority of the KOIs are indeed exoplanets, but identifying which ones aren't will require detailed follow-up.


What telescope is this, and how does it work?

This is indeed the VLT, or rather one of the four 8.2m Unit Telescopes (UT in ESO parlance). Looking at the optical layout (found here along with a description of the telescope), the small mirror in the centre of your screenshot is M3, the tertiary mirror. M1 is the primary (in the background), and M2 is out of the frame towards the top- the white supports you see in the foreground hold it above M1.

One advantage of this design is that you can have two large instrument platforms on the vertical tilt axis of the telescope - this way, you don't have to dangle several tons of instrument from the support structure (Nasmyth focus in the image).


Virial shocks appear when gas falls in toward a galaxy, and the shocked gas stays hot so the pressure in the post-shock gas can support the shock and keep it away from the galactic disk.
 The alternative is if the post-shock gas cools and loses pressure support, and the (non-virial) shock collapses right down to the galactic disk.  The significance is that if gas falls directly onto the high-density disk, it has different effects than if it meets a low-density shock well away from the galaxy.  There's a lot of information at https://academic.oup.com/mnras/article/345/1/349/984798


Is it possible to see an image with radio telescope of something not so far away in detail?

In brief: you cannot resolve details of an object on a scale less than roughly a wavelength or two.  The distance to the object only marginally affects this problem.

So take a look at the wavelengths your radiotelescope is capable of receiving and compare with dimensions of interesting "features" on a human body.

Further, as the comments pointed out, unless the human is either emitting radio-wavelength energy in significant amounts (unlikely) or is being "painted" by a large emitting source (as active radar systems do), you won't get any signal in the first place.


Nothing "escapes" a BH - in the sense that a signal originating inside the event horizon remains forever inside. If something is observed moving away from the BH, then it was generated outside the event horizon. If it was generated inside, it would never be observed at all, forever and ever.

Gravity itself does not "escape" a BH - and neither does "not escape". Gravity is simply a characteristic of the metric of spacetime. If spacetime is warped in a certain way, gravity can be measured to exist. A BH is simply a very powerful distortion of spacetime, nothing more, nothing less. It is generated by a concentration of mass/energy, which warps spacetime, and then that concentration becomes trapped by this distortion that it has produced.

In that sense, gravity is simply part of the BH, because gravity is spacetime being warped, and because a BH is essentially just that - warped spacetime. A BH's gravitational field is part of the BH itself, extending to infinity (but getting weaker with distance). It doesn't "escape" because there's nothing there in the process of escaping.

It's like having a plastic bag tied into a knot to keep water inside, and someone asks "so how does the plastic escape the knot?" The plastic does not "escape" the knot, the knot is part of the plastic.

This all becomes easier to understand when you realize that gravity is not a thing, it's just an effect of spacetime being distorted.

EDIT: I think what you were really asking was - can a BH have its own magnetic field? The answer is yes.

A BH can have 3 characteristics: mass, spin (rotation), and electric charge (a.k.a. the no-hair theorem). All other characteristics of the matter falling in it are lost, except these three. If you drop a proton into a neutral BH, then the BH acquires a charge equal to one proton, and that's a measurable electric field.

Now consider a spinning BH with an electric charge, the Kerr-Newman metric. You have a charge, and you have spin. That means you have magnetism. So, yes, a BH can have a magnetic dipole. However, the rotation axis and the magnetic dipole axis must be aligned - a BH cannot be seen as "pulsing". Again, no signal from inside the event horizon can be observed outside.

However, you should not imagine the electric (or magnetic, same thing) field as "escaping" the BH. It does not escape. What happens is, when the charges were swallowed by the BH, the lines of electric field remain "glued" to the BH, which then acquires a charge. Those lines of electric field have existed forever, they don't "escape" anything, and continue to exist after the charge is trapped by the BH.

Note: electric fields and magnetic fields are one and the same. One could appear to be the other, depending on the motion of the observer.


Why does a mirror bent 'like a potato chip' allow space telescopes to be smaller and have a wider field of view?

Freeform optics are a response to the specific challenge of cramming a telescope in a very limited space. A traditional instrument would have all optics symmetrical and aligned on the same axis. It would waste a lot of space within the cubesat. Also, traditional designs tend to be much longer than they are wider; they don't fit well in a cube; it is very hard to make classic instruments that are as short as they are wide.

But with freeform optics you could bounce light in a few directions within the cube. You'd still achieve a decent focal length, and you would use all the volume available to you.

