How to estimate the inclination of a galaxy between edge on and face on?

If you tilt a circular object, then along one axis the length will appear to be shortened, while along a perpendicular axis the apparent size will be unchanged.  So you can use that to determine the tilt angle (inclination).
If we define $0^\circ$ to be face-on, then when tilted at an angle $i$ away from face-on, the shorter axis will appear to be decreased in length by a factor of $\cos i$.  Thus, assuming that the galaxy is circular in cross-section, you can determine its inclination from $\cos i = \frac{\rm short\ axis}{\rm long\ axis}$.
This should work for spiral galaxies, but won't work for elliptical galaxies, which aren't necessarily circular in projection when viewed face-on.


How can the Advanced Composition Explorer (ACE) warn of incoming solar storms?

Solar flare X-rays travel at light speed, but the solar wind from CME's travels at merely 600 to 2000 kilometers per second. It's the wind that causes aurora and solar storms, depending on the orientation of its magnetic filed.
Here's a SOHO view of a CME. It's clear the gas is moving slower than light. You can tell if the earth is in the path of an ejection by how near to symmetrically the gas cloud expands around the sun.

So, to answer your question, the satellite measures magnetic field direction, and can give 15 minutes warning because the CME gas is moving much slower than light.


What is the minimum size to be a natural satellite?

As long as the satellite is way smaller than the object it is orbiting, the mass does not make any difference to its orbit.

This is because even while gravitational force depends on product of masses of both objects, the acceleration of orbiting object is this force divided by its mass. Therefore any change to its mass will keep its acceleration and thus velocity and thus orbit about the same as before (as long as there are no significant changes to barycenter the two objects are orbiting; gravitational waves do change things, but only for very extreme cases such as close black hole and/or neutron star binaries)

This will get more complicated if there are more than one satellite orbiting the body, and stability in these cases is often hard to determine (aspects like orbital resonances and phase do matter in these cases)

EDIT: added explanation why mass does not affect orbit


Why aren't brown dwarfs the dark matter?

Two reasons.

We know from looking at galaxy rotation curves and the motion of galaxies in clusters and from gravitational lensing, that the amount of "dark matter" is some 30% of the density of the universe. But on the other hand, estimates of the abundances of deuterium, helium, tritium and lithium produced in the big bang indicate that only 5% of the density of the universe can be in the form of normal matter (e.g atoms, protons, neutrons etc). As brown dwarfs are made of these things, they can only contribute to this small percentage and cannot be responsible for most of the dark matter (the same is true for anything made of normal, "baryonic matter").
Astronomers have counted how many brown dwarfs there are, both in the local neighbourhood and in star clusters (see for example Andersen et al. 2008;  Kirkpatrick et al. 2011; Burningham et al. 2013; . In other words we can see brown dwarfs - they emit most of their light in the infrared part of the spectrum, and surveys like SDSS, 2MASS and WISE have uncovered them in their thousands. It turns out that there is about 1 brown dwarf for every 4 more massive stars. So, although numerous, they contribute a very tiny fraction of the normal matter in our Galaxy. In addition, microlensing experiments towards the Galactic bulge and Magellanic clouds suggest that although there are lots of brown dwarfs elsewhere in our Galaxy (e.g. Alcock et al. 2000; Novati et al. 2008), there are not enough to make much of a contribution to dark matter, and so the results from our solar neighbourhood appear representative of the Galaxy as a whole.


How do we differentiate between a star or planet and noise introduced due to equipment?

You can differentiate between planet (Did you mean asteroid? Planets are quite bright.) and noise quite easily. Just take three pictures of the sky and they will be probably quite noisy. If some spot changes its position on all three pictures, then this is some near object (astronomically near). You can additionally verify it if its trajectory is line or not. If it is not, then it is probably just noise, but that isn't always the case.
Differentiating between a star and noise is harder. You have to take more pictures and hope that the noise will cancel out.
About flux density: this is the measure for light over some surface and it is proportional to the number of photons that hit the sensor.


A black hole that doesn't take in matter?

This is the full quote

  If, for example, the Sun were replaced by a black hole of equal mass,
  the orbits of the planets would be essentially unaffected. A black
  hole can act like a "cosmic vacuum cleaner" and pull a substantial
  inflow of matter, but only if the star it forms from is already having
  a similar effect on surrounding matter

It's badly written, but what it's basically saying is that it doesn't matter the form of the massive object that's being orbited by a planet or other object.  A stable orbit would remain stable, so, if our sun was to collapse into a black hole and nothing else changes, Earth's orbit would remain the same.  It would get dark and cold, but the orbit would be unchanged.

The misconception is, if the sun became a black hole than Earth would be sucked into the sun.   That's 100% false and that's all they're saying.

Gravity is a function of mass and distance.   Black holes have very high gravity, but a big part of the reason for that is because when they form out of dying stars, they become very small, a few solar masses squeezed into only about 10 miles across so the distance to the center of mass gets very small.

For a planet to be in any danger from a black hole, it would probably need to be inside the Roche Limit, perhaps only a few million miles away, which is several times closer than Mercury is to the sun, for example.   The safe distance, of-course, varies with the density and solidity of the orbiting object.

Here's a Q on Roche Limits and Black holes if interested.


Is the Celestron astromaster 130 EQ Model able to see nebulas and galaxies?

The question isn't whether you can see deep-sky objects but how many.
Given a reasonably dark observing site, the Orion nebula and the Andromeda galaxy are visible in 7x50 binoculars.
A 130mm objective should be enough to see most of the nebulae and galaxies in the Messier catalogue.
Once you get the hang of it, you may be able to see a few NGC objects too, but a bigger scope will show more of them.


How to calculate a phase shift in the process of deriving Baade-Wesselinke distance?

It is admirable that you are tackling this type of problem as a high school student, but I really think that some of these topics are out of your grasp currently, partly because of the highly technical nature, but also because of your lack of access to the necessary data.

For one thing, you say you have V-R colors as a function of time, but all these equations were calibrated to use V-K colors. From your linked paper, you're going to need equation 2 for your process to calculate $p$ and equation 3 for getting $\theta_0$. Unfortunately, these equations assume V-K colors. You can redo the analysis done in this paper to calculate $p$ and this paper to get $\theta_0$ for V-R colors, but it's not going to be easy.

For another, you're going to need measurements of the angular size as a function of time (and your cited paper suggests that only exists for a handful of Cepheids and was measured using the VLTI) as well as the RV as a function of time. The $\delta \theta$ that you reference is described exactly in the paper as "a  phase shift  between  the  radial  velocity  curve  and  the  angular  diameter changes measured either interferometrically or via the SB relation." This means if you have $RV(t)$ and $\Delta \theta(t)$, you can determine a phase shift between the two curves, presumably by fitting some equation of the form $f(t) = a\ \mathrm{sin}(bt+c)$ to both equations and saying that $\delta \theta = c_{RV} - c_{\Delta \theta}$.

The analysis you're trying to do requires very specific data and it seems that the relevant papers on this subject have already done the analysis on all stars in which all necessary data existed. You cannot easily apply this process to any Cepheid without significant work.


How far is the nearest black hole from our solar system? Is there an official list?

According to table 2 of Corral Santana (2015) (The BlackCAT catalogue of stellar mass black holes in X-ray transients - which other than gravitational waves, is how they are found), the nearest known black hole is the primary of the V616 Mon system (otherwise known as A0620-003) at a distance of $1.06 \pm 0.10$ kpc ($3460 \pm 390$ ly).

Black holes can only be found at present when they are in close binary systems with other stars (or black holes). This is the tiny minority of an estimated Galactic population of 100 million black holes.

The nearest black hole is likely invisible to us at present. A statistical estimate can be made (see my answer here) that the nearest will be within 20 pc or so, depending on the parent population and detailed physics of supernovae.

The online BlackCAT catalogue.


I was wondering if it were possible to have a Universe that were isotropic but NOT homogeneous OR a Universe that were homogeneous but NOT isotropic.

Most people will be happy with the UC Berkeley definition that says homogeneous means "looks the same at every location" and isotropic means "looks the same in every direction". And some will know that as per Ned Wright's article, these attributes aren't quite the same:

He says "the figure above shows a homogeneous but not isotropic pattern on the left and an isotropic but not homogeneous pattern on the right".  However as far as I know pictures like this just don't apply to our universe scattered with galaxies.

  My spidey sense is telling me that it is possible that a Universe could be isotropic but not homogeneous.

My spidey sense is telling me some guy 46 billion light years away might say the universe is neither isotropic nor homogeneous. Because when he looks up, half the night sky is black or something.

  Surely, by default, if a Universe is homogenous then it follows that it has to be isotropic?

I agree with the gist of that. IMHO if an observer sees a homogeneous universe, he sees an isotropic universe too. Yes, one can find hypothetical scenarios wherein the universe is homogeneous but not isotropic. But they're only hypothetical. And let's not forget that it's only an assumption. If you lived in a forest would you assume that the world was covered in trees? Which look the same at every location and in every direction? It isn't a particularly scientific  assumption. For all you know some guy lives near the edge of the forest. I think you're better off saying we just don't know.


Theoretically, what is the biggest optical telescope that may exist?

It's complicated.

Until late-20th century, we've tried to make bigger and bigger monolithic telescopes. That worked pretty well up to the 5 meter parabolic mirror on Mount Palomar in California in the 1940s. It kind of worked, but just barely, for the 6 meter mirror on Caucasus in Russia in the 1970s. It did work, but that was a major achievement, for the twin 8.4 meter mirrors for the LBT in Arizona in the 2000s.

We've learned eventually that the way to go is not by pouring larger and larger slabs of low-expansion glass. It is generally accepted that somewhere just below 10 meters diameter is about as large as possible for monolithic mirrors.

The way to go is by choosing to make smaller mirror segments (1 meter to a few meters in diameter each) and combining those into a tiled mirror. It's somewhat harder to carve the asymmetrical parabolic (or hyperbolic, or elliptic, or spherical) reflecting curved surface in a segment like that, but it's far easier to manage thermal and cooling issues when you have to deal with smaller solid objects.

Each segment is mounted in an active mirror cell, with piezo actuators that very precisely control its position. All segments must combine into a single smooth surface with a precision better than 100 microns (much better than that in reality). So now you have a large array of massive objects, dynamically controlled via computer, each with its own vibration modes, each with its own source of mechanical noise, each with its own thermal expansion motions, all of them "dancing" up and down a few microns on piezo elements.

Is it possible to orchestrate a very large system like that? Yes. The 100 meter OWL was considered feasible technically. From the perspective of keeping the mirrors aligned, an even larger structure should be doable; the computer-controlled actuators should overcome most vibrations and shifts up to quite large distances.

