Sky Glow calculations I am leading a project for my High School NASA Club, and am looking for some advice. The goal of our project is to use a Raspberry Pi to control a camera that takes photos of the sky, and uses the photo to calculate sky glow, and create a map of where it in the sky is there the most sky glow. After calculating sky glow the device would add it to a plot that shows how sky glow varies throughout the year, and creates a nightly twitter post rating the night for Astronomy. I currently have working code that when given a picture, it will convert down to greyscale, and then find the brightness map, and find an average sky glow. The part that I am stuck on is in figuring out what kind of camera to use. I don't know if the RPi camera module would work for taking photos of the stars, or if I would be better suited using a USB webcam with it. I am not sure what factors are important in this decision at all. Also, once the camera is set up, what would be the best process for calibrating it to get values that agree most closely with accepted ones? I couldn't decide if this should go on the Astronomy, Raspberry Pi, or Photography Stack Exchange, but I chose this one since sky glow may be something astronomers calculate a lot. Thank you for your help.

I would recommend the PiNoir Camera. Since it has no IR-filter attached, it's perfectly made for nightly observations. If you need the Raspberry just to calculate Skyglow in general, you could try to collect all light and bundle it to a beam, then measure its intensity.

What also came to mind is: When calculating the rating of the night, or a point in the sky, consider moon phases in your calculations. A bright and full moon will significantly raise average sky-glow and vice versa.


Time measurement outside solar system? The sun is used to determine and measuring time in our solar system. What if we were to travel outside the solar system i.e. to interstellar space, or other galaxies and other celestial bodies (out of our solar system). How would we keep a tab on time then?

Units of time are actually quite arbitrary.

Their relationship to astronomical features is historical:

it's what people used when counting rotations and revolutions of the planet were the best tools to hand. Nowadays, they are legacy values: the "true" units are other things that happen to be feasible to measure using modern tools (wavelengths of light, resonances of atoms), with particular values chosen and scaled to match the old values for continuity.

And other units derived from them were purely matters of arithmetical convenience. Factors of 12 and 60 were common because they make it easy to do arithmetic in thirds and quarters, and when doing fractions, they reduce easily. So a value like a "second" is absolutely and utterly arbitrary; it applies to outer space exactly as well as it does here.

The second's arbitrary value is the fundamental unit of the metric system, that that's what scientists use to communicate with each other. The numbers can be scaled for their convenience: as long as they're labeled, one system is as good as any other. (Especially with computers to do the work.)

Astronomers try to measure rotations and revolutions of distant bodies, and when speaking of them in public they might use "day" or "year" to help people visualize it. But when speaking to each other, they know what the technical terms mean, and avoid the ambiguity of layperson speak.

  So in the end, it's all just "seconds", scaled up or down. It may be
  converted to something else for public consumption or even with each
  other, but when it comes down to brass tacks they find it's easiest to
  all use the same unit, even if it's completely arbitrary.

(It is worth noting that some physicists use a rather unfamiliar set of units call "natural units", under which important constants like the speed of light and the pull of gravity work out to precisely 1. These are convenient when you're pushing around equations, but require a fair bit of effort to translate into clocks and rulers when you want to build an actual experiment.)

  Another approach can be Through an atomic clock.

Modern atomic clocks use atoms to determine the passing of time. So you would measure it using that.

  For example 1 second is defined as --- the duration of 9192631770
  periods of the radiation corresponding to the transition between the
  two hyperfine levels of the ground state of the caesium 133 atom This
  mean that if you can make a laser shine on a ceasium 133 atom and by
  counting the periods of the light emitted you will now how many
  seconds have passed.

As a interesting side note sins our orbit around the sun is not constant it is sometimes necessary by the scientific community to prolong a year by a second or so every now and then to make clocks match the moment of the earth. There called leap seconds


Direction Names Within a Galaxy Are there commonly used names for objective directions within a spiral galaxy?

Central and peripheral describe objective directions, because the observer's frame of reference doesn't affect whether something is closer or farther from the middle.  "Clockwise" or "counterclockwise" don't work, because they depend on which side you're viewing from.

Are there names that can describe one object's location relative to another object, either in the galaxy's direction of spin or against it?  Or words to describe whether something is farther or closer out along a particular arm of a spiral galaxy?

The galactic coordinate system is usually used for galaxies. It is based on the galactic plane — approximately where the galactic disc is.

It uses galactic latitude (for the y-value of the objects) and galactic longitude (x and z values, based on the disc).

There is also a supergalactic coordinate system used for superclusters.

Its equator is an invisible line called the supergalactic plane, which is the preferred distribution of galaxies in a supercluster.


I have a Model 36050 telescope My Model 36050 telescope came with the H6mm, H20mm and the green filtered moon lenses. It has a 360mm focal length and a 50mm diamiter.  What can be seen with it?

With this scope you will be able to see quite a few things. Unfortunately, you won't be able to resolve much detail with some objects.

At 50mm aperture, the light gathering ability is not fantastic. That being said, some brighter objects will be easily seen. Your max power (zoom) is approximately 2*aperture, and the power is your objective focal length (360) / eyepiece focal length. Say you have a 6mm eyepiece in, your power (zoom) will be 360/6 = 60

PLANETS

Mercury and Venus will be easily seen with your telescope, at the highest power, you may be able to resolve some of the phases of Venus.

Mars will be visible as a reddish round blob, with hazy edges.

Jupiter will be visible along with the 4 Galilean moons (Io, Ganymeade, Europa, Callisto), although cloud bands will be very hard to resolve. Perhaps if you are at a dark sky site with perfect seeing, you may be able to make out a faint cloud band, but you will likely be looking at a large bright dot, with 4 smaller ones near it. A handy app for your phone called 'Moons of Jupiter' will allow you to tell which ones you are looking at.

Saturn will be a nice sight as you will be able to see the rings. This is always a spectacular sight for newcomers to astronomy. Although it will be rather small (even at maximum power), it will still be a nice sight.

Uranus and Neptune will likely be too dim for you to see. In a dark sky site in great conditions, you may be able to make out a tiny dim dot for each, but that is unlikely.

DSO's (Deep Sky Objects)

For DSO's, your best bet is to go for clusters and nebula at around mag +6.5-+7 max. Any dimmer than that and you will struggle. This means you will be able to see such globular clusters as the great cluster in Hercules, the Andromeda galaxy and the Orion nebula (these 3 were picked as they are the brightest in their categories) although they will look like just fain smudges.

Stars

Some of the brighter Carbon stars are a nice sight as they are often pleasing shades of red, which distinguish them from background stars. Binary stars are also a popular sight. There are a lot of good lists of binary stars online, although you will want to go for ones with a large separation between them, so that you will be able to see both. Mizar and Alcor are a good starting point for binary stars, these are easy to locate in the handle of the 'Big Dipper' in Ursa Major. If you want a bit more from your binary stars, you could look for some colour contrasting binaries, such as Albiero in Cygnus, which is a lovely binary star with one blue and one gold star.

Some open star clusters may be nice to go for, such as the Beehive cluster in Cancer, or the Pleiades in Taurus.

The best thing to do is to download a few apps, get yourself a star chart and get to a local dark sky sight and see what you can see. There are hundreds of objects you will be able to see. You could also look for a local Astronomy group based in your location for further guidance and also a chance to look through some other telescopes too!

Oh and let us not forget the Moon! That is always a beautiful sight!


How to search SIMBAD using identifiers? I'm trying to get some data about the moon from SIMBAD (specifically, radiation spectrum in certain wavelengths), and I encountered an unexpected problem: I can't find the Moon.
In SIMBAD, I can search using several different methods. Since I'm looking for a specific object in the sky, I thought the best method would be just to search for the moon by its name, using "Query by identifiers".
It may sound silly, but the problem is I don't know what is the nomenclature of the moon, nor how to find it. Using the obvious guesses ("Moon", "Luna", etc.) gave me nothing. What am I missing here? Does identifier mean something else than what I thought?
P.S. This problem also occurred to me while searching for the Sun, which signals to me that I'm probably missing something critical here...

If you look on the main SIMBAD page, you will see the following in the box titled "Content":

The SIMBAD astronomical database provides basic data, cross-identifications, bibliography and measurements for astronomical objects outside the solar system.

(emphasis mine)
The Moon and the Sun are inside the Solar System, thus they are out-of-scope for SIMBAD.


How massive must a rocky body be to liquify adequately to eliminate internal mass concentrations that could disrupt natural satellite orbits? Maybe a slightly weird question, and maybe more appropriate for some future geophysics SE, but I feel like this is what questions like Do moons have moons? are really driving at, and it just almost gets answered there.

How massive must a rocky body be to liquify adequately to eliminate internal mass concentrations that could disrupt natural satellite orbits?

I understand geothermal energy to be a function of both gravitation and fissionable material in the planetary body, so an ideal answer would at least state its assumptions in regards to the amount of energy from fissionable material present.

Planets and moons are never going to be completely homogenized such that their radial density is perfectly spherically symmetric. They're always going to have mass concentrations (mascons) with regions of more mass than average. These mascons, due to the extra mass, cause gravitational anomalies of higher gravity in those local regions. These mascons exist for all bodies, be it the Earth (e.g., Hawaii), the Moon (e.g., Mare Imbrium), or Mars (e.g., Isidis Planitia).

Your question is to ask, how massive must a body be such that it can liquify itself and eliminate these mass concentrations so as to not disturb satellite orbits. This is specifically an issue in regard to orbiting the Moon because its Hill sphere is so small that only very low orbits can exist, but the mascons in the Moon make most of these unstable.

I think though, you're asking the wrong question. As I said above, the Earth itself has mascons which are in fact caused by its liquid interior! The region around and below Hawaii has a slightly higher than average mass caused by a mantle plume, effectively an upwelling of less dense material from the core-mantle boundary. Aside from creating the mascon, it was this mantle plume which also created Hawaii, but that's beside the point.