Since light is reflected from mirrors at angles different from normal, you cannot use the traditional symmetric shapes such as parabolic, spherical, etc. You need to basically take a paraboloid and squish it in one direction so that it works about the same like a parabolic mirror (I'm simplifying), but at an angle of reflection of, say, 45 degrees.

In such an instrument you could have multiple "potato chip" mirrors, as in the diagram above. You have to design the instrument as a whole; computer simulations will adjust the shape of each mirror until the performance of the whole instrument is close to a classic straight design.

As far as I can tell, the manufacturing precision is such that freeform optics are only usable at long wavelengths such as infrared, where less precise optics can be used. But technology improves all the time. It also depends on how much aberration you can tolerate in your image.

For usage from ground level this is less useful, unless you absolutely need a telescope in a very small form factor for some reason. Classic optics are still preferred when space and shape are not restricted.


How many arc seconds is Sirius "spiralling motion" around Sirius B?

From Sirius A to Sirius B on the sky is around 3 arcsec on average.

The separation between Sirius A and Sirius B actually varies from 3 to 11 arcseconds b/c of their orbital motion over a long period of time (Here).

I believe you're referring to that. There's a lot of information (and even some graphs and diagrams) here: Here. The information on that website is sort of vague, but I think it'll be helpful in visualization.

Hope that helps!


This should really be a comment, but... you can usually check your calculations using a program like Stellarium (screenshot above).

From the latitude/longitude and right ascension/declination you give, it appears you are viewing Aldebaran from Chicago, IL at 7:20am UTC on 2017-12-27 (since Chicago is on Central Time and not observing Daylight Saving Time, which is 6h behind UTC).

As you can see from the above, the hour angle is about 3h16m19s (or 49.16 degrees), the azimuth (which is what I think you actually wanted instead of hour angle) is about 254 degrees, and the altitude is about 41 degrees.

The process for converting right ascension and declination to azimuth and elevation isn't difficult, and I considered posting a general answer, but I wasn't sure what format to use. Most programming languages have an astronomy library (which will be more accurate), and there are plenty of sites that already walk through how we get the formula, not to mention my https://astronomy.stackexchange.com/a/14508/21 which gives sourced formulas.


The colour of a meteor depends mostly on its chemical composition emission spectrum. To simplify magnesium meteors are noted to have an emission spectrum which results in blue to green colour. Changing factors like adding additional metals and the speed of the meteor could results in a cyan meteor compared to a blue one.

Source:
NASA on meteors


WOW! signal calculation of point source intensity

The plot is pretty straight forward.   6EQUJ5 puts the peak signal between Q and U, closer to U.   The letters represent two digit numbers, 1-9, then A-Z.  A being 10, Z being 35.   U was the highest signal strength received equivalent to a 30.

A plot fits rather neatly.

Wikipedia has a nice summary of the event.  The location is discussed here with some margin for error due to there being two listening devices and the design being, as I understand it, somewhat less than ideal.

While the signal remains unexplained, it was some 30 times stronger than baseline, but 3s and 4s were fairly common.  It was only 7-10 times stronger than signal strengths they received regularly.   I lean towards it being a natural but unexpected event, but there's not sufficient data to close the case, so to speak.


Could Tabby's star be beamed by a neutron star's jet causing irregular star spots?

I think this scenario is unlikely.  There is no evidence Tabby's star is near a neutron star.  Astronomers are monitoring Tabby's star very closely and there has not been any detection of the radio emission, gamma rays, or x-rays you would expect to see from a neutron star.  Also something as heavy as a nearby neutron star would have a lot of gravity probably show up in radial velocity measurements of the star.  Globs of dust, broken up comets, or more mundane stellar variability are a far more likely explanations for the weird dips seen in the star's light-curve.


What are some formulas that are associated with the Schmidt corrector?

The specific purpose of a Schmidt Corrector plate is to have an equal but opposite spherical aberration to the primary mirror they compensate for. So any formula you try to come up with will depend on the primary.

In fact the information on SCT's shows how it is made using physical forms, not using formulae.

From revolvy.com:

  A thin glass disk with a perfectly polished accurate flat form was placed on a heavy metal pan. The upper edge of the pan was ground at a precise angle or bevel based on the coefficient of elasticity of the particular type of glass plate that was being used. The glass plate was sealed to the ground edge of the pan, then a vacuum pump was used to exhaust the air until a particular negative pressure had been achieved. This caused the glass plate to warp slightly. The exposed side was then ground and polished to a perfect flat. When the vacuum was released, the plate sprang back until its bottom surface was again plane, while the upper surface had the correct figure.