Like you said, the real limits are financial. The complexity of such a system increases with the square of the diameter, and with complexity comes cost.

The entire discussion above was about "filled aperture" telescopes: given a round shape of a certain diameter, it is filled with mirror segments. For a given aperture, this design captures the largest amount of light.

But the aperture does not have to be filled. It can be mostly empty. You could have a few reflecting segments on the periphery, and the center would be mostly void. You'd have the same resolving power (you would see the same small details), it's just that the brightness of the image would decrease, because you're capturing less light total.

This is the principle of the interferometer. The twin 10 meter segmented Keck mirrors in Hawaii can work as an interferometer with a baseline of 85 meters. This is effectively equivalent to a single 85 meter aperture in terms of resolving power, but obviously not in terms of image brightness (amount of light captured).

The US Navy has an interferometer in Arizona with mirrors placed on 3 arms in a Y shape, each arm 250 meters long. That gives the instrument a baseline (equivalent aperture) of several hundreds of meters.

U of Sydney has a 640 meter baseline interferometer in the Australian desert.

Interferometers cannot be used to study very faint objects, because they can't capture enough light. But they can produce very high resolution data from bright objects - e.g. they are used to measure the diameter of stars, such as Betelgeuse.

The baseline of an interferometer can be made extremely large. For terrestrial instruments, a kilometers-wide baseline is very doable now. Larger will be doable in the future.

There are talks about building interferometers in outer space, in orbit around Earth or even bigger. That would provide a baseline at least in the thousands of kilometers. That's not doable now, but seems feasible in the future.


Pluto's orbit overlaps Neptune's, does this mean Pluto will hit Neptune sometime?

No, Pluto is a so called resonant trans-neptunian object; the orbital period of Pluto is almost exactly 3:2 (1.5) times that of Neptune. This means that every time Pluto nears perihelion and is therefore closest to the Sun and also closest to the orbit of Neptune, Neptune is always at a specific angle (50° according to Wikipedia) in front or behind Pluto.
(See for instance Figure 3 and surrounding text in Jewitt, D., Morbidelli, A., & Rauer, H. (2007). Trans-Neptunian Objects and Comets. Springer.)


Why is the core of a gas giant supported by electron degeneracy pressure instead of nuclear fusion?

The test to see whether degeneracy pressure is going to be significant is to compare $kT$ with the Fermi energy $E_F$

The Fermi energy is the energy level up to which all energy states would be occupied in a completely degenerate fermion gas. It is given by (for non-relativistic conditions)
$$ E_F = \frac{h^2}{2m}\left(\frac{3}{8\pi}\right)^{2/3} n^{2/3},$$
where $m$ and $n$ are the mass and number density of the fermions (in this case, electrons).

If $E_F \gg kT$, then the gas can be considered completely degenerate and the electrons exert a non-relativistic degeneracy pressure. If $E_F \sim 10 kT$ the electrons are partially degenerate.
In both these circumstances, the pressure exerted by the electrons is much higher than they would have in a perfect gas, because a large fraction of electrons occupy high energy (and momentum) states.

Degeneracy can be achieved either by having a low temperature or a high number density of fermions. In a white dwarf, electrons are completely degenerate because the number density of electrons is extremely high. In a gas giant, the electron density is nowhere near as high, but then the temperature is a lot lower and so they reach a state of partial degeneracy.

The pressure of a partially degenerate gas is higher than that of a perfect gas at the same density and temperature. More importantly, the pressure is only very weakly dependent on temperature. As a gas giant radiates away its potential energy, it is able to do so whilst only contracting very slightly and the core does not become hot enough to initiate nuclear fusion.

Note that degeneracy pressure and thermal pressure are not two different things. Both arise merely as a consequence of particles having a distribution of momenta and obeying (in this case) Fermi-Dirac statistics to determine how they occupy the possible energy states. The expression $P = nkT$ is simply a convenient approximation for fermions that holds only when $E_F \ll kT$ and the quantum nature of the particles is not apparent.


What is the degree of ionization is the solar photosphere?

The degree of ionization in the photosphere varies with depth of course, but overall it is small.   Table 1 of the Bilderberg Continuum Atmosphere  (Solar Physics, 3, 5, 1968) gives the pressure and the electron pressure at various optical depths in a comprehensive model.  The ratio of the pressure gives the Ne/Ntotal.  At optical depth = 1 at 5000 Angstroms, Pgas = 1.412E5 (cgs) and Pe = 6.239E1.  The ratio is 4.4E-4. The hydrogen ionization, also given in the table is 4.07E-4.  The model assumes Helium abundance of 0.1 which is off a bit from the modern number of 0.087.


There are night sky images in high resolution available from ESO, NASA and potantially others. They are already projected, openly licensed, and can directly be used, e.g. as skyboxes or -domes in computer graphics.
https://www.eso.org/public/images/eso0932a/
https://svs.gsfc.nasa.gov/3895


Note that they are related via the reddening law, i.e.
\begin{equation*}R_\lambda = \frac{A_\lambda}{E(B-V)}\end{equation*}
where $A_\lambda$ is the extinction at wavelenght $\lambda$, and $E(B-V)$ is the  color excess for $B-V$. In literature is commonly used $E(B-V)$, while in practice is usually needed $A_\lambda$, to switch back and forth from apparent magnitude $m_\lambda$ to absolute magnitude $M_\lambda$ with
\begin{equation*}m_\lambda=M_\lambda+A_\lambda+(m-M)_0\end{equation*}
using the absolute distance modulus $(m-M)_0$.

Back to your question, classical references for the color excess of galactic globular clusters are the Harris catalog and the Schlegel maps (I would prefer the second). A nice solution is the Python module dustmaps (Green 2018); in the documentation you find all the available maps (here).

Finding $R_\lambda$ is more tricky. You can search in the literature for tabulated values (e.g. Tables 14 and 15 in Sirianni et al. 2005), otherwise a rough-but-still-working approach is to take a $R_\lambda$ vs. $\lambda$ plot (e.g. the one in Fitzpatrick 1999) and to extrapolate by-hand $R_\lambda$ at the effective wavelength of your filter. Note that usually are presented several reddening laws, distinguished by $R_V$. This is because they're not universal and depends on the environment; a reasonable choice for GGCs is $R_V=3.1$ .


How a star's rotation affects a star in the main sequence?

This is a well-studied problem. The effect of rotation on the structure of a low-ish mass star (like the Sun) is summarised by Eggenberger (2013).

Such stars are never observed to rotate so fast that the rotation plays any significant role in their hydrostatic equilibrium, however rotation does play a role by causing additional mixing in the star.

This is important for two reasons: (i) it inhibits the gradual diffusion of helium towards the core, this slightly lowers the opacity in the core and raises it in the envelope (compared with a non-rotating star). This results in a slightly higher luminosity and a slightly higher surface temperature. (ii) More importantly, the additional mixing brings additional hydrogen into the core and this increases the lifetime on the main sequence.

The effects on stars of a solar mass are however unlikely to be very significant in practice, because these stars efficiently lose angular momentum via a magnetised wind during their early lives and the effects of rotation are unlikely to be significant even at several times the solar rotation rate.

The effects on more massive stars can be more severe. These can spin at an appreciable fraction of their break-up rate for a large part of their lives and do not lose angular momentum as efficiently as lower mass stars (they do not have magnetised winds). The effects are described in the canonical paper of Meynet & Maeder (2000); they are more pronounced than for lower mass stars and more complicated because of the radiative envelope and uncertainties in the rotation-dependence of the significant mass-loss.

The hydrostatic effects of rotation are expected to be important early in the main sequence and contribute to a slightly lower surface temperature. At later times the dominant effects are caused by changes in the mixing and diffusion near the core and in the envelope as for the lower mass stars, resulting in higher luminosities and hotter temperatures. The main sequence lifetimes can be extended by 30% due to the additional mixing of fresh hydrogen fuel into the core.


How to create 3D animation of particles position vs time in python?

After that I want to make it a bit complicated. I have the stars size and mass as well. I was wondering if I can change the size of the circles in the plot according to the radius and use different Colors for different mass range?

That part has been addressed in previous questions and answers here, I remember reading maybe a year ago (+/- 0.9 years).

Here is something you may find helpful reprinted from my answer in Stack Overflow. It's based on astronomer and data scientist Jake VanderPlas' blogpost from 2013.

For your needs keep the dots but not the lines. What I recommend is that you start from this and get as far as you can, then ask a question in Stack Overflow showing what you've tried and explaining what you still need.

  Here is the Lorenz attractor both in 3D and animated. The script is in the following link (along with many goodies) in Jake VanderPlas' Pythonic Perambulations. You can learn a lot by going line-by-line through the script - it's an elegant use of matplotlib objects.

  https://jakevdp.github.io/blog/2013/02/16/animating-the-lorentz-system-in-3d/

  I added these two lines just before return in the animate function, and then used ImageJ to import the "image stack" and save the "animated GIF":

fname = "Astro_Jake_" + str(i+10000)[1:]
fig.savefig(fname)

  Note: For OSX it seems to be necessary to set blit = False in animation.FuncAnimation.

  Here is a minimal, simplified example of plotting lines in 3D based on the above:

def lorentz_deriv((x, y, z), t0, sigma=10., beta=8./3, rho=28.0):
    """Compute the time-derivative of a Lorentz system."""
    return [sigma * (y - x), x * (rho - z) - y, x * y - beta * z]

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from scipy.integrate import odeint as ODEint

x = np.linspace(0, 20, 1000)
y, z = 10.*np.cos(x), 10.*np.sin(x) # something simple

fig = plt.figure()
ax = fig.add_subplot(1,2,1,projection='3d')
ax.plot(x, y, z)

# now Lorentz
times = np.linspace(0, 4, 1000)

start_pts = 30. - 15.*np.random.random((20,3))  # 20 random xyz starting values

trajectories = []
for start_pt in start_pts:
    trajectory = ODEint(lorentz_deriv, start_pt, times)
    trajectories.append(trajectory)

ax = fig.add_subplot(1,2,2,projection='3d')
for trajectory in trajectories:
    x, y, z = trajectory.T  # transpose and unpack
    # x, y, z = zip(*trajectory)  # this also works!
    ax.plot(x, y, z)

plt.show()

I wrote this before I knew of PIL. Apparently you can use PIL to generate your GIFs instead of using an external program like I did.

See

Creating GIFs in Python Using Pillow (PIL Fork)
https://pythonprogramming.altervista.org/create-an-animated-gif-with-pil/
https://pythonprogramming.altervista.org/png-to-gif/

and have fun!