The point here is that all bodies have mascons, whether they have liquid interiors like the Earth or primarily solid interiors like the Moon. The difference between orbiting the Earth and the Moon though, is the distance you can do it at. The reason mascons are such a problem for orbits around the Moon is that you have to orbit so close such that the effect of these mascons compared to the general field is a noticeable perturbation on the orbit. If you could orbit the Moon farther out, they wouldn't be a problem. The opposite applies for the Earth. Most satellites around the Earth are very far away such that the Earth's mascons don't have any noticeable effect. If you orbited close enough (and likely that would require orbiting inside the atmosphere which presents lots of other problems), the mascons would become a problem just as they are for lunar orbits.

In fact, shown below is what the Earth would look like if you magnified all the various mass/gravity anomolies (such a shape is known as the Earth's geoid). All those irregularities would perturb a satellite if it had to orbit the Earth as close as it would need to orbit the Moon.

I know I kind of side-stepped your question, but I hope you'll agree that the question isn't really meaningful. All bodies have mascons to some degree or another, you can't get away from that.


Will Starlink deface the night sky? There is some discussion (and there are some questions on this site) about the impact of Starlink on observational astronomy. I am not an astronomer, but I am amazed by the beauty an immensity of the night sky.
Must I be concerned, from an aesthetical point of view, that Starlink will change the night sky that I and countless people before me have admired? For example, will there be less stars visible to the naked eye, or will the night sky be defaced by a myriad of artificial light sources?
For background: Starlink is a commercial project by SpaceX, a private company, that seeks to launch thousands of small satellites into the earth's orbit.

Satellites just add moving lights to the sky, they do not obscure stars. However, some may find that disrupting their view of what a sky should look like.
The visual magnitude of starlink satellites is about 5.92 at zenith, and usually darker (but also sometimes brighter). The new darker satellites are about 0.77 magnitudes fainter. This is in the lower range for naked eye vision under good conditions.
Light pollution is likely to make the satellites invisible to most people, since they already cannot see much of the sky. You need to be outside suburban skies on the Bortle scale to be able to notice them. So if you are concerned with the aesthetics of the sky, city light pollution is a far, far greater problem.


Stellar Classification Table and Star RNG I found a table that was based on data from 2003 for Stellar Classification, and I can't get to the given data source. I plan on using the table as a basis for a program that spits out random star systems with stars +-5% of the given data.

I know type R was renamed to C so I am already throwing that out. Is there anything else I should ignore from this table or add?

Or this table is crap and should not be used as there are glaring errors.

I found a table that was based on data from 2003 for Stellar Classification, and I can't get to the given data source ...

The website that was the source of the data was last active  in 2006. I've replaced the broken URLs with working URLs. I wouldn't recommend relying on data that old, there's likely a newer source.

Original (404) Text and URLs:

  Credit:

  The table data was obtained from SpaceGear.Org's SCTable version 3.0 09/23/2003. The following data source was obtained from their SCTable web page:

Original Text and repaired URLs:

  Credit:

  The table data was obtained from SpaceGear.Org's SCTable version 3.0 09/23/2003. The following data source was obtained from their SCTable web page:

The 3rd links says that they derived the data from:

  Absolute magnitude was derived from the standard Hertzsprung-Russell (HR) diagram. The book Stars and Their Spectra by James B. Kaler was used as a resource for deriving the absolute magnitude for each possible luminosity class. Classifications N, S, WC, WN, C, and R are mapped as follows:

N and S are equivalent to M.
WC and WN are equivalent to O.
C spans G4 to M9.
R spans G4 to M0.

  Temperature was derived from the relationship between luminosity class and spectral class. The book Stars and Their Spectra by James B. Kaler and Mitchell Charity's work on star colors was used as a resource for deriving temperature for Main Sequence (V), Supergiant (Ia), Giant (II), and Dwarf (D) luminosity classes. Luminosity classes Ia0, Ib, III, IV, and VI are mapped as follows:

Temperatures of Ia0 and Ib are equivalent to Ia.
Temperatures of II is equivalent to III.
Temperatures of IV and VI are equivalent to V.

  Color was derived from Mitchell Charity's work on star colors. Each designated color is defined with at least one field (R, G, or B) at full saturation (value equals 255). Actual color values for display would be attenuated across all color components based on the star's apparent magnitude in a given situation.

  Mass was derived from many sources including Astrophysical Formulae by Kenneth R. Lang. Many rows are heavily interpolated from known data on star mass.

See how I fixed all the URLs so they work correctly. If there is more from that webpage that you wanted sources for then you can use the same method. Since I was able to get to the raw CSV data that should be easier to use in your program.

See also Wikipedia's webpage on Stellar Classification.

An example of a website providing similar data (from 2015) is: "The Mag 5 Star Catalog" or see: "The HYG Database", his latest update to GitHub is a couple of years ago.


When was the diameter of Titan first measured? Titan is enshrouded in a thick opaque cloud of methane : with a telescope, you can't see the moon's surface. Because of this, from a distance, you can only see the diameter of the moon + its atmosphere, not the diameter of the moon itself.
When was the diameter of Titan first measured and how?

Discovery of Titan, 1655: Unknown diameter.
Dollfus, 1970: 4,850$\pm$300km (1). Measured by Filar micrometer (2) and diskmeter / double-image micrometer (3).  (Apparently a summary of earlier measurements, currently trying to find print copy)
NASA SP-340, 1974: Summary of above techniques, propose settling on 5,000km diameter until it can be measured by stellar occultation (the same process used to determine the size and shape of Ultima Thule). (4)
Elliot, 1975: Measured the diameter by limb darkening / lunar occultation (i.e. by how Titan passed by Saturn).
Calculated diameter somewhere between 5,132$\pm$47km and 5,832$\pm$53km.  (5)
Pioneer 11, 1979: Only 5,800km found in references, appears to still be using the 1975 calculation. (6)
Voyager 1, 1980: Same as above.
Cassini, 2004: The Cassini probe had radar which was able to "see" right through Titan's atmosphere.  Although the first flyby was on July 2, 2004, it appears that the radar was not used until the second of the 45 planned flybys (7,8).
This paper from 2009 combined data from all of Cassini's passes at that point.  This allowed the authors to build a model of the actual surface over enough of Titan to determine its size (~5150km) and that it is slightly oblate, like the Earth.  In terms of actually measuring the surface, this is probably the best answer.
(1): Surfaces and interiors of planets and satellites, Dollfus, Audouin, 1970.  Page 129 is apparently the specific table.
(2): Lunar Occultation of Saturn. I. The Diameters of Tethys, Dione, Rhea, Titan, and lapetus, Elliot, J.L., 1975.


Locations of farthest matter Regarding matter that has the highest redshifts, do we see such matter in every general direction we look (relative to Earth: up down left right front back)?

It's not possible to see photons from "beyond" the cosmic microwave background (CMB), because (assuming our cosmology is generally correct) the CMB is the result of recombination of electrons with protons in the early universe. Photons from before that era were absorbed in the plasma.

Essentially then, the earliest matter we can see (in photonic terms) is this glowing hot hydrogen, but massively red-shifted so it looks like hydrogen at 3K.

We see the CMB in every direction.


How to find the source of the light? There are millions of galaxies out there in the entire universe. When a telescope is set to a direction to to identify the stellar objects, it can capture all lights from all the objects. How do we identity that the particular light is emitted from one star/galaxy and not the other one?

Even if we precisely point to one star, there are chances that other luminous objects interfere with the light ray.

The positions of many galaxies are recorded, so astronomers can point their telescopes to the exact point in the sky that they wish to observe. In addition,the relatively nearby galaxies can be recognised by their appearance,& these are the galaxies most intensively studied. Professional astronomers have memorised most of the stars you will see in the night sky, and the constellations (Sagittarius, Ursa Major, Orion etc) guide them to their target. Those galaxies near the limit of observation are of lttle interest except when something unusual & very luminous happens within them,such as a supernova or gamma ray burst.


What are the Time Zones on Mercury? I normally post on worldbuilding, but this question is merely about facts of a planets orbit and spin. I am trying to figure out how time periods work on Mercury to possibly make a worldbuilding guide for the planet. I recently heard on an educational video that the eccentricity of Mercuries orbit can cause the sun to temporarily become still on mercury, and then move in the opposite direction, then become still again, and move along it's normal course in the sky. The possible effects of this are that certain regions of mercury could experience a "season" where the son only goes partway through the sky before turning around and setting creating something of a cycle of "short days" and "long days". I imagine because of mercury's mathematically synchronized orbit, these short days can only be experience on small bands of the planet, I would imagine there is only one circular band going around the world along a line of longitude that is like this.
What I am trying to understand is how different parts of mercury experience the day and not simply the length of mercuries day in absolute terms. I am wondering how a hypothetical person that would be following the "evening" and "morning" sections of mercury would experience the backwards movement of the sun over the course of the mercurian year. I am wondering if this backwards movement of the sun causes one side of the planet to experience less daylight than the other side, but I will leave the practical effects of this on colonization for a worldbuilding SE.
Simply put: I am trying to understand how different regions of mercury experience the day/night cycle and the way the day/night cycle travels over the planet. This is ultimately a question of what mercuries natural time zones are and the length (and irregularity) of the day-night cycle on the different parts of the planet.

Partial answer about the retrograde motion only:
Yes, the Sun has yearly retrograde motion as seen from Mercury
I've written a simple script to generate the orbital motion of Mercury based on its eccentricity. I've used reduced units with $GM = a = 1$ and period $T = 2 \pi$.
Mercury's rotational period is 2/3 of its orbital period.
I've plotted the rotational angle of mercury minus the angle of the Sun as seen from Mercury's position. To address retrograde motion, it's not necessary to consider from where on Mercury you are looking,
There are tiny dips below zero every year.