If, however, you do wish to work out the SCT parameters from formulae, this page on telescope-optics.net has the fairly complex info you need to use this formula:

$$z=\frac{(\rho^4-\Lambda\rho^2)d^4}{4(n^\prime-n)R^3}=\frac{(\rho^4-\Lambda\rho^2)D}{512(n^\prime-n)F^3}$$

It's not an easy page to work through - but for someone who hasn't needed complicated maths for some years I could just about handle it.


There's a subtle misunderstanding there, which is rarely made clear in discussions of the Big Bang and the current model of the universe.

Big Bang did not explode to fill the universe. Matter and energy has always filled the entire universe pretty uniformly, but 13+ GYears ago, the universe was very small and its energy density was very high. But it was still completely filled. Our theories combined with our observations don't tell us whether it was infinite or finite, just that whatever it was, it was was very dense and very energetic.

From that point -- what came before, if anything is utterly beyond the current scope of physics and is pure speculation -- space itself started to expand very, very rapidly and since the mass-energy was conserved, the density of mass-energy rapidly decreased.  This hasn't stopped, though it has slowed.  Our best measurements to date suggest that it will never stop and probably is speeding up some.

When we look at the CMB, we're looking back in time to about 380,000 years after the Big Bang, to the point where the universe had cooled so that it was no longer a plasma and became transparent to light. Prior to that point the plasma was essentially opaque and there is no way to use electromagnetic radiation to see earlier.

We see the CMB 13 GYears away in every direction, but we're not seeing to the "edge" of the universe (if there is one). We're seeing to the most distant point from which light has has time to get to us travelling at lightspeed. So 13 GYears back is 13 GLightyears out.  There's stuff (probably stars and planets and galaxies and all the stuff we know and love) further away, but we can't see it because light hasn't had time to get here from there.

Basically, the universe that we can in principle observe is a 26 GLightyear bubble in a much larger universe. (BTW, all these measurements are light-travel distances. When you take GR into account it gets more complicated, but those complications are not critical to this discussion.  See Observable Universe, Distance Measures and Comoving and Proper Distances for typically Wikipedian incomplete-and-confusing but still useful articles.)

The one thing based on observation that we can say is that the universe is probably at least ten times the diameter of what we can observe. Assuming General Relativity and assuming that the larger universe is about as homogeneous as what we can see, the fact that we observe our bit of it to be very flat means that the edge or the radius of curvature or whatever makes the whole thing finite has to be at least ten times further away -- otherwise we'd see curvature in our own observable patch of space.

So the problem is that the part of the universe that we can see is observed to be too small a part of the whole shebang to make any kind of estimate of how big the entire universe is.  (It's a whole lot like trying to survey an area of a few square miles accurately enough to estimate the size of the Earth. There's so much noise in the measurements due to local topography, that all you can do is say "It's a lot bigger than a few miles.")

Caveat: If I understood you correctly, your question is about the size of the whole universe, not just the visible universe.  If I'm wrong, let me know and I'll frantically re-write this!

Note that even the parts which rely on observation necessarily rely on those observations filtered through the lenses of theory. I've tried to use only the parts of cosmology which seem to be solid and to avoid the wild speculation that produces popular science books and TV specials.  But it's still quite possible that it will all be overturned as we come to learn more. (I'll be surprised, but that's happened more than once!)


Could dark matter be considered a medium?

I don't see any reason why dark matter couldn't be considered a medium just as air or water are, but this does not mean that your conclusion is correct.

Dark matter does not interact with photons, however the reason we know dark matter exists is because of the measurable effect of its gravity.  Galaxies have a tendency to clump around it.

Light travels unimpeded at its standard rate between galaxies where we don't detect any dark matter, so interesting question though it is, I'm afraid the answer is no.


What is the farthest reported distance from observer from which a solar system body has occulted a star?

Wikipedia lists some of the most farthest objects from the sun in the solar system. The third farthest object "Eris" was the subject of one of the most distant stellar occultations observed from Earth. From this occultation, its diameter, density and albedo was calculated.
On 6 November 2010, Eris occulted a faint star (V $\sim$ 17.1) which was observed from Chile. The following data was recorded from the occultation:

Radius, RE = 1163 $\pm$ 6 km
Density, $\rho$ = 2.5 $\pm$ 0.05 g/cm3
Albedo, $p_V$ = $0.96 ^{+0.09} _{-0.04}$
Distance, ~98 AU

Eris is now near its aphelion (at 97.6 AU), and is the most remote body presently observed in the solar system. With an orbital period of almost 560 years, it will take more than 240 years from now to reach its perihelion, at 37.8 AU.