An example of a GIF with stars in it from the first blogpost (Nyan Cat):


I can't see any detail through my telescope

The planets appears to be smaller than the Moon, so it is more difficult to see the details on them. These are some of the factors that determine how good the image will be.

Steadiness of the scope. You mentioned that it was windy, so I assume that the scope was vibrating a lot. To get a good view, you need a scope that does not vibrate when you do not touch it. (A rule of thumb is that the vibration should dampen out in 1 or 2 seconds when you are done focusing or moving the scope to track an object.)

Steadiness of the air. If you look over a paved road on a hot day, you can see how the air is shimmering. Now imagine magnifying that shimmering air by 50x or 100x, and that is what you are dealing with when using an astronomical telescope. The steadiness of the air will limit the size of details that can be seen. (Astronomers refer to this as the "seeing". The shimmering of the air is generally not as bad as my example of the road, but looking through 20 miles of atmosphere and magnifying it make it seem as bad.) The telescope can also create air currents if the scope is warmer than the air. Waiting 15 to 60 minutes (depending on the design and size of the scope) to let the telescope cool down to the air temperature can help with the air currents inside the tube.

Viewing an object when it is high in the sky will help with the seeing because you are looking through less atmosphere (and less distortion). Viewing an object when low in the sky will usually be disappointing. (Move to a location where the object is higher in the sky! :-)

Quality of the mirror A telescope mirror needs to be ground and polished to a high degree of accuracy. Without someone performing a test on your scope, it is hard to know if the figure is excellent, acceptable, or poor. (I have not ground a mirror myself, but I have read that it is not hard to make a good mirror. Or course, it is even easier to make a poor mirror!)

Quality of the eyepieces The eyepieces are just as important as everything else. Some eyepieces provided with small telescopes are of low quality. If the diameter of the eyepiece barrel is about 1 inch (25 mm) or has the letter "H" in front of the eyepiece size (like H20mm), the quality is likely to be poor. 1.25 inch (32 mm) diameter eyepieces are generally better quality.

Magnification A rule of thumb is that the maximum useful magnification is 50 times the diameter of the optics in inches (or 2 times the diameter of the optics in mm). For 114 mm, that gives a useful magnification of 220x. You will find that lower magnifications are used more often because the image gets worse at higher magnification, especially when the seeing is poor. 50x to 150x should be enough to give good views of Jupiter and Saturn.

Collimation A Newtonian reflector has two mirrors. If the mirrors are not aligned properly, the image quality will deteriorate. (The alignment is known as collimation.) The way to check the collimation is to look at a bright star high in the sky. When in focus, the star should be a sharp point surrounded closely by a faint ring. When you un-focus the image slightly, the star will be a larger dot, and the faint rings (diffraction rings) will be a little larger and easier to see. The rings should look like a bulls eye around the dot of the star. The image should be the same on either side of the focus.

If the image is not round, or if the image changes from inside focus to outside focus, either the quality of the mirrors is poor, or the mirror is warped in the mount (too much force in the holding mechanism is bending the mirror!) or the collimation is off. The warping of the mirror can be corrected by not applying as much force on the clips that hold it in the cell. The collimation can be adjusted to correct for problems with the alignment. See How to Align Your Newtonian Reflector Telescope on the Sky & Telescope website for example. A poor quality mirror cannot be easily corrected unless you know someone with the skill to re-polish it.


I don't know if this is what most astronomers use, but it is certainly a method that could be used. It also makes for some interesting photos.

Gravitational lensing was one of the great predictions of Einstein's theory of general relativity, and was actually one of the first pieces of evidence for it - see the solar eclipse of 1919. Gravitational lensing is the bending of light by a massive object. Because mass and energy warp space-time, we know that if you can calculate how much the light is bent, you can calculate the mass of the body bending it. Wikipedia gives the following formula:

$$\theta=\frac{4GM}{rc^2}$$

where $M$ is mass, $r$ is the distance the object is from the light, and $G$ and $c$ are, of course, constants. So if you can measure how much light is bent by an object, you can calculate its mass. It's hard, and you would need good conditions for it to work, but it could work, nonetheless.


What are these faint streaks on the moon?

i would rather call the streaks you are referring to as ridges.

the surface of the moon is ofcourse not a perfect sphere and the same way the earth had ups and downs, there are ridges on the moon too.

you say that most of them are diagonal - top-left to bottom-right but on the right hand corner of the image, there are ones vertical ones. So, there is no specific orientation for these "streaks".

on the other hand, if there were streaks around a crater such that concave part of the streaks point towards the crater, they are due to shock waves travelling outwards from the point of impact.


Eyepieces and accessories recommendations for a Starblast 4.5-inch scope in a city as well as darker suburbs, and kid friendly?

The one thing I would add is an eyepiece which provides the widest possible field for your telescope.
With 1.25" eyepieces this would be an eyepiece with a focal length from 20mm to 26mm and an apparent field of view between 62° and 70°, such as:

Explore Scientific 26mm 62°
APM UFF 24mm 65°
Tele Vue Panoptic 24mm 68°
Pentax XW 20mm 70°

If you want a wider field of view then you have choose a lower focal length - 17.5mm and 76° with a Baader Morpheus, for example, and 16mm and 85° with a Masuyama. These two will give a slightly smaller true field of view.
Your zoom + 2x Barlow should cover all other situations, so I wouldn't go and get any more eyepieces right away.
Later on you might want to buy a fixed focal length eyepiece: maybe you find yourself often using the zoom at 9mm with the 2x Barlow, so you might get a dedicated 4.5mm eyepiece to use instead of the zoom for those situations.

Use the handy FoV calculator at astronomy.tools to compare eyepieces and see the visible area given by each one.
You'll see that all of the eyepieces show roughly the same real area of sky (the 24mm/68° shows the most, and a 40mm Plössl with 43° would show more still).
Where they differ is in the apparent field of view and the size of the image when you look through the eyepiece.
Your Orion zoom has a FoV of about 40° at its 21mm setting. A 24mm eyepiece with a FoV of 68° will give you a low power (about 20x magnification) view which is twice as wide as that given by your zoom.

You mention viewing under light polluted skies as well as darker skies, and the wish to see nebulae and galaxies.
Light pollution and galaxies are incompatible, unless you use image enhancement technology of some kind (night vision, live stacking, or full-on astrophotography).
Except for the biggest and brightest galaxy - Andromeda, M31 (the core of which should still be visible beneath all but the worst inner-city skies) - your search for galaxies will be a frustrating one.
Even under dark skies, seeing galaxies with the naked eye is not easy, requiring the use of averted vision and memory in order to see any detail.

Some nebulae are visible from the city. The Orion nebula, M42, is very bright and always worth a look, no matter where you are or how many times you've seen it before. In the summer the Triffid nebula can be seen in the southern skies, and the Swan (aka Omega) nebula comes out with the help of a filter.
There are also some bright planetary nebulae which you can see from the city. The Saturn nebula comes to mind as one I've seen under light pollution.
Generally, though, the best sights for the city astronomer are:

The inner planets and Jupiter and Saturn
The moon
Bright star clusters
Bright double stars and carbon stars

(Disclaimer: my urban skies could be a lot worse, and my telescope could be a lot smaller, so what I can see won't necessarily be the same as what you can see)

Also, you could get yourself a couple of books. A paper star atlas is a nice supplement to an electronic one, especially for those trips to darker skies.
Turn Left At Orion is a classic book which is frequently recommended. Nightwatch is another.
Get a red light as well, to preserve your night vision in the dark.


Find the height of the shooting star from the Earth's ground

This is basically a trigonometry question.  I wasn't sure from the question if the apparent location from your friend's position was six degrees higher or lower.  I chose higher for the example, but you could change it.

You have two triangles with the same altitude, but different bases and angles.

$$\text{altitude} = x \tan(81^{\circ}) = (x + 20km) \tan (75^{\circ})$$  Do you see how to calculate it from there?


What abbreviations, letters, or symbols do we use to denote highly degenerate stars?

Neutron stars and black holes do not have assigned spectral types since they do not have a measurable optical/IR spectrum - which is the basis for assigning a spectral type.

There are many classifications for systems containing black holes and neutron stars. They are not related to the letter-based spectral types of "normal" stars or white dwarfs, but more to do with their X-ray, radio or inferred physical properties (e.g low- or high-mass X-ray binary systems).


I'm not absolutely sure if it is correct, but I use information from this site: http://star-www.st-and.ac.uk/~fv/webnotes/chapt9a.htm

And I convert equatorial coordinates (RA = α, Dec = δ) to ecliptic coordinates (λ, β)  using this formula.

  sin(β) = sin(δ) cos(ε) - cos(δ) sin(ε) sin(α)

  cos(λ) cos(β) = cos(α) cos(δ)

  ε = Axial tilt (23,5°)

I know that due to precession only λ changes (β remains same for ecliptic coordinates). So I add the angle of precession to the λ. Then I convert the new λ and β back to the equatorial coordinates.

  sin(δ) = sin(β) cos(ε) + cos(β) sin(ε) sin(λ)

  cos(λ) cos(β) = cos(α) cos(δ)


What's the rationale behind the false colours in solar observation photographs?

Some of the images shown in your question are not colored or colorized, it is not a false color; it's the real color. The calculations that are made to produce the false colors that are used on colorized image datasets are shown in the aia_lct.pro file, where the exact RGB values are provided.

Where a false color is chosen it is done so (for SDO and some other images) using the AIA Standard Color Tables (which are generated in the aia_lct.pro file).

Information on the "VSO's Guide to SDO Data Analysis" (.PDF), on page 46, describes use of the subroutines in the sdo/aia/idl/calibration/aia_colors.pro source code file from http://www.heliodocs.com/ or view "Installing Solarsoft" for ready to run versions and more info.

See Section 6.6.1 "Using AIA Standard Scaling and Colors" for additional information about the aia_colors.pro function. The Introduction of that Guide contains additional information and links about the SDO mission.