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

def deriv(X, t):
    x, v = X.reshape(2, -1)
    acc = - x * ((x**2).sum())**-1.5
    return np.hstack((v, acc))

halfpi, pi, twopi, fourpi, sixpi = [f*np.pi for f in (0.5, 1, 2, 4, 6)]
to_degs, to_rads = 180/pi, pi/180

# Mercury
# https://space.stackexchange.com/questions/43213/what-is-the-analytical-closed-form-solution-of-the-two-body-problem-to-verify-it/43237#43237
# https://solarsystem.nasa.gov/planets/mercury/by-the-numbers/

e = 0.20563593
a = 1
peri = (1 - e) * a
v_peri = np.sqrt(2/peri - 1/a)

X0 = np.array([peri, 0, 0, v_peri])
times = np.linspace(0, sixpi, 1001)
dt_years = (times[1] - times[0]) / twopi

answer, info = ODEint(deriv, X0, times, full_output=True)
xy = answer.T[:2]
x, y = xy
r = np.sqrt((xy**2).sum(axis=0))

theta_rot = np.mod((3/2) * times, twopi)
sun = np.zeros(2)[:, None] - xy
theta_sun = np.arctan2(sun[1], sun[0])

if True:
    plt.figure()
    plt.subplot(2, 1, 1)
    angle = theta_rot - theta_sun
    dangle = np.mod(angle[1:] - angle[:-1]+halfpi, twopi)-halfpi
    dangle_dt = dangle / dt_years
    angle = np.mod(angle+halfpi, twopi)-halfpi
    plt.plot(times/twopi, angle)
    plt.ylabel('angle (rad)')
    plt.subplot(2, 1, 2)
    plt.plot(times[1:]/twopi, dangle_dt)
    plt.plot(times[1:]/twopi, np.zeros_like(dangle_dt), '-k', lw=0.5)
    plt.ylabel('dangle/dt (rad/year)')
    plt.xlabel('years')
    plt.show()

if True:
    plt.figure()
    plt.subplot(3, 1, 1)
    plt.plot(x, y)
    plt.plot(x[:1], y[:1], 'ok', ms=2)
    plt.plot([0], [0], 'oy', ms=8)
    plt.gca().set_aspect('equal')
    plt.subplot(3, 1, 2)
    plt.plot(times/twopi, x)
    plt.plot(times/twopi, y)
    plt.plot(times/twopi, r, '-k')
    plt.subplot(3, 1, 3)
    plt.plot(times/twopi, theta_rot)
    plt.plot(times/twopi, theta_sun)
    plt.show()


Radio telescope targeting How do you target a radio telescope on the precise object you wish to observe? You can point it in the general direction but how do you get the information from the exact point in the sky that you are investigating?
This seems self evident with an optical telescope but not with a radio telescope.

Large radio telescope have pretty good pointing accuracy:

the individual dishes of the VLA are accurate to about 10 arcsec.
the giant Lovell telescope at Jodrell Bank has a similar accuracy.

The second critical parameter is the beam width. Beam width depends highly on frequency:
beamwidth = wave length/dish diameter
When you use several dishes in an interferometer, you can increase their effective accuracy and decrease their collective beam width.
A bit more info on the mechanics of the Lovell telescope:

A control computer calculates the required drive rates to follow each radio source. The drive motors are servo-controlled, so there is a continuous check that the correct rate has been achieved. The position of the telescope is constantly monitored and fed back to control computer to ensure that the telescope is pointing correctly.
For good tracking the pointing accuracy should be about a twentieth of the resolution. Since the resolution is proportional to the wavelength being received (see below), it follows that the pointing accuracy is more critical at shorter wavelengths. The control computer is able to correct for pointing errors caused by the telescope bowl sagging under its own weight as it moves up and down. In this way the pointing errors can be kept to about 10 arcsec.

So, servo motors and presumably, calibration are what enables this accuracy.
The Lovell telescope has 2 elevation drive motors with gearboxes on each side of the dish. These can be driven so that one motor pulls the dish, and the dish drags along the second motor. This eliminates gear backlash in the system: the two gear trains are "wound up" in opposite directions. This means they can make adjustments without gear backlash interfering. (source: a video presentation that runs in the visitor center at Jodrell Bank)
Further reading: the story of Jodrell Bank, and a Radio Electronics article.


What if there was an asteroid belt with the mass of a planet? In the past, the asteroid belt was thought to be the remains of a shattered planet. Of course, that was all pseudoscience, the mass of the asteroid belt being 4% of the moon. I've been wondering, what would happen if this did occur. What would an asteroid belt with around 50% of the Earth's mass be like? How often would asteroids hit earth? Would it influence Mars?

The mass of a planet? There probably was, billions of years ago, and perhaps still is, but the debris of this planet-in-the-making has been shattered and dispersed by impacts. We know from fragments which have struck the Earth that the planet which was forming there had grown large enough to have a molten interior where the heavy metals, mostly iron and nickel, had sunk to the centre and differentiated from the rocky mantle. A massive impact must have shattered this forming planet about 4 or 5 billion years ago, and the debris has become so dispersed that it is difficult to assess how much there originally was. Some was kicked out of the asteroid belt completely, the rest is still there but much of it so finely shattered and dispersed that we can't see it (the asteroid belt covers an enormous area, much larger than the orbit of the Earth-moon system).

The resulting absence of a body with substantial gravitational field to attract comets & other objects has enabled them to pass on by, perhaps to impact the Earth and inner planets. Bodes Law says here should be a planet there, but whether it would have been a Mars-sized planet or not, nobody knows. The differentiation of nickel-iron from the rocky material becomes hard to explain without a fairly substantial planetesimal there to separate them, and in our solar system at least, Bodes Law seems fairly reliable.


What is the nature of "rock vapor" in this description of the formation of the Moon? The NPR News item MacArthur Fellow And Planetary Scientist Sarah Stewart Discusses How The Moon Was Formed and audio podcast begins:

Ari Shapiro, Host: Sarah Stewart likes to think about what happens when planets collide. She uses two actual cannons to simulate those massive impacts. Here's one firing in her lab at UC Davis.
Unidentified Person:  Firing in three, two, one.

Soundbite of Cannon FIring

Ari Shapiro, Host: Her work earned her a spot in this year's class of MacArthur Fellows. Many of us call it the genius grant. For years, experts have thought that Earth's moon formed after a large collision knocked off a bunch of rock. Stewart told me her research suggests a different story.
Planetary Scientist Sarah Stewart: During planet formation, when two bodies collide, there's so much energy released that most of these bodies are vaporized. That means that a rocky planet like Earth is mostly rock vapor.
Shapiro: What is rock vapor, and what does that have to do with our moon?
Stewart: Rock vapor is taking the rocks that we stand on and heating it up to the point where it becomes a gas. And when that occurs, the Earth becomes much larger because vapor is much less dense. And it extends out into this enormous object hundreds of times larger than the Earth today. And we proposed that our moon grows within the rock vapor of the Earth after a giant impact.
Shapiro: So the moon actually came from the Earth.
Stewart: The moon grows within the rock vapor of the Earth. And that gives the moon the same chemistry as the Earth.

We don't learn about "rock vapor" in Earth Science class, but I know it's got to be a lot hotter than the lava we see in the news. A significant fraction of Earth's crust is SiO₂ based and it's boiling point is roughly 3,000°C, and I have a hunch the temperatures involved here are much much higher than that. The kinetic energy associated with say a relative velocity of 40 km/s is roughly 8 eV per AMU, over 130 eV for every oxygen atom for example.
So does "rock vapor" start out as a highly ionized "rock plasma" with almost no covalent bonds remaining, or does most of the energy of the original impactor get transferred to a much greater mass of Earth?
Is there a good place to read about her and her students' MacArthur grant-getting research described in the podcast?

Planetary Scientist Sarah Stewart's research is on the formation of the moon, not, as far as I can tell, as much on the chemical composition and precise temperature of the atmosphere after impact, so I don't know if plasma is all that relevant to her work, but I think she'd have to model and account for total energy and temperature, similar to what you did in your question.

If I understand you correctly, you want to know what Earth's atmosphere was like, lets say a weeks or a few weeks or maybe a year after the giant impact.   Dr. Stewart's team has a word for this type of planet, a Synestia

Plasma temperature is tricky for 2 reasons.  One, there's not a specific temperature where gas becomes plasma.  Unlike melting points of boiling points, which happen at specific temperatures (and specific pressure for boiling points), the plasma phase of matter is closer to a dimmer switch that turns on gradually than a specific plasma point.  Similar to the temperature where fusion happens, individual electrons are unpredictable, so heating a gas, it will turn into a plasma gradually.

It's also possible for a rock vapor, take SiO2 as a baseline, to retain it's double bonds as a gas and at the same time, be a low level plasma, emitting some electrons, so it can be both rock vapor and a plasma.   That's not possible for water, for example, because those bonds are too weak.  Water splits into Hydrogen and Oxygen a couple thousand degrees lower than when the individual molecules begin to enter the plasma state.

Another problem is pressure.  The center of the Earth is plasma temperature (low level Plasma but it's in that temperature range), but people generally don't call that state of matter a plasma.

I think your 40 km/s estimate is too high, because Theia was thought to be a Trojan object before it collided with Earth, so the collision rate should be not much more than escape velocity, maybe 12 or 13 km/s because they shared the same orbit.

40 km/s meteor collisions on Earth happen because they approach at a different inclination, where the orbital directions are much less lined up, that's how you get 11 to 70 km/s for meteors, but Theia was probably on the low side of that, perhaps 14 or 15 km/s tops depending on it's eccentricity - if I can make a bad guess.

I'm sure that plasma temperature happens during giant impacts.  But the temperature is highest where the two objects collide, so initially, the highest temperature corresponds with the highest pressure.   After the impact you have the explosive rebound, because collisions of this magnitude are more like large explosions than anything else and after that you can model where the temperature goes as the Earth begins to settle.

Models would have to account for how the heat moves around and through the planet, how much heat ends up burred vs goes into rock vapor, heat of vaporization, heat lost due to expansion of rock vapor, how quickly heat radiates away (I would think it would be highly opaque, so radiation would be somewhat slow),

Bigger factors would be how much is lost in rebound and ejected material and how much is transferred to angular momentum.  There's also the uncertainty on how massive Theia was.  I think later estimates put it at 1/2 to 1/3rd the mass of Mars.