Reference:

Sicardy, B., Ortiz, J., Assafin, M. et al. A Pluto-like radius and a
high albedo for the dwarf planet Eris from an occultation. Nature
478, 493–496 (2011). https://doi.org/10.1038/nature10550
(PDF)


Could population III stars develop directly into population I stars?

No they couldn't. Population I stars contain elements like strontium, barium, gold, lead etc. that are not formed (much) in type II (core collapse) supernovae. They are also much richer in iron, nickel, manganese etc. than are stars formed from gas enriched only by supernovae.

Population I stars (like the Sun) can be termed third generation stars because they contain material that has been through at least two stars - but not just two high mass stars that exploded as supernovae.

Iron-peak elements are mainly formed and disseminated by type Ia supernova that are exploding white dwarfs. White dwarfs are the remnants of low-mass stars with long lives.

Many heavy elements (strontium, barium, lead etc) are formed by the s-process in intermediate mass stars that also have (relatively) long lives and do not explode as supernovae. Furthermore, this neutron capture requires iron-peak nuclei to act as seeds, so these stars must in turn have formed out of material already enriched with iron-peak nuclei.

It is also now thought that other elements (silver, gold, osmium etc) are formed by the merger of neutron stars, where there must be a significant delay between the initial supernovae that produced the neutron stars and the subsequent merger by orbital decay.


It would be good to specify what "majority of radiation" means. If it means "radiative energy", then this premise is surely wrong. To find the energy emitted per wavelength, take the Planck function $B(T, \lambda)$ at a certain temperature $T$ and multiply with the wavelength $\lambda$. You will find that the short-wavelength radiation always dominates for a black-body.
Wien's law gives you the peak of the Planck function. Radiation at all other wavelengths exists.

Note that objects predominantly loose most of their energy, if photons and matter are in equilibrium. If this equilibrium is disturbed, or particle accelerations play a role (like in the presence of magnetic fields), a whole zoo of non-thermal processes can also lead to important radiative energy loss.
Non-thermal processes in space can be radio emission from gyrating particles, line radiation, Masers, Bremsstrahlung, Compton radiation... Those processes then do not show a Planck spectrum, but other characteristic spectra.

Edit: I've added a plot of the Planck function at two different temperatures. In a linear plot (on the left) it might look as if there's a cut-off, but in a logarithmic plot (on the right), one sees what mathematical analysis already betrays: The function is defined everywhere. It just becomes really small on the high-energy end, as a commentator pointed out correctly.


Are there are any photographs online that approximate what the Milky Way looks like to the unaided eye?

It may be difficult to find proper photos that emulate this, both because given the tools and the subject, most people would opt for longer exposures, and also because of the general difficulty in emulating how our brain processes visual information into an image, especially in low-light or high contrast situations like this.  I think the photo at the bottom of this article may be a decent recreation of it, though it's a bit fuzzy and not so wide. Note: There are star trails forming in the image, which is indicative of it being a longer exposure. The image (unless cropped) doesn't appear to be from a wide-angle lens - I would guess somewhere in the 35-50mm range.  Perhaps a 10 secondish exposure, but my judgement is more based on result than technique.

What you can see will depend a lot upon the conditions of the night. You're probably not expecting a lot of light pollution, but you also have to keep in mind that your eyes will adjust to light as well.  A cell phone or a full moon can be enough to prevent you from being able to see well.  If conditions are perfect - little to no light pollution, no light from cars, flashlights, etc, a new moon, low humidity/no clouds, etc, and the milky way is in a good position, you will be able to pick out the galaxy quite well. You should be able to notice both the high concentration of stars as well as the 'milky' lighter shade caused by the many stars too dim or distant to appear bright.


When was the term "orbit" (in the modern sense) first used and by whom?

Searching for the definition in dictionary.com you can find this under the word origin and history section:

  Late 14c., "the eye socket," from Old French orbite or directly from
  Medieval Latin orbita, transferred use of Latin orbita "wheel track,
  beaten path, rut, course, orbit Astronomical sense first
  recorded 1690s in English; it was in classical Latin, revived in
  Gerard of Cremona's translation of Avicenna.

I believe that this is what you are looking for.


if the big bang only expanded the universe when and how did it originate?

"before" is not the right word. It suggests that there was an arrow of time which allowed to trace events. But time was created only with the big bang, there was no before.

We don't know where the matter came from, or what caused the Big Bang. One hypothesis suggests that it was caused by the collision of branes in other universes, but there's no way to verify this, so in a strict sense it isn't science. Many researchers think we'll never have an answer this.