Here is the Description of the colors NASA uses for different wavelengths (using the AIA Standard Color Tables):

  view-source:https://sdo.gsfc.nasa.gov/data/aiahmi/

Telescopes

<select name="telescope" id="telescope" class="form-control">
    <option value="aia_0171">AIA 171 (gold)</option>
    <option value="aia_0193">AIA 193 (bronze)</option>
    <option value="aia_0304">AIA 304 (red)</option>
    <option value="aia_0211">AIA 211 (purple)</option>
    <option value="aia_0131">AIA 131 (teal)</option>
    <option value="aia_0335">AIA 335 (blue)</option>
    <option value="aia_0094">AIA 094 (green)</option>
    <option value="aia_1600">AIA 1600 (yellow/green)</option>
    <option value="aia_1700">AIA 1700 (pink)</option>
    <option value="hmib">HMI Magnetogram (gray)</option>
    <option value="hmibc">HMI Colorized Magnetogram (colorized)</option>
    <option value="hmii">HMI Intensitygram (gray)</option>
    <option value="hmiic">HMI Intensitygram (orange)</option>
    <option value="hmiihi">HMI Intensitygram High Cadence (orange)</option>
    <option value="hmiif">HMI Intensitygram Flat (orange)</option>
    <option value="hmid">HMI Dopplergram (gray)</option>
    <option value="COMP211193171">AIA Composite 211, 193, 171</option>
    <option value="COMP304211171">AIA Composite 304, 211, 171</option>
    <option value="COMP094335193">AIA Composite 094, 335, 193</option>
    <option value="COMPHMI171">Composite AIA 171, HMI Magnetogram</option>
</select>

Here is a webpage showing the images in your question in various sizes, including a 48 hour movie for each.

The full Electromagnetic Spectrum:

The Visible Spectrum, including Fraunhofer Line Letters and Wavelength Numbers:

The Fraunhofer lines are typical spectral absorption lines. Absorption lines are dark lines, narrow regions of decreased intensity, that are the result of photons being absorbed as light passes from the source to the detector. In the Sun, Fraunhofer lines are a result of gas in the photosphere, the outer region of the sun. The photosphere gas is colder than the inner regions and absorbs light emitted from those regions.

Here is the wavelength's frequency for each Element of interest:

Each Filter accepts a narrow band which has a specific use and specific properties. Filters are more expensive when the bandwidth is very narrow, single frequency Filters are unnecessary and unaffordable.

Different frequency ranges of the spectrum are used to resolve different aspects of the object being observed, similarly different bandwidths are used to further enhance those features.

Listing the bandwidths for every filter and the expected results would be exhaustive, one website that gives such info for each of their filters is Daystar where their filter comparison page indeed confirms that the Sun can be viewed in black and white using a monochrome sensor. It's when you want to combine multiple channels that use of color becomes necessary.

Here is an example of the differences between bandwidths for the Hydrogen Alpha Filter (a deep-red visible spectral line in the Balmer series with a wavelength of 656.28 nm):

Filters with 0.8 ångström bandwidths will reveal prominences in high contrast
Filters with 0.7 ångström bandwidths will reveal prominences in high contrast and occasionally, surface texture. Prominences are generally larger with .7Å filters than in narrower bandwidths
Filters with 0.6 ångström bandwidths will reveal some surface contrast as well as prominences.  A .6Å filter can be a good compromise for those having a difficult decision
Filters with 0.5 ångström bandwidths will reveal better surface contrast as well as great prominences
Filters with 0.4 ångström bandwidths will reveal excellent surface contrast as well as fine chromosphere detail
Filters with 0.3 ångström bandwidths will reveal superior surface contrast above any other filter. Views are generally somewhat dimmer.  Expect "Pencil-Thin" details on surface and prominence features

It is bandwidth that is used to derive contrast, not coloring.

The SDO (Solar Dynamics Observatory) is a sun-pointing semi-autonomous spacecraft that allows nearly continuous observations of the Sun. Unlike the human eye which uses rods and cones, the SDO uses 3 instruments:

The AIA (Atmospheric Imaging Assembly) images the solar atmosphere in multiple wavelengths to link changes in the surface to interior changes. Data includes images of the Sun in 10 wavelengths every 10 seconds.
The EVE (Extreme Ultraviolet Variability Experiment) measures the Sun's extreme ultraviolet irradiance with improved spectral resolution, "temporal cadence", accuracy, and precision over preceding measurements made by TIMED SEE, SOHO, and SORCE XPS.
The HMI (Helioseismic and Magnetic Imager) studies solar variability and characterizes the Sun's interior and the various components of magnetic activity. HMI produces data to determine the interior sources and mechanisms of solar variability and how the physical processes inside the Sun are related to surface magnetic field and activity.
The Magellan craft used Synthetic Aperture Radar uses radio waves and not visible light to image map objects. Magellan synthetic aperture radar mosaics from the first cycle of Magellan mapping are mapped onto a rectangular latitude-longitude grid to create this image. Data gaps are filled with Pioneer Venus Orbiter altimetric data, or a constant mid-range value. Simulated color is used to enhance small-scale structure. The simulated hues are based on color images recorded by the Soviet Venera 13 and 14 spacecraft. The image was produced by the Solar System Visualization project and the Magellan science team at the JPL Multimission Image Processing Laboratory.

Similarly SDO data outside the visible spectrum is colorized to allow invisible portions of the electromagnetic spectrum (and Magnetometer recordings) to be visualized by the human eye in a readily digestible image rather than serving the general public a large dataset of numbers.
The visible spectrum is the portion of the electromagnetic spectrum that is visible to the human eye. Electromagnetic radiation in this range of wavelengths is called visible light or simply light. A typical human eye will respond to wavelengths from about 390 to 700 nm. In terms of frequency, this corresponds to a band in the vicinity of 430–770 THz.

The spectrum does not, however, contain all the colors that the human eyes and brain can distinguish. Unsaturated colors such as pink, or purple variations such as magenta, are absent, for example, because they can be made only by a mix of multiple wavelengths. Colors containing only one wavelength are also called pure colors or spectral colors.

  Q: How is the convention of which temperature or wavelength gets which colour established?

  Is it simply so that scientists can immediately tell what temperature or wavelength a picture shows, or are the colours optimized to show particular features?

The colors are not optimized to show particular features or provide contrast. The aia_lct.pro file is where the exact RGB values are calculated, see top of this answer.

  Q: And why is there a white ring or circle in the sunshade in some SOHO imagery, and why different sizes compared to the disc in the two photos above?

The coronagraph blocks out the Sun with an occulting disk to reveal the much fainter structures in the Sun's atmosphere. Sometimes multiple images are combined to remove the disk and provide a unoccluded image.

The Large Angle and Spectrometric Coronagraph (LASCO) in the Solar and Heliospheric Observatory (SOHO) doesn't use the latest technology, sometimes there are disturbances in the images.

There are two kinds of disturbances that repeatedly occur:

Blackouts and Whiteouts, in broken lines, circle-like shapes, or over the whole picture. They are caused by the electronics box. There has never been a firmware update, since it was judged as too sensitive changing the flight-software.
Black and white pixels, occurring in patterns, without pattern or alone. Those "missing blocks" are telemetry dropouts, caused by radio interference or a disturbance in the data transfer to Goddard Space Flight Center.

[The section below might benefit from some more editing if the OP requests it.]

  AIA 193

  This channel highlights the outer atmosphere of the Sun - called the corona - as well as hot flare plasma. Hot active regions, solar flares, and coronal mass ejections will appear bright here. The dark areas - called coronal holes - are places where very little radiation is emitted, yet are the main source of solar wind particles.

  Where: Corona and hot flare plasma

  Wavelength: 193 angstroms (0.0000000193 m) = Extreme Ultraviolet

  Primary ions seen: 11 times ionized iron (Fe XII)

  Characteristic temperature: 1.25 million K (2.25 million F)

  AIA 304

  This channel is especially good at showing areas where cooler dense plumes of plasma (filaments and prominences) are located above the visible surface of the Sun. Many of these features either can't be seen or appear as dark lines in the other channels. The bright areas show places where the plasma has a high density.

  Where: Upper chromosphere and lower transition region

  Wavelength: 304 angstroms (0.0000000304 m) = Extreme Ultraviolet

  Primary ions seen: singly ionized helium (He II)

  Characteristic temperature: 50,000 K (90,000 F)

  AIA 171

  This channel is especially good at showing coronal loops - the arcs extending off of the Sun where plasma moves along magnetic field lines. The brightest spots seen here are locations where the magnetic field near the surface is exceptionally strong.

  Where: Quiet corona and upper transition region

  Wavelength: 171 angstroms (0.0000000171 m) = Extreme Ultraviolet

  Primary ions seen: 8 times ionized iron (Fe IX)

  Characteristic temperature: 1 million K (1.8 million F)

  AIA 211

  This channel (as well as AIA 335) highlights the active region of the outer atmosphere of the Sun - the corona. Active regions, solar flares, and coronal mass ejections will appear bright here. The dark areas - called coronal holes - are places where very little radiation is emitted, yet are the main source of solar wind particles.

  Where: Active regions of the corona

  Wavelength: 211 angstroms (0.0000000211 m) = Extreme Ultraviolet

  Primary ions seen: 13 times ionized iron (Fe XIV)

  Characteristic temperature: 2 million K (3.6 million F)

  AIA 131

  This channel (as well as AIA 094) is designed to study solar flares. It measures extremely hot temperatures around 10 million K (18 million F), as well as cool plasmas around 400,000 K (720,000 F). It can take images every 2 seconds (instead of 10) in a reduced field of view in order to look at flares in more detail.

  Where: Flaring regions of the corona

  Wavelength: 131 angstroms (0.0000000131 m) = Extreme Ultraviolet

  Primary ions seen: 20 and 7 times ionized iron (Fe VIII, Fe XXI)

  Characteristic temperatures: 10 million K (18 million F)

  AIA 335

  This channel (as well as AIA 211) highlights the active region of the outer atmosphere of the Sun - the corona. Active regions, solar flares, and coronal mass ejections will appear bright here. The dark areas - or coronal holes - are places where very little radiation is emitted, yet are the main source of solar wind particles.

  Where: Active regions of the corona

  Wavelength: 335 angstroms (0.0000000335 m) = Extreme Ultraviolet

  Primary ions seen: 15 times ionized iron (Fe XVI)

  Characteristic temperature: 2.8 million K (5 million F)

  AIA 094

  This channel (as well as AIA 131) is designed to study solar flares. It measures extremely hot temperatures around 6 million Kelvin (10.8 million F). It can take images every 2 seconds (instead of 10) in a reduced field of view in order to look at flares in more detail.

  Where: Flaring regions of the corona

  Wavelength: 94 angstroms (0.0000000094 m) = Extreme Ultraviolet/soft X-rays

  Primary ions seen: 17 times ionized iron (Fe XVIII)

  Characteristic temperature: 6 million K (10.8 million F)

  AIA 1600

  This channel (as well as AIA 1700) often shows a web-like pattern of bright areas that highlight places where bundles of magnetic fields lines are concentrated. However, small areas with a lot of field lines will appear black, usually near sunspots and active regions.

  Where: Transition region and upper photosphere

  Wavelength: 1600 angstroms (0.00000016 m) = Far Ultraviolet

  Primary ions seen: thrice ionized carbon (C IV) and Continuum

  Characteristic temperatures: 6,000 K (11,000 F), and 100,000 K (180,000 F)

  AIA 1700

  This channel (as well as AIA 1600) often shows a web-like pattern of bright areas that highlight places where bundles of magnetic fields lines are concentrated. However, small areas with a lot of field lines will appear black, usually near sunspots and active regions.

  Where: Temperature minimum and photosphere

  Wavelength: 1700 angstroms (0.00000017 m) = Far Ultraviolet

  Primary ions seen: Continuum

  Characteristic temperature: 6,000 K (11,000 F)

  AIA Composite 211, 193, 171

  This image combines three images with different, but very similar, temperatures. The colors are assigned differently than in the single images. Here AIA 211 is red, AIA 193 is green, and AIA 171 is blue. Each highlights a different part of the corona.

  AIA Composite 304, 211, 171

  This image combines three images with quite different temperatures. The colors are assigned differently than in the single images. Here AIA 304 is red (showing the chromosphere), AIA 211 is green (corona), and AIA 171 is dark blue (corona).

  AIA Composite 094, 335, 193

  This image combines three images with different temperatures. Each image is assigned a color, and they are not the same used in the single images. Here AIA 094 is red, AIA 335 is green, and AIA 193 is blue. Each highlights a different part of the corona.

  [Webpage had no Info Popup for: AIA 171 & HMIB]

  HMI Magnetogram

  This image comes from HMI, another instrument on SDO. It shows the magnetic field directions near the surface of the Sun. White and black areas indicate opposite magnetic polarities, with white showing north (outward) polarity and black showing south (inward) polarity.

  Where: Photosphere

  Wavelength: 6173 angstroms (0.0000006173 m) = Visible (orange)

  Primary ions seen: Neutral iron (Fe I)

  Characteristic temperature: 6,000 K (11,000 F)

  HMI Colorized Magnetogram

  This image comes from HMI, another instrument on SDO. It shows the magnetic field directions near the surface of the Sun. White and black areas indicate opposite magnetic polarities, with white showing north (outward) polarity and black showing south (inward) polarity.

  Where: Photosphere

  Wavelength: 6173 angstroms (0.0000006173 m) = Visible (orange)

  Primary ions seen: Neutral iron (Fe I)

  Characteristic temperature: 6,000 K (11,000 F)

  [Webpage had no Info Popup for: HMI Intensitygram - colored]

  [Webpage had no Info Popup for: HMI Intensitygram - Flattened]

  [Webpage had no Info Popup for: HMI Intensitygram]

  [Webpage had no Info Popup for: HMI Dopplergram]

  [Webpage had no Info Popup for: HMI Intensitygram - High Cadence]

  EVE/ESP and EVE/MEGS-P

  This plot shows time series of 5 strong EUV emission lines, showing how bright the flare is emitting at these various emissions. Also shown is a Dark value, which is a detector that is blocked from seeing the Sun, which shows energetic particles from the Sun that can penetrate the EVE instrument and cause false counts. This Dark diode will increase during solar storms.

  Where: Solar Irradiance (full Sun)


Is there a reason two interstellar objects were found in the solar system in such a short period of time?

The biggest contributor to these two observations happening so close to each other is capability and interest.

Our ability to observe such objects is ever increasing.
Once we saw one, our interest increased and we looked harder ie. lots more funding for the necessary observations to be made that would identify such objects.
We saw the next one soon after.

There is unlikely to be anything special here, just the politics of scientific funding.

It may be that these were two objects that coincidentally passed through the solar system at nearly the same time, but we are going to need to make many more observations of such events before making such a determination.


Massive degraded impact crater on Mercury - chemical evidence?

There is still some research to do to definitively answer this question. Having said that though, according to research published in the paper "Evidence for geochemical terranes on Mercury: Global mapping
of major elements with MESSENGER’s X-Ray Spectrometer (Weider et al. 2015), the high Mg region (HMR) could be due to the consequences of an impact event - the authors from the article suggest that:

  therefore, that the HMR may also be an impact related
  feature, and that its high Mg/Si ratio is evidence of
  mantle excavation during a large impact event early in Mercury’s
  history (>4.1 Ga).

and later, partly resurfaced by subsequent volcanism and smaller impact events.

The authors elaborate further that the mechanism behind it could be

  If an ancient large impact at the HMR site penetrated into a
  vertically differentiated mantle (with additional lateral variations
  possible) and reached a deep, otherwise unsampled layer (i.e., the
  lherzolitic source proposed by Charlier et al. 2013*), the HMR composition
  would be the signature of high-degree partial melting of
  that mantle source.

Additional resource cited:

Charlier et al. 2013, Phase equilibria of ultramafic compositions on Mercury and the origin of the compositional dichotomy, Earth and Planetary Science Letters


Is there a more modern/updated paper than Hurley's 2000 SSE?

My guess would be no because, as of this writing, from the paper link you provided you can see there are 949 citations to that paper with a good chunk of them in this year. You could always contact the author(s) to discuss it. Perhaps your effort could lead to an update.


No, "dark fluid" is not a synonym for dark matter.
The Daily Mail article derives from a press release by the EPFL and an article in Science magazine, neither of which mention the word 'fluid'.
We don't know what dark matter is yet. There are several theories. One of them compares dark matter to a fluid:

Normally speaking, subatomic particles exchange momentum when they interact. Consequently, galactic collisions should also lead to interaction by dark matter particles and by studying it, we would have a better idea of what dark matter really is.
To verify whether dark matter is made up of particles, the researchers explored two possible outcomes. The first, was that particles of dark matter often interact, but don't exchange much momentum. The second was that they seldom interact, but exchange a lot of momentum.
If the first case was true, dark matter would slow down after the collision, since the greater frequency of the interactions would have drag-like effect. If the second case was proven to be true, then dark matter would likely be dispersed into space.
Surprisingly, the study discovered that the dark matter particles from two colliding galaxies, simply pass through each other, suggesting that dark matter particles do not interact with themselves, but have a more fluid-like character instead.

(emphasis mine). So they're not saying dark matter is exactly like a fluid, they're saying dark matter has some properties that are comparable to the behavior of fluids.


A planet's apparent position in the sky is determined by orbital geometry alone.
The tilt of its axis only affects its orientation as we see it in a telescope.

It's easier to think of the solar system in terms of ecliptic coordinates.
The Sun's geocentric latitude is always near 0°, as is the Earth's heliocentric latitude.
A planet's orbital inclination is measured relative to the ecliptic, and its maximum excursion in latitude is related to that inclination.
Each planet crosses the ecliptic northward at its ascending node and southward at its descending node.

For example, Saturn's orbit has a semimajor axis of 9.6 au and is inclined 2.5° to the ecliptic.
Its heliocentric latitude runs that far north or south of the ecliptic over its 29.5 year period.
Earth's orbit brings us 1 au closer at opposition, so Saturn's geocentric ecliptic latitude can reach ±2.8°.

Mercury's orbit has a semimajor axis of 0.4 au and is inclined 7°, so its heliocentric ecliptic latitude range is ±7°.
At inferior conjunction ~0.6 au away, its geocentric ecliptic latitude range is about ±5°.

The transformation from ecliptic to equatorial coordinates is a 3D rotation by the 23.4° obliquity, not a simple sum.
Also a planet's orbit and the Earth's axis may be tilted in different directions.
Saturn's ascending node is 114° longitude east of the J2000 vernal equinox; its orbital inclination and our axial tilt are that far out of phase.
A JPL HORIZONS ephemeris for 29 years of Saturn oppositions and conjunctions gives geocentric declinations in a range of ±22.6°.

Saturn's 26.7° axial tilt relative to its orbital plane does not affect its declination at all.
The effect we do see is that at different points in its orbit, its rings appear open to the north, open the south, or edge-on and slanted east or west, as shown here.

Since you mentioned the solar equator, it's inclined 7.2° to the ecliptic.
The planets do cross it, but their declinations are not related to it either.


In fact there is no official definition of a satellite.

Issac Asimov makes a good argument that the Moon is a planet since its orbit is convex around the Sun for its entire orbit, unlike the other moons of the solar system which are concave when in opposition and convex (to the Sun) when in conjunction.

Additionally, the idea that a body is a satellite of another body based on the barycenter being inside one body is flawed. Consider moving the Moon further or closer from/to the Earth. In which direction would you move it to make the two a "binary system" as opposed to a primary/moon system. In fact, though naively one would consider two closer bodies to be a binary system, moving the Moon closer puts the barycenter closer to the center of the Earth. Yet moving the moon further away moves the barycenter outside the Earth. Would you consider the Earth/Moon system as a binary system if the Moon were 30% further from the Earth?

Thus, we see that all proposed definitions for a satellite are flawed and have corner cases which make a precise definition contentious. This is the same problem as we had when trying to define a planet. No definition will nicely fit all the corner cases that Nature will give us.

Note that this answer was translated into English by my father, who is the one typing it. I'm eight years old and don't use SE unsupervised. A good portion of this information I learned just now because of this question, so thank you to the OP.


Which is the largest planet ever observed?

Yes, there is a limit. Anything with a mass larger than about 13 times that of Jupiter would be called a brown dwarf (a failed star), though whether such an object would consist entirely of gas, or had a rocky/icy core as is probable for most giant planets, is not presently observable. Any larger than about 75 Jupiter masses and we would just call it a star. The exact definition of what a planet is (especially, the 13 Jupiter mass boundary) is still disputed.

Of the bonafide planets that have been detected and confirmed, the catalogue at exoplanets.org lists Kepler-435b as the one with the largest measured radius (although its radius error bar overlaps with that of other planets). The quoted radius is $1.99 \pm 0.18$ times that of Jupiter.

Most giant planets have very similar radii for masses between about 0.5 and 10 times that of Jupiter. The reason for this is that they are largely supported by electron degeneracy pressure. The diversity in the radii (between about 0.7 and 2 times that of Jupiter) of such planets is not yet fully understood.

The plot below shows mass vs radius for "planets". The smaller (probably rocky/icy) planets do show a trend of increasing radius with mass (the solid line is where a theoretical relationship for rocky/icy planets has been used to estimate the mass from the radius). The gas giants above about 0.5 Jupiter masses show no trend and a small scatter.


How to make a star chart from entries in a Bright Star Catalogue?

Here is a tentative answer based on your comment that you want to make a star chart. I'm not an expert in this but I used answers to RA/dec to Alt/Az program or method which link to http://www.stargazing.net/kepler/altaz.html

You have the star's position on the celestial sphere, convert it to decimal degrees:

     item              original             conversion                  decimal degrees
Declination:      -33°  31′   46″    -33 - 31/60. -  46/3600.         =   -33.529444
Right Ascension   00h  08m  03.5s   (  0 +  8/60. + 3.5/3600.) x 15   =     2.014583

You have the time $UTC$ and longitude $LON$, convert it to $LST$ (local sidereal time). From the question Local Sidereal Time and @DavidHammen's answer:

$$LST = 100.46 + 0.985647 d + LON + 15 UT $$

where

$LST$ is local sidereal time in degrees
$d$ is the number of days from J2000, including the fraction of a day
$UT$ is the universal time in decimal hours
long is your longitude in decimal degrees, East positive.

You have $RA$ and $LST$, get $HA$ (Hour Angle)

$$HA = LST - RA$$

You have $DEC$ and $HA$, get altitude and azimuth

$$ALT = \arcsin\left( \sin(DEC) \sin(LAT) + \cos(DEC) \cos(LAT) \ cos(HA) \right)$$

$$AZ = \arccos \left( \frac{\sin(DEC) - \sin(ALT) \sin(LAT)}{\cos(ALT) cos(LAT)} \right)$$

From there you have to decide how you want to plot altitude and azimuth on your map.

If you want to plot it on a circle of radius $R$ then use

$$X = R (1 - ALT/90) \cos(AZ)$$

$$Y = R (1 - ALT/90) \sin(AZ)$$

assuming $ALT$ is in degrees. In this map "top" or $X, Y = 0, R$ will be north and $X, Y = R, 0$ will be East. You will have to decide if that's the best way to plot it or if you want to mirror East-West.


The universe is expanding on a large scale. But locally things are always messy.

Locally, galaxies are not set in stone, they move relative to each other, and the directions are random. If they're moving towards each other fast enough, then they will collide.

Also, there's gravity. Some galaxies are bound to each other by gravity, and that will tend to pull them together.

As to why galaxies move at all, relative to each other - well, things in this universe have kinetic energy, and it's distributed randomly. Being distributed randomly, all kinds of scenarios are possible - things running away from each other, zooming past each other, bumping into each other, etc.

It's a messy and random universe, and the order of expansion becomes apparent only on the largest scale.


Can objects enter hydrostatic equillibrium through processes other than the influence of gravity?

Strictly speaking (as far as I know), hydrostatic equilibrium applies whenever a fluid balances external body forces with the pressure gradient.  From Wikipedia:

  In continuum mechanics, a fluid is said to be in hydrostatic equilibrium or hydrostatic balance when it is at rest, or when the flow velocity at each point is constant over time. This occurs when external forces such as gravity are balanced by a pressure gradient force.

I think the concept happens to be most frequently used in areas where gravity is the external force, but it could in principle be anything else. So, though I stand to be corrected, I think a droplet isolated in space long enough could be said to be in hydrostatic equilibrium, even though the most relevant force is the surface tension, rather than gravity.


How fast would an observer have to travel in order to be able to study an entire star's life in one human lifetime?

We want it so that $1 \text{ year at the speed of x% of c} = \dfrac{10^{10}}{80} \text{ stationary years}$. The Lorentz factor (in terms of c) is $\gamma=\dfrac{1}{\sqrt{1-v^2}}$. Solving for $v$ with $\gamma =  \dfrac{10^{10}}{80},$ we get  $v=0.999999999999999968c = 299792457.999... \text{m}\cdot\text{s}^{-1}$. This is definitely impossible, as the energy requirements are astronomically large: $$5572282108577709241 \text{J}\cdot c^{-2} \cdot 125000000 =696535263572213655125000000 \text{J} = 7.75 \cdot 10^9 \text{ kg of mass converted to energy}$$
This is definitely unrealistic, and is a probably impossible feat to complete. It would be better to study more and more stars instead of exploiting time dilation to view a single star for billions of years.


Does a photon honor the causality principle?

Instead of saying "space has no meaning because it's collapsed into one single point" it's a lot easier if you say that a photon is not at rest in any frame. Then all of the paradoxes you mention go away.

It's easy to use the Lorentz transformations (the mathematical core of special relativity) to show that it's impossible to transform a frame moving at $v<c$ to $c$, or vice versa. (Actually, the Lorentz transformations were developed / discovered a decade or so before Einstein proposed the theory of special relativity. His great insight was to realise that those transformations made more sense if you abolish the hypothesis of the luminiferous aether, and treat space and time as a united geometric structure).

In relativity theory, it simply doesn't make sense to talk about "the point of view of a photon". A photon doesn't have a point of view because it's not geometrically possible for an inertial reference frame to have a velocity of $c$ relative to any other inertial reference frame.

The geometrical structure of spacetime proposed by relativity theory has important implications for causality because no causal influence can travel faster than $c$. Two key concepts here are light cones, and the relativity of simultaneity. There are plenty of questions on this topic on the Physics stack, so I won't go into the details here, but very briefly, if a photon leaves spacetime event A and is absorbed at spacetime event B, then all observers will agree that A occurred before B, although different observers may measure different spatial distances between A & B and different time durations between A & B, but they will all agree that the (local) speed of the photon was $c$. We say that the spacetime interval between A & B is light-like (or null).

If spacetime events A & B have greater space separation than time separation, eg in my frame, B happens 1 second after A, but the distance between them is 2 light-seconds, then we say that the spacetime interval between them is space-like. No causal influence can travel between them, and different observers will disagree on whether A happened before B, or vice versa, and for some observers A & B occur simultaneously. Here's a nice diagram from the Wikipedia article on the relativity of simultaneity:

  Events A, B, and C occur in different order depending on the motion of the observer. The white line represents a plane of simultaneity being moved from the past to the future.


Conversion of UBVRIJHK magnitudes to luminosity?

Luminosities cannot be calculated from magnitudes without knowing the distance to the source (and perhaps something about extinction).

I assume therefore that you are talking about absolute magnitudes in those filter band passes.

Luminosity can also not be arrived at from a single absolute magnitude measurement in the way you suggest in your post, since for a source of a given luminosity, how much of that luminosity emerges in a specific band is entirely dependent on the spectrum of the source.

OK, but now I see what you want - you want to divide up the bolometric luminosity of the Sun between the individual bands using the absolute magnitude of the Sun in those bands. But the luminosity doesn't divide up neatly like that, partly because the bands overlap and partly because the bandpasses are not top-hat functions so it's difficult to see what a "luminosity in the U-band" actually means other than an integration of the intrinsic stellar spectrum (which you don't have) over the (normalised) bandpass sensitivity profile.

$$ L_{\rm band} = \int f_{\nu} S_{\nu} d\nu$$

The formulae you quote contain the absolute magnitudes of the Sun in those bands. That is the first number in the brackets. If you subtract the bolometric magnitude from this, then that gives you (minus) the bolometric correction. The problem is that the UBV etc. magnitude system and the bolometric magnitude system do not share the same zeropoint. The bolometric correction in the V band is about zero, so your formulae would have the entire luminosity of the Sun emerging in the V band. The zeropoints of the UBV etc. system are set by defining an A0 star (like Vega) to have an absolute magnitude of zero in all pass bands.

What you can do is convert the magnitudes into some sort of mean flux density over the bandpass and then just multiply this by the "effective width" of the filter. This is very approximate and essentially assumes the spectrum is flat across the filter bandpass.

Page 100 of Zombeck (1992) gives the flux density zeropoints in UBVRIJHK corresponding to magnitude 0.0. So take your magnitude and scale the zeropoint flux accordingly (by $10^{-0.4m}$) and then multiply the flux density by the effective bandpass width. Finally to convert an observed flux to a luminosity, multiply by $4\pi d^2$, where $d=10$ pc, if you are dealing with absolute magnitudes.

By "PARSEC" I assume you are talking about the stellar evolution models produced by the Padova group. In which case a basic output of the model is the stellar luminosity. Any magnitudes that are provided by the models are calculated afterwards by using a particular set of bolometric corrections (the relationship between absolute bolometric magniude and absolute magnitude in a particular passband).

As regards the specific columns in the tables, you will likely have to go to the original papers that describe the models. I cannot comment further without knowing exactly which models they are or where you got them.


Are there any to-scale diagrams of the TRAPPIST-1 system?

As BMF already pointed out, you won't really see any planets in such overview diagrams for TRAPPIST-1, either. At least not with more than a few pixels per planet.

As I did not find a good model online, I would recommend looking at "Games" like Universe Sandbox or SpaceEngine. For the latter, older versions are even free.

Then you can generate screenshots like these:

Inner 3 planets to scale in front of a white background:

The whole system with the habitable zone:

And a double eclipse from TRAPPIST-1d:


Angular diameter of the Sun's reflection from the ocean, seen from Sun-Earth L1?

$d_o$ and $h_o$ are the distance and diameter of the Sun and $d_i$ and $h_i$  of the respective images. The Earth's radius is $r$ and $f=\frac{r}{2}$ is the focal length.

The mirror equation $$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$$ in the approximation $d_o \gg f$ gives $d_i \approx f$. Then, using the magnification equation

$$h_i = \frac {d_i}{d_o}h_o \approx \frac {f}{d_o}h_o.$$

If $S$ is the distance to the satellite then the angular size is

$$\frac{h_i}{S} \approx \frac{h_o}{d_o}\frac{f}{S} = \frac{h_o}{d_o}\frac{r}{2S} \approx \frac{1.4\times10^6}{150\times10^6}\frac{6.4\times10^3}{2\times 1.6\times 10^6}\approx 1.8 \times 10^{-5}$$

where distances are in km. (We have also ignored that the image is $\frac{R}{2}$ closer to the satellite than is the Earth's center.)

On a more descriptive note, the reflection on perfectly smooth water would look like a very bright spot about 29 km in diameter. It is blurred out (due to sea waves) to about ten times that diameter on the image. It would be interesting to calculate the brightness (per area) of the (unblurred) spot compared to the brightness of the sunlit Earth's surface.


James K has provided a short answer "no". The longer answer is that because Mercury and Venus are in closer orbits to the Sun than Earth's, they will always be in roughly the same area of the sky as the Sun, whereas the other planets aren't so restricted. Since something is generally only visible if it's above the horizon, it follows that if the Sun is below the horizon, Mercury and Venus will either be close to or below the horizon.

The very furthest that Venus gets from the Sun (its "elongation") is 47.8°. If Venus is at its peak elongation around the time of the summer solstice, its angle above the horizon around sunrise (or sunset, depending on which side of the Sun Venus is on at the time) will be maximised. Add to this that since Venus is very bright at this elongation, it's actually visible even when the Sun is low above the horizon; and it remains visible for about three hours after sunset (or, if it's on the Sun's other side, it rises above the horizon about three hours before sunrise). In winter, on the other hand, the ecliptic is substantially lower and Venus's elevation at sunrise/sunset is therefore much less.

If the Sun is more than 47.8° below the horizon, then Venus must also be below the horizon. You can never see Mercury or Venus in "deep dark night" (e.g. around midnight), whereas Mars, Jupiter and Saturn are readily visible (and quite bright!) if they happen to be above the horizon at midnight. [NB I'm excluding the situation of summer at extreme latitudes, since that isn't "deep dark night"].

This diagram on Wikipedia provides a simple representation of elongation and angular separation, and shows the difference in how the "inner" and "outer" planets can appear in the sky.


Taken at face value it's not correct that points on the equator are visible for exactly six months.

For example, Orion is on the celestial equator. It is true that Orion will be above the horizon for half the day (12 hours), but during the summer when it is above the horizon, so is the sun, so it can't be seen.

But Orion is visible from  Northern Europe between the middle of August (when it rises shortly before the sun) and the middle of April (when it sets shortly after dusk) That is rather more than six months. Exact dates can't be given as visibility during twilight depends on factors like the clarity of the atmosphere, and stars don't wink off as the sun rises, they fade out.

On the other hand, if you always observe at the same time then points on the celestial equator will be above the horizon for six months of the year, but just being above the horizon doesn't make it visible, when close to the horizon the thicker atmosphere (and more prosaically, trees and buildings) will make a constellation effectively obscured.

It is true that the north celestial pole is always above the horizon, as are many of the constellations around it. From my location that includes Cassiopeia, Ursa minor, most of Ursa major.

More generally, for Northern Hemisphere observers, the further north a point is, the longer it will remain visible.


Is there really a blood wolf moon this month?

On January 20th, 2019, there will be a total eclipse of the moon. Since Rayleigh scattering of sunlight passing through the earth's atmosphere and then hitting the moon during a total lunar eclipse makes the moon look reddish, total lunar eclipses are sometimes given the (portentous and slightly silly) name of "blood moon". (This seems to be the result of vague New Testament references to lunar eclipses being promoted by a couple of apocalyptic fundamentalists in the last decade; see here for more.)

In addition, there are various folk traditions which give special names to the (first) full moon in each month; it's apparently an old Anglo-Saxon tradition to call the first full moon in January the "Wolf Moon". Since total lunar eclipses necessarily happen during full moons, you could thus (somewhat) legitimately call this a "blood wolf moon". (Some discussion of other "full moon names", including the "harvest moon", can be found here.)

Finally, there is a recent trend of referring to a full moon that happens near to lunar perigee (when the moon makes its closest approach to the earth and appears somewhat larger in the sky) as a "super moon"; this term was apparently coined by the astrologer Richard Nolle in 1979. This sort of thing happens three or four times a year.

So you could call the upcoming lunar eclipse on January 20th a "blood wolf super moon" (or "super blood wolf moon", or whatever), if you really wanted to....

(Personally, I don't think that's nearly enough in the way of unnecessary and portentous adjectives. I'm holding out for "blood wolf demon vampire super extreme power moon".)


What is the explanation for rapid inflation just after the big bang?

With normal matter, the strength of gravity depends on the stress-energy tensor, which, in an isotropic homogeneous universe, has trace of $\rho + 3p$.  Note that the pressure from relativistic particles (photons and neutrinos) enhances the mass density in slowing down expansion.   In inflation theory, it is assumed that there was a scalar field (perhaps) in a small section of the universe that began in a false vacuum state. That is, the field's energy density remained roughly constant as the universe began to inflate, and therefore the total energy of this field grew as the volume of the universe grew.  The field's pressure was negative and greatly overwhelmed the inward force of the density of normal matter and pressure. It is this negative pressure that rapidly expanded this section of the universe by many factors.
Energy was conserved because work was done by the gravitational field as it pressed inward against the field's outward pressure; the gravitational potential became very negative, while the energy density in the field became very large.  Eventually, the field decayed away, heating the particles, which resulted in the traditional Big Bang conditions of a hot expanding universe.
Throughout this theory, standard physics is expected to hold with just the addition of this new scalar field.  It is introduced because it helps to explain why the entire visible universe today is so homogeneous even though there are subregions that were previously thought to have not been in causal contact until quite recently.  It also explains why the universe is very nearly flat.  It converted a small section of the universe to extremely large radii and thus small curvature.
You can read about this from one of its inventors here


What is the physical interpretation of the derivative of the emitted and observed wavelength?

I think you can understand that factor as follows:

When photons lose energy, they're spread out over a larger wavelength range. Since there is a fixed number of photons, that means that the number of photons per observed wavelength bin decreases by a factor of (1+z), and hence the flux density, which is really what $f_\lambda$ is, decreases by this factor.


Not an answer, but I thought this was a good slice of a picture of the Moon's orbit around the sun.

Source:  http://www.wired.com/2012/12/does-the-moon-orbit-the-sun-or-the-earth/


What is "class 4 light pollution" and could I estimate my own area's class visually?

Only the original poster can answer what was meant, but one system for rating the darkness of the night sky is the Bortle Scale.

https://en.m.wikipedia.org/wiki/Bortle_scale


Which star did Uranus occult in 1977 when the planet's rings were discovered?

From wikipedia, the rings of uranus

  The definitive discovery of the Uranian Rings [...] use[d] the occultation of the star SAO 158687[...]

The star SAO 158687, also cataloged as HD 128598 is a magnitude 8.7, orange dwarf star in Libra. There seems to be nothing special about the star, except that one day, a planet passed in front of it.


Books describing first-hand experience of astrophysicists finding planets without skipping actual maths

The discovery and recapture of Ceres, and the brilliant discovery and use by Gauss of the Normal distribution, the method of least squares, and a method that would now be thought of as a type of fast fourier transform to predict the position of the planet to within 0.5 degrees, based on just 6 observations is one of the great achievements of "pen and paper astronomy". Gauss, though we now think of him as a pure mathematician, held the post of Professor of Astronomy in Göttingen.

Discovery of the First Asteroid, Ceres: Historical Studies in Asteroid Research by Clifford Cunningham, ISBN 978-3-319-21777-2 tells this story in some detail, but ouch its not cheap! nearly £100 on dead trees.


How does the center of a black hole can contain so much mass?

The notion of "centre of the black hole" is misleading. It treats the spacetime in and around the black hole as if it were normal flat spacetime, with no gravitational distortion.  In fact the spacetime is highly curved. So curved, in fact, that "space" behaves more like "time".  Instead of "where is the singularity" you should ask "when is the singularity."

The answer to this question is always "in the future".  There are no singularities in our past.  This is significant, since when we say "there is a pen on the table" we mean that the pen was there a few nano-seconds ago, because of the time light takes to reach us. Everything that we say "exists" actually exists in our past.  Singularities are a "when" that have a "before" but don't have an "after". Nothing can have a singularity in its past, because there is no future after the singularity.

So in this sense, singularities don't exist. We can't study them because they are always in the future (due to the extreme bending of spacetime). If we fall into a black hole we will (rapidly) reach the singularity but there is no way to report back about it, since the singularity is surrounded by an event horizon. Some of us may (if we are unlucky) fall into a black hole. For these people the singularity is in their future. But it is not possible for a singularity to be observed, as it can never be in the past of any timeline.

So a black hole behaves "as if" all the mass was concentrated in one point. But don't think of the singularity as a "thing" in the centre of the black hole, but think of it as a "when", and always remember "this is weird, and analogies and metaphor are never a complete explanation".


How can a naked singularity be possible?

Nobody likes the idea of naked singularities, as they would have a toxic effect on causality. If a singularity existed that was not separated from us by an event horizon, then not only would the future not be predictable, but the past would not be fixed. Like the grandfather paradox, it wouldn't make sense, so it cant exist. The trouble is that GR doesn't implicitly rule them out.

If a black hole is spinning fast enough, or has enough charge then it seems a naked singularity could form. Naively you may think that the centrifugal force in the first case, and the electromagnetic force in the second are sufficient to overcome gravity.

I particular if $G^2M^4/c^2 < J^2$, where J is angular momentum, M is the mass, G is the gravitation constant, and c is the speed of light. The there will be a naked singularity. For a small, stellar mass black hole would need an angular momentum of $10^{42} \mathrm{kg m^2 s^{-1}}$ to lose its event horizon


How does the gravity of one planet affect the velocity of other planets?

The most famous example of this lead to the discovery of Neptune.   Planetary orbits were neatly worked out by Kepler and Newton, so they had a good idea of how fast planets should be moving.   Uranus wasn't moving as fast as it should have, and that lead to the conclusion that there was an 8th planet.

Assuming the two planets orbit in a similar plane and don't cross, the inner planet moves faster so it's always catching up to and passing the outer planet.   When it's behind the planet and catching up, it moves faster, when it's ahead, it moves slower.   The effect is, as you stated, very very small but in the case of Uranus, with an orbital period of 84 years, so it spends a couple decades being accelerated by Neptune until it passes it, then it spends a similar amount of time being slowed down by it.   The observed speeding up and slowing down over years was enough to lead to a calculation of the approximate location of, and shortly after, the discovery of Neptune.

The 3-body problem is complicated and the math involved is somewhat advanced, but you can do a rough approximation of acceleration and altered distance from the expected orbit using Newton's laws of gravitational acceleration g=GM/r^2 over roughly a fraction of the orbital period.  It's surprisingly small, but over a couple months (for Earth) or a couple decades (for Uranus) it works out to about what you might expect.  A noticeable but tiny fraction of the orbit.

Over time, these variations balance out, so the orbits remain largely unchanged over thousands, millions and even billions of years, unless the planets happen to fall into resonance.  Resonance can lead to the gradual destabilizing of orbits.


Under Earth's gravitational pull, how dense would an object have to be to reach its Schwarzschild Radius?

Earth's gravity is quite weak. It manages to compress air (luckily for us), but that's about it. Since the Earth is not a black hole it can never compress something else to be a black hole.

That being said, lets consider some object and assume something compresses it to a black hole.

The Schwarzschild radius is defined as
$$
R_s = \frac{2GM}{c^2}.
$$
Now lets say the object is a sphere of radius $R_s$, then its uniform density is
$$
 \rho = \frac{M}{ \frac{4\pi}{3} R_s^3 }.
$$
Using this to get rid of the mass term leaves us with
$$
R_s = \frac{2G \rho \frac{4\pi}{3} R_s^3}{c^2}
$$
Inverting this expression gives the black hole density for a given radius
$$
\rho = \frac{3 c^2}{8 \pi G R_s^2}.
$$

Notice that if you want a really small black hole, say $R=1$ m, the density has to be about $1.6\times10^{26}$ kg/m$^3$. That's about 20 times the mass of Earth stuffed in a cubic meter, so quite a lot.

But if the black hole can be as large as we please, then the density can get extremely low. A great example of this are supermassive black holes, which have about the same density as water (that's $10^3$ kg/m$^3$).


About how many revolutions has the Earth made around the sun?

Since the very early solar system, there have not been large-scale movements of the planets.

In the early solar system, while the planets were still embedded in a protoplanetary disk, there were large movements. (notably the hypothesised "Grand Tack" of Jupiter) However once Jupiter reached it's current position of 5.2AU it remained there, and the Earth formed then too.

There have not been any significant changes to the Earth's orbit since the Earth formed, otherwise there would have been significant heating or cooling. So the average year length was and is about 30 000 000 seconds, and Earth has made about 4.5 billion orbits of the sun since then.


Generally bright points of light moving across the sky at a constant speed are satellites. Meteors are much faster and usually have a trail.
If you saw two or more close together and moving at the same speed they were probably some of the Starlink satellites launched by SpaceX.
The sudden vanishing is usually because they either moved into Earth's shadow, or the angle of the reflecting surfaces moved so that they didn't reflect light to you as an observer.


Is there a stellar database that indicates how long ago stars in our Galaxy formed?

There isn't really a database as you request. Finding the ages of stars is difficult. Only one star has an accurately known age - the Sun. That comes from radioisotope dating of meteorites. For other stars we must rely on models to a greater or lesser extent and we can only estimate an age if the star has a mass or is in a phase of its evolution where things are changing rapidly enough to give some handle on how old it is. For stars at the mass of the Sun or a little bit bigger, one can use evolution in the Hertzsprung-Russell diagram. Stars become more luminous as they burn through their core hydrogen and precise measurements can give an age to about $\pm 1$ Gyr. These are the data that Snaith et al. (2015) use in their paper.

So what Snaith et al. do is they attempt to constrain the star formation history of our Galaxy (or at least stars in the disk of our Galaxy) by making a Galactic Chemical Evolution model that predicts how the abundances of silicon and iron change with time and compare the results of their model with the observed silicon and iron abundances in this set of solar-type stars with reasonably well-estimated ages. Iron and silicon are used because they provide differing constraints. Iron is produced during the evolution of relatively low-mass ($1.5 - 4 M_{\odot}$) stars that live their long(ish) lives (1-10 Gyr), become white dwarfs and then some of the white dwarfs explode as type Ia supernovae. Silicon is produced by massive stars with short lives ($<0.1$ Gyr) and spread promptly into the interstellar medium by core-collape type II supernovae.

A representative lot from Snaith's paper is shown below. The star forming rate is shown in plot (a) and the match of the model to the observational data (they are trying to match the solid green points here) on chemical abundance is shown in the other plots.

They test a number of models and different assumptions, but it appears to be quite a robust result (found in other studies also) that there was a burst of star formation 10-12 billion years ago, a lull at 8-9 billion years ago and then a more constant, lower rate over the last 7 billion years. The very low rate in their model over the last 2 billion years is unlikely to be true and is probably an artefact of the age uncertainties in their data points (e.g. stars with age of $1 \pm 1$ billion years) and we certainly see plenty of star formation in the disk of the Galaxy today.

Contrary to what you say in your question this does (almost) represent the age distribution of stars in the Galaxy. That is because the vast majority of stars (90+%) are of a solar mass or lower. Such stars have main sequence life times of 11+ billion years, such that almost every one of them that was born is still a main sequence star now.

There will be a small reduction in the numbers of the oldest stars, because a small fraction of those ($<10$%) will have lived and died and also because a small fraction of the oldest stars may have escaped the Galaxy entirely (due to various kinematic processes that cause their orbital speeds to change). There is also an effect whereby if you take a volume close to the Sun, the oldest stars will be under-represented because they are distributed more broadly around the disc midplane for the same reason.

But by an large, the upper left plot in the picture is roughly what we expect the age distribution of stars to be in the Galactic disk.

There are a small population of very old stars ($12+$ bilion years), perhaps 1%, that are distributed more widely in the Galactic halo.


Can Eight Retrograde (seen from Earth) planets occur?

EDIT: I used the DE431 ephemeris to get the results below. However, if you visit HORIZONS (https://ssd.jpl.nasa.gov/horizons.cgi) and create an ephemeris for Pluto (actually Pluto's barycenter, object 999), the results include this warning:

  The Horizons output log may report using DE431 as the source of target
  or center data, but it is nevertheless using DE433 as an override if
  Pluto is involved. This is to maintain ephemeris consistency for the
  New Horizons  spacecraft encounter at Pluto.

Therefore, my calculations may be inaccurate, above and beyond the normal inaccuracy of predicting planetary data in the past and future.

There are 2 times in the past 15,000 years and 5 times in the next 15,000 years where all 8 planets are in retrograde as observed from Earth.

The table below lists the times when this happens:

The last column is the ephemeris time, the number of seconds since '2000-JAN-01 12:00:00 UTC', which NASA refers to as "the epoch".
The first line in each pair shows the start time of when all 8 planets are in retrograde, and the final planet to enter retrograde to make this true.
The second line in each pair shows the end time when at least one planet is no longer in retrograde, and the planet whose retrograde period has ended.

B.C. 12435-AUG-10 07:52 JUPITER STARTS RETROGRADE -455482152466.038696
B.C. 12435-AUG-24 17:12 MERCURY ENDS RETROGRADE -455480909251.797852

B.C.  4139-MAR-23 15:02 MARS STARTS RETROGRADE -193692373029.630005
B.C.  4139-MAR-28 21:57 MERCURY ENDS RETROGRADE -193691916114.978455

A.D.  7662-APR-27 19:33 MARS STARTS RETROGRADE 178685595220.195740
A.D.  7662-MAY-12 05:19 NEPTUNE ENDS RETROGRADE 178686840014.478882

A.D.  8807-JAN-07 02:02 MERCURY STARTS RETROGRADE 214808680981.726929
A.D.  8807-JAN-20 01:05 JUPITER ENDS RETROGRADE 214809800771.851685

A.D. 10575-FEB-14 21:34 VENUS STARTS RETROGRADE 270604748086.593933
A.D. 10575-FEB-24 18:53 NEPTUNE ENDS RETROGRADE 270605602468.037781

A.D. 13679-NOV-06 16:03 VENUS STARTS RETROGRADE 368580427477.379639
A.D. 13679-NOV-19 01:05 MERCURY ENDS RETROGRADE 368581496795.683594

A.D. 16364-MAR-11 19:25 MERCURY STARTS RETROGRADE 453290109945.561157
A.D. 16364-MAR-13 08:53 MARS ENDS RETROGRADE 453290244824.895996

Notes and caveats:

I wrote https://github.com/barrycarter/bcapps/tree/master/ASTRO/bc-retrograde.c to print out retrograde times for all planets.
The file https://github.com/barrycarter/bcapps/tree/master/ASTRO/all-retrogrades.txt.bz2 shows start and end retrograde times for all 8 planets. NOTE: the lines indicating that planets leave retrograde at "17191-MAR-01 00:00" is incorrect-- that's simply the maximum time for which I could make calculations.
I wrote https://github.com/barrycarter/bcapps/tree/master/ASTRO/bc-moving-backwards.pl to determine which planets were in retrograde at any given time, and used the results to find the times above where all 8 planets were in retrograde.
I spot checked my results against https://www.calsky.com/ -- note that I am computing "ecliptic retrograde", when a planet's ecliptic longitude is decreasing, not "equatorial retrograde", when a planet's right ascension is decreasing.
As always, I invite people to double check my results, and remind everyone not to use my results for anything critical.


Does the Time of Sunset (Sunrise) Change Over Time Given a Fixed Location and Consistent Atmospheric Conditions?

Shalom! Strictly speaking, yes, the time of sunset does change, but the amount of change is imperceptible to humans over a lifetime. The change is due to the slowly varying obliquity of the Earth on its axis, the slowly varying eccentricity of Earth’s orbit around the Sun, and other such cyclic or secular changes.
Other than that, you’ll notice a slight change from one year to the next, as the year doesn’t last exactly 365 days. This is a four-year cycle, and the difference is around one minute, I would say, on average, depending on the time of the year—for example, when there is less of a difference in sunrise/sunset times from one day to the next, the difference between the same date of two successive years is lesser than for dates when there is a large difference between one day and the next.


What that ESA (European Space Agency) page titled Hot gas sloshing in a galactic cauldron that you link to describes are called WHIM (Warm–Hot Intergalactic Medium). They are not interstellar medium, but intergalactic medium gas. The difference in density is huge, with interstellar medium density at an average of $\rho ∼ 1\ ppcm$ (one proton per cubic centimeter), but the density of these WHIM being even a few orders of magnitude lower at $\rho ∼ 10^{−6}−10^{−5}\ ppcm$, or roughly 1 to 10 protons per cubic meter (NASA's Chandra X-ray Observatory quotes average density of 6 protons per cubic meter).

What is interesting about WHIM is that they are absolutely huge. We're talking of distances that extend across clusters of galaxies (so stretching multiple millions of light years), which means that even as tenuous as they are, account for a large portion of the baryonic matter of the Universe:

  Such matter is predicted to account for a sizable fraction ($∼ 50\%$)
  of all the baryons in the local        ($z < 1$ [redshift in the infrared
  spectrum]) Universe, and it is thus considered the best candidate to
  host the baryons seen at high redshift and missing from the low
  redshift census.

So now about their heat emissions, and why are they detected in the X-ray range in the first place (the ESA's article mentions the photograph featured there was taken by the ESA’s XMM-Newton X-ray observatory):

  Electrons and baryons in the WHIM are shock-heated during their infall
  in the dark matter LSS [Large–Scale Structures] potential well, and settle in
  filamentary/sheet-like structures surrounding LSSs.

I've added a few clarifications in quotes encapsulated in square brackets, but what this means is that parts of these WHIM interact with AGN (Active Galactic Nucleus) as the galaxies pass by, and the X-ray emissions of AGN excite Baryonic matter to a temperature $T ∼ 10^5−10^7 K$.

Quote sources:

The Cosmic Baryon Budget, Fukugita et al. 1998
The Cosmic Energy Inventory, Fukugita & Peebles 2004

Additional reading:

X-ray Discovery Points to Location of Missing Matter, NASA
Last “Missing” Normal Matter Is Found, Sky & Telescope
Studying the WHIM Content of the Galaxy Large–Scale Structures along the Line of Sight to H 2356-309 (PDF), L. Zappacosta et al., Harvard-Smithsonian Center for Astrophysics