Another way to look at this question is, after formation, the synestia would have layers, similar to any gas giant, though gas giant layers aren't well understood, we could use the sun as an example.  There might be convection, conduction and condensation layers, and perhaps lapse rate could be applied, and obviously gravity would be lower with the material more spread out.   You might also have layers where the pressure was sufficient that different types of matter would form, like, hot enough to be liquid but enough pressure to be a solid, similar to Earth's core.

All that said, trying to calculate the lower atmospheric temperature of this theoretical, recently formed synestia is a little bit more math than I want to do, and I'd probably get it wrong anyway even if I did the math.  But it seems entirely reasonable that the lower atmosphere was at plasma temperatures if much of the upper atmosphere was at rock-vapor temperature.  But if you can get a temperature model, that would be a step in the direction of a plasma model.   I'd guess the low level plasma temperature for rock vapor would begin somewhere in the 5,000 or 6,000 C range, but it's a hard thing to look up as different compounds have different plasma temperatures.  There's even some cold plasma, like florescent bulbs work on that property, but they require an electric field.

I don't know if my long "I don't know if there was plasma" counts as an answer, but it's a fun question and I thought I'd give it a shot.   I like her idea a lot and I've read a few articles that indicated problems with the more traditional giant impact models, so she may end up being right.


apparent size of an astronomical body What's the name of the measure of apparent (visible) size of a body? e.g. the Sun and the Moon appear to be roughly the same size, despite the Sun being vastly bigger - because the Moon is much closer. What's this apparent size called?

It's angular diameter (a.k.a angular size or apparent size). This XKCD comic strip shows in an interesting way the angular size of some of the most remarkable objects in the sky:


Third rotational angle in equatorial coordinates Right ascension and declination identify an object's position in the sky, but if we are looking at something through a telescope, there is a third angle of interest, namely how our view is rotated relative to the point we are looking at.
We could could think of this angle as a sort of celestial roll that goes with RA (yaw), and D (pitch).
Is there a standard name for this "roll" angle as a third element with RA and D?
What is its reference direction?  (Orthogonal to the direction given by RA and D, in the plane containing that direction and the north pole?)
And would increasing the angle turn our view clockwise, or counterclockwise?
Thanks,
Doguleez

This angle is dependent on the specific positioning of the eyepiece on your telescope. For models on which the tube can be rotated or custom-positioned, this means there are even more possibility—technically, an infinity.
However, no matter which telescope you use, the angle of rotation of the field of view will not change as objects move across the sky due to the Earth’s rotation if you are using an equatorial mount. Of course, using an altazimuth mount would make this angle change with time.
Usually, angles increase eastwards at the eyepiece. See for example https://en.wikipedia.org/wiki/Position_angle
Clear skies!


Exposure details for this now (in)famous image from Lowell Observatory? IAU's announcement 19035 IAU Statement on Satellite Constellations, 03 June 2019 links to 19035a Trails made by Starlink satellites which contains the image shown below.
Question: Is it possible to track down the exact start time and exposure duration for this image, and which telescope was used?

An image of the NGC 5353/4 galaxy group made with a telescope at Lowell Observatory in Arizona, USA on the night of Saturday 25 May 2019. The diagonal lines running across the image are trails of reflected light left by more than 25 of the 60 recently launched Starlink satellites as they passed through the telescope’s field of view.
Although this image serves as an illustration of the impact of reflections from satellite constellations, please note that the density of these satellites is significantly higher in the days after launch (as seen here) and also that the satellites will diminish in brightness as they reach their final orbital altitude.
Credit: Victoria Girgis/Lowell Observatory

I was the one operating the telescope at the time! I was using our 11 inch telescope fitted with a 1 MP MallinCam camera. The image was taken at 9:21 pm with a 25 second exposure.


If an Asteroid was to strike the Earth, would it affect the Earth's rotation? If an Asteroid was to strike the Earth, would it affect noticeably the Earth's rotation, and if so, how large would this Asteroid have to be?

To have a noticeable effect the impactor needs to be BIG.

Most questions about "what would happen if ... hits" can be answered by the "Earth impact effects program" (http://impact.ese.ic.ac.uk/)

Here are calculations for a 100km stony asteroid...

A brute like this would have a good chance of wiping out most complex life on the planet. There has been nothing like this in the last 4 billion years (or so)... It could cause the length of the day to change by "up to 2.42 seconds"

As gerrit said, it would be the last of our worries.


Special Relativity as Applied to an Interstellar Starship? I'm once again confused by the twin paradox. Let's say I am on an interstellar starship flying at 0.6c from a star 30 light years away from Earth to a star 50 light years away from Earth and 40 light years from each other. How would you estimate the difference between the time on the ship when you reach the destination versus the time a person on Earth thinks it should be? Now let's say I travel back to the original star (In both circumstances, I accelerate at 1g until reaching 0.6c.) I would assume that I could use the twin paradox formula to determine the difference in elapsed time on the ship from that measured by a clock left at the first star.

Now, let's say I'm traveling in the same manner (1g acceleration until reaching 0.6c and 1g deceleration for the same duration until arrival) beginning at Earth, traveling to 16 stars and returning to Earth. I am thinking that the time difference between the ship clock (date) and Earth time would be based on the distance and acceleration/ deceleration segments of each leg of the journey independent of the components of the travel vectors in the direction (toward or away from the Earth). Am I right??

Would Earth send messages to the ship based on conventional measurements (ie Physics 101)? I would want to determine the difference between the time on Earth when the message is sent and the ship time when the message is received, considering of course the time of transit of the message (at light speed).

In the first example, you certainly could use the calculation of special relativity to measure the difference in the time shown on a clock which was making the high speed journey and one which was at rest. The details may make the calculation trickier, but we can use the flat minowsky space of the clock at rest to do the calculation, so it is relatively straightforward. [see what I did there?]. It would be a factor of about 1.6 so the journey that would take a little over 67 years would, on board ship, seem to take 67*1.6 = 41 years. (I have ignored the time spent accelerating and decelerating, in order to keep the maths simple. Earth would observe the ship leaving 30years after it did (light distance) and arriving after 67 years, however, since the light must travel further, the ship would appear to arrive after 67+50-30 years = 87 years.

For the multiple journey the calculations are just longer. The direction of travel doesn't affect time dialation, only the speed. So it would again be straightforward to calculate the total time dialation.

It would be quite possible for Earth to communicate with the ship. But those the messages would not arrive for up to 50 years. Messages would be certainly out of date.


Does Mars' "Northern ocean" have another name? What about the "Inland Sea"? The March 15, 2019 Phys.org article A NASA spacecraft may have explored the edges of an early Mars sea in 1997 shows a stylized map of an area on Mars where the Mars Pathfinder landed and uses colors to indicate where bodies of surface water might have once existed.

They are labeled as

Northern ocean
Inland Sea

and I noticed they are capitalized. Do these reflect proper nouns and names of possible bodies of water that are recognized by planetary scientists or Martian cartographers (or should I have said Mars cartographers?)? Do these possible bodies of water have other names as well?

According to the Wikipedia article on the Mars ocean hypothesis, the ocean is sometimes called the Paleo-Ocean, and sometimes called "Oceanus Borealis" (a.k.a. "Northern Ocean" for those who feel that quidquid Latine dictum sit, altum videtur.)

The inland sea in the article is located in the Simud Interior Basin (part of the Simud Valles). The Nature article describing the findings (Rodriguez et al. 2019) doesn't appear to give the sea a name, although it does use the term "SIB sea" at one point.


When did it take place, and what was the cause of the first blackhole in the universe? When did it take place, and what was the cause of the first back hole in the universe? Does this time precede stars collapsing as a result of old age?

We don't know for sure.

It is possible that random fluctuations in the density of matter of the early universe created black holes that could be much smaller than regular black holes formed by stellar collapse. These would have formed in the moments after the big bang.

Otherwise the first stars formed about 400 million years after the big bang, and some of these could form black holes after their (short) lives. Stars over 250 solar masses (which could form in the early universe) would produce large black holes when they collapsed.


Is there a website that shows equivalent views through different telescopes? I'm trying to buy a beginners telescope. I have a Celestron TravelScope, but would like something better. When I look at other models, many users have posted images from their viewfinders. Is there a site that collates all of these, to give an idea of what I can expect from a given scope (e.g. is Mars just a red dot, can I make out Saturn's rings, what's the detail on the moon like, etc)?

Deep sky watch has an article that covers this
http://www.deepskywatch.com/Articles/what-can-i-see-through-telescope.html
It includes sample image of what can be seen in, for example, a medium telescope (150mm objective) with 180x magnification


Can "rogue" supermassive black holes be made this way? Could two galaxies (one big and one small)intersect at a velocity to allow the smaller black hole to escape but not the galaxy around it?

I assume you're asking about central supermassive black holes (SMBHs, one per galaxy), not stellar-mass black holes.

The answer is yes, but what actually happens is the two SMBHs have to merge first, and then the resulting combined SMBH can sometimes be ejected from the combined (merged) galaxy.

[Edited to add: Since you've updated the question with a series of diagrams, I should state explicitly that the scenario suggested by the diagrams -- stars in smaller galaxy merge into big galaxy, but SMBH continues on almost unaffected -- is not physically possible. Most of the stars from the smaller galaxy will not end up in the center of the big galaxy, but because of dynamical friction, the SMBH will.]

This NASA press release from 2017 describes the discovery of a quasar apparently ejected from a recently merged galaxy. I'll go ahead and quote their description of the suggested mechanism (this possibility has been suggested by theoretical studies going back at least ten or fifteen years):

  According to their theory, two galaxies merge, and their black holes
  settle into the center of the newly formed elliptical galaxy. As the
  black holes whirl around each other, gravity waves are flung out like
  water from a lawn sprinkler. The hefty objects move closer to each
  other over time as they radiate away gravitational energy. If the two
  black holes do not have the same mass and rotation rate, they emit
  gravitational waves more strongly along one direction. When the two
  black holes collide, they stop producing gravitational waves. The
  newly merged black hole then recoils in the opposite direction of the
  strongest gravitational waves and shoots off like a rocket.

Since most massive galaxies -- including those that have undergone major mergers in the past -- have a SMBH in their center, the gravitational recoil usually isn't strong enough to eject the SMBH; instead, the SMBH loses energy to the stars in the inner part of the merged galaxy via dynamical friction, and settles back into the center. But it appears that sometimes there's enough of a kick to allow the SMBH to escape.

Another possibility is that if two galaxies merge and their SMBHs form a binary, and then another galaxy (with its own SMBH) merges before the previous two SMBHs have actually merged, then you can have a three-body interaction between the late-arrival SMBH and the binary SMBH, which could result in one of the SMBHs being ejected. But this requires the right timing, and probably doesn't happen very often.


Why are blue stars the hottest? I know that blue stars are the hottest and youngest (according to Universe Today). But what makes them blue, instead of other colors like purple?

Our eyes and brains make them blue.

Stars are (close to) a black body, they emit a broad band of radiation. The hottest stars will emit most of their radiation in the ultraviolet but they will still be very bright in visible light.

As they get hotter, they emit even more ultraviolet, and more visible light but the mixture of wavelengths in visible light doesn't change much.

This mixture is interpreted by our brains as being a sort of light blue. No matter how hot the star gets, it will still be emitting lots of visible light in a mixture that looks "light blue".  It will never look purple.

Purple can only be seen if either a single short-wavelength visible light is being emitted, or a mixture of short and long wavelengths (red and blue). Neither of the mixtures can occur in light emitted from a black body.

So stars can appear red, orange, yellow, white or light blue, but never green or purple.


Why is Saturn the only large planet without any trojans? Every planet apart from Mercury and Saturn has trojans in its L4 and/or L5 points.
Mercury is easy enough to explain: It is small, has an eccentric orbit that precesses, and any trojans it has would be heavily perturbed by the Sun's gravity.
However Saturn doesn't have any of those complications. The answer given to this question is that Jupiter's gravity perturbs any trojans at Saturn's L4 and L5 points.
But if Mars, which is much less massive than Saturn and comes closer to Jupiter, can have multiple trojans then there's no reason why Saturn shouldn't.

They are pulled out of stable orbits by Jupiter.
The details are in https://ui.adsabs.harvard.edu/abs/2012jsrs.conf..225B/abstract
Full text https://syrte.obspm.fr/jsr/journees2011/pdf/baudisch.pdf

The planet Jupiter is solely responsible for the hole of instability for short time integrations
($T < 10^{7}$) compared to the age of the planetary system. On the long term scale this planet also destabilizes the whole region around the Saturnian libration points. If we find in the future Trojans of Saturn,
these Trojans could only be captured asteroids, in orbits in the 1:1 MMR for a short time


Is the size and heat of a star important in the development of a solar system? How do the size and heat of stars affect development of solar systems and what kinds of planets can develop? Is it possible to orbit supermassive stars? Furthermore, would star's unleashing violent winds such as wolf-rayet stars be candidates for solar system formations?

I think the quick answer to this is no, at least, any solar-system around a Wolf-Rayet or similarly large star wouldn't have time to develop.   I would think that large stars have solar-systems cause I see no reason why they wouldn't, but the stars don't last long enough for their solar-systems to develop much.   The formation period, including bombardment period(s) and time for the planets to cool down from the heat of formation takes a significant amount of time, perhaps a couple hundred million years minimum.

Our solar system's late heavy bombardment occurred between about 4.1 and 3.8 billion years ago, ending some 700 million years after our solar-system formed and Earth didn't cool down enough to have a magnetic field until about 3.5 billion years ago, roughly 1 billion years after formation.   Granted, that's a sample size of one, so numbers could vary quite a bit, but I have a hard time imagining that planet's could cool down and complete most of their early bombardment in the lifespan of a very large star.

For developed solar-systems where the planets had time to cool and the solar-systems had time to mostly coalesce and get past early-bombardment, I don't think there's much of a window beyond maybe 2 solar masses  (a 2 solar mass star would have roughly 1/16th the lifespan of our sun, or a bit less than a billion years in main sequence), based on an exponential ratio to the 4th power of the star's mass.  Source.

Now, as to whether a solar-system could survive a Wolf-Rayet star's enormous ejections, that's another question.   I'm not 100% sure but large planets are pretty resilient, so my guess is yes, but it's just a guess.

Another interesting question is gravity vs luminosity.   Our solar-system the furthest planet Neptune, about 30 astronomical units from the Sun, and, presumably a planet could orbit a fair bit further out than that and be in a stable orbit, but lets go with 30 AU.

A star with 10 times the mass of our sun during it's main sequence has roughly 3,000 times the luminosity (using the exponential relationship of 3.5 Mass to Luminosity - see here)

The gravity for 10 solar masses, gravity reduces by the square of the distance, so square root of 10, a Neptune equivalent orbital velocity would be about 95 Astronomical Units, so, at 95 astronomical units from a star with 3,000 times the luminosity of our sun, the energy received from that star at that distance would be roughly 1/3rd the energy the Earth gets from the sun.

Too much larger than that, somewhere around 20 solar masses or, not much more than that, the star gets too hot to have much of a solar-system within it's theoretical Goldilocks zone.   Not that that matters much, cause the short lifespan of the star kind of answers the question anyway, but my guess would be, if planets survive around a Wolf-Rayet star, even the outer parts of such a solar-system should be hot unless they were enormously distant, like a couple hundred AUs.   There might be no ice or water planets around a star of that size.


Poisson's equation-like formula in axis-symmetric Galactic potential in terms of Oort constants I am tackling this problem, from a 2016 Cambridge Astrophysics Tripos past paper:
Let $\Phi(R, z)$ be the axi-symmetric Galactic potential. At the Solar location, $(R, z) = (R_0, 0)$, prove that
$$\frac{\partial^{2} \Phi}{\partial z^{2}}=4 \pi G \rho_{0}+2\left(A^{2}-B^{2}\right)$$
where $G$ is the gravitational constant, $\rho_0$ is the density in the Solar neighborhood and $A$ and $B$ are Oort’s constants.
What I do is:
Rewrite equation we want to prove
Note:
$A^2-B^2=-\frac{v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}$
Using definiton of $A$ and $B$. (Which are: $\left.A \equiv \frac{1}{2}\left(\frac{v_{c}}{R}-\frac{d v_{c}}{d R}\right)\right|_{R_{0}}$ and $B \equiv-\left.\frac{1}{2}\left(\frac{v_{c}}{R}+\frac{d v_{c}}{d R}\right)\right|_{R_{0}}$).
Rewrite what we want to prove:
$$
\bbox[5px,border:3px solid green]{
\frac{\partial^{2} \Phi}{\partial z^{2}}=4 \pi G \rho_{0}-2\frac{v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}
}
$$
Use Poisson's equation:
$$\nabla^2\Phi=4\pi G\rho_0$$
Expand LHS of Poisson:
$$\frac{\partial^2}{\partial R^2}\Phi+\frac{\partial^2}{\partial z^2}\Phi=4\pi G \rho_0$$
Use equation of motion for circular orbit:
$$\frac{v_c^2}{R}=\frac{\partial}{\partial R}\Phi$$
Differentiate once more wrt $R$:
$$\frac{\partial^2}{\partial^2 R}\Phi=\frac{\partial}{\partial R}\frac{v_c^2}{R}=\frac{2v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}-\frac{1}{R^2}v_c^2$$
Collect results
Substitute result into Poisson's equation expanded form:
$$\frac{2v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}-\frac{1}{R^2}v_c^2+\frac{\partial^2}{\partial z^2}\Phi=4\pi G\rho_0$$
Rearrange:
$$
\bbox[5px,border:3px solid green]{
\frac{\partial^2}{\partial z^2}\Phi=4\pi G\rho_0-2\frac{v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}+\bbox[5px,border:3px solid red]{\frac{1}{R^2}v_c^2}
}$$
Conclude
Compare with what we wanted to prove (ie the two green-boxed equations). We have an extra term (red boxed) which shouldn't be there. How do I fix the proof?

Poission equaiton:
$$\nabla^{2} \Phi=4 \pi G S_{0}$$
In cylindrical polar coordinates:
$$\frac{1}{r} \frac{d}{dr}\left(r \frac{d \Phi}{d r}\right)+\frac{d^{2}}{d z^{2}} \Phi=4 \pi G \rho_{0}$$
We also know from $\ddot{\vec{r}}=-\nabla\Phi$ that $\ddot{r}-r \dot{\phi}^{2}=-\frac{\partial\Phi}{\partial r}$, so for circular orbit, we have: $r\dot{\phi}^2=\frac{\partial \Phi}{\partial r}$. This means that we can write $v_c=r^2\dot{\phi}^2=r\frac{\partial \Phi}{\partial r}$
Rewrite Poisson above:
$$\frac{d^{2}}{d z^{2}} \Phi=4 \pi G \rho_{0}-\frac{1}{r} \frac{d}{d r}\left(v_{c}^{2}\right) = 4 \pi G \rho_{0}- \frac{2v_c}{r}\frac{dv_c}{dr}$$
Keeping in mind that  $A^2-B^2=-\frac{v_c}{R}\frac{\mathrm{d}v_c}{\mathrm{d}R}$, rewrite RHS:
$$\frac{d^{2}}{d z^{2}} \Phi=4 \pi G \rho_{0}+2(A^2-B^2)$$
which is what we wanted.


Why do our clocks not go out of sync with the day over a year? Our clocks use 24 hours for a day, our planet takes 24 hours to rotate completely and our planet revolves around the sun in 365 days.
Imagine at the start of the year, the side of Earth facing Sol is 'Side A' and the dark side is 'Side B'. The side of the sun that Earth sees is 'Side A' and the opposite side is 'Side B'.
After half a year, wouldn't 'Side B' of Earth be facing 'Side B' of Sol, and thus the day/night cycle would be out of sync by 12 hours?
I've been trying to get my head around this problem for ages and I can't seem to figure it out. It seems like the angle that the sun hits the Earth must change by about a degree per day until it's gone around 360 degrees. Would that not mean the clocks offset by 1/365th of 24 hours every day over a year?

You are correct, and in fact the clocks are offset. But most people never notice this because of the way days are defined.

If you measure the "day" length by the stars (instead of the sun) you will see it's not 24 hours at all. It's around 23 hours 56 minutes. (Check this by going out at night, picking a bright star, and noting the exact time it passes due south. Then observe it the following night and note it passes 4 minutes earlier.)  By the way this shortened "day" is the sidereal day.

Each day, the Earth shifts along in its orbit (by about a degree, as you correctly observed in the question). So if you go out and point at the Sun at Noon every day, you are pointing in a slightly different direction. (I'll also neglect the Solar Analemma at this point too as it's just too confusing.)

Really, the "day" of 24 hours is not related to the true rotation period of Earth at all; loosely speaking it's just the average time between sunrises, or sunsets, or noons.


Translating a zenith position to the nadir What would be the easiest way to translate a zenith position to a nadir on the celestial sphere?
For example I have a position from the azimuthal grid in Stellarium; 3h44m32.64s/+50°50'16.2"
which was obtained based on my location at +50°54'13.9" -2°34'19.1" and a time of ; 15/03/2019 16:24:20.

In Astronomical Algorithms, Second Edition, Jean Meeus shows us how to convert local coordinates to equatorial and vice versa.
Formulas 13.5 and 13.6, p. 93:
$$ \mathrm{tan}\ A = \frac {\mathrm{sin}\ H}{\mathrm{cos}\ H\ \mathrm{sin}\ \phi - \mathrm{tan}\ \delta\ \mathrm{cos}\ \phi} $$
$$ \mathrm{sin}\ h = \mathrm{sin}\ \phi\ \mathrm{sin}\ \delta + \mathrm{cos}\ \phi\ \mathrm{cos}\ \delta\ \mathrm{cos} H $$
Unnumbered formulas, p. 94:
$$ \mathrm{tan}\ H = \frac {\mathrm{sin}\ A}{\mathrm{cos}\ A\ \mathrm{sin}\ \phi + \mathrm{tan}\ h\ \mathrm{cos}\ \phi } $$
$$ \mathrm{sin}\ \delta = \mathrm{sin}\ \delta\ \mathrm{sin}\ h - \mathrm{cos}\ \phi\ \mathrm{cos}\ h\ \mathrm{cos}\ A $$
where $ \alpha $ is the right ascension in degrees, $ \delta $ is the declination in degrees, $ \phi $ is the latitude (+ North, $ - $ South) in degrees, $ H $ is the hour angle measured westwards from South in degrees, $ A $ is the azimuth in degrees, and $ h $ is the height of the object in degrees. Please note that Meeus measures azimuths from the South heading East (90°) then North (180°) and West (270°).
With these formulas, you can find the required position of the point you want to know about.
In order to get the nadir, I would just put $ h = -90° $. Another way would be to find the position of the object at zenith, add/remove 180° to its right ascension, and change the sign of its declination.
Hope this helps!
Clear skies.


How did Astronomers deduce that the Sun was not a ball of fire? Its common knowledge that people used to think that the sun is a ball of fire or molten metal, but when did science start to prove otherwise?

I think it's maybe not the case that there was a moment when the astronomy community conclusively rejected the ball-of-fire hypothesis; astronomers simply accumulated more and more evidence against it. If you want to put a rough date on it, you could put your finger somewhere in the middle of the 19th century, as by then, other ideas had taken hold.

Back in the classical period, Anaxagoras had proposed that the Sun was a heap of molten metal. I don't know whether this was widely accepted by his contemporaries. The idea of the Sun as a ball of metal or fire certainly persisted for some time, though perhaps largely for lack of any better ideas. We didn't even understand oxygen and combustion until the work of Lavoisier and others in the late 18th century, so detailed calculations were presumably out of the question for a millennium or two after Anaxagoras. I don't know when calculations of how long combustion could sustain the Sun were first done, but it appears to have been not more than several decades after the theory of combustion was developed.

Why? Well, we can say that by the middle of the 19th century, the predominant explanation for the Sun's luminosity was not the burning of coal but instead gravitational potential energy. By the 1860s, it was widely known that chemical reactions could only power the Sun for a few thousand years. We also now had a potentially viable alternative: a decade earlier, Hermann von Helmholtz had begun exploring the idea that gravitational contraction of some sort, by what we now call the Kelvin-Helmholtz mechanism, was the source of energy, with gravitational potential energy being transformed into heat$^{\dagger}$. Around the same time, Lord Kelvin suggested that meteors falling into the Sun provided the necessary energy, a similar mechanism to Helmholtz's. I believe astronomers continued with the contraction hypothesis through the turn of the century - I've seen an article written around 1900 to that effect.

However, during the early and mid- 1900s, quantum theory and nuclear physics were being developed, and the work of Eddington, Bethe and others would lay the groundwork for our current understanding of solar energy production. Previous models (including, finally, Kelvin-Helmholtz contraction) were now known to be insufficient because they allowed the Sun to shine for only thousands or millions of years, and geologists had established that Earth itself was much older than this. Fusion, on the other hand, allows the Sun to survive for billions of years - a timescale that matches up well with the age of the Earth. We also knew that hydrogen and helium were the dominant constituents of the Sun and other stars; while Wollaston and Fraunhofer had performed the first solar spectroscopy observations in the early 1800s, the true composition of the Sun was not accepted for more than a century, when Cecilia Payne made a detailed study of spectral lines.

$^{\dagger}$ While this does produce heat in various bodies, including T Tauri stars, it is not significant in most stars beyond that stage.


Are Saturn's rings stable? Saturn's rings contain many moonlets that shape the rings of Saturn. The structures in the rings of Saturn around moonlets are similar to those in protoplanetary disks around newly formed planets, which makes me wonder if the material in the rings will follow a similar fate and form into larger bodies.

Will the material within the rings of Saturn eventually coalesce into moons, or are the rings of Saturn stable enough to last billions of years?

Most of Saturn's rings are inside it's Roche limit, which means they will never clump together.   Tidal forces prevent this from happening.

small objects that are already together can withstand the tidal forces.    A sufficiently large body inside a planet's Roche limit should break apart by tidal forces.   That may be how the rings formed in the first place, or they may have formed by collision.

Saturn's rings are thought to be relatively young, if 100 million years qualifies as young, but it is for objects inside the solar system.    Being young doesn't tell us how long they will last.   I think the 300 million year estimate may be accurate enough, originally mentioned in the comments, but until that happens, the rings will remain, just grow smaller over time and they'll never coalesce into moons.

As a footnote, Saturn's thinner outer rings or gossamer rings are outside the Roche limit, and those might, one day, form into moons, or they may be too thin for that to happen.  A certain density is probably required for moon formation.


Observatories that allow public access Are there any observatories with high powered telescopes that also allow public access?
Astronomy is a VERY expensive addiction, because no matter what telescope you get, you always want the next model up.   If we can find an observatory that allows the public to see through their scope in America or Canada, we can make our vacation plans accordingly.  Kids want to see Pluto.  Scopes that powerful are "slightly" over our budget.

Amateur astronomers regularly broadcast their viewing sessions over the Internet on the LiveSkies network. Pluto is within the viewing capabilities of most of the equipment used to broadcast but it just looks like a dim star that can be seen to move over long periods of time.
https://www.liveskies.org/
Scroll down to the  View As A Guest Link then click on a live broadcast. There are no schedules. People can broadcast from anywhere but the majority broadcast from North America. If it is dark and clear somewhere in North America someone may be on line.
Many amateur astronomy clubs have their own observatories that are open to the public on occasions. Google amateur astronomy clubs. There is probably a group near you.
Professional observatory time is a valuable commodity. Observatories with "high powered" equipment and skies dark enough for good viewing will have limited if any public access.


How do star densities work? If we take a look at stars more massive than the Sun, their densities vary a lot. UY Scuti is an extremely low-density star that's only 8.5x more massive than the Sun, but is 1000-2000x its size. r136a1, on the other hand, is 256x more massive than the Sun, but is only 30x its size. Neither of them are main-sequence.

The Sun has a lower mass than both of them, yet it's also smaller than both of them.

It doesn't seem like mass makes a big difference. Both reach hydrostatic equilibrium at completely different sizes. r136a1 has a much stronger gravitational pull than UY Scuti, but it should also have much more radiation pressure, right?

So how can we determine the density of a star? Is there a formula?

The mean density of the star is really only defined by the formula $\bar\rho=M/V=3M/4\pi R^3$.  The radius of a star is a generally a very complicated function of a star's other properties.  When we determine the radius in stellar models, it's only because we've solved equations that describe the structure of the whole star, and read off the value at what we define as the surface. So no simple formula in general.

That said, one can derive the approximate functional dependence for stars of various evolutionary states through the principle of homology. i.e. assuming that stars of a certain type are just rescaled versions of each other. Glancing at my old course notes, on the upper main sequence, where stars burn hydrogen principally through the CNO cycle and have radiative envelopes dominated by electron-scattering opacity, we derived $R\propto M^{15/19}$. The same principle (but with different assumptions about the star) is used to determine the location of the Hayashi track for pre-main-sequence stars, along which $R\propto M^{-7}T^{49}$.  Particular formulae can be found for different types of star but the relationships between $M$ and $R$ vary wildly.

Neither the two stars you mentioned are typical main-sequence stars. R136a1 is a Wolf-Rayet star, which is basically a star that has blasted away most of its hydrogen envelope.  Mass-radius relations are usually strongly dependent on mean molecular weight, which is higher without hydrogen, so the relations break down (or, rather, would have to be derived separately). But usually higher mean molecular weight gives a more compact star. UY Scuti has probably finished burning hydrogen in its core and has moved off the main sequence. So again, it'll follow a different relation.


Are galaxies moving in each direction with same speed? All galaxies are moving away from each other (us) because of the expansion of the universe.

Are they moving with same speed in each direction?

Are the galaxies moving towards each direction with same speed ?

Short version : Locally, No.  On the cosmic scale, yes-ish.

The motions of galaxies are more complex than this questions suggests.

Let's start with the Milky Way.  From our point of view stationary.

But the Milky Way is part of a group of galaxies (the local group).  They essentially move in their mutual gravitational fields.  Even that field is complicated by the gravitational field of dark matter halos, which are invisible (unobservable directly).

But the local group is not the only group out there.  The local group is part of an even larger grouping known as the Virgo Supercluster.  Again motions of these groups are complex and again dark matter plays a part.

The local group contains 54 galaxies, and the super cluster contains something on the order of 100 such groups.  That's about 5400 galaxies in a fairly complex motion.  There's no one easy direction to spot here.

But the larger scale universe is thought governed by what is known as the FLRW metric, which (with dark matter, dark energy and some adjustments) fits the overall expansion of the universe we see.

It's only on that scale (essentially the entire universe excluding the nearest bits) that you see the kind of pattern you mean.  So the local bits are maybe 5400 galaxies, but the rest of the universe is very roughly two trillion galaxies.  So our local part is on our scale vast and unimaginable, but on the scale that FLRW applies it's a microscopic dot lost in the averages.

FLRW is a result of using General Relativity to model the expansion of the universe.  It does predict (when you match it to our universe's conditions) that from any point everything else will be moving away from you (and we now think accelerating away).

There is an apparent contradiction between these two extreme (local group and universe away from that).  The resolution of this is that FLRW ignores local grouping and irregularities and assumes a nice continuous distribution (galaxies don't even feature on this grand scale).  On that local scale gravitational fields (currently) are large enough to overcome the expansion, hence our closest galaxy (Andromeda) is approaching us, not moving away.  But the further away you get, the larger the overall effects predicted by FLRW become and dominate "normal" gravitational effects.

FLRW does not say that they all move away at the same speed.  It's distance dependent - further away = faster moving.  So not the same speed, but from every viewpoint in the universe the same pattern of faster when further away.


Does Earth revolves around Milky Way? As we know Earth rotates around Sun and Sun around Milky Way but then Sun must have some velocity/angular momentum (I don't know that much physics terms).

So as Earth revolves around Sun, it must adjust its orbit as Sun also revolves.

Is it so? Or am I missing something?

Earth is bound gravitationally to the Sun. So it is orbiting together with the Sun around the center of the Milky Way.

This takes one galactic year.

Hence your presumption is correct.

Seen from a non-rotating system relative to the Milky Way barycenter, Earth's orbit isn't an ellipse, but ressembles a distorted helical trajectory. Distorted, since the orbit around the Sun isn't perpendicular to the orbit around the center of the Milky Way.


Is it possible to tell the time of day by shadows on the photo I have a photo where objects and their shadows (e.g. trees, people) are visible, the Sun is not visible directly, but it is clearly neither dusk nor dawn.

Let's assume that coordinates of this place are known more or less precisely so that I can find it on google maps.

I also know the date when this photo was taken but don't know the time of day. Is it possible to estimate it from the photo?

I have a debate with a friend of mine about this.
My friend states that a reasonable enough estimate is to calculate the Sun's elevation above the horizon by

$$
\text{elevation} = \arctan \frac{\text{object's height}}{\text{shadow's length}}
$$

and then use one the myriad websites on the internets to look up the time for that date and location.

I doubt it because

The length of a shadow on the photograph depends on the relative position of the Sun, the object and the camera.
Most importantly, the slope of the ground can't be estimated neither from the photo, nor from google maps. This picture is a good example.

Who's right? Am I missing something?

Also, it would be interesting to understand the role of uncertainties in this calculation. Assuming that the surface is reasonably smooth and the slope is close to zero, what uncertainties (location, object's height, shadow's length, the slope, camera position) would play the major role? Does it depend on the time of year?

In short: If you know length of shadow and object height (assuming vertical object casting shadow of an horizontal surface), you can tell time if you know date and you can tell date if you know time. If you know direction of shadow (e.g. relative to north) you can tell date and time without knowing any of them.

The most tricky part is to measure objects and shadows from a photograph - because you need their real length and perspective distorts object and shadow lengths -. To do this you need to make some assumptions about the geometry of the objects in the photograph. In fact, if you know the geometry of the settings, after some geometrical calculation you can find the Sun position and therefore date and time.

For example, with an image like this one and a detailed map, you can find date, time, position of camera and even focal length of lens - with a lot of measurement and geometry calculation, of course.

However, if the only objects casting shadows in your image are trees and people, measuring their exact position and size would be impossible and date and time estimations will be quite rough, specially if the image doesn't show any other identifiable and measurable features - and even worse if you can't tell the north from features in the photo. In general, finding date and time from shadows works better with buildings, specially with sharp shaped buildings with known orientation.


What is the degree of ionization is the solar photosphere? I am wondering how many free electrons per baryon and how many free electrons per atom there are in the solar photosphere.  This number depends on the abundances of the various atoms found in the photosphere as well as each atom's ionization potential.

The book "Introduction to Stellar Astrophysics" by Boehm-Vitense, vol. 2, 1997 reprint, on page 76 claims that one out of every $10^4$ hydrogen atoms are ionized, so hydrogen is mostly neutral in the photosphere; however, there are other heavier elements with lower ionization potentials.  Boehm-Vitense goes through and calculates the fraction of ionization of iron but calculations for only two species present in the photosphere can hardly be considered comprehensive.

Whether you calculate these numbers yourself or provide a summary of a reference (with a link to that reference if possible) that has more details about ionization in the photosphere it will be appreciated.

My guess is that, since hydrogen has such a low degree of ionization, the number of electrons per baryon and per atom will be small, since hydrogen is the majority constituent of the 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.


What are the large round dark "holes" in this NASA Hubble image of the Crab Nebula? I came across this image of the Crab Nebula taken from NASA Hubble telescope. What are the large round "holes" and how are they formed?

I think my deleted answer to your previous question covers this well, so I'll add it here.
These two spots are known as the east and west bays of the Crab Nebula. They appear to be the result of a torus partially encircling a section of the nebula. The pulsar's magnetic field interacts with the gas and dust in the torus, which blocks synchrotron radiation from gases in the nebula. It has been conjectured that the torus is the result of a $\sim0.1M_{\odot}$ ejection of material from a circumstellar disk prior to the supernova, based on studies of the proper motion of gas on the edges of both bays.
References

Fesen et al. (1992)
Hester et al. (1995)

We can also rule out that this is the result of a defect in Hubble, because these features have been observed in different wavelengths, including infrared, visible, ultraviolet, and x-ray (though not in radio or gamma-ray wavelengths):

Image courtesy of Wikipedia user Hunster under the Creative Commons Attribution-Share Alike 3.0 Unported license.
The infrared, ultraviolet, and x-ray images come from the Spitzer Space Telescope, the SWIFT observatory, and the Chandra observatory, respectively.


Open problems in astronomy that an amateur (with a PhD in some other field) would have a chance of solving? What are some open problems in astronomy that an amateur would have a chance of solving? Suppose the amateur has a PhD in some other field, owns a basic telescope, a set of filters, diffraction gratings, cameras, and happens to know a lot about machine learning, signal processing, spectral estimation, statistics & design of experiments, and basic physics and chemistry.

Are there better tags available (such as "research")?

If you have good knowledge of software development and pattern recognition, there are several problems that you could assist in solving. Much of observational astronomy requires long time series data, and removing the noise from this data. I have just left the field where some colleagues are trying to develop some software to use image subtraction techniques to isolate individual stars in the centre of clusters. The centre of the cluster is typically more dense and harder to get clear measurements of each individual star to analyse.

Pattern recognition would be particularly useful in pipeline analysis, where a generic pipeline is used on large amounts of data to 1: find the types of stars one is interested in; and 2: extracting some interesting information on these stars. Machine learning techniques could also be used to assist in developing the general pipelines for more specific interests.

I am happy to put you in touch with a few people that could provide you with some specific problems that you might be able to assist in.


To what extent are structure formations unexplained? I hear conflicting claims about our understanding of the structure formation in our universe. I heard that ΛCDM explaines well the large scale structure of the universe i.e.

  Comparison of the model with observations is very successful on large
  scales (larger than galaxies, up to the observable horizon)

and then I read the opposite on nasa.gov:

  However, we still do not understand how these structures condensed
  out of the 'primordial soup'. The formation of structure appears much
  quicker than the current theories would have it!

So what is Lambda-CDM actually good at and what are its potential weaknesses, if there is any?

The info you reported are not opposite.
They refers to different scales on structure formation, and are both true.
As you said, ΛCDM model is very useful to explain large scale structures, as galaxies, clusters, and larger.
This means that the model fits really well to the observables, first of all the CMB map and its anisotropy.

What the model can NOT explain is the original formation of the structures, at the observed rate (take a look at this).
We know the age of universe (within this model), and we know the age of the observed galaxies. Then it results that galaxies have formed much faster then in a "simple" ΛCDM scenario. We can not explain how the primordial structures started to collapse and to grow in what now we see as galaxies.
To account for this observation we need to invoke dark matter primordial fluctuations. But this is an assumption, and does not give us the original spectrum of the fluctuations, neither it explain how it is formed. And this is a parameter that the ΛCDM model can not account for, at least at this stage.


Why can't moon light (reflected sun light) turn the sky blue? Does turning the colour of the sky blue need more luminous light? Does it depend on luminosity or some other factors are also responsible for this phenomenon? Why can't the moon light turn the sky blue even a little bit (at least the area near the disc).

Thanks

The simple answer is that it does, but it's not bright enough to be visible to the naked eye. Earth's atmosphere scatters the moon light just like sunlight.

The full moon (like the sun) fills about 1/2 of 1 degree of the sky, the entire sky being 180 degrees, give or take, so the full moon fills less than 1 part in 100,000 of the night sky, so there simply isn't enough blue light to be visible over the brighter stars even with the brightest full moon. Our eyes are very good at seeing variations in brightness, but not that good. . . . and, for what it's worth, the night sky has always appeared to have a dark bluish tint to me, but that might just be my brain playing tricks on me because logically I know it's there. I'm not sure whether it's actually visible.

With a good sized telescope, moonlight scattering acts as a form of light pollution. Telescope users know that you get better visuals when there's no moon.

Source.


Is the universe always expanding Currently we observe that the universe is expanding. And this expansion even goes faster and faster.

Why do we know (or presume) that this expansion started slowly at a singularity (i.e. big bang) and is getting faster and faster?

Did a serious scientist ever presumed that the universe might be oscillating, i.e. like a breathing bubble? Currently the universe is expanding but who knows, perhaps in 1 Million year we will observe a shrinking universe and 3 Million year later again expanding - provided the humans will still exist at this time which I doubt.

Of course, this would open many new questions like: How and when did it start? What is the frequency and amplitude? But let's keep this question off side.

It will always expand (probably).

The mass and energy in the universe curves space, and it seems that there is exactly the right amount of mass and energy to keep the universe expanding forever.

Had you asked this question 20 years ago the answer would have been "we don't know". Recent discoveries of "Dark Energy" suggest that the expansion of the universe is not slowing down, but speeding up. This would mean that the universe is destined to expand forever, at a faster and faster rate.

The notion that there will be a "Big Crunch" followed by another "Big Bang" was seriously considered by multiple scientists: among them Willem de Sitter and George Gamow. There are problems with this hypothesis: It is not well supported by observations of the rate of expansion. Moreover, there would need to be way to reduce the entropy of the big crush, as entropy can only increase. And moreover any "pre-big bang state" may be hidden behind the singularity at the start of the (current) universe and so fundamentally hidden from us.


What may be the period and speed of revolution of the star that is closes to galactic centre? The Sun moves at the speed of 220km/s around the galactic center in a circular orbit. The sun is located at about 30,000 light years from the galactic center. The Sun orbits around the center of the Milky Way once every 225 million years. The period of time is called a cosmic year.

I am curious about the speed at which the star closest to galactic center revolves around galactic center. Has this been calculated or approximated. Also what is the time it takes to complete one revolution for the star around galactic center or what is the cosmic year length for the star that is closest to galactic Center .

I have taken reference from http://www.bibliotecapleyades.net/ciencia/ciencia_astrosciences07.htm

Star S2 reaches maximum velocities of 5,000 km/s according to this ESO page.

The orbital period is given as a little over 15 years and this paper gives a peri-center (closest approach to the black hole) of 17 light hours.

There's another star in the area called S0-102 which has a shorter period, but its orbit is less eccentric - so its closest approach to the black hole is further out at 36 light hours. Its maximum velocity will probably be less than S2 but I haven't been able to find a figure.


Does the cosmic microwave background recede at the speed of light? Does the cosmic microwave background recede at the speed of light?  Is it possible it recedes even faster because of the stretching of space?

Well obviously the CMBR arrives (towards us, not receding) at us at the speed of light as it is a form of electromagnetic radiation and local velocities are not affected by expansion.

What I guess you may mean though is the surface of last scattering, which is the surface from which the CMBR that we receive at a given instant was emitted in the early Universe. This surface of course is no longer emitting CMBR in the present cosmological time and it is not a fixed (comoving) surface either as which surface we receive the CMBR from changes with time, however we can give a recession velocity to the comoving set of points which make up the surface of last scattering at any given instant.

In the standard model of cosmology, the Lambda-CDM model, the instantaneous surface of last scattering is receding from us at approximately 3.2c (i.e. a recession velocity 3.2 times the speed of light).


Star class according to initial mass Studying the stars classes for my homework (school) i learned that stars are classified by the spectral types (Harvard) and luminosity class (MKK) ..

So the sun is a star class G2V

But can not understand that clasificacion (far in the future) changes to MIII when his main sequence ends and begins to burn helium

What is the relationship there?
Would not it be more logical to classify stars according to their initial mass and any change that is made in the classification is related to the lifetime of the star?
sub classify them according to:
Fusion element: Deuterium, Hydrogen, Helium, etc.
And Remaining time fusing current element?

We are glued to the story?

The classification of stars using spectral class is a very useful classification when considering the properties of (the atmosphere of) a star at that moment. If you consider the different stages in human development (embryo, fetus, infant, toddler, etc...), for instance, here one person also continuously changes its class.
So it is not strange/uncommon/forbidden to change a classification of a thing whether it is a human or a star.

It also does not mean that other classifications are not useful. It depends on how you want to use the classification. In literature, concepts such as low-mass star, or Sun-like star are also used.

Classification according to the evolutionary stage (i.e. the type of fusion going on) of a star is also very common.

See the image below from ATNF in which you can see how stars of different zero-age main-sequence mass move along a colour-magnitude diagram. If we follow the track of a star of one solar mass then we get the following subsequent evolutionary stages:

MS - Main sequence stars (spectral type V) fuse hydrogen into helium in the core. When the hydrogen in the core is depleted, the star becomes an RGB star.
RGB - Red Giant Branch stars (V->II/III) fuse hydrogen into helium in a shell around the core while the core is inert helium. At some point (Helium flash) the core temperature becomes high enough to start fusing helium into carbon and oxygen, the star then becomes an HB star.
HB - Horizontal Branch stars (II/III) fuse helium into carbon/oxygen in the core. A shell with inert helium surrounds the core, which in turn is surrounded by a hydrogen burning shell.
AGB - Asymptotic Gian Branch stars (I) have a carbon/oxygen core with several inert or fusion shells around it.

These types (MS/RGB/HB/AGB) are classes in a very useful and common classification scheme.

In the end it just depends on what you want to do with your classification scheme.

NB. The classification of the Sun changes from G2 to M because the outer envelope and atmosphere expand enormously, causing the atmosphere to cool down. A cooler atmosphere means that the Sun changes from a yellow (6000K) to a red (4000K) star. The wavelength, and therefore colour, at which most energy is transmitted depends on the temperature of the star's atmosphere.


Which celestial object in the solar system has the fastest tangential velocity at its surface? Consider all of the naturally-occurring objects in the solar system : in the reference frame of the object's center (ie, ignoring the orbital velocity around the sun or, in the case of satellites, their planet) and considering an average point on their surface : which object has the fastest tangential velocity?

Note - I am not asking about the rotational velocity ($\omega$), but the tangential velocity, which for a solid spherical body would be simply $\vec v=\vec \omega\times r $. Since the solar system is typically not made of up homogeneous solid spheres, I can't just look up the tables of rotational velocities and radii of the objects.

I'm looking to be surprised, so everything goes : planets, their satellites and any of the other known objects (comets, asteroids, etc) - the only condition should be that this velocity should be more or less constant (I'm not sure if bodies in highly-elliptical orbits have severe changes in rotational velocity as they approach the sun).

Hands down it's Jupiter.  It has both the largest radius and faster rotational velocity of the planets, and smaller objects will not be able to compete with that huge radius.  Jupiter's rotational velocity is approximately 13.4 km/s (with some variation due to wind currents, as it is a gas), which can be determined both from Doppler shifts of reflected sunlight, and also from watching the motion of equatorial cloud features.  Saturn is only 10.9 km/s, as it has a longer rotation period and a smaller radius.  None of the other planets can compete with Jupiter, and smaller objects would need prohibitively short rotation periods to have any chance.  For example, the Earth would need to have a day that was about half an hour long to compete with Jupiter.  Even a molecule that spins millions of times a second would need a size on the order of a cm to compete with Jupiter.


How much star can star fusion fuse if star fusion can fuse hydrogen? During stellar nucleosynthesis a star will convert a portion of its hydrogen atoms into helium.  Is there a constant of how many tons of hydrogen is converted based on the mass of the star?  Is it based on a percentage of the star's mass? Or can it vary? A Nasa blurb stated that 600 million tons of hydrogen per second are consumed by our own sun. Is there a mathematical formula that can tell you how much any star would consume and convert to helium?

There's a rough equation based on a star's mass but the age and metalicity of the star are factors too.   As large stars grow older they grow hotter on the inside and as they grow hotter, they fuse hydrogen faster, but they also become less massive over their lifetimes and have less and less hydrogen over time, so there's two opposing factors.

A rough equation is found here:

http://www.astronomynotes.com/evolutn/s2.htm

  a star's luminosity increases with mass as $(\text{the star's mass})^p$. The
  value of the exponent p varies between 3 and 4. For the rare massive
  stars (M* > 30 Msun), p = 3 and for the more common low-mass stars (M*
  < 10 Msun), p = 4

Loosely speaking, the luminosity is roughly equivalent to the rate of hydrogen fusion though the hydrogen fusion takes place inside the star and it takes some time for the energy to make it's way to the outside of the star and radiate away, the age of the star is far greater than the time it takes the newly created energy to escape, that it's still a pretty good ratio direct, luminosity to hydrogen fusion rate.

A star with 2 solar masses is about (4th power) 16 times as luminous, which means it fuses hydrogen about 16 times as fast.   When you get into significantly larger stars, the 4th power rule gradually becomes the 3rd power.

But as far as age being a factor, our Sun during it's main sequence will be about twice (or a bit more than twice) as luminous when it's 8 or 9 billion years old as it was when it was 1 billion years old, which corresponds to a gradual doubling of hydrogen fusion rate.  This happens (basically) because the core retains heat and more heat makes fusion more likely to happen, even as the amount of hydrogen is significantly lower in the later stage of it's main sequence than the early stage.


Do planetary rings always follow rotational axis? In solar system,most ring systems seem almost lies exactly on the equator of planet, is it possible that the plane of a ring system is not lie on the equator of the planet?

Nearly always.

The asymmetric gravity of a spinning (and hence oblate) planet will induce tides that can pull small moons and ring particles into orbit around its equator in a few million years.

If you had a moon that was for some reason not orbiting around the equator, and it broke up (for example by passing the Roche limit) then the resulting ring would not be around the equator at first. However it is unlikely that a moon would be in such an orbit in the first place.

Very large moons (such as the the Moon) and those that are captured minor planets (Triton) can be in an inclined orbit. Triton is large, and was captured from the Kuiper belt. There has not been enough time for Triton to correct its orbit. Triton orbits backwards too. Such moons are comparatively far from their host planets, and so not at risk of being broken up to form a ring.