Roche limit - calculation with centrifugal forces

The roche limit is a calculation of when pieces of a satellite will be pulled off the satellite. Any force that affects the whole satellite equally can be ignored, as this will affect the motion of the satellite as a whole but not (directly) whether pieces are pulled off.

If you consider an object which maintains a constant orientation, but rotates about another body, the paths each part of the object take differ only by a translation in space, and so each part will experience the same centrifugal force. So, the only rotation that needs to be considered is the rotation of the body around its own axis.

In the special case of a tidally locked satellite in a circular orbit, the satellite's rotation around the planet, and the satellite's rotation around its own axis, are the same rotation, just differing in what axis we are considering the rotation to be around.

In the wikipedia article's calculations, the centrifugal force for the rigid body case is considered around the satellite's own axis, but for the fluid body case it is considered as a rotation about the primary. This is an equally valid calculation for the special case of a tidally locked satellite in a circular orbit.


A black dwarf is a white dwarf that has cooled to the point that it is no longer emitting light. However it takes a long time for a white dwarf to cool down. The exact timescale is rather uncertain, but at least hundreds of billions of years, perhaps hundreds of trillions of years.

The upshot is that that no black dwarfs exist in the universe today. The Universe is only about 14 billion years old, so there hasn't been time for any black dwarfs to form.

The paper A Dying Universe: The Long Term Fate and Evolution of Astrophysical Objects discusses brown dwarfs, black dwarfs, and other objects that may exist in an aging universe.


What is the size of the image sensor in the largest optical telescopes?

The current largest digital CCD camera is that of the Vera C. Rubin Observatory1 which has a whopping 3.2 gigapixels. The previous largest features on the Panoramic Survey Telescope & Rapid Response System (Pan-STARRS), and has a resolution of ~1.4 gigapixels.[1]
Based on the spec sheet provided by the Vera Rubin Observatory, the LSST camera has a resolution of roughly 0.2 arcseconds per 10 $\mu$m pixel. It is about 5 feet (1.52 meters) wide and weighs over 6000 lbs (2721 kilos, $2.6\times10^9$ dyn).[2] The actual photosensitive portion of the camera is ~64 cm ($4\times10^{34}$ Planck lengths) across.
Cameras of this size often have pointing, calibration and recording issues. These sources of error are usually extremely well understood, but it's still interesting to see. Pan-STARRS has a detailed list on their data site[3]. These issues include:

Randomly missing data that gets filled in later
Pointing errors related to astrometric positions of their targets
Registration issues near the celestial pole resulting in poor photometry

1formerly known as the Large Synoptic Sky Survey or LSST. The acronym has now been repurposed: What is the LSST now? Where does LSST end and Vera C. Rubin Observatory begin?


For stars around 13th magnitude, are their entries in various catalogues usually linked, or does one do this manually?

If you search by position in the CDS Simbad database by position it will give you entries in all three of these catalogues. If you search by name, then it will look for matched objects in these three catalogues. I am unsure what the matching algorithm is - it must cross-correlation on the position and look for objects within an appropriate matching radius, taking account of the proper motion of the object and epochs of the various catalogues.

This will not be foolproof, but the density of 13th magnitude objects on the sky is low enough that false matches are highly unlikely.


What complementary observation campaigns have looked at Kepler's field of view?

The Kepler Field had been covered in the near IR by the 2MASS catalog since that covered the entire northern sky.  There was also a Smithsonian Institute program called the Kepler Input Catalog program that observed the Kepler field with the 48-inch telescope at the Whipple Observatory on Mount Hopkins, Arizona in the SDSS g, r, i, and z bands plus a custom filter centered on the Mg b lines.  Also many individual objects studied by Kepler were observed in follow up studies.  So, yes this is a well studied region of the sky now.  Now, however, the Pan-Starrs survey has imaged most of the whole sky with several optical filters.

After the reaction wheel failure on Kepler, it was repurpose as the K2 project.  In this project, it observes a different part of the ecliptic plane every 3 months.  Catalogs of stars and galaxies in each of these locations are created ahead of time with the EPIC survey.  In addition, there is a SN/AGN search ongoing with K2 that includes simultaneous groundbased observations using various wide field imaging telescopes around the world such as the SkyMapper in Australia.  A workshop on this will be held in February.


Is it proper to refer to objects beyond Neptune as "Kuiper Belt Objects?"

Yes there is a difference, Kuiper Belt Objects (KBOs) are a subset of Trans Neptunian Objects (TNOs). Other subsets are Oort-cloud objects (OCOs) and scattered disk objects (SDOs). These are not KBOs but they are TNOs. See for instance this wikipedia